DE09

DIGITALS ELECTRONICS

TYPICAL QUESTIONS & ANSWERS
PART - I

OBJECTIVE TYPE QUESTIONS
Each Question carries 2 marks. Choose correct or the best alternative in the following: Q.1 The NAND gate output will be low if the two inputs are (A) 00 (B) 01 (C) 10 (D) 11 Ans: D The NAND gate output will be low if the two inputs are 11 (The Truth Table of NAND gate is shown in Table.1.1) X(Input) Y(Input) F(Output) 0 0 1 0 1 1 1 0 1 1 1 0 Table 1.1 Truth Table for NAND Gate Q.2 What is the binary equivalent of the decimal number 368 (A) 101110000 (B) 110110000 (C) 111010000 (D) 111100000 Ans: A The Binary equivalent of the Decimal number 368 is 101110000
(Conversion from Decimal number to Binary number is given in Table 1.2) 368 184 --- 0 92 --- 0 46 --- 0 23 --- 0 11 --- 1 5 --- 1 2 --- 1 1 --- 0 0 --- 1 Table 1.2 Conversion from Decimal number to Binary number 2 2 2 2 2 2 2 2 2

1

DE09 Q.3 The decimal equivalent of hex number 1A53 is (A) 6793 (B) 6739 (C) 6973 (D) 6379

DIGITALS ELECTRONICS

Ans: B The decimal equivalent of Hex Number 1A53 is 6739 (Conversion from Hex Number to Decimal Number is given below) 1 A 5 3 Hexadecimal 16³ 16² 16¹ 16° Weights (1A53)16 = (1X16³) + (10 X 16²) + (5 X 16¹) + (3 X 16º) = 4096 + 2560 + 80 + 3 = 6739 Q.4

(734)8 = ( )16
(A) C 1 D (C) 1 C D Ans: D (734)8 = (1 D C)16 0001 │ 1101 │ 1100 1 D C (B) D C 1 (D) 1 D C

Q.5

The simplification of the Boolean expression ABC + A BC is (A) 0 (B) 1 (C) A (D) BC Ans: B

( )( )

The Boolean expression is ABC + A BC is equivalent to 1 = (A+ A )(B+ B )(C+ C ) = 1X1X1 = 1 Q.6 The number of control lines for a 8 – to – 1 multiplexer is (A) 2 (B) 3 (C) 4 (D) 5

(ABC )+ (ABC ) = A + B + C + A + B + C = A + B + C + A + B + C

( ) ( )

Ans: B The number of control lines for an 8 to 1 Multiplexer is 3 (The control signals are used to steer any one of the 8 inputs to the output) Q.7 How many Flip-Flops are required for mod–16 counter? (A) 5 (B) 6 (C) 3 (D) 4 Ans: D The number of flip-flops is required for Mod-16 Counter is 4. 2

DE09

DIGITALS ELECTRONICS (For Mod-m Counter, we need N flip-flops where N is chosen to be the smallest number for which 2N is greater than or equal to m. In this case 24 greater than or equal to 1)

Q.8

EPROM contents can be erased by exposing it to (A) Ultraviolet rays. (B) Infrared rays. (C) Burst of microwaves. (D) Intense heat radiations. Ans: A EPROM contents can be erased by exposing it to Ultraviolet rays (The Ultraviolet light passes through a window in the IC package to the EPROM chip where it releases stored charges. Thus the stored contents are erased).

Q.9

The hexadecimal number ‘A0’ has the decimal value equivalent to (A) 80 (B) 256 (C) 100 (D) 160 Ans: D The hexadecimal number ‘A0’ has the decimal value equivalent to 160 ( A 0 161 160 = 10X161 + 0X160 = 160)

Q.10

The Gray code for decimal number 6 is equivalent to (A) 1100 (B) 1001 (C) 0101 (D) 0110 Ans: C The Gray code for decimal number 6 is equivalent to 0101 (Decimal number 6 is equivalent to binary number 0110) + 0 1 + 1 + 0

0 Q.11

1

0

1

The Boolean expression A.B + A.B + A.B is equivalent to (A) A + B (B) A.B (C) A + B (D) A.B Ans: A The Boolean expression A .B + A. B + A.B is equivalent to A + B ( A .B + A. B + A.B = B( A + A ) + A. B = B + A. B {Q ( A + A ) = 1} = A + B {Q (B + A. B ) = B + A}

Q.12

The digital logic family which has minimum power dissipation is

3

DE09 (A) TTL (C) DTL (B) RTL (D) CMOS

DIGITALS ELECTRONICS

Ans: D The digital logic family which has minimum power dissipation is CMOS. (CMOS being an unipolar logic family, occupy a very small fraction of silicon Chip area) Q.13
The output of a logic gate is 1 when all its inputs are at logic 0. the gate is either (A) a NAND or an EX-OR (B) an OR or an EX-NOR (C) an AND or an EX-OR (D) a NOR or an EX-NOR

Ans: D The output of a logic gate is 1 when all inputs are at logic 0. The gate is either a NOR or an EX-NOR . (The truth tables for NOR and EX-NOR Gates are shown in fig.1(a) & 1(b).) Input A B 0 0 0 1 1 0 1 1 Output Y 1 0 0 0 Input A B 0 0 0 1 1 0 1 1 Output Y 1 0 0 1

Fig.1(a) Truth Table for NOR Gate

Fig.1(b) Truth Table for EX-NOR Gate

Q.14

Data can be changed from special code to temporal code by using (A) Shift registers (B) counters (C) Combinational circuits (D) A/D converters.

Ans: A Data can be changed from special code to temporal code by using Shift Registers. (A Register in which data gets shifted towards left or right when clock pulses are applied is known as a Shift Register.) Q.15
A ring counter consisting of five Flip-Flops will have (A) 5 states (B) 10 states (C) 32 states (D) Infinite states.

Ans: A A ring counter consisting of Five Flip-Flops will have 5 states. Q.16
The speed of conversion is maximum in (A) Successive-approximation A/D converter. (B) Parallel-comparative A/D converter. (C) Counter ramp A/D converter. (D) Dual-slope A/D converter.

4

DE09

DIGITALS ELECTRONICS Ans: B The speed of conversion is maximum in Parallel-comparator A/D converter (Speed of conversion is maximum because the comparisons of the input voltage are carried out simultaneously.)

Q.17

The 2’s complement of the number 1101101 is (A) 0101110 (B) 0111110 (C) 0110010 (D) 0010011

Ans: D The 2’s complement of the number 1101101 is 0010011 (1’s complement of the number 1101101 is 0010010 2’s complement of the number 1101101is 0010010 + 1 =0010011) Q.18
The correction to be applied in decimal adder to the generated sum is (A) 00101 (B) 00110 (C) 01101 (D) 01010

Ans: B The correction to be applied in decimal adder to the generated sum is 00110. When the four bit sum is more than 9 then the sum is invalid. In such cases, add +6(i.e. 0110) to the four bit sum to skip the six invalid states. If a carry is generated when adding 6, add the carry to the next four bit group . Q.19
When simplified with Boolean Algebra (x + y)(x + z) simplifies to (A) x (B) x + x(y + z) (C) x(1 + yz) (D) x + yz

Ans: D When simplified with Boolean Algebra (x + y)(x + z) simplifies to x + yz [(x + y) (x + z)] = xx + xz + xy + yz = x + xz + xy + yz (Q xx = x) = x(1+z) + xy + yz = x + xy + yz {Q (1+z) = 1} = x(1 + y) + yz = x + yz {Q (1+y) = 1}] Q.20
The gates required to build a half adder are (A) EX-OR gate and NOR gate (B) EX-OR gate and OR gate (C) EX-OR gate and AND gate (D) Four NAND gates.

Ans: C The gates required to build a half adder are EX-OR gate and AND gate Fig.1(d) shows the logic diagram of half adder.
A B S

C

Fig.1(d) Logic diagram of Half Adder 5

DE09

DIGITALS ELECTRONICS

Q.21

The code where all successive numbers differ from their preceding number by single bit is (A) Binary code. (B) BCD. (C) Excess – 3. (D) Gray.

Ans: D The code where all successive numbers differ from their preceding number by single bit is Gray Code. (It is an unweighted code. The most important characteristic of this code is that only a single bit change occurs when going from one code number to next.) Q.22
Which of the following is the fastest logic (A) TTL (B) ECL (C) CMOS (D) LSI

Ans: B ECL is the fastest logic family of all logic families. (High speeds are possible in ECL because the transistors are used in difference amplifier configuration, in which they are never driven into saturation and thereby the storage time is eliminated. Q.23
If the input to T-flipflop is 100 Hz signal, the final output of the three T-flipflops in cascade is (A) 1000 Hz (B) 500 Hz (C) 333 Hz (D) 12.5 Hz.

Ans: D If the input to T-flip-flop is 100 Hz signal, the final output of the three Tflip-flops in cascade is 12.5 Hz {The final output of the three T-flip-flops in cascade is Frequency 100 (T) = = 3 =12.5Hz} 2N 2 Q.24
Which of the memory is volatile memory (A) ROM (B) RAM (C) PROM (D) EEPROM

Ans: B RAM is a volatile memory (Volatile memory means the contents of the RAM get erased as soon as the power goes off.) Q.25
-8 is equal to signed binary number (A) 10001000 (C) 10000000

(B) 00001000 (D) 11000000

Ans: A - 8 is equal to signed binary number 10001000
6

DE09

DIGITALS ELECTRONICS
(To represent negative numbers in the binary system, Digit 0 is used for the positive sign and 1 for the negative sign. The MSB is the sign bit followed by the magnitude bits. i.e., - 8 = 1000 1000 ------- ----------------Sign Magnitude ------- ---------------

Q.26

DeMorgan’s first theorem shows the equivalence of (A) OR gate and Exclusive OR gate. (B) NOR gate and Bubbled AND gate. (C) NOR gate and NAND gate. (D) NAND gate and NOT gate

Ans: B DeMorgan’s first theorem shows the equivalence of NOR gate and Bubbled AND gate (Logic diagrams for De Morgan’s First Theorem is shown in fig.1(a)
A

A B

Y

Y

B

Fig.1(a) Logic Diagrams for De Morgan’s First Theorem

Q.27

The digital logic family which has the lowest propagation delay time is (A) ECL (B) TTL (C) CMOS (D) PMOS

Ans: A The digital logic family which has the lowest propagation delay time is ECL (Lowest propagation delay time is possible in ECL because the transistors are used in difference amplifier configuration, in which they are never driven into saturation and thereby the storage time is eliminated). Q.28
The device which changes from serial data to parallel data is (A) COUNTER (B) MULTIPLEXER (C) DEMULTIPLEXER (D) FLIP-FLOP

Ans: C The device which changes from serial data to parallel data is demultiplexer. (A demultiplexer takes in data from one line and directs it to any of its N outputs depending on the status of the select inputs.) Q.29
A device which converts BCD to Seven Segment is called (A) Encoder (B) Decoder (C) Multiplexer (D) Demultiplexer

7

DE09

DIGITALS ELECTRONICS Ans: B A device which converts BCD to Seven Segment is called DECODER. (A decoder coverts binary words into alphanumeric characters.)

Q.30

In a JK Flip-Flop, toggle means (A) Set Q = 1 and Q = 0.

(B) Set Q = 0 and Q = 1. (C) Change the output to the opposite state. (D) No change in output. Ans: C In a JK Flip-Flop, toggle means Change the output to the opposite state. Q.31
The access time of ROM using bipolar transistors is about (A) 1 sec (B) 1 msec (C) 1 µsec (D) 1 nsec.

Ans: C
The access time of ROM using bipolar transistors is about 1 µ sec.

Q.32

The A/D converter whose conversion time is independent of the number of bits is (A) Dual slope (B) Counter type (C) Parallel conversion (D) Successive approximation.

Ans: C The A/D converter whose conversion time is independent of the Number of bits is Parallel conversion. (This type uses an array of comparators connected in parallel and comparators compare the input voltage at a particular ratio of the reference voltage). Q.33
When signed numbers are used in binary arithmetic, then which one of the following notations would have unique representation for zero. (A) Sign-magnitude. (B) 1’s complement. (C) 2’s complement. (D) 9’s complement.

Ans: A Q.34
The logic circuit given below (Fig.1) converts a binary code y1y 2 y 3 into

8

DE09 (A) Excess-3 code. (C) BCD code. Ans: B Gray code as X1=Y1, X2=Y1 XOR Y2 , For Y1 Y2 Y3 0 0 0 0 0 1 0 1 0 0 1 1 Q.35

DIGITALS ELECTRONICS (B) Gray code. (D) Hamming code.

X3=Y1 XOR Y2 XOR Y3 X1 X2 X3 0 0 0 0 0 1 0 1 1 0 1 0

The logic circuit shown in the given fig.2 can be minimised to

(A)

(B)

(C)

(D)

Ans: D As output of the logic circuit is Y=(X+Y’)’+(X’+(X+Y’)’)’ (X+Y’)’=X’Y Using DE Morgan’s Now this is one of input of 2nd gate. F=(A+X’)’=A’X=[(X’Y)’.X] =[(X+Y’)X]=X+XY’=X(Y’) =X Q.36
In digital ICs, Schottky transistors are preferred over normal transistors because of their (A) Lower Propagation delay. (B) Higher Propagation delay. (C) Lower Power dissipation. (D) Higher Power dissipation.

Ans: A Lower propagation delay as shottky transistors reduce the storage time delay by preventing the transistor from going deep into saturation. Q.37
The following switching functions are to be implemented using a Decoder: f1 = ∑ m(1, 2, 4, 8, 10, 14 ) f 2 = ∑ m(2, 5, 9, 11) f 3 = ∑ m(2, 4, 5, 6, 7 ) 9

DE09

DIGITALS ELECTRONICS
The minimum configuration of the decoder should be (A) 2 – to – 4 line. (B) 3 – to – 8 line. (C) 4 – to – 16 line. (D) 5 – to – 32 line.

Ans: C 4 to 16 line decoder as the minterms are ranging from 1 to 14. Q.38
A 4-bit synchronous counter uses flip-flops with propagation delay times of 15 ns each. The maximum possible time required for change of state will be (A) 15 ns. (B) 30 ns. (C) 45 ns. (D) 60 ns.

Ans: A
15 ns because in synchronous counter all the flip-flops change state at the same time.

Q.39

Words having 8-bits are to be stored into computer memory. The number of lines required for writing into memory are (A) 1. (B) 2. (C) 4. (D) 8.

Ans: D
Because 8-bit words required 8 bit data lines. 1 LSB is added to the 2

Q.40

In successive-approximation A/D converter, offset voltage equal to D/A converter’s output. This is done to

(A) (B) (C) (D) Ans: B Q.41

Improve the speed of operation. Reduce the maximum quantization error. Increase the number of bits at the output. Increase the range of input voltage that can be converted.

The decimal equivalent of Binary number 11010 is (A) 26. (B) 36. (C) 16. (D) 23.

Ans: A
4 3 2 1 11010 = 1 X 2 + 1 X 2 + 0 X 2 + 1 X 2 = 26

Q.42

1’s complement representation of decimal number of -17 by using 8 bit representation is (A) 1110 1110 (B) 1101 1101 (C) 1100 1100 (D) 0001 0001

Ans: A (17)10 = (10001)2 In 8 bit = 00010001 1's Complement = 11101110
10

DE09 Q.43
The excess 3 code of decimal number 26 is (A) 0100 1001 (B) 01011001 (C) 1000 1001 (D) 01001101

DIGITALS ELECTRONICS

Ans: B
(26)10 in BCD is ( 00100110 ) BCD Add 011 to each BCD 01011001 for excess – 3

Q.44

How many AND gates are required to realize Y = CD+EF+G (A) 4 (B) 5 (C) 3 (D) 2

Ans: D To realize Y = CD + EF + G Two AND gates are required (for CD & EF). Q.45
How many select lines will a 16 to 1 multiplexer will have (A) 4 (B) 3 (C) 5 (D) 1

Ans: A
4 In 16 to 1 MUX four select lines will be required to select 16 ( 2 ) inputs.

Q.46

How many flip flops are required to construct a decade counter (A) 10 (B) 3 (C) 4 (D) 2

Ans: C
Decade counter counts 10 states from 0 to 9 ( i.e. from 0000 to 1001 ) Thus four FlipFlop's are required.

Q.47

Which TTL logic gate is used for wired ANDing (A) Open collector output (B) Totem Pole (C) Tri state output (D) ECL gates

Ans: A
Open collector output.

Q.48

CMOS circuits consume power (A) Equal to TTL (C) Twice of TTL

(B) Less than TTL (D) Thrice of TTL

Ans: B As in CMOS one device is ON & one is Always OFF so power consumption is low. Q.49
In a RAM, information can be stored (A) By the user, number of times. 11

DE09 (B) By the user, only once. (C) By the manufacturer, a number of times. (D) By the manufacturer only once. Ans: A
RAM is used by the user, number of times. The hexadecimal number for (95.5)10 is (A) (5F.8)16 (B) (9A.B)16

DIGITALS ELECTRONICS

Q.50

(C) Ans: A

(2E.F)16
(95.5)10 = (5F.8)16
Integer part 16 95 16 5 15

(D) (5A.4 )16

Fractional part 0.5x16=8.0

0

5

Q.51

The octal equivalent of (247 )10 is (A) (252)8

(C) (367 )8 Ans: C (247)10 = (367)8
8 8 8 247 30 3 0

(B) (350 )8 (D) (400)8

7 6 3

Q.52

The chief reason why digital computers use complemented subtraction is that it (A) Simplifies the circuitry. (B) Is a very simple process. (C) Can handle negative numbers easily. (D) Avoids direct subtraction.

Ans: C Using complement method negative numbers can also be subtracted. Q.53
In a positive logic system, logic state 1 corresponds to (A) positive voltage (B) higher voltage level (C) zero voltage level (D) lower voltage level

Ans: B
12

DE09

DIGITALS ELECTRONICS
We decide two voltages levels for positive digital logic. Higher voltage represents logic 1 & a lower voltage represents logic 0.

Q.54

The commercially available 8-input multiplexer integrated circuit in the TTL family is (A) 7495. (B) 74153. (C) 74154. (D) 74151.

Ans: B MUX integrated circuit in TTL is 74153. Q.55
CMOS circuits are extensively used for ON-chip computers mainly because of their extremely (A) low power dissipation. (B) high noise immunity. (C) large packing density. (D) low cost.

Ans: C Because CMOS circuits have large packing density. Q.56
The MSI chip 7474 is (A) Dual edge triggered JK flip-flop (TTL). (B) Dual edge triggered D flip-flop (CMOS). (C) Dual edge triggered D flip-flop (TTL). (D) Dual edge triggered JK flip-flop (CMOS).

Ans: C
MSI chip 7474 dual edge triggered D Flip-Flop.

Q.57

Which of the following memories stores the most number of bits (A) a 5M × 8 memory. (B) a 1M × 16 memory. (D) a 1M × 12 memory. (C) a 5M × 4 memory.

Ans: A 5Mx8 = 5 x 220 x 8 = 40M (max) Q.58
The process of entering data into a ROM is called (A) burning in the ROM (B) programming the ROM (C) changing the ROM (D) charging the ROM

Ans: B The process of entering data into ROM is known as programming the ROM. Q.59
When the set of input data to an even parity generator is 0111, the output will be (A) 1 (B) 0 (C) Unpredictable (D) Depends on the previous input

Ans: B
In even parity generator if number of 1 is odd then output will be zero.

13

DE09 Q.60

DIGITALS ELECTRONICS
The number 140 in octal is equivalent to (A) (96 )10 . (B) (86 )10 . (C) (90 )10 . (D) none of these.

Ans: A (140)8 = (96)10 1 x 82 + 4 x 8 + 0x 1 = 64 + 32 = 96 Q.61
The NOR gate output will be low if the two inputs are (A) 00 (B) 01 (C) 10 (D) 11

Ans: B, C, or D O/P is low if any of the I/P is high Q.62
Which of the following is the fastest logic? (A) ECL (B) TTL (D) LSI (C) CMOS

Ans: A Q.63
How many flip-flops are required to construct mod 30 counter (A) 5 (B) 6 (C) 4 (D) 8

Ans: A Mod - 30 counter +/- needs 5 Flip-Flop as 30 < 25 Mod - N counter counts total ' N ' number of states. To count 'N' distinguished states we need minimum n FlipFlop's as [N = 2n] For eg. Mod 8 counter requires 3 Flip-Flop's (8 = 23) Q.64
How many address bits are required to represent a 32 K memory (A) 10 bits. (B) 12 bits. (C) 14 bits. (D) 16 bits.

Ans: D 32K = 25 x 210 = 215, Thus 15 address bits are required, Only 16 bits can address it. Q.65
The number of control lines for 16 to 1 multiplexer is (A) 2. (B) 4. (C) 3. (D) 5.

Ans: B As 16 = 24, 4 Select lines are required. Q.66
Which of following requires refreshing? (A) SRAM. (B) DRAM. 14

DE09 (C) ROM. Ans: B Q.67 (D) EPROM.

DIGITALS ELECTRONICS

Shifting a register content to left by one bit position is equivalent to (A) division by two. (B) addition by two. (C) multiplication by two. (D) subtraction by two.

Ans:C Q.68
For JK flip flop with J=1, K=0, the output after clock pulse will be (A) 0. (B) 1. (C) high impedance. (D) no change.

Ans: B Q.69
Convert decimal 153 to octal. Equivalent in octal will be (A) (231)8 . (B) (331)8 .

(C) (431)8 . Ans: A (153)10 = (231)8 8 8 8 153 19 2 1 3 2

(D) none of these.

Q.70

The decimal equivalent of (1100)2 is (A) 12 (C) 18

(B) 16 (D) 20

Ans: A (1100)2 = (12)10 Q.71
The binary equivalent of (FA )16 is (A) 1010 1111 (C) 10110011

(B) 1111 1010 (D) none of these

Ans: B (FA)16 = (11111010)10 Q.72
The output of SR flip flop when S=1, R=0 is (A) 1 (B) 0 (C) No change (D) High impedance 15

DE09

DIGITALS ELECTRONICS Ans: A As for the SR flip-flop S=set input R=reset input ,when S=1, R=0, Flip-flop will be set.

Q.73

The number of flip flops contained in IC 7490 is (A) 2. (B) 3. (C) 4. (D) 10.

Ans: A Q.74
The number of control lines for 32 to 1 multiplexer is (A) 4. (B) 5. (C) 16. (D) 6.

Ans: B The number of control lines for 32 (25) and to select one input among them total 5 select lines are required. Q.75
How many two-input AND and OR gates are required to realize Y=CD+EF+G (A) 2,2. (B) 2,3. (C) 3,3. (D) none of these.

Ans: A Y=CD+EF+G Number of two input AND gates=2 Number of two input OR gates = 2 One OR gate to OR CD and EF and next to OR of G & output of first OR gate. Q.76
Which of following can not be accessed randomly (A) DRAM. (B) SRAM. (C) ROM. (D) Magnetic tape.

Ans: D Magnetic tape can only be accessed sequentially. Q.77
The excess-3 code of decimal 7 is represented by (A) 1100. (B) 1001. (C) 1011. (D) 1010.

Ans: D An excess 3 code is always equal to the binary code +3

Q.78

When an input signal A=11001 is applied to a NOT gate serially, its output signal is (A) 00111. (B) 00110. (C) 10101. (D) 11001.

Ans: B As A=11001 is serially applied to a NOT gate, first input applied will be LSB 00110.
16

DE09

DIGITALS ELECTRONICS

Q.79

The result of adding hexadecimal number A6 to 3A is (A) DD. (B) E0. (C) F0. (D) EF.

Ans: B Q.80
A universal logic gate is one, which can be used to generate any logic function. Which of the following is a universal logic gate? (A) OR (B) AND (C) XOR (D) NAND

Ans: D NAND can generate any logic function. Q.81
The logic 0 level of a CMOS logic device is approximately (A) 1.2 volts (B) 0.4 volts (C) 5 volts (D) 0 volts

Ans: D CMOS logic low level is 0 volts approx. Q.82
Karnaugh map is used for the purpose of (A) Reducing the electronic circuits used. (B) To map the given Boolean logic function. (C) To minimize the terms in a Boolean expression. (D) To maximize the terms of a given a Boolean expression.

Ans: C Q.83
A full adder logic circuit will have (A) Two inputs and one output. (B) Three inputs and three outputs. (C) Two inputs and two outputs. (D) Three inputs and two outputs.

Ans: D A full adder circuit will add two bits and it will also accounts the carry input generated in the previous stage. Thus three inputs and two outputs (Sum and Carry) are there. Q.84
An eight stage ripple counter uses a flip-flop with propagation delay of 75 nanoseconds. The pulse width of the strobe is 50ns. The frequency of the input signal which can be used for proper operation of the counter is approximately (A) 1 MHz. (B) 500 MHz. (C) 2 MHz. (D) 4 MHz.

Ans: A

17

DE09

DIGITALS ELECTRONICS
Maximum time taken for all flip-flops to stabilize is 75ns x 8 + 50 = 650ns. Frequency of operation must be less than 1/650ns = 1.5 MHz.

Q.85

The output of a JK flipflop with asynchronous preset and clear inputs is ‘1’. The output can be changed to ‘0’ with one of the following conditions. (A) By applying J = 0, K = 0 and using a clock. (B) By applying J = 1, K = 0 and using the clock. (C) By applying J = 1, K = 1 and using the clock. (D) By applying a synchronous preset input.

Ans: C Preset state of JK Flip-Flop =1 With J=1 K=1 and the clock next state will be complement of the present state. Q.86
The information in ROM is stored (A) By the user any number of times. (B) By the manufacturer during fabrication of the device. (C) By the user using ultraviolet light. (D) By the user once and only once.

Ans: B Q.87
The conversation speed of an analog to digital converter is maximum with the following technique. (A) Dual slope AD converter. (B) Serial comparator AD converter. (C) Successive approximation AD converter. (D) Parallel comparator AD converter.

Ans: D Q.88
A weighted resistor digital to analog converter using N bits requires a total of (A) N precision resistors. (B) 2N precision resistors. (C) N + 1 precision resistors. (D) N – 1 precision resistors.

Ans: A Q.89
The 2’s complement of the number 1101110 is (A) 0010001. (B) 0010001. (C) 0010010. (D) None.

Ans: C 1’s complement of 1101110 is = 0010001 Thus 2’s complement of 1101110 is = 0010001 + 1 = 0010010 Q.90
The decimal equivalent of Binary number 10101 is (A) 21 (B) 31 (C) 26 (D) 28 18

DE09

DIGITALS ELECTRONICS

Ans: A 1x24 + 0x23 +1x22 +0x21 + 1x20 = 16 + 0 + 4 + 0 + 1 = 21. Q.91
How many two input AND gates and two input OR gates are required to realize Y = BD+CE+AB (A) 1, 1 (B) 4, 2 (C) 3, 2 (D) 2, 3

Ans: A There are three product terms, so three AND gates of two inputs are required. As only two input OR gates are available, so two OR gates are required to get the logical sum of three product terms. Q.92
How many select lines will a 32:1 multiplexer will have (A) 5. (B) 8. (C) 9. (D) 11.

Ans: A For 32 inputs, 5 select lines will be required, as 25 = 32. Q.93
How many address bits are required to represent 4K memory (A) 5 bits. (B) 12 bits. (C) 8 bits. (D) 10 bits.

Ans: B For representing 4K memory, 12 address bits are required as 4K = 22 x 210 = 212 (1K = 1024 = 210) Q.94
For JK flipflop J = 0, K=1, the output after clock pulse will be (A) 1. (B) no change. (C) 0. (D) high impedance.

Ans: C J=0, K=1, these inputs will reset the flip-flop after the clock pulse. So whatever be the previous output, the next state will be 0. Q.95
Which of following are known as universal gates (A) NAND & NOR. (B) AND & OR. (C) XOR & OR. (D) None.

Ans: A NAND & NOR are known as universal gates, because any digital circuit can be realized completely by using either of these two gates. Q.96
Which of the following memories stores the most number of bits (A) 64K × 8 memory. (B) 1M × 8 memory. 19

DE09 (C) 32M × 8 memory.

DIGITALS ELECTRONICS (D) 64 × 6 memory.

Ans: C 32M x 8 stores most number of bits 25 x 220 = 225 (1M = 220 = 1K x 1K = 210 x 210) Q.97
Which of following consume minimum power (A) TTL. (B) CMOS. (C) DTL. (D) RTL.

Ans: B CMOS consumes minimum power as in CMOS one p-MOS & one n-MOS transistors are connected in complimentary mode, such that one device is ON & one is OFF.

20

DE09 PART – II

DIGITALS ELECTRONICS

NUMERICALS
Q.1
Convert the octal number 7401 to Binary.

(4)

Ans: Conversion of Octal number 7401 to Binary: Each octal digit represents 3 binary digits. To convert an octal number to binary number, each octal digit is replaced by its 3 digit binary equivalent shown below. 7 4 0 1

111 100 000 001 Thus, (7401)8 = (111100000001)2

Q.2

Find the hex sum of (93)16 + ( DE )16 .

(4) Ans:
Hex Sum of (93)16 + (DE)16 Convert Hexadecimal numbers 93 and DE to its binary equivalent shown below:93 → 10010011 DE → 11011110 ---------------101110001 → 171 ----------------Thus (93)16 + (DE)16 = (171)16

Q.3

Perform 2’s complement subtraction of (7 )10 − (11)10 .

(4)

Ans: 2’s Complements Subtraction of (7)10 – (11)10 First convert the decimal numbers 7 and 11 to its binary equivalents. (7)10 = (0111)2 (11)10 = (1011)2 in 4-bit system Then find out the 2’s complement for 1011 i.e.,
1’s Complement of 1011 is 0100 2’s Complement of 1011 is 0101 So, (7)10 – (11)10 = 0111 0101 --------1100 ---------

21

DE09

DIGITALS ELECTRONICS
Since there is no carry over flow occurring in the summation, the result is a negative number, to find out its magnitude, 2’s Complement of the result must be found. 2’s Complement of 1100 is 0011 1 -------0100 -------Here the answer is (-4)10 (or) in 2’s complement it is 1100. What is the Gray equivalent of (25)10 .

Q.4

(2)

Ans: Gray equivalent of (25)10 The binary equivalent of Decimal number 25 is (00100101)2 1. The left most bit (MSB) in gray code is the same as the left most in binary 2. Add the left most bit to the adjacent bit 3. Add the next adjacent pair and so on., Discard if we get a carry. 0+0+1+0+0+1+0+1

0

0

1

1

0

1

1

1

Gray Number

Q.5

Evaluate x = A .B + C A .D using the convention A = True and B = False.

( )

(4)

Ans: Evaluate x = A .B + C A.D

( )

= A B + C ( A + D ) (Since A.D = A + D by using Demorgan’s Law) = A .B + C . A + C . D By using the given convention, A = True = 1; B = False = 0 =1 .0 + C.1+ C. D = 0 + 0 + C. D = C. D

Q.6

Simplify the Boolean expression F = C(B + C)(A + B + C).

(6)

Ans: Simplify the Boolean Expression F = C (B +C) (A+B+C) F = C (B+C) (A+B+C) = CB + CC [(A+B+C)] = CB + C [(A+B+C)] (Q CC = C) = CBA + CBB + CBC + CA + CB + CC = ABC + CB + CB + CA + CB + CC (Q CBB =CB & CBC = CB) = ABC + CB + CA + C (Q CB+CB+CB = CB; CC = C) = ABC + BC + C (1+A) = ABC + BC + C (Q 1+A = 1) = ABC + C (1+B)
22

DE09
= ABC + C (Q 1+B = 1) = C (1+AB) = C {Q (1+AB)=1}

DIGITALS ELECTRONICS

Q.7

Simplify the following expression into sum of products using Karnaugh map F(A, B, C, D) = ∑ (1,3,4,5,6,7,9,12,13) (7)

Ans: Simplification of the following expression into sum of products using Karnaugh Map: F(A,B,C,D) = Σ (1,3,4,5,6,7,9,12,13) Karnaugh Map for the expression F(A,B,C,D) = Σ (1,3,4,5,6,7,9,12,13) is shown in Fig.4(a). The grouping of cells is also shown in the Figure.

The equations for (1) is A B; (2) is C D; (3) is A D; (4) is B C Hence, the Simplified Expression for the above Karnaugh map is F(A,B,C,D) = A B+ C D+ A D+B C = A (B + D) + C ( B + D)

Q.8

Simplify and draw the logic diagram for the given expression F = ABC + ABC + ABC + A BC + A BC .

(7)

Ans:
Simplification of the logic expression F = ABC + AB C + A B C + A BC + A B C F = ABC + AB C + A B C + A BC + A B C 23

DE09

DIGITALS ELECTRONICS
F = A + B + C + ( A + B )C + A B C + A ( B + C ) + A B C (Q ABC = A + B + C and AB = A + B by using Demorgan’s Law) = A + B +C + A C + B C + A BC + A B + AC + A B C = A + A C+ B + B C + C + A C + A B C +A B + A B C = A (1+C)+ B (1 + C) + C (1 + A) + A B C + A B + A B C = A + B + C + A B C + A B + A B C {Q (1+C) = 1 and (1+A) = 1} = ( A + A B ) + B (1 + AC) + C (1+ A B) = ( A + B )+ B + C {Q ( A + A B ) = ( A + B ); (1+AC) = 1 and (1+ A B) =1} F = ( A + B + C ) (Q B + B = B ) The logic diagram for the simplified expression F = ( A + B + C ) is given in fig.5(a)
A _ A

_ B B _ C C

_ _ _ F =A +B+ C

Fig.5(a) Logic diagram for the expression F = ( A + B + C )

Q.9

Determine the binary numbers represented by the following decimal numbers. (i) 25.5 (ii) 10.625 (iiii) 0.6875

(6)

Ans: (i) Conversion of decimal number 25.5 into binary number: Here integer part is 25 and fractional part is 0.5. First convert the integer part 25 into its equivalent binary number i.e., divide 25 by 2 till the quotient becomes 0 shown in table 2(a)
Quotient 25 2 12 2 6 2 3 2 1 2 12 6 3 1 0 Remainder 1 0 0 1 1

Table 2(a)
24

DE09

DIGITALS ELECTRONICS
So, integer part (25)10 is equivalent to the binary number 11001. Next convert fractional part 0.5 into binary form i.e., multiply the fractional part 0.5 by 2 till you get remainder as 0 0.5 X2 -----1.0 Remainder

1 (Quotient) The decimal fractional part 0.5 is equivalent to binary number 0.1. Hence, the decimal number 25.5 is equal to the binary number 11001.1

(ii) Conversion of decimal number 10.625 into binary number: Here integer part is 10 and fractional part is 0.625. First convert the decimal number 10 into its equivalent binary number i.e., divide 10 by 2 till the quotient becomes 0 shown in table 2(b)
Quotient 10 2 5 2 2 2 1 2 5 2 1 0 Table 2(b) Remainder 0 1 0 1

So, the integer part 10 is equal to binary number 1010. Next convert the decimal fractional part 0.625 into its binary form i.e., multiply 0.625 by 2 till the remainder becomes 0 0.625 0.250 0.50 X2 X2 X2 -------------------1.250 0.50 1.0 (Remainder)

1

0

1

(Quotient)

So, the decimal fractional part 0.625 is equal to binary number 0.101. Hence the decimal number 10.625 is equal to binary number 1010.101.

(iii)Conversion of fractional number 0.6875 into its equivalent binary number: Multiply the fractional number 0.6875 by 2 till the remainder becomes 0 i.e.,
25

DE09
0.6875 X2 ----------1.3750 0.3750 X2 --------0.75 0.75 X2 -------1.5 0.5 X2 -----1.0

DIGITALS ELECTRONICS

(Remainder)

1

0

1

1

(Quotient)

So, the decimal fractional number 0.6875 is equal to binary number 0.1011.

Q.10

Perform the following subtractions using 2’s complement method. (i) 01000 – 01001 (ii) 01100 – 00011 (iii) 0011.1001 – 0001.1110

(8)

Ans: (i) Subtraction of 01000-01001: 1’s complement of 01001 is 10110 and 2’s complement is 10110+ 1 =10111. Hence 01000 = 01000 - 01001 = +10111 (2's complement) ------------------------11111 (Summation) ------------------------Since the MSB of the sum is 1, which means the result is negative and it is in 2's complement form. So, 2's complement of 1111 =00001= (1)10. Therefore, the result is – 1. (ii) Subtraction of 01100-00011: 1’s complement of 00011 is 11100 and 2’s complement is 11100 + 1 = 11101. Hence
01100 = 01100 – 00011 = + 11101 (2's complement) -------------------------------------------------1 01001 = + 9 Ignore -------------------------------------------------If a final carry is generated discard the carry and the answer is given by the remaining bits Which is positive i.e., (1001)2 = (+ 9)10 (iii) Subtraction of 0011.1001 – 0001.1110: 1’s complement of 0001.1110 is 1110.0001 and its 2’s complement is 1110.0010. 0011.1001 = 0011.1001 - 0001.1110 = + 1110.1011 (2’s complement) ------------------------------------------1 0001.101I = + 1 .68625 Ignore 26

DE09

DIGITALS ELECTRONICS
If a final carry is generated discard the carry and the answer is given by the remaining bits which is positive i.e., (0001.1011)2 = (+ 1.68625)10

Q.11

Simplify the expressions using Boolean postulates

(9)

(i) X Y + XYZ + X (Y + X Y ) (iii) XY + XZ + X Y Z (XY + Z) Ans: (i) X Y + XYZ + X (Y + X Y )

(ii) Y = (A + B)( A + C)(B + C)

= X Y + XYZ + X (Y + X Y ) = X (Y + YZ ) + X (Y + X Y )

= X (Y + Z ) + X (Y + X )
(Because Y + YZ = Y + Z and Y + X Y = Y + X ) = X Y + XZ + XY + XX = X Y + XZ + XY + X (Because XX=X) = X Y + XZ + X (1 + Y )

= X Y + XZ + X (Because (1+Y=1) = ( X + Y )( X + Z ) + X (Because XY = X + Y ) = X X + X Z +Y X +YZ + X = X + X Z +Y X +YZ + X = X (1 + Z + Y ) + Y Z + X = X +YZ + X = (X + X ) + Y Z = 1 + Y Z (Because X + X = 1) = 1 = 0 (Because 1 + Y Z =1) (ii) Y = (A + B)( A + C)(B + C) Y = (A + B)( A + C)(B + C) = (A A + AC + B A + BC) (B + C) = (AC + B A + BC) (B + C) = ABC + BB A + BBC + ACC + B A C + BCC = ABC + B A + BC + AC + B A C + BC = ABC + AC + B A + B A C + BC =AC (B+1) + B A + BC ( A +1) = AC + B A + BC = AC + B A + BC (A + A ) = AC + B A + BCA + BC A = AC(1 + B) + B A (1 + C)
27 (Because X X = X )

(Because A A = 0) (Because BB = B) (Because BC + BC = BC) (Because B + 1 = 1 and A + 1 = 1) (Because A + A = 1)

DE09

DIGITALS ELECTRONICS
= AC + B A {Because (1 + B) = 1 and (1 + C) =1} (iii) XY + XZ + XY Z (XY + Z) = XY + XZ + XY Z (XY + Z) = XY + XZ + XXY Y Z + X Y Z Z = XY + XZ + X Y Z (Because Y Y = 0 & ZZ = Z) = XY + X + Z + XY Z (Because XZ = X + Z ) = X + XY + Z + XY Z = X + X (Y +Y Z) + Z = X + X (Y +Z) + Z (Because Y +Y Z = Y +Z) = X + X Y (Z+ Z ) + XZ + Z (Because Z+ Z =1) = X + X Y Z + XY Z + XZ + Z = X + XZ (1+ Y) + Z (1+XY) = X + XZ + Z (Because 1+ Y = 1 &1+XY = 1) = X + XZ) + Z =( X + Z) + Z (Because X + XZ = X + Z = X +( Z + Z ) = X +1 (Because Z + Z = 1) =1 (Because X +1 = 1)

Q.12

Minimize the logic function Y( A, B, C, D) = ∑ m(0,1,2,3,5,7,8,9,11,14) . Use Karnaugh map. Draw logic circuit for the simplified function. (9)

Ans: Fig. 4(a) shows the Karnaugh map. Since the expression has 4 variables, the map has 16 cells. The digit 1 has been written in the cells having a term in the given expression. The decimal number has been added as subscript to indicate the binary number for the concerned cell. The term ABC D cannot be combined with any other cell. So this term will appear as such in the final expression. There are four

groupings of 4 cells each. These correspond to the min terms (0, 1, 2, 3), (0, 1, 8, 9), (1, 3,5,7) and (1, 3, 9, 11). These are shown in the map. Since all the terms (except 14) have been included in groups of 4 cells, there is no need to form groups of two cells.

28

DE09

DIGITALS ELECTRONICS

The simplified expression is Y (A,B,C,D) = ABC D + A B + B C + B D+ A D Fig.4 (b) shows the logic diagram for the simplified expression Y (A,B, C, D) = ABC D + A B + B C + B D+ A D
A B C D _ A B C D

_ _ A B

_ A D

Y

_ _ B C

_ B D

Fig.4(b) Logic diagram for Y

Q.13

Simplify the given expression to its Sum of Products (SOP) form. Draw the logic circuit for the simplified SOP function Y = (A + B) A + AB C + A B + C + AB + ABC (5)

(

)

(

)

Ans:
Simplification of given expression Y = (A + B) (A + AB ) C + A (B + C ) + A B + ABC in some of products (SOP) form:29

DE09

DIGITALS ELECTRONICS
Y = (A + B) (A + AB ) C + A (B + C ) + A B + ABC =(A + B) (A + AB ) C + A (B + C ) + A B + ABC =(A + B) (A + A + B )C + A (B + C ) + A B + ABC =(A + B) (1+ B )C + A (B + C ) + A B + ABC (Because A + A = 1) = (A + B) (C+ B C) + A B + A C + A B + ABC = (A + B) (C+ B C) + A B + A C + A B + ABC =AC + A B C + BC + B B C + A B + A C + A B + ABC = AC + AC( B + B) + BC + 0 + A B + A C + A B (Because B B = 0) = AC + AC+ BC+ A B + A C (Because B + B = 1) = AC+ BC+ A B + A C (Because AC + AC = AC) = C (A+ B) + A (B + C )
A B C C A +B C (A + B)

_ A

Y
_ _ A(B + C) _ B +C

B _ C

Fig.4(c) Simplified Logic Circuit Q.14 Design a 8 to 1 multiplexer by using the four F( A, B, C, D) = ∑ m(0,1,3,4,8,9,15) . variable function given by (10)

Ans: Design of 8 to 1 Multiplexer: This is a four-variable function and therefore we need a multiplexer with three selection lines and eight inputs. We choose to apply variables B, C, and D to the selection lines. This is shown inTable 8.1. The first half of the minterms are associated with A' and the second half with A. By circling the minterms of the function and applying the rules for finding values for the multiplexer inputs, the implementation shown in Table.8.2. The given function can be implemented with a 8-to-1 multiplexer as shown in fig.8(a). Three of the variables, B, C and D are applied to the selection lines in that order i.e., B is connected to s2, C to s1 and D to s0. The inputs of the multiplexer are 0, 1, A and A’. When BCD = 000,001 & 111 output F = 1 since I0 & I8 = 1 for BCD(000), I1 = 1and I9 =1 respectively. Therefore, minterms m0 = A’ B’ C’ m1 = A’ B’ C, m8 = A’, B’, C’ and m9 = A’ B’ C produce a 1 output. When BCD = 010, 101 and 110, output F = 0, since I2, I5 and I6 respectively are equal to 0.

30

DE09 Minterm 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 A 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 B 0 0 0 0 1 1 1 1 0 0 0 0 1 1 1 1 C 0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1

DIGITALS ELECTRONICS D 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 F 1 1 0 1 1 0 0 0 1 1 0 0 0 0 0 1

Table .8.1 Truth Table for 8-1 Multiplexer

Table 8.2 Implementation Table for 8 to 1 MUX

31

DE09

DIGITALS ELECTRONICS

1

I I

0 1 2 3 4 5 6

0

I I I I I

8 X1 MUX

Y

F

A

I 7

S
2

S

1

S

0

B C D

Fig.8(a) Logic circuit for 8-to-1 Multiplexer

Q.15 Convert the decimal number 82.67 to its binary, hexadecimal and octal equivalents. Ans: (i)Conversion of Decimal number 82.67 to its Binary Equivalent Considering the integer part 82 and finding its binary equivalent
2 2 2 2 2 2 2 82 41 20 10 5 2 1 0 Remainder ----- 0 (LSB) Remainder ----- 1 Remainder ----- 0 Remainder ------0 Remainder ----- 1 Remainder ---- 0 Remainder ---- 1 (MSB)

(6)

The Binary equivalent is (1010010)2

32

DE09
Now taking the fractional part i.e., 0.67 Fraction Fraction X 2

DIGITALS ELECTRONICS

0.67 0.34 0.68 0.36 0.72 0.44 0.88 0.76

1.34 0.68 1.36 0.72 1.44 0.88 1.76 1.52

Remainder New Fraction 0.34 0.68 0.36 0.72 0.44 0.88 0.76 0.52

Integer

1 0 1 0 1 0 1 1

It is seen that, it is not possible to get a zero as remainder even after 8 stages. The process continued further on an approximation can be made and the process is terminated here. The binary equivalent is 0.10101011 Therefore, the binary equivalent of decimal number 82.67 is (1010010.10101011)2 (ii)Conversion of the binary equivalent of decimal number 82.67 into Hexadecimal: The binary equivalent of decimal number 82.67 is (1010010.10101011)2 Convert each 4-bit binary into an equivalent hexadecimal number i.e. 0101 0010 .1010 1011 5 2 A B Therefore, the hexadecimal equivalent of decimal number 82.67 is (52.AB)16 (iii)Conversioin of the binary equivalent of decimal number 82.67 into Octal number: The binary equivalent of decimal number 82.67 is (1010010.10101011)2 Convert each 3-bit binary into an equivalent octal number i.e. 001 010 010 1 2 2 .101 010 110 . 5 2 6

Therefore, the Octal equivalent of decimal number 82.67 is (122.526)8 Q.16 Add 20 and (-15) using 2’s complement. (4)

33

DE09

DIGITALS ELECTRONICS

Ans: Addition of 20 and (-15) using 2’s Complement:
2 2 2 2 2 20 10 5 2 1 0 Remainder ----- 0 (LSB) Remainder ----- 0 Remainder ----- 1 Remainder ------0 Remainder ----- 1(MSB) 2 2 2 2 2 16 8 4 2 1 0 (20)10 = 1 0 1 0 0 (-16)10 = 0 1 1 1 1(1’s Complement) +1(2’s Complement) --------------------10000 --------------------Therefore, 20 = 1 0 1 0 0 -16 = 1 0 0 0 0 ----------------------------1 0 0 1 0 0 Remainder ----- 0 (LSB) Remainder ----- 0 Remainder ----- 0 Remainder ------0 Remainder ----- 1(MSB)

(16)10 = 1 0 0 0 0

(Neglect) -------------------------------Since the MSB of the sum is 0, which means the result is positive i.e +4

Q.17 Add 648 and 487 in BCD code. Ans: Addition of 648 and 487 in BCD Code: 6 4 8 = 0110 0100 1000 4 8 7 = 0100 1000 0111 -------------------------------------------------1010 1100 1111

(4)

10 12 15 -------------------------------------------------In the above problem all the three groups are invalid, because the four bit sum is more than 9. In such cases, add +6(i.e. 0110) to the four bit sum to skip the six invalid states. If a carry is generated when adding 6, add the carry to the next four bit group i.e.

34

DE09
6 4 8 = 0110 0100 1000 4 8 7 = 0100 1000 0111 -------------------------------------------------1010 1100 1111 0110 0110 0110 1 11 1 1 1 11 -----------------------------------------------------0001 0 0 0 1 0 0 1 1 0 1 0 1 3 5 1 1 ------------------------------------------------------Addition of 648 and 487 in BCD Code is 1135.

DIGITALS ELECTRONICS

Q.18 Prove the following Boolean identities. (i) XY + YZ + Y Z = XY + Z (ii) A.B + A.B + A.B = A + B Ans: (i) Prove the Boolean Identity XY + YZ + Y Z = XY + Z L.H.S = XY + YZ + Y Z = XY(Z+ Z ) + YZ + Y Z (Q Z + Z = 1) = XYZ + XY Z + YZ + Y Z = YZ(1+X) + XY Z +Y Z = YZ + XY Z +Y Z (Q 1+X = 1)
= Z (Y+ Y ) + XY Z = Z + XY Z (Q Y+Y =1) = Z + XY(Q Z + XY Z = Z + XY) = R.H.S (Hence Proved)

(4)

(ii) Prove the Boolean Identity A B + A B + A B = A + B R.H.S = A + B = A (B + B ) + B (A + A ) (Q B + B = 1 & A + A = 1) = A (B + B ) + B (A + A ) = AB + A B + B A + B A
= A B + A B + B A ( A B + A B = A B) = L.H.S (Hence Proved)

Q.19 For F = A.B.C + B.C.D + A.B.C , write the truth table. Simplify using Karnaugh map and realize the function using NAND gates only. (10) Ans: Simplification of Logic Function F = A B C + B C D + A B C

35

DE09

DIGITALS ELECTRONICS

(i)The Truth Table is given in Table 4.1
Inputs A 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 B 0 0 0 0 1 1 1 1 0 0 0 0 1 1 1 1 C 0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1 D 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 Output (F) 0 0 0 0 0 1 1 1 0 0 0 0 0 1 1 1

Table 4.1 (ii) The Karnaugh Map is shown in fig.4(a). The simplified expression is F = BC + BD

(iii) The NAND-NAND Realization is shown in fig.4(b)

36

DE09
__ BC

DIGITALS ELECTRONICS

B C

B D

__ BD

______ __ __ F = BC . BD = BC + BD

Fig. 4(b) NAND-NAND Realization

Q.20 Determine the analog output voltage of 6-bit DAC (R-2R ladder network) with Vref as 5V when the digital input is 011100. (10) Ans: For 6-bit R-2R DAC ladder network, the output voltage is given by V V0 = R a n−1 2 n −1 + a n − 2 2 n− 2 + ⋅ ⋅ ⋅ ⋅ + a1 21 + a 0 2 0 2n Given Data : VR = 5V, n = 6, a5=0, a4=1,a3=1,a2=1,a1=0,a0=0 5 V0 = 6 a 5 2 5 + a 4 2 4 + a3 2 3 + a 2 2 2 + a1 21 + a 0 2 0 2 5 V0 = 6 0 × 2 5 + 1 × 2 4 + 1 × 2 3 + 1 × 2 2 + 0 × 2 1 + 0 × 2 0 2 28 V0 = 5 × = 2.1875 V 64

(

)

(

)

(

)

Q.21 Solve the following equations for X (i) 23.610 = X 2 (ii) 65.53510 = X16 Ans: (i) Solve the equation 23.610 = X2 for X 23.610 = X2 In order to find X, convert the Decimal number 23.610 into its Binary form. First take the decimal integer part 23 to convert into its equivalent binary form
2 2 2 2 2 23 11 ------- 1 5 --------1 2 --------1 1 --------0 0 --------1

(6)

Hence 2310 = 101112 Next take the decimal fractional part 0.6 to convert into its equivalent binary form. 37

DE09

DIGITALS ELECTRONICS

Fraction 0.6 0.2 0.4 0.8 0.6 0.2 0.4

Fraction X 2 1.2 0.4 0.8 1.6 1.2 0.4 0.8

Remainder new fraction 0.2 0.4 0.8 0.6 0.2 0.4 0.8

Integer 1 0 0 1 1 0 0

It is seen that it is not possible to get a zero as remainder even after 7 stages. The process can be continued further or an approximation can be made and the process terminated here. The binary equivalent is 0.1001100. Hence 23.610 = 10111.10011002. (ii) In order to find X, convert the Decimal number 65.535 into its equivalent Hexadecimal form. First taking the integer part 65 to convert into its equivalent Hexadecimal form. 16 16 65 4 ---- 1 0 ---- 4

Hence 6510 = 4116 Next take the decimal fractional part 0.6 to convert into its equivalent binary form. Fraction 0.535 0.56 0.96 0.36 0.76 0.16 0.56
Fraction X 16

8.56 8.96 15.36 5.76 12.16 2.56 8.96

Remainder new fraction 0.56 0.96 0.36 0.76 0.16 0.56 0.96

Integer 8 8 15 (F) 5 12(C) 2 8

It is seen that it is not possible to get a zero as remainder even after 7 stages. The process can be continued further or an approximation can be made and the process terminated here. The Hexadecimal equivalent is 0.88F5C28. Hence 65.53510 = 41.88F5C2816.

Q.22 Perform the following additions using 2’s complement (i) -20 to +26 (ii) +25 to -15

(5)

Ans:
38

DE09

DIGITALS ELECTRONICS (i) First convert the two numbers 20 and 26 into its 8-bit binary equivalent and find out the 2’s complement of 20, then add -20 to +26. 20 = 0 0 0 1 0 1 0 0 (8-bit binary equivalent of 20) 20 = 1 1 1 0 1 0 1 1 (1’s complement) +1 ------------------------------20 = -20 = 1 1 1 0 1 1 0 0 (2’s complement of 20) +26 = 0 0 0 1 1 0 1 0 (8-bit binary equivalent of 26) ----------------------------Addition of -20 to +26 = +6 = 0 0 0 0 0 1 1 0 ----------------------------Hence -20 to +26 = (6)10 = (0110)2. (ii) First convert the two numbers 25 and 15 into its 8-bit binary equivalent and find out the 2’s complement of 15, then add +25 to -15. 15 = 0 0 0 0 1 1 1 1 (8-bit binary equivalent of 15) 15 = 1 1 1 1 0 0 0 0 (1’s complement) +1 -----------------------------15 = -15 = 1 1 1 1 0 0 0 1 (2’s complement of 15) +25 = 0 0 0 1 1 0 0 1 (8-bit binary equivalent of 25) -----------------------------Addition of -15 to +25 = +10 = 0 0 0 0 1 0 1 0 ------------------------------Hence -15 to +25 = (10)10 = (1010)2.

Q.23 (i) Convert the decimal number 430 to Excess-3 code: (ii) Convert the binary number 10110 to Gray code:

(6)

Ans: (i) Excess 3 is a digital code obtained by adding 3 to each decimal digit and then converting the result to four bit binary. It is an unweighted code i.e., no weights can be assigned to any of the four digit positions. 4 3 0 +3 +3 +3 ----------------------------7 6 3

0111 0110 0011 (Excess-3 Code) -------------------------------(ii) The rules for changing binary number 10110 into its equivalent Gray code are, the left most bit (MSB) in Gray code i.e., 1 is the same as the left most bit in binary and

39

DE09

DIGITALS ELECTRONICS add the left most bit (1) to the adjacent bit (0) then add the next adjacent pair and discard the carry. Continue this process till completion. + + + +

1

0

1

1

0

1 1 1 0 1 Hence Gray equivalent of Binary number 10110 is 11101. Q.24 Verify that the following operations are commutative but not associative (i) NAND (ii) NOR (6)

Ans: (i) Commutative Law is AB = BA . To verify whether the NAND operation is Commutative or not, prepare truth table shown in Table No.3.1
A 0 0 1 1 B 0 1 0 1 Table No.3.1 AB 1 1 1 0 BA 1 1 1 0

From the Table No.3.1, we observe that the last two columns are identical, which means AB = BA Associative Law is A.( B.C ) = ( A.B ).C To verify whether the NAND operation is Associative or not, prepare truth table shown in Table No.3.2

A 0 0 0 0 1 1 1 1

B 0 0 1 1 0 0 1 1

A.( B.C ) 0 1 1 1 0 1 1 1 0 0 1 0 0 0 1 1 Table No.3.2

C

( A.B ).C 1 0 1 0 1 0 1 1

From the Table No.3.2, we observe that the last two columns are not identical, which means A.( B.C ) ≠ ( A.B ).C

40

DE09

DIGITALS ELECTRONICS (ii) Commutative Law is A + B = B + A . To verify whether the NOR operation is Commutative or not, prepare truth table shown in Table No.3.3
A 0 0 1 1 B A+ B 0 1 1 0 0 0 1 1 Table No.3.3 B+ A 1 0 0 1

From the Table No.3.3, we observe that the last two columns are identical, which means A+ B = B+ A Associative Law is A + ( B + C ) = ( A + B ) + C To verify whether the NOR operation is Associative or not, prepare truth table shown in Table No.3.4

A + ( B + C ) ( A + B) + C 0 0 0 0 0 0 0 1 1 0 0 1 0 1 1 0 1 1 1 0 1 0 0 0 1 1 0 1 0 0 1 1 0 0 1 1 1 1 0 0 Table No.3.4 From the Table No.3.4, we observe that the last two columns are not identical, which means A.+ ( B + C ) ≠ ( A + B ) + C

A

B

C

Q.25 Prove the following equations using the Boolean algebraic theorems: (i) A + A .B + A . B = A + B (ii) A BC + A B C + AB C + ABC = AB + BC + AC Ans: (i)

(5)

Given equation is A + A .B + A. B = A + B L.H.S. = A + A .B + A. B = (A + A. B ) + A .B = A (1+ B ) + A .B = A + A .B (Q 1+ B =1) = (A + A ) (A + B) = (A + B) (Q A + A = 1) = R.H.S Hence Proved 41

DE09

DIGITALS ELECTRONICS (ii) Given equation is A BC + A B C + AB C + ABC = AB + BC + AC L.H.S = A BC + A B C + AB C + ABC = A BC + A B C + AB C + ABC = A BC + A B C + AB (C + C ) = A BC + A B C + AB (Q C + C = 1) = A BC + A (B + B C) = A BC + A (B + C) (Q B + B C = B + C) = A BC + A B + AC = C (A + A B) + A B + AC = C (A + B) + A B + AC (Q A + A B = A + B) = AC + BC + AB + AC = AB + BC + AC (Q AC + AC = AC) = R.H.S Hence Proved

Q.26 A staircase light is controlled by two switches one at the top of the stairs and another at the bottom of stairs (5) (i) Make a truth table for this system. (ii) Write the logic equation is SOP form. (iii) Realize the circuit using AND-OR gates. Ans: A staircase light is controlled by two switches S1 and S2, one at the top of the stairs and another at the bottom of the stairs. The circuit diagram of the system is shown in fig.4(a).

0 S 1

1

0 S

1
2

L

BU LB SU PPLY

ON = 1 OFF = 0

Fig.4(a) Circuit diagram (i)
The truth table for the system is given in truth table 4.1

S1 S2 L 0 0 0 0 1 1 1 0 1 1 1 0 Table 4.1 (ii) The logic equation for the system is given by L = S1 S2 + S1 S 2 42

DE09

DIGITALS ELECTRONICS

(iii) Realization of the circuit using AND-OR gates is shown in fig 4(b)
_ S . S 2 1 1 2

S S

L

_ S . S 2 1

Fig.4(b) Logic Diagram for the system Q.27 Minimize the following logic function using K-maps and realize using NAND and NOR gates. F( A, B, C, D) = ∑ m(1,3,5,8,9,11,15) + d( 2,13) (9) Ans: Minimization of the logic function F(A, B, C, D) = ∑ m(1,3,5,8,9,11.15) + d(2,13) using Kmaps and Realization using NAND and NOR Gates (i) Karnaugh Map for the logic function is given in table 4.1

The minimized logic expression in SOP form is F = A B C + C D + B D + AD The minimized logic expression in POS form is F = (A + B + C ) ( C +D) ( B +D) (A+D)

(ii) Realization of the expression using NAND gates:
The minimized logic expression in SOP form is F = A B C + C D + B D + AD and the logic diagram for the simplified expression is given in fig.4(c)

43

DE09
A B C D _ A+B+C

DIGITALS ELECTRONICS

_ C+D

F
_ B+D

_ _ A+D

Fig.4(c) Logic Diagram (iii) Realization of the expression using NOR gates:
The minimized logic expression in POS form is F = (A + B + C ) ( C +D) ( B +D) (A+D) and the logic diagram for the simplified expression is given in fig.4(d)
A B C D ________ _ _ A+B+C

____ _ C+D _____ _ B+D

F

____ A+D

Fig.4(d) Logic Diagram Q.28 Design a 4 to 1 Multiplexer by using the three variable function given by F( A, B, C) = ∑ m(1,3,5,6) (7) Ans: Design of 4 to 1 Multiplexer by using the three variable function given by F(A,B,C) = ∑ m(1,3,5,6) The function F(A,B,C) = ∑ m(1,3,5,6) can be implemented with a 4-to-1 multiplexer as shown in Fig.7(a). Two of the variables, B and C are applied to the selection lines in that order, i.e., B is connected to S1 and C to S0. The inputs of the multiplexer are 0, I, A, and A'. When BC = 00, output F = 0 since I0 = 0. Therefore, both minterms m0 = A' B' C' and m4 = A B' C' produce a 0 output, since the output is 0 when BC = 00 regardless of the value of A. When BC = 01, output F = 1, since I1 = 1. Therefore, both minterms m1 =A' B'C and

44

DE09

DIGITALS ELECTRONICS m5 = AB'C produce a 1 output, since the output is 1. when BC = 01 regardless of the value of A. When BC = 10, input I2 is selected. Since A is connected to this input, the output will be equal to 1 only for minterm m6 = ABC', but not for minterm m2 = A' BC', because when A' = I, then A = 0, and since I2 = 0, we have F = 0. Finally, when BC = 11, input I3 is selected. Since A' is connected to this input, the output will be equal to 1 only for minterm m3 = A' BC, but not for m7 = ABC. This is given in the Truth Table shown in Table No 7.1

Minterm 0 1 2 3 4 5 6 7

A 0 0 0 0 1 1 1 1

B 0 0 1 1 0 0 1 1

C 0 1 0 1 0 1 0 1

F 0 1 0 1 0 1 1 0

Table 7.1 Truth Table

Fig.7(a) Implementation Table

45

DE09

DIGITALS ELECTRONICS

0 1 A
A'

I I I I

0

1 2 3

4 x1 MUX

Y

F

S

1

S0

B C

Fig.7(b) Logic Diagram of 4X1 Multiplexer Q.29 Find the conversion time of a Successive Approximation A/D converter which uses a 2 MHz clock and a 5-bit binary ladder containing 8V reference. What is the Conversion Rate? (4) Ans: Given data: Frequency of the clock (F) = 2 MHZ Number of bits (n) = 5 n 5 (i) Conversion Time (T) = = = 2.5 µ sec clockrate 2 X 10 6 1 1 = (ii) Conversion Rate = = 400,000 conversions/sec T 2.5 X 10 −6 Q.30 A 6-bit R-2R ladder D/A converter has a reference voltage of 6.5V. It meets standard linearity. Find (i) The Resolution in Percent. (ii) The output voltage for the word 011100. (4) Ans: Given Data Number of Bits (n) = 6 Reference Voltage (VR) = 6.5 V For R-2R Ladder D/A Converter,
1 1 1 = 6 = = 1.59 % 2 − 1 2 − 1 63 (ii)The Output Voltage (VO ) of 6-bit R-2R Ladder D/A Converter for the word 011100 is given by (i)The Resolution in Percent is given by n VO =

VR a n−1 .2 n −1 + a n − 2 .2 n − 2 + − − − − + a1 21 + a 0 2 0 n 2

[

]

46

DE09
VO =

DIGITALS ELECTRONICS
6 .5 0.2 6−1 + 1X 2 5−1 + 1X 2 4−1 + 1X 2 3−1 + 0 X 2 2−1 + 0 X 2 0 6 2 6 .5 4 VO = 2 + 23 + 2 2 26 VO = 2.84 V.

[

]

[

]

Q.31

Convert 2222 in Hexadecimal number.
Ans:

(4)

16 16

2222 138 8 0

14 10 8

=8AE

Q.32

Subtract –27 from 68 using 2’s complements.
Ans: 68-(-27)=68-(-27)using 2’s complement 2’s complement representation of 68=01000100(64+4) 2’s complement representation of - (-27) = 00011011 =+ 27 11100101 =-27 in 2’s complement Now add 68 and 27 68 01000100 -(-27) +00011011 95 010111111

(6)

Which is equal to +95
Q.33

Divide (101110 )2 by (101)2 .
Ans: 101

(4)

101110 101 000110 101 001 Quotient -1001 Remainder -001

1001

Q.34

Prove the following identities using Boolean algebra: (i) (A + B) A + AB C + A B + C + AB + ABC = C(A + B) + A B + C .

(

)

(

)

(

)

(ii) A(A ⋅ B) ⋅ B(A ⋅ B) = A ⊕ B .
47

DE09

DIGITALS ELECTRONICS

(iii) AB + A + AB = 0 .
Ans: (i) (A+B)(A+A’B’)C+A’(B+C’)+A’B+ABC =C(A+B)+A’(B+C’) LHS (A+B)(A+A’+B’)C+A’B+A’C’+A’B+ABC = (A+B)(1+B’)C+A’B+ A’C’+ABC as (A+A’=1) = (A+B).1.C+A’B+ A’C’+ABC = AB+AC+A’B+ A’C’+ABC = ABC+AB+ABC+AC+A’B+A’C’ AB(C+1)+AC(B+1) +A’B+ A’C’ = AB+AC+A’B+ A’C’ = C(A+B) + A’(B+C’) = RHS Hence Proved (ii) A( A.B).B( A.B) = A ⊕ B

(9)

Let us take X = A( A.B) Y = B ( A.B) So we have X .Y = A ⊕ B Also X = A( A.B )
= A( A′ + B′) -------3

(iii)

By using DeMorgan’s Law (AB)’=A’+B’ X = (A(A’+B’))’=(AA’+AB’)’=(AB’)’=(A’+B) ------1 Now Y = (B(AB)’)’=[B(A’+B’)]’= [A’B+BB’]’=(A’B)’=(A+B’) ------2 Now Combining X & Y from 1 & 2 above, we have L.H.S in 3 as : ((A+B’)(A’+B))’ =[AA’+BB+A’B’+AB]’ =(AB+A’B’)’ =A XOR B = RHS Hence Proved ((AB)’+A’+AB)’=0 LHS ( AB + A + AB ) = (1 + A′)′ since AB + AB = 1 = 1′ since 1 + A′ = 1 = 0 = RHS Hence Proved

Q.35

A combinational circuit has 3 inputs A, B, C and output F. F is true for following input combinations A is False, B is True A is False, C is True A, B, C are False A, B, C are True 48

DE09

DIGITALS ELECTRONICS

(i) (ii) (iii) (iv)
Ans: (i)

Write the Truth table for F. Use the convention True=1 and False = 0. Write the simplified expression for F in SOP form. Write the simplified expression for F in POS form. Draw logic circuit using minimum number of 2-input NAND gates. (7)

Making the truth table A B C 0 0 0 0 0 1 0 1 0 0 1 1 1 0 0 1 0 1 1 1 0 1 1 1 A is false b is true For both value of c F is true. (ii) Simplified expression for F can be found by K-map

F 1 1 1 1 0 0 0 1

In SOP Form F = A’+BC (iii) Simplified expression for F in POS form I. In POS Form MINIMIZE ZEROS F’=AB’+AC’ II. F=A’+BC taking complement twice F’=( A’+BC)’=(A.(BC)’) F”=F=(A.(BC)’)’ (iv) Logic circuit by using minimum number of 2-input NAND gates

Q.36

Minimise the logic function

F (A, B, C, D ) = Π M (1, 2, 3, 8, 9, 10, 11,14) ⋅ d (7, 15) Use Karnaugh map. Draw the logic circuit for the simplified function using NOR gates only. (7)

49

DE09 Ans:

DIGITALS ELECTRONICS

F=∏M(1,2,3,8,9,10,11,14).d(7, 15)

F’=B’D+B’C+AC+AB’ By Complementing F F=(B’D+B’C+AC+AB’)’ = [(B’D)’(B’C)’(AC)’(AB’)’]’ = (B+D’)(B+C’)(A’+C’)(A’+B) Taking complement twice and without opening the bracket F=[(B+D’)+(B+C’)’(‘A’+C’)+(A’+B)]’ The logic circuit for the simplified function using NOR gates

Q.37

The capacity of 2K × 16 PROM is to be expanded to 16 K × 16. Find the number of PROM chips required and the number of address lines in the expanded memory. (4)
Ans: Required capacity =16k x 16 Available chip (PROM) =2k x 16 The no of chip =16k x 16 =8 2k x 16 In the chip total word capacity = 2 x 210

50

DE09

DIGITALS ELECTRONICS

Thus the address line required for the single chip = 11 In the expanded memory the word capacity 16k = 214 Now the address lines required are 14. Among then 11 will be common and 3 will be connected to 3 x8 decoder.
Q.38

Perform following subtraction (i) 11001-10110 using 1’s complement (ii) 11011-11001 using 2’s complement
Ans: (i ) 11001 - 10110 1' s Compliment of 10110 = 01001 11001 + 01001 -----------------100010 Add 1 and ignore carry. Ans is 00011 = 3. (ii) 11011 – 11001 = A – B 2's complement of B = 00111 11011 + 00111 100010 Ignore carry to get answer as 00010 = 2.

(8)

Q.39

Reduce the following equation using k-map Y = A BC + ACD + A B + ABCD + A BC
Ans:
Y = A B C + AC D + A B + A BC Y = A BC D + A BC D + A BC D + ABC D + A BCD + A BC D

(8)

+ A BC D + A BC D + ABC D + A BC D + A BCD

Q.40 Write the expression for Boolean function

51

DE09

DIGITALS ELECTRONICS

F (A, B, C) = ∑m (1,4,5,6,7) in standard POS form.
Ans:

(8)

f ( A, B,C ) = ΣM (1,4,5,6,7) in standard POS form F = m1 + m4 + m5 + m6 + m7 F = Σm(1,4,5,6,7) = ∏ M(0,2,3) = M0 M2 M3 = (A+B+C)(A+ B +C)(A+ B + C )
Q.41

Design a 32:1 multiplexer using two 16:1 multiplexers and a 2:1 multiplexer.
Ans:

(8)

To design a 32 X 1 MUX using s3 s2 s1 s0 I0 I15 s3 16 X1 MUX s2 s1 s0 2X1 MUX f

I16 I31

16X1 MUX Select line M

Two 16 X 1 MUX & one 2 X 1 There are total 32 input lines and one O/P line. The 2 X 1 MUX will transmit one of the two I/P to output depending upon its select line M. For M = 0 upper MUX ( I0 – I15 ) will be selected and M = 1 lower MUX ( I16 – I31 ) will be selected.
Q.42

Implement the following function using a 3 line to 8 line decoder. S (A,B,C) = ∑ m(1,2,4,7) C (A,B,C) = ∑ m ( 3,5,6,7)

(8)

Ans: S (A,B,C) = m (1,2,4,7) C (A,B,C) = m (3,5,6,7) These are full adder's output as sum (S) and carry ( C ). We know that 3 to 8 line decoder generates all the minterms from 0 to 7. In the decoder shown in the figure, Do correspond to minterm mo, and so on. So by ORing appropriate outputs of the decoder we can implement these functions.

52

DE09

DIGITALS ELECTRONICS

Q.43

Perform the following operations using the 2’s complement method: (i) 23 – 48 (ii) – 48 – 23
Ans: (i) 23 - 48 add them

(4)

23 - (- 48) 71 (ii) – 48 - 23 = - 48 + (-23) -48 =11010000 -23 =11101001 1 1 0 1 1 1 0 0 1 = -71 Carry is discarded
Q.44

010111 + 010000 100111

Prove the following Boolean identities using the laws of Boolean algebra: (i) (A + B)(A + C ) = A + BC (ii) ABC + A BC + ABC = A (B + C )
Ans: (i) (4)

(A+B)(A+C)=A+BC LHS AA+AC+AB+BC=A+AC+AB+BC OR A((C+1)+A(B+1))+BC OR A+A+BC OR A+BC = RHS Hence Proved (ii) ABC+AB’C+ABC’=A(B + C) LHS AC(B+B’)+AB(C+C’) OR AC+AB OR A(B+C)= RHS Hence Proved 53

DE09 Q.45

DIGITALS ELECTRONICS

The Karnaugh map for a SOP function is given below in Fig.1. Determine the simplified SOP Boolean expression. (5)

Ans:

Q.46

A certain memory has a capacity of 4K × 8 (i) How many data input and data output lines does it have? (ii) How many address lines does it have? (iii) What is its capacity in bytes?

(5)

Ans: (i) available capacity =4Kx8 = 210 x210 x 8 = 212x8 As in the 4Kx8 ,the second number represents the number of bits in each word so the number of data input lines will be 8(also the data output lines) . (ii) It has total 4K (2 12) address line which are required to address 212 locations. (iii) Its capacity in bytes is 4K bytes. Q.47

A 5-bit DAC produces an output voltage of 0.2V for a digital input of 00001. Find the value of the output voltage for an input of 11111. What is the resolution of this DAC? (6)
Ans: For the Digital output of 00001 Output voltage is =0.2 volt =Resolution The output=.2x31=15.5volts Resolution=(0.2volt)/(15.5v)x100=1.290

54

DE09 Q.48

DIGITALS ELECTRONICS

An 8-bit successive approximation ADC has a resolution of 20mV. What will be its digital output for an analog input of 2.17V? (4)
Ans: Resolution =20mv Analog input =2.17v Equivalent value=(2.17)/(2.17)=108.5 Equivalent Binary value=1101100.1

Q.49

A microprocessor uses RAM chips of 1024 × 1 capacity.
(i) (ii)

How many chips will be required and how many address lines will be connected to provide capacity of 1024 bytes. How many chips will be required to obtain a memory of capacity of 16 K bytes. (5)

Ans: ( i ) Available chips = 1024 x 1 capacity Required capacity = 1024 x 8 capacity
No. of Chips= 1024X8 =8 1024X1

Number of address lines are required = 10 (i.e. 1024 = 210 ) As the word capacity is same ( 1024 ) so same address lines will be connected to all chips. ( ii )

No.Of Chips Required =
Q.50

16X1024X8 = 128 1024X1

Find the Boolean expression for logic circuit shown in Fig.1 below and reduce it using (6) Boolean algebra.

Ans: Y = (AB)’ + (A’ + B)’ = A’ + B’ + AB’ Using Demorgan’s Theorem. = A’ + B’(1+A) = A’ + B’ Since 1+A=1 Q.51

Implement the following function using 4-to-1 multiplexer. Y(A, B, C ) = ∑ (2,3,5,6) 55

(8)

DE09 Ans:

DIGITALS ELECTRONICS

Y(A,B,C)=∑(2,3,5,6) Let us take B,C as the select bits and A as input. To decide the input we write. Y = A’BC’+A’BC+AB’C+ABC’ = 0 if B=0, C=0 = A if B=0, C=1 = 1 if B=1, C=0 = A’ if B=1, C=1 The corresponding implementation is shown in the figure. Thus

0 A 1 A’

4x1 MUX

Y

B
Q.52

C
(8)

Design a mod-12 Synchronous up counter.

Ans: Design a mod 12 synchronous counter using D-flipflops. I state table Present state Next state Required D Inputs A B C D A B C D DA DB DC

DD 1 0 1 0 1 0 1 0 1 0 1 0

0 0 0 0 0 0 0 0 1 1 1 1

0 0 0 0 1 1 1 1 0 0 0 0

0 0 1 1 0 0 1 1 0 0 1 1

0 1 0 1 0 1 0 1 0 1 0 1

0 0 0 0 0 0 0 1 1 1 1 0

0 0 0 1 1 1 1 0 0 0 0 1 56

0 1 1 0 0 1 1 0 0 1 1 0

1 0 1 0 1 0 1 0 1 0 1 0

0 0 0 0 0 0 0 1 1 1 1 0

0 0 0 1 1 1 1 0 0 0 0 1

0 1 1 0 0 1 1 0 0 1 1 0

DE09

DIGITALS ELECTRONICS

First draw the state table having present state, next state and required flip-flop input to give the transition. D flip flop gives the output same as the next state itself. Then solve by using K maps to find out DA DB DC DD for all states. Unused states are 1100,1101,1110,1111 they can be treated as don’t care conditions from the table. Draw Karnaugh-maps for DA, DB, DC and DD as follows and obtain Boolean expressions for them.

57

DE09

DIGITALS ELECTRONICS

58

DE09

DIGITALS ELECTRONICS

Logic diagram for mod-12 Synchronous up-counter

Q.53

Find how many bits of ADC are required to get an resolution of 0.5 mV if the maximum full scale voltage is 10 V. (8)
Ans: Resolution=.5mv Full scale output=+10v %resolution =(5mv)/10x100=0.05% No of bits =Log2(2x1000) = 20

Q.54

Convert the decimal number 45678 to its hexadecimal equivalent number.
Ans: (45678)10=(B26E)16

(4)

59

DE09

DIGITALS ELECTRONICS

16

45678

16 16 16

2854 178 11 0

14 6 2 11

E 6 2 B

(45678)10=(B26E)16
Q.55

Write the truth table of NOR gate.
Ans: A 0 0 1 1

(4)

B 0 1 0 1

F 1 0 0 0

Q.56

Design a BCD to excess 3 code converter using minimum number of NAND gates. Hint: use k map techniques. (8)
Ans:

First we make the truth table BCD no ABCD 0000 0001 0010 0011 0100 0101 0110 0111 1000 1001

EXCESS-3 NO WXYZ 0011 0100 0101 0110 0111 1000 1001 1010 1011 1100

Then by using K maps we can have simplified functions for w, x, y, z as shown below:

60

DE09

DIGITALS ELECTRONICS

61

DE09

DIGITALS ELECTRONICS

NAND gate implementation for simplified function

62

DE09

DIGITALS ELECTRONICS

W = BD + AD + AB’ + BC By complementing twice we get W = ((BD + AD + AB’ + BC)’)’ = ((BD)’ . (AD)’ . (AB’)’ . (BC)’)’ X = BC’D + B’D + B’C By complementing twice we get X = BC’D + B’D + B’C = ((BC’D)’ . (B’D)’ . (B’C)’)’ Y = C’D’ + CD = ((C’D’)’ + (CD)’)’ Z = D’

Logic diagram for BCD to excess 3 code converter by using minimum number of NAND gates

Q.57

With the help of a suitable diagram, explain how do you convert a JK flipflop to T type (4) flipflop.

63

DE09

DIGITALS ELECTRONICS Ans: Given flip flop is JK flip flop and it is required to convert JK into T. First we draw the characteristic table of T flip flop and then relate the transition with excitation table of JK flip flop.

Now we solve K maps for J and K by considering T and Q(t) as input.

Logic diagram convert a JK flipflop to T type flipflop. Q.58

A number of 256 x 8 bit memory chips are available. To design a memory organization of 2 K x 8 memory. Identify the requirements of 256 x 8 memory chips and explain the details. (8)
Ans: Chips available=256x8 Required capacity=2048x8 Number of chips=(2048x8)/(256x8)=8=(256=28) Address lines required for 2048x8chip=11(2048=211) Thus the size of the decoder=3x8

64

DE09

DIGITALS ELECTRONICS

Q.59

Convert (177.25)10 to octal.
Ans: (177.25)10 = ( )8 First we take integer part

(8)

8 8 8

177 22 2 0

1 6 2

Thus (177)10 = (261)8

Now as 0.25 x 8 = 2.00 and 0.00 x 8 = 0 Thus (0.25)10 = (0.2)8 Therefore, Thus (177.25)10 = (261.2)8

Q.60

Perform the following subtraction using 1’s complement (i) 11001 – 10110 (ii) 11011 - 11001
Ans:

(8)

(i) 11001 – 10110 = X – Y X = 11001 1’s complement of Y = 01001 Sum = 1 00010 End around carry = 1 So X-Y = 00011 (ii) 11011 – 11001 = X – Y X = 11011 1’s complement of Y = 00110 Sum = 1 00001 End around carry = 1 So X-Y = 00010
Q.61

Prove the following identities (i) A B C + A B C + A B C + A B C = C (ii) A B + A B C + A B + A B C = B + A C
Ans: (i) LHS = A’B’C’ + A’BC’ + AB’C’ + ABC’ = A’C’ (B’ + B) + AC’ (B’ + B) = A’C’ + AC’ [as B’+B = 1] = C’ (A’ + A) = C’ [as A’+A =1]

(8)

65

DE09

DIGITALS ELECTRONICS

= RHS. Hence Proved
(ii) LHS = AB + ABC + A’B + AB’C = B + AC = B (A + A’) + AC (B + B’) = B + AC [as B + B’ = A + A’ = 1] = B + AC = RHS. Hence Proved Q.62

Find the boolean expression for the logic circuit shown below.

(8)

Ans:

Output of Gate-1 (NAND) = (AB)’ Output of Gate-2 (NOR) = (A’+B)’ Output of Gate-3 (NOR) = [(AB)’ + (A’+B)’]’ Now applying De-Morgans law, (X+Y)’ = X’Y’ and (XY)’ = (X’+Y’) [(AB)’ + (A’+B)’]’ = [(AB)’]’ [(A’+B)’]’ = (AB) (A’+B) = AA’B + ABB = ABB = AB.
Q.63

Reduce the following equation using k-map Y = BCD+A B CD+A BCD+A BCD+A BCD
Ans:

(8)

Multiplying the first term by (A+A’) Y = A’BC’D’ + ABC’D’ + A’BC’D + ABC’D + A’BCD + ABCD = ∑ (4,12,5,7,15,13) = BC’ + BD

66

DE09

DIGITALS ELECTRONICS

Q.64

Implement the following function using 8 to 1 multiplexer Y(A, B, C, D ) = ∑ (0,1,2,5,9,11,13,15)

(8)

Ans: We will take three variables B,C & D at selection lines and A as input. Now there are eight inputs and they can be 0,1,A or A’ depending on the Boolean function.

I0 A’ A 0 8 A’ Now, the realization is:

I1 1 9 1

I2 2 10 A’

I3 3 11 A

I4 4 12 0

I5 5 13 1

I6 6 14 0

I7 7 15 A

67

DE09

DIGITALS ELECTRONICS

I0 = A’ I1 = 1 I2 = A’ I3 = A I4 = 0 I5 = 1 I6 = 0 I7 = A

8x1 MUX

Y

B

C

D

Select Lines

Q.65

(i)

How many 128 × 8 RAM chips are required to provide a memory capacity of 2048 bytes. (ii) How many lines of address bus must be used to access 2048 bytes of memory. How many lines of these will be common to each chip? (iii) How many bits must be decoded for chip select? What is the size of decoder? (8) Ans: (i) Available RAM chips = 128 x 8 Required memory capacity = 2048 x 8 Number of chips required = (2048 x 8) / (128 x 8) = 16. (ii) Chips available are of 128 x 8 in size. It means that total 128 (27) locations are there and each location can store 8 bits. Thus the total number of address lines required to access 128 locations is 7. As seven address lines can address 27 locations. These seven lines are common to all chips. Now to access 2048 locations, we require 11 address lines, as 2048 = 211 (iii) These higher order lines will be applied to decoder input. The number of inputs to the decoder will be 11 - 7 = 4. The size of the decoder will be 4x16. These 16 decoder outputs will be connected to the chip select input of individual chips.
D0 To chip select input of chip-1

A10 A9 A8 A7

4 x 16 Decoder
D15 To chip select input of chip-16

A6
~

A0

To all chips

68

DE09 Q.66

DIGITALS ELECTRONICS

How many bits are required at the input of a ladder D/A converter, if it is required to give a resolution of 5mV and if the full scale output is +5V. Find the %age resolution. (8) Ans: First we find out the ratio of Full scale output to Resolution = 5V / 5 mV = 1000. Now number of bits = log2 1000 = 10. Percentage Resolution = 5 mV / 5 V * 100 = 0.1%

Q.67

A 6-bit Dual Slope A/D converter uses a reference of –6V and a 1 MHz clock. It uses a fixed count of 40 (101000). Find Maximum Conversion Time.
Ans The time T1 given by T1 = 2 N TC where N = no. of Bits, Tc = time period of clock pulse Given N = 6, TC = 1/ 1MHz = 1 µs. Therefore T1 = 26 X 10 -6 s = 64 µs.

(4)

Q.68

A 2-digit BCD D/A converter is a weighted resistor type with E R = 1 Volt, with R = 1MΩ , R f = 10KΩ . Find resolution in Percent and Volts. (5)
Ans Resolution = 1/22 = 0.25 volts. As the resolution is determined by number of input bits of D/A converter; For example two bit converter has 22 (4) possible output levels, therefore its resolution is 1 part in 4 In percent it will be ¼ X 100 = 25% In volts, it will be 0.25 volts.

69

DE09 PART – III

DIGITALS ELECTRONICS

DESCRIPTIVES
Q.1 Ans:

Distinguish between min terms and max terms. Distinguish between Minterms and Maxterms:

(6)

(i) Each individual term in standard Sum Of Products form is called as minterm whereas each individual term in standard Product Of Sums form is called maxterm. (ii) The unbarred letter represent 1’s and the barred letter represent 0’s in min terms, whereas the unbarred letter represent 0’s and the barred represent 1’s in maxterms. (iii) If a system has variables A, B, C then the minterms would be in the form ABC, whereas the maxterm would be in the form A+B+C. (iv) The minterm designation for three variable expression be Y=∑m (1, 3, 5, 7) Where the capital ∑ represents the product and m stands for minterms. Decimal number 1 corresponds to binary number 001 or A B C Decimal number 3 corresponds to binary number 011 or A BC Decimal number 5 corresponds to binary number 101 or A B C Decimal number 7 corresponds to binary number 111 or ABC. Whereas the Maxterm designation for three variable expression be Y=∏M (0, 1, 3, 4) Where the capital ∏ represents the product and M stands for maxterms. Decimal 0 means binary 000 and term is A+B+C Decimal 1 means binary 001 and term is A+B+ C Decimal 3 means binary 011 and term is A+ B + C Decimal 4 means binary 100 and term is A +B+C

Q.2

What are universal gates. Construct a logic circuit using NAND gates only for the expression x = A . (B + C). (7)
Ans: Universal Gates: NAND and NOR Gates are known as Universal gates. The AND, OR, NOT gates can be realized using any of these two gates. The entire logic system can be implemented by using any of these two gates. These gates are easier to realize and consume less power than other gates. Construction of a logic circuit for the expression X = A (B + C) using NAND gates is Shown in fig.4 (b)

70

DE09
___ AB

DIGITALS ELECTRONICS

A B

A C

___ AC

_________ ____ ____ (AB) (AC) = AB + AC

Fig.4(b) Logic Diagram for the expression X = A (B + C) Q.3

Mention the various IC logic families.

(7)

Ans: Various IC Logic Families: Digital IC’s are fabricated by employing either the Bipolar or the Unipolar Technologies and are referred to as Bipolar Logic Family or Unipolar Logic Family I Bipolar Logic Families: There are two types of operations in Bipolar Logic Families 1. Saturated Logic Families 2. Non-saturated Logic Families 1. Saturated Logic Families: In Saturated Logic, the transistors in the IC are driven to saturation. (i) Resistor-Transistor Logic (RTL). (ii) Direct-Coupled Transistor Logic (DCTL) (iii) Integrated-Injection Logic (I²L) (iv) Diode -Transistor Logic (DTL) (v) High-Threshold Logic (HTL) (vi)Transistor-Transistor Logic (TTL) 2. Non-saturated Logic: In Non-saturated Logic, the transistors are not driven into saturation. (i) Schottky TTL (ii) Emitter Coupled Logic (ECL) II Unipolar Logic Families: MOS devices are Unipolar devices and only MOSFETs are employed in MOS logic circuits. The MOS logic families are (i) PMOS (ii) NMOS, and (iii) CMOS

while in PMOS only p-channel MOSFETs are used and in NMOS only n-channel MOSFETs are used, in complementary MOS (CMOS), both P and N channel MOSFETs are employed and are fabricated on the same silicon chip.

Q.4

What is a half-adder? Explain a half-adder with the help of truth-table and logic diagram. (10) Ans: Half Adder: A logic circuit for the addition of two one-bit numbers is referred to as an half-adder. The addition process is illustrated in truth table shown in Table 6.1. Here A and B are the two inputs and S (SUM) and C (CARRY) are two outputs. 71

DE09

DIGITALS ELECTRONICS

A 0 0 1 1

B 0 1 0 1

S 0 1 1 0

C 0 0 0 1

Table 6.1 Truth Table for Half Adder

From the truth table, we obtain the logical expressions for S and C outputs as S = A B+A B C = AB The logic diagram for an Half-adder using gates is shown in fig.6(a)
A B S

C

Fig.6(a) Logic Diagram for an Half-adder

Q.5

Using a suitable logic diagram explain the working of a 1-to-16 de multiplexer.(7)
Ans: Working of a 1-to-16 Demultiplexer: A demultiplexer takes in data from one line and directs it to any of its N outputs depending on the status of the selected inputs. If the number of output lines is N (16), the number of select lines m is given by 2m = N.i.e., 24 = 16. So, the number of select lines required for a 1-to-16 demultiplexer is 4. Table 7.1 shows the Truth Table of 1-to-16 Demultiplexer. The input can be sent to any of the 16 outputs, D0 to D15. If DCBA = 0000, the input goes to D0. If DCBA = 0001, the input goes to D1 and so on. Fig.7(a) shows the logic diagram of a 1-to-16 demultiplexer, consists of 8 NOT gates, 16 NAND gates, one data input line(G), 4 select lines (A,B,C,D) and 16 output lines (D0, D1, D2 ------D16). The 8 NOT gates prevent excessive loading of the driving source. One data input line G is implemented with a NOR gate used as negative AND gate. A low level in each input G1 and G 2 is required to make the output G high. The output G of enable is one of the inputs to all the 16 NAND gates. G must be high for the gates to be enabled. If the enable gate is not activated then all sixteen de multiplexer outputs will be high irrespective of the state of the select lines A,B,C,D.

72

DE09

DIGITALS ELECTRONICS

Demultiplexer Input 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

Selection Lines

Logic Function

Demultiplexer Outputs D0 D1 D2 D3 D4 D5 D6 D7 D8 D9 D10 D11 D12 D13 D14 D15 0 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0

DCB A 0 0 0 0 0 0 0 1 0 0 1 0 0 0 1 1 0 1 0 0 0 1 0 1 0 1 1 0 0 1 1 1 1 0 0 0 1 0 0 1 1 0 1 0 1 0 1 1 1 1 0 0 1 1 0 1 1 1 1 0 1 1 1 1

DC B A D C BA DC B A DC BA DCB A DCB A DC B A DC B A
DC B A DC B A DC B A D

CB A

DCB A DCB A DCBA DCBA

1 1 1 1

Table 7.1 Truth Table of 1-to-16 Demultiplexer

73

DE09

DIGITALS ELECTRONICS

G _ A _ B D _ C 0 _ D A

_ _ _ _ (A B C D )

_ _ _ D (A B C D ) 1 B _ _ _ (A B C D ) 2 _ _ (A B C D ) 3 C _ _ _ (A B C D ) 4

1 Data Input

_ G

D

G

_ G2

D

A A
_ A

D

D 5 S E L E C T L I N E S

_ _ (A B C D )

B B
_ B D 7 D 6

_ _ (A B C D )

_ (AB C D)

C C
_ C

D D 8

_ _ _ (AB C D)

D 9

_ _ (A B C D )

D D
_ D D

_ _ (A B C D ) 10

_ C

_ D (A B C D ) 11 _ _ (A B C D ) 12

D

_ B

D 13

_ (A B C D )

_ A

_ D (A B C D ) 14

A

B

C

D (A B C D) 15 D

Fig.7(a) Logic Diagram of 1-to-16 De multiplexer

74

DE09 Q.6 .

DIGITALS ELECTRONICS

With relevant logic diagram and truth table explain the working of a two input EX-OR gate. (7)
Ans: Two-Input EX-OR Gate: An Exclusive-OR (EX-OR) gate recognizes words which have an odd number of ones. Fig.7(b) shows the logic diagram of an EX-OR gate and Fig.7(c) shows the symbol of an EX-OR Gate. The upper AND gate gives an output A B and the lower AND gate gives an output A B .
_ A A _ A B

_ _ Y =A B+ A B _ B _ A B

B

Fig.7(b) Logic Diagram of EX-OR Gate
A B

Y

Fig.7(c) Symbol of EX-OR Gate

Therefore, the output equation becomes Y = A B + A B = A EX-OR B = A ⊕ B If both A and B are low, the output is low. If either A or B (not both) are high (and the other is low), the output is high. If both A and B are high, output is low. Thus the output is 1 when A and B are different. Table 7.2 shows the Truth Table for EX-OR gate.
A B Y(A B+A B)

0 0 1 1

0 1 0 1

0 1 1 0

Table 7.2 Truth Table of EX-OR Gate Q.7

With the help of clocked JK flip flops and waveforms, explain the working of a three bit binary ripple counter. Write truth table for clock transitions.
(14)

Ans: 3-Bit Binary Ripple Counter: In Ripple Counters, all the Flip-Flops are not clocked simultaneously and the flip-flops do not change state exactly at the same time. A 3-bit

75

DE09

DIGITALS ELECTRONICS

Binary Counter has maximum of 23 states i.e., 8 states, which requires 3 Flip-Flops. The word Binary Counter means a counter which counts and produces binary outputs 000,001,010--111.It goes through a binary sequence of 8 different states (i.e, from 0 to 7). Fig.8(a) shows the logic circuit of a 3-bit Binary Ripple Counter consisting of 3 Edge Triggered JK flip-flops. As indicated by small circles at the CLK input of flipflops, the triggering occurs when CLK input gets a negative edge. Q0 is the Least Significant Bit (LSB) and Q2 is the Most Significant Bit (MSB). The flip-flops are connected in series. The Q0 output is connected to CLK terminal of second flip-flop. The Q1 output is connected to CLK terminal of third flip-flop. It is known as a Ripple Counter because the carry moves through the flip-flops like a ripple on water. Working: Initially, CLR is made Low and all flip-flops Reset giving an output Q = 000. When CLR becomes High, the counter is ready to start. As LSB receives its clock pulse, its output changes from 0 to 1 and the total output Q = 001. When second clock pulse arrives, Q0 resets and carries (i.e., Q0 goes from 1 to 0 and, second flip flop will receive CLK input). Now the output is Q = 010. The third CLK pulse changes Q0 to 1 giving a total output Q = 011. The fourth CLK pulse causes Q0 to reset and carry and Q1 also resets and carries giving a total output Q = 100 and the process goes on. The action is shown is Table 8.1.The number of output states of a counter are known as Modulus (or Mod). A Ripple Counter with 3 flip-flops can count from 0 to 7 and is therefore, known as Mod-8 counter. Counter State 0 1 2 3 4 5 6 7 Q2 0 0 0 0 1 1 1 1 Q1 0 0 1 1 0 0 1 1 Q0 0 1 0 1 0 1 0 1

Table.8.1 Counting Sequence of a 3-bit Binary Ripple Counter

76

DE09
HIGH

DIGITALS ELECTRONICS
Q Q Q

0

1

2

J CLOCK PULSES

0

Q
0

J

1

Q

1

J2

Q

2

FF
K 0 Cr

0

FF
K 1 Cr

FF
1 K 2 Cr

2

CLEAR

Fig.8(a) Logic Diagram of 3-Bit Binary Ripple Counter

Ripple counters are simple to fabricate but have the problem that the carry has to propagate through a number of flip flops. The delay times of all the flip flops are added. Therefore, they are very slow for some applications. Another problem is that unwanted pulses occur at the output of gates.

1 CLOCK PULSES 0 1

2

3

4

5

6

7

8

9

10

Q

1 0 0 1

Q

1 0 1

Q

2

0 T ime

Fig.8(b) Timing Diagram of 3-bit Binary Ripple Counter

The timing diagram is shown in Fig.8(b). FF0 is LSB flip flop and FF2 is the MSB flip flop. Since FF0 receives each clock pulse, Q0 toggles once per negative clock edge as shown in Fig. 8(b).The remaining flip flops toggle less often because they receive negative clock edge from preceding flip flops. When Q0 goes from 1 to 0, FF1 receives a negative edge and toggles. Similarly, when Q1 changes from 1 to 0, FF2 receives a negative edge and toggles. Finally when Q2 changes from 1 to 0, FF3 receives a negative edge and toggles. Thus whenever a flip flop resets to 0, the next higher flip flop toggles. This counter is known as ripple counter because the 8th clock pulse is applied, the trailing edge of 8th pulse causes a transition in each flip flop. Q0 goes from High to Low, this causes Q1 go from High to Low which causes Q2 to go from High to Low which causes Q3

77

DE09

DIGITALS ELECTRONICS

to go from High to Low. Thus the effect ripples through the counter. It is the delay caused by this ripple which result in a limitation on the maximum frequency of the input signal.

Q.8 Using D-Flip flops and waveforms explain the working of a 4-bit SISO shift register.

(14)

Ans: Serial In - Serial Out Shift Register: Fig.9(a) shows a 4 bit serial in - serial out shift register consisting of four D flip flops FF0 , FF1 , FF2 and FF3. As shown it is a positive edge triggered device. The working of this register for the data 1010 is given in the following steps.
DATA IN D
0

Q
0

D

1

Q
1

D

2

Q
2

D3

Q
3

FF

0

FF

FF

1

2

FF

3 _ Q3

CLOCK

Fig.9(a) Logic Diagram of 4-bit Serial In – Serial Out Shift Register
1 C LK 0 1 0 2 3 4

D ATA

1

Q

0

Q
1

Q
2

Q
3

Fig.9(b) Output Waveforms of 4-bit Serial-in Serial-out Register

1. 2. 3. 4. 5.

Bit 0 is entered into data input line. D0 = 0, first clock pulse is applied, FF0 is reset and stores 0. Next bit 1 is entered. Q0 = 0, since Q0 is connected to D1, D1 becomes 0. Second clock pulse is applied, the 1 on the input line is shifted into FF0 because FF0 sets. The 0 which was stored in FF0 is shifted into FF1. Next bit 0 is entered and third clock pulse applied. 0 is entered into FF0, 1 stored in FF0 is shifted to FF1 and 0 stored in FF1 is shifted to FF2. Last bit 1 is entered and 4th clock pulse applied. 1 is entered into FF0, 0 stored in FF0 is shifted to FF1, 1 stored in FF1 is shifted to FF2 and 0 stored in FF2 is shifted to FF3. 78

DE09

DIGITALS ELECTRONICS

6. 7. 8. 9. 10.

This completes the serial entry of 4 bit data into the register. Now the LSB 0 is on the output Q3. Clock pulse 5 is applied. LSB 0 is shifted out. The next bit 1 appears on Q3 output. Clock pulse 6 is applied. The 1 on Q3 is shifted out and 0 appears on Q3 output. Clock pulse 7 is applied. 0 on Q3 is shifted out. Now 1 appears on Q3 output. Clock pulse 8 is applied. 1 on Q3 is shifted out. When the bits are being shifted out (on CLK pulse 5 to 8) more data bits can be entered in.
(14)

Q.9

With the help of R-2R binary ladder, explain the working of a 4-bit D/A converter

Ans: R-2R Ladder network method: In a R-2R ladder network method of digital to analog conversion, irrespective of number of bits of the DAC only two convenient values of resistors are needed in the ratio of 1:2 as depicted in fig 10(a). An R-2R Ladder Network based on constant reference current. In the circuit of fig 10(a) points G are actual ground and points G' are virtual ground. Therefore the potential at all the Gs and G’s is zero. Between ground (actual or virtual) and node A there are two resistors each of value 2R in parallel. Therefore this resultant resistance between ground and node A is R and the current through each of the 2R resistance connected to node A must be same. Let us say this current is I. Then the current flowing from A to B through the resistor R is 2I. Then the total resistance from ground to node B through the node A becomes 2R. Also the resistance directly connected between ground and B is also 2R. So between the node B and ground there are two equal resistances in parallel each of value 2R. Therefore, the resultant resistance is R and the current approaching to node B from both sides must be equal. Since current approaching from the side of node A is 2I, therefore the current approaching to the node B from the resistor 2R under it must also be 2I. Hence the total current approaching the node C from the side of node B is 4I. On the basis of the same logic the current approaching to node D from the side of node C must be 8I and the current approaching it form the 2R resistor under node D should also be 8I.
D I ref R 8I

C

R 4I

B

R 2I

A I

= 16 I

I ref 2R

8I

2R

4I

2R

2I

2R

I

2R

-V G'

D

3

D G G'

D
2

1

D G

0

G G'

G G' I out Virtual Ground Line G ' G

R

f

V o +

Fig.10(a) R-2R Ladder Network D/A Converter

79

DE09

DIGITALS ELECTRONICS

Whenever any of the bit or bits of the digital input word D3 D2 D1D0 is high, the corresponding transistor switch is ON i.e. connected to virtual ground and the current of that vertical branch of the ladder comes from the output, otherwise the current of the vertical branch comes directly from the actual ground without any effect on the output. Hence the output current (lout) gives the analog current value corresponding to the digital input word. This analog current gets converted to the analog voltage Vo. An R-2R 4-Bit Ladder Network DAC based on reference voltage: An R-2R 4-bit Ladder Network D/A Converter is shown in fig. 10(b)
V LOW D
0 ref

H IGH D
1

D

2

D

3

R 2R 2R 2R 2R

f

2R A

R B

R C

R D

R

V

+

o

Fig.10(b) R-2R 4-bit Ladder Network D/A Converter Proof: Step 1: If the digital value to be converted to analog value is 0001 i.e.D0 is on the high side connected to Vref while D1, D2, and D3 are connected to ground. Then the circuit redrawn as shown inFig.10(c).
Rf X 2R V ref 1

X B

2

X R C

3

X R D

4

A

R

R

V o +
2R 2R 2R 2R

X'

1

X'

2

X'

3

X'

4

Fig.10(c) R-2R Ladder Network D/A Converter when D0 is connected to Vref and D1,D2,D3 are connected to ground

Applying Thevenin’s theorem at X1,X1’ , the circuit of fig.10(c) becomes the equivalent circuit shown in fig.10(d)

80

DE09

DIGITALS ELECTRONICS
R X X1 R B

f

2

R

C

R

D

R

V R
+ o

+
2R 2R 2R

V ref /2

X'

1

X'

2

Fig.10(d) R-2R Ladder Network D/A Converter when Thevenin’s Theorem applied at X1 and X1’ Again Applying Thevenin’s Theorem at X2,X2’, then the circuit of fig.10(d) becomes the equivalent circuit shown in fig.10(e).
Rf X X2 R C

3

R

D

R

V R
+ o

+
2R 2R

V ref /4

X'

2

X'

3

Fig.10(e) R-2R Ladder Network D/A Converter when Thevenin’s Theorem applied at X2 and X2’ Again Applying Thevenin’s Theorem at X3,X3’ the circuit of fig.10(e) becomes the equivalent circuit shown in fig.10(f):
R X X3 R D
4 f

R

V R
+ o

+
2R

V ref /8

X'

3

X'

4

Fig.10(f) R-2R Ladder Network D/A Converter when Thevenin’s Theorem applied at X3 and X3’

81

DE09

DIGITALS ELECTRONICS

Once Again applying Thevenin’s theorem at section X4, X4’ the circuit of fig.10(f) finally becomes the equivalent circuit shown in fig.10(g).
Rf Rf R

V R
+ o

2R V ref / 16

V o +

+
V
ref

/ 16

Fig.10(g) R-2R Ladder Network D/A Converter when Thevenin’s Theorem applied at X4 and X4’ Step 2: If D1 is high (connected to Vref) and D0, D2, D3 are all low (connected to ground), then the circuit becomes:
R X 2R V re f 1 f

X R C

2

X R D

3

B

R

V o +
2R 2R 2R

X'

1

X'

X'

2

3

Fig.10(h) R-2R Ladder Network D/A Converter when D1 is connected to Vref and D0,D2,D3 are connected to ground Applying Thevenin’s Theorem, three times and reducing the circuit each time at sections X1, X2, X3 we finally get the circuit as shown in fig.10(i).
R
f

2R V ref /8

V o +

Fig.10(i) Equivalent Circuit when D1 is connected to Vref D0,D2,D3 are connected to ground

82

DE09

DIGITALS ELECTRONICS

Step 3: Repeating the same exercise of D2 High and other bits Low, we get the finally reduced Circuit as shown in fig.10(j).
R
f

2R V ref /4

V

+

o

Fig.10(j) Equivalent Circuit when D2 is connected to Vref D0,D1,D3 are connected to ground Step 4: Repeating the same for D3 High and other bits Low, we can reduce the circuit shown in fig.10(k).
R
f

2R V ref /2

V

+

o

Fig.10(k) Equivalent Circuit when D3 is connected to Vref D0,D1,D2 are connected to ground Step 5: Compiling the reduced circuits of the above four steps by applying Superposition Theorem, then the network of fig 10(g),10(i),10(j),10(k) becomes the equivalent circuit shown in fig.10(m).
2R V ref R

/ 16 /8 /4

f

D 0 D D D
1 2

2R 2R 2R

V

ref

V

V ref +

o

V

ref

/2

3

Fig.10(m) Equivalent circuit by applying Superposition Theorem for the circuits of fig.10(g),10(i),10(j),10(k)

83

DE09

DIGITALS ELECTRONICS

Hence the derived equivalent circuit of the R-2R ladder network proves that the bits of the input digital word D3, D2, D1, D0 receive the applied voltages as per their binary weights and we get the corresponding analog value at Vo. Therefore,
D D D  D .R f  3 + 2 + 1 + 0  2R 4 8 16   2 Vref D D  D VO = .R f  n−1 + n − 2 + ..... + n0  2R 4 2   2 VO = Vref

If Rf is also selected equal to 2R, then D D  D VO = Vref . n −1 + n− 2 + ..... + n0  4 2   2 VO is independent of the numerical values of R. Thus any convenient value of R & 2R can be taken for the design of the D/A converter. The maximum output analog voltage is nearly equal to Vref. The actual values of R-2R resistors influence only the maximum current handled by the op-amp. Voltage resolution of n-bit ladder network DAC is Vref/2n
Q.10

With relevant diagram explain the working of master-slave JK flip flop.

(9)

Ans: Master-Slave J-K FLIP-FLOP: A master-slave J-K FLIP-FLOP is a cascade of two SR FLIP-FLOPS. One of them is known as Master and the other one is slave. Fig.11(a) shows the logic circuit. The master is positively clocked. Due to the presence of inverter, the slave is negatively clocked. This means that when clock is high, the master is active and the slave is inactive. When the clock is low, the master is inactive and the slave is active. Fig.11(b) shows the symbol. This is a level clocked Flip-Flop. When clock is high, any changes in J and K inputs can affect S and R outputs. Therefore, J and K are kept constant during positive half of clock. When clock is low, the master is inactive and J and K inputs can be allowed to be changed. The different conditions are Set, Reset, and Toggle. The race condition is avoided because of feedback from slave to master and the slave being inactive during positive half of clock. (i) SET State: Assume that Q is low (and Q is high). For high J, low K and high CLK, the Master goes to SET state giving High S and Low R. Since Slave is inactive, Q and Q do not change. When CLK becomes Low, the Slave becomes to

Set state giving High Q (and low Q ).
(ii) RESET State: At the end of Set State Q is High (and Q low). Now if J is low, K is

high and CLK is high, the Master Resets giving Low S and High R. Q and Q do not change because Slave is inactive. When CLK becomes Low, the Slave becomes active and resets giving Low Q (and High Q ). 84

DE09

DIGITALS ELECTRONICS

(iii) Toggle State: If both J and K are High, the Slave copies the Master. When CLK is High, the Master toggles once. Then the Slave toggles once when CLK is low. If the Master toggles into Set state, the slave copies the Master and toggles into Set state. If the Master toggles into Reset state, the slave again copies the Master and toggles into Reset state. Since the second FLIP-FLOP simply follows the first one, it is referred to as the slave and the first one as the master. Hence, this configuration is referred to as masterslave(M-S) FLIP-FLOP. Truth Table of JK Master Slave Flip-Flop in Table 11.1 shows that a Low PR and Low CLR can cause race condition. Therefore, PR and CLR are kept High when inactive. To clear, we make CLR Low and to preset we make PR Low. In both cases we change them to High when the system is to be run. Low J and Low K produce inactive state irrespective of clock input. If K goes High, the next clock pulse resets the Flip-Flop. If J goes High by itself, the next clock pulse sets the Flip-Flop. When both J and K are High, each clock pulse produces one toggle.

PR S J Q C LK

MASTER

SLAVE
_ Q

K C LR

R

___ CLK

Fig.11(a) Logic Diagram of Master-Slave J-K FLIP-FLOP

Pr

J M-S J-K FF

Q

C LK

K

_ Q

Cr

Fig.11(b) Logic Symbol of Master-Slave J-K FLIP-FLOP

85

DE09

DIGITALS ELECTRONICS

Inputs

Output

PR 0 0 1 1 1

CLR 0 1 0 1 1

CLK X X X X

J X X X 0 0

K X X X 0 1

Q Race Condition 1 0 No change 0

1 1

1 1

1 1

0 1

1 Toggle

Table 11.1 Truth Table of JK Master-Slave Flip-Flop Q.11

Compare the memory devices RAM and ROM.

(5)

Ans: Comparison of Semi-conductor Memories ROM and RAM The advantages of ROM are: 1. It is cheaper than RAM. 2. It is non-volatile. Therefore, the contents are not lost when power is switched off. 3.It is available in larger sizes than RAM. ' 4. It's contents are always known and can be easily tested. 5. It does not require refreshing. 6. There is no chance of any accidental change in its contents. The advantages of RAM are: 1. It can be updated and replaced. 2. It can serve as temporary data storage.

3. It does not require lead time (as in ROM) or programming time (as in PROM). 4. It does not require any programming equipment
Q.12

State and prove Demorgan’s laws.

(5)

Ans: De Morgan's Theorems: (i) Statement of First Theorem: A + B = A . B Proof: The two sides of the equation A + B = A B is represented by logic diagrams shown in fig.3 (a) & 3(b)

86

DE09

DIGITALS ELECTRONICS
_ A _ _ A .B

A

A B

A +B

_____ A+B

_ B B

Fig.3(a) Logic diagram for A + B

Fig.3(b) Logic diagram for A B

The equality of the logic diagrams of fig.3 (a) & 3(b) is proved by the truth table shown in table 2(c)
Inputs A 0 0 1 1 Intermediate Values A+B A 0 1 1 1 1 0 1 0 Table 2(c) (ii) Statement of second theorem: AB = A + B Proof: The two sides of the equation AB = A + B is represented by the logic diagrams shown in fig.3(c) & 3(d).
_ A A _ _ A +B

B 0 1 0 1

B 1 0 1 0

Outputs A+ B 1 0 0 0

A B 1 0 0 0

A B

A .B

____ A .B

_ B B

Fig.3(c) Logic diagram for AB

Fig.3(d) Logic diagram for A + B

The equality of the logic diagrams of fig.3(c) & 3(d) is proved by the truth table shown in table 2(d)
Inputs A B 0 0 0 1 1 0 1 1 Intermediate Values A .B A B 0 1 1 0 1 0 0 0 1 1 0 0 Table 2(d) Outputs A + B A.B 1 1 1 1 1 1 0 0

Q.13

Discuss in detail, the working of Full Adder logic circuit and extend your discussion to explain a binary adder, which can be used to add two binary numbers. (14) 87

DE09

DIGITALS ELECTRONICS Ans: Full-Adder: A half-adder has only two inputs and there is no provision to add a carry from the lower order bits when multibit addition is performed. For this purpose, a third input terminal is added and this circuit is used to add An, Bn, and Cn-1, where An and Bn are the nth order bits of the numbers, A and B respectively and Cn-1 is the carry generated from the addition of (n-1)th order bits. This circuit is referred to as fulladder and its truth table is given in Table 5.1

Inputs An 0 0 0 0 1 1 1 1 Bn 0 0 1 1 0 0 1 1 Cn-1 0 1 0 1 0 1 0 1 Sn 0 1 1 0 1 0 0 1

Outputs Cn 0 0 0 1 0 1 1 1

Table 5.1 Truth Table of a Full-Adder

The K-maps for the outputs Sn and Cn are given in Fig.5(a) and Fig.5(b) respectively and the minimized expressions are given by Sn = An Bn C n −1 + An Bn Cn-1 + An Bn C n −1 + An Bn Cn-1 Cn = An Bn + Bn Cn-1 + An Cn-1
An Bn An Bn

An Bn

AnAnn Bn B 1

____ Cn-1 Cn-1 1

1 1
Fig.5(a) K-map for Sn

Fig.5(b) K-map for Cn

88

DE09

DIGITALS ELECTRONICS

The logic diagrams for the Sn and Cn are shown in fig.5(c) & fig.5(d).
A n Bn C n-1

Sn

Fig.5(c) NAND-NAND Realization of Sn
A n B n Bn C n-1 Cn

A n C

n-1

Fig.5(d) NAND-NAND Realization of Cn Binary Adder: The full adder forms the sum of two bits and a previous carry. Two binary numbers of n bits each can be added by means of Binary Adder. If A = 1011 and B = 0011, whose sum is S = 1110. When pair of bits is added through a full-adder, the circuit produces a carry to be used with the pair of bits one significant position higher. This is shown in Table 5.2 The bits are added with full-adders, starting from the Least Significant Position (subscript 1), to form the sum bit and carry bit. The input carry C1 in the Least Significant position must be 0. The value of Ci+1 in a given significant position is the output carry of the full-adder. This value is transferred into the input carry of the full-adder that adds the bits one higher significant position to the left. The sum bits are thus generated starting form the rightmost position and are available as soon as the corresponding previous carry bit is generated

Subscript i 4 3 2 1 Full-Adder Input Carry 0 1 1 0 Ci Z Augend 1 0 1 1 Ai X Addend 0 0 1 1 Bi Y Sum 1 1 1 0 Si S Output Carry 0 0 1 1 Ci+1 C Table 5.2 Truth Table for Binary Adder 89

DE09

DIGITALS ELECTRONICS

A Binary Parallel Adder is a digital function that produces the arithmetic sum of two binary numbers in parallel. It consists of full-adders connected in cascade, with the output carry from one full-adder connected to the input carry of the next full-adder. Fig.5(e) shows a 4-bit Binary Parallel Adder. The augend bits of A and the addend bits of the B are designated by subscript numbers from right to left, with subscript 1 denoting the low-order bit. The carries are connected in a chain through the full-adders. The input carry to the adder is C1 and the output carry is C5. The S outputs generate the required sum bits.
B
4

A

4

B

3

A

3

B

2

A

2

B

1

A

1

C
5

FU LL AD DER

C

4

FU LL AD DER

C3

FU LL AD DER

C2

FU LL AD DER

C

1

S

4

S

3

S

2

S

1

Fig.5(e) 4-bit Binary Parallel Adder using Full-Adders Q.14

What is a decoder? Draw the logic circuit of a 3 line to 8 line decoder and explain its working. (7)
Ans: Decoder: A Decoder is a combinational logic circuit that converts Binary words into alphanumeric characters. Thus the inputs to a decoder are the bits 1, 0 and their combinations. The output is the corresponding decimal number. It converts binary information from n input lines to a maximum of 2n unique output lines. If the n-bit decoded information has unused or don't-care combinations, the decoder output will have less than 2n outputs. Working: The logic circuit of a 3 line to 8 line decoder is shown in fig.6 (a). The three inputs (x, y, z) are decoded into eight outputs (from D0 to D7), each output representing one of the minterms of the 3-input variables. The three inverters provide the complement of the inputs, and each one of the eight AND gates generate one of the minterms. A particular application of this decoder is a binary-to-octal conversion. The input variables may represent a binary number, and the outputs will then represent the eight digits in the octal number system. However, a 3-to-8 line decoder can be used for decoding any 3-bit code to provide eight outputs, one for each element of the code. The operation of the decoder may be verified from its input-output relationships shown in Table 6.1.The table shows that the output variables are mutually exclusive because only one output can be equal to 1 at any one time. Consider the case when X=0, Y=0 and Z=0, the output line D0 (X’, Y’, Z’) is equal to 1 represents the minterm equivalent of the binary number presently available in the input lines.

90

DE09 Inputs X Y Z D0 1 D1 D2 D3 Outputs D4

DIGITALS ELECTRONICS

D5

D6

D7

0 0 0 0 1 1 1 1

0 0 1 1 0 0 1 1

0 1 0 1 0 1 0 1

0
1

0 0
1

0 0 0
1

0 0 0 0
1

0 0 0 0 0
1

0 0 0 0 0 0
1

0 0 0 0 0 0 0
1

0 0 0 0 0 0 0

0 0 0 0 0 0

0 0 0 0 0

0 0 0 0

0 0 0

0 0

0

Table 6.1 Truth Table of 3-to-8 line Decoder

D = X' Y' Z'
0

D = X' Y' Z
1

D = X' Y Z'
2

Z

D = X' Y Z
3

Y

D = X Y' Z'
4

X

D = X Y' Z
5

D = X Y Z'
6

D =XYZ
7

Fig. 6(a) Logic Circuit of 3-to-8 line Decoder Q.15

What is an encoder? Draw the logic circuit of Decimal to BCD encoder and explain its working. (7) 91

DE09

DIGITALS ELECTRONICS Ans: Encoder: An Encoder is a combinational logic circuit which converts Alphanumeric characters into Binary codes. It has 2n (or less) input lines and n output lines. An Encoder may be Decimal to Binary, Hexadecimal to Binary, Octal to BCD etc. Decimal to BCD Encoder: This encoder has 10 inputs (for decimal numbers 0 to p) and 4 outputs for the BCD number. Thus it is a 10 line to 4 line encoder. Table 6(a) lists the decimal digits and the equivalent BCD numbers. From the table, we can find the relationship between decimal digit and BCD bit. MSB of BCD bit is Y3. For decimal digits 8 or 9, Y3 = 1. Thus we can write OR expression for Y3 bit as Y3 = 8 + 9 Similarly , Bit Y2 is 1 for decimal digits 4,5,6 and 7. Thus we can write OR expression Y2 = 4 + 5 + 6 + 7 Y1 = 2 + 3 + 6 + 7 Yo = 1 + 3 + 5 + 7 + 9

The logic circuit for the expressions (Y0, Y1, Y2, Y3) is shown in fig. 6(b). When a High appears on any of input lines the corresponding OR gates give the BCD output. For e.g., if decimal input is 8, High appears only on output 3 (and LOW on Y0, Y1 , Y2), thus giving the BCD code for decimal 8 as 1000. Similarly, if decimal input is 7, then High appears on outputs Y0, Y1, Y2 (and LOW on Y3), thus giving BCD output as 0111.
Decimal Digit Y3 Y2 Y1 Y0 BCD Code

0 1 2 3 4 5 6 7 8 9

0 0 0 0 0 0 0 0 1 1

0 0 0 0 1 1 1 1 0 0

0 0 1 1 0 0 1 1 0 0

0 1 0 1 0 1 0 1 0 1

92

DE09
1

DIGITALS ELECTRONICS

Y ( LSB )
0

2

3 Y
1

4 5 6 7 Y
2

8 9

Y ( MSB )
3

Fig.6(b) Logic diagram for Decimal to BCD Encoder Q.16 What is a flip-flop? What is the difference between a latch and a flip-flop? List out the application of flip-flop. (4) Ans: Flip-Flop: A flip-flop is a basic memory element used to store one bit of information. Both Flip-flops and latches are bistable logic circuits and can reside in any of the two stable states due to a feedback arrangement. The main difference between them is in the method used for changing the state. Applications of Flop-Flops: (1) Bounce elimination switch (2) Parallel Data Storage in Registers (3) Transfer of Data from one bit to another. (4) Counters (5) Frequency Division Q.17 Draw the circuit diagram of a Master-slave J-K flip-flop using NAND gates. What is race around condition? How is it eliminated in a Master-slave J-K flip-flop. (10) Ans: Logic Diagram of Master-Slave J-K Flip-Flop using NAND Gates: Fig.7(a) shows the logic diagram of Master-Slave J-K Flip-Flop using NAND gates.

93

DE09

DIGITALS ELECTRONICS

G J

PR
3M

G

1M

Q

M

S

G
S

3S

G

1S

Q C LK

MASTER

SLAVE
_ Q

K G
4M

G C LR

2M

_ Q

R
M

S

G

G
4S

2S

___ CLK

Fig.7(a) Logic Diagram of Master-Slave J-K FLIP-FLOP The Race-around Condition: The difficulty of both inputs 1 (S = R = I) being not allowed in an S-R Flip-Flop is eliminated in a J-K Flip-Flop by using the feedback connection from outputs to the inputs of the gates. In R-S Flip-Flop, the inputs do not change during the clock pulse (CK = 1), which is not true in J-K Flip-Flop because of the feedback connections. Consider that the inputs are J = K = 1 and Q = 0 and a pulse as shown in Fig. 7(b) is applied at the clock input. After a time interval t equal to the propagation delay through two NAND gates in series, the output will change to Q = 1. Now we have J = K = 1 and Q = 1 and after another time interval of t the output will change back to Q = O. Hence, for the duration tp of the clock pulse, the output will oscillate back and forth between 0 and 1. At the end of the clock pulse, the value of Q is uncertain. This situation is referred to as the race-around condition.The race-around condition can be avoided if tp < t < T. However, it may be difficult to satisfy this inequality because of very small propagation delays in ICs. A more practical method of overcoming this difficulty is the use of the master-slave (M-S) configuration.
T RAILING (NEGAT IVE) EDGE LEADING (POSIT IVE) EDGE

t 0

P

T

Fig.7 (b) a Clock Pulse

A master-slave J-K Flip-Flop is a cascade of two S-R Flip-Flops with feedback from the outputs of the second to the inputs to the first as illustrated in Fig.7(a). Positive clock pulses are applied to the first Flip-Flop and the clock pulses are inverted before these are applied to the second Flip-Flop. When CK=1, the first Flip-Flop is enabled and the outputs QM and QM respond to their inputs J and K according the Table 7.1. At this time, the second FlipFlop is inhibited because its clock is LOW ( CK = 0). When CK goes LOW ( CK = 1), the first Flip-Flop is inhibited and the second Flip-Flop is enabled, because now its clock is HIGH ( CK = 1). Therefore, the outputs Q and Q Follow the outputs QM and Q M respective (second and third rows of Table 7.1). Since the second Flip-Flop simply follows the first one, it is referred to as the Slave and the first one as the Master. Hence, this 94

DE09

DIGITALS ELECTRONICS

configuration is referred to as Master-Slave Flip-Flop. In this circuit, the inputs to the gates G3M and G4M do not change during the clock pulse, therefore the Race-around condition does not exist. The state of the Master-Slave Flip-Flop changes at the negative transition (trailing end).
Inputs PR 0 0 1 1 1 CLR 0 1 0 1 1 CLK X X X X J X X X 0 0 K X X X 0 1 Output Q Race Condition 1 0 No change 0

1

1

1

0

1

1

1

1

1

Toggle

Table 7.1 Truth Table of JK Master-Slave Flip-Flop Q.18 What is a demultiplexer? Discuss the differences between a demultiplexer and a decoder. (4) Ans: Demultiplexer: It is a logic circuit that accepts one data input and distributes it over several outputs. A demultiplexer has one data input, m select lines, and n output lines, whereas a decoder does not have the data input but the select lines are used as input lines. Q.19 What is a shift register? Can a shift register be used as a counter? If yes, explain how? (4)

Ans: Shift Register: A register in which data gets shifted towards left or right when clock pulses are applied is known as a Shift Register. A shift register can be used as a counter. If the output of a shift register is fed back to serial input, then the shift register can be used as a Ring Counter. Q.20 What are synchronous counters? Design a Mod-5 synchronous counter using J-K Flip-Flops. (10)

Ans:

95

DE09

DIGITALS ELECTRONICS Synchronous Counters: The term synchronous means that all flip-flops are clocked simultaneously. The clock pulses drive the clock input of all the flip-flops together so that there is no propagation delay. Mod-5 Counter Synchronous Counter: The Mod-5 Synchronous Counter have five counter states. The counter design table for this counter lists the three flip-flop and their states (0 to 5 states), as shown in table 9(a), the six inputs required for the three flip-flops. The flip-flop inputs required to step up the counter from the present to the next state have been worked out with the help of the excitation table shown in the table.
Input pulse A Count 0 1 2 3 4 5(0) 0 1 0 1 0 0 0 0 1 1 0 0 0 0 0 0 1 0 1 X 1 X 0 X 1 X 1 X 0 1 X X 0 X X 0 1 X 0 0 0 1 X X X X X 1 B C JA KA JB KB JC KC Counter States Flip-Flop Inputs

Table 9(a) counter Design Table for Mod-5 Counter

A flip-flop: The initial state is 0. It changes to 1 after the clock pulse. Therefore JA should be 1 and KA may be 0 or 1 (that is X ). B flip-flop: The initial state is 0 and it remains unchanged after the clock pulse. Therefore JB should be 0 and KB may be 0 or 1 (that is X) C flip-flop: The state remains unchanged. Therefore Jc should be 0 and KC should be X.The flip-flop input values are entered in Karnaugh maps shown in Table 9(b) [(i) (ii) (iii) (iv) (v) and (vi)] and a boolean expression is found for the inputs to the three flip-flops and then each expression is simplified. As all the counter states have not been utilized, X’s (don't) are entered to denote un-utilized states. The simplified expressions for each input shown under each map. Finally, these minimal expressions for the flip-flop inputs are used to draw a logic diagram for the counter, which is shown in fig.9 (b).

96

DE09

DIGITALS ELECTRONICS

(i) Map for JA JA =

(ii) Map for KA KA = 1

C

(iii) Map for JB JB = A

(iv) Map for KB KB = A

(v) Map for JC
JC = AB

(vi) Map for KC
KC = 1

Table 9(b) Karnaugh Maps for MOD-5 Synchronous Counter

A

B

C

J CLOCK PULSES 1

A

A

J

B

B

JC

C

FF
KA

A _ A

FF
KB

B _ B 1 K
C

FF

C _ C

Fig.9 (b) Logic Diagram of MOD-5 Synchronous Counter

97

DE09

DIGITALS ELECTRONICS

Q.21 With the help of a neat diagram, explain the working of a successive approximation A/D converter. (14) Ans: Successive Approximation ADC: This is the most widely used A/D converter. As the name suggests, the digital output tends towards analog input through successive approximations. In Successive Approximation ADC, the comparison with the input analog voltage is done in descending order starting from maximum voltage. Fig.10 (a) shows the block diagram of SA A/D converter. The main components are Op-amp Comparator, Control Logic, SA register and D/A converter. It uses Digital to Analog converter as a feedback element. The control logic is the most important part of Successive Approximation Converter, as this decides the next step to be taken. The ring counter provides timing waveform to control the operation of the converter. The Digital to Analog Converter unit, n bit register and ring counter are all reset by the first pulse from the ring counter. The ring counter containing a single one sets the MSB of the Digital to Analog Converter to 1 and the other to 0

Ring Counter

C lock

------------------

Control logic

Vo

---------

+

-

V

in

V

n - bit register f a
----------

0 1

S/ H

-------------

a

a n -2 a

n -1

Analog i nput

D/A converter

Fig.10 (a) Block Diagram of Successive Approximation A/D Converter

The basic operating principle of Successive Approximation Converter is that the voltage ouput of DAC corresponding to MSB is compared by the comparator with the input voltage and if the voltage is less, the bit 1 is retained. If the voltage is more, it is reset to 0 and counter moves to next position. Similar decisions are made at each bit position until the nearest value is reached. 98

DE09

DIGITALS ELECTRONICS

Assume that the MSB of a unipolar 6 bit converter produces 10 V output and we have to measure an analog output voltage of 8.2 V. Each bit divides the voltage by 2 so that the voltages for the 6 bits from MSB downwards is
Bits 5 4 3 2 1 0

Voltage

10

5

2.5

1.25

0.625

0.3125

MSB

LSB

The operation of SA A/D converter is shown in Table No.10(a). Let the analog input be 8.2 V. The SA register is first set to zero. Then 10 is placed in MSB. This is fed to D/A converter whose output goes to comparator. Since the analog input (8.2 V) is greater than D/A output (i.e.,10 V), the MSB is set to one. Then 1 is placed in bit next to MSB (i.e., 1 is placed in second position). Now the output of D/A is 5 V. Since analog input is less than 5 V, it is reset to 0. Next 0 is placed in third position. Now the D/A output is (5+2.5=7.5V) which is less than analog input. Therefore, this 0 bit is retained and 0 is placed in the next bit (i.e., fourth position). Now the D/A output is (7.5+1.25=8.75), which is more than analog input. Therefore, the 1 bit is placed in fifth position. Now the D/A output is (8.75+0.625=8.125) which is less than analog input, it is reset to 0. Now 0 is placed in LSB producing a D/A output of (8.125+0.3125=8.4375) which is more than analog input. Therefore, LSB is set to one. The various steps and voltages are tabulated in Table No.10 (a).
Step Register DAC Output Comparator decision w.r.t 8.2V. Start 2 3 4 5 6 100000 010000 011000 011100 011010 011011 10 5 5+2.5=7.5 7.5+1.25=8.75 7.5+0.625=8.125 8.125+0.3125=8.4375 High Low Low High Low High

Table 10(a)

The D/A converter waveform is shown in fig.10 (b)

99

DE09
10 9 8 7 6 5 4 3 2 1 0

DIGITALS ELECTRONICS

0

1

1

0

1

0

Fig.10 (b) Output Waveform of D/A Converter Features: (i) It is one of the most widely used ADC (ii) Its conversion time is very next only to Flash or Parallel ADC (iii) SACs have fixed value of conversion time that is not dependent on the value of analog input voltage. (iv) Data can be taken out either in serial or in parallel. (v) During the period of comparison the input analog voltage should be held constant and so the input to comparator is through a Sample Hold circuit. Q.22 Difference between static and dynamic RAM. Draw the circuits of one cell of each and explain its working. (10) Ans: Differentiation between Static RAM and Dynamic RAM: Static RAMs store ones and zeros using conventional FLIP-FLOPs. whereas, the memory cells of dynamic RAMs are basically charge storage capacitors with driver transistors.The presence or absence of charge in a capacitor is interpreted as Logic1 or 0. Static RAMs do not require refreshing because there is no problem of charge leaking-off in FLIP-FLOPs whereas Dynamic RAMs require periodic charge refreshing to maintain data storage because the charge stored on capacitors leak-off with time. Static RAMs are slower but easier to drive than dynamic memories, which generally require clock signals in addition to extra power supplies whereas dynamic circuits usually require externally generated clock voltages, Advantages of Static RAMs over Dynamic RAMs: (i)Higher speed of operation (faster)i.e, lower access –time. (ii)Does not require refreshing. Advantages of Static RAMs over Dynamic RAMs: (i)Higher number of bits storage on a given silicon chip area. i.e, Higher packaging density. (ii)Lower power consumption.

100

DE09

DIGITALS ELECTRONICS Static RAM Cell: A RAM memory cell consisting of two cross-coupled MOS inverters is shown in Fig.11 (a). It is addressed by setting AX and AY to 1. When AX = 1, the cell is connects to the data and data line. When AY = 1, T7 and T8 are ON. To write into the cell, set W = 1, T9 becomes ON. If data input is 1, the voltage at node D will correspond to level 1 making T3 ON and level at D will be 0. On the other hand, if the data input is at logic 0, then T3 will be OFF and D would be at 1. To read the state of the FLIP-FLOP, we set R = 1.This connects the data output to D . Thus, the complement of the data level written into the cell is read at data output.
A
X

V

DD

To other cells with same A

X

V To other cells with same A Y T
2

GG

T

4

To other cells with same A ___ D ata line

Y

D ata line

T

5

T
D D

6

T

1

T

3

Write (W) T D ata Input

R ead (R) T

7

8

T

T
9

A

10

Y

____ D ata Output

Fig.11(a) Logic Diagram of a Static MOS RAM Cell Dynamic RAM Cell: A dynamic cell uses four transistors in place of the six used in a static cell. This reduces the silicon chip area and results in saving of power. The circuit of a 4transistor dynamic MOS RAM cell is shown in Fig.11 (b).The state of the cell is stored on the stray capacitances C1 and C2, whose presence is essential. The cell is addressed by making AX = AY =1. In one state of the cell, the voltage across C1 is large and T1 is ON. Correspondingly, C2 has zero voltage and T2 is OFF. In the other state, the voltages on C1 and C2 and the conducting states of T1 and T2 are reversed. For writing into the cell, we set W = 1 and for reading from the cell we set R = 1.It is necessary to refresh the cell periodically, otherwise the charge stored on the capacitors leak off. The refreshing operation is accomplished by allowing brief access from the supply voltage VDD to the cell. This is done by making AX = 1 and the refresh terminal voltage corresponding to 1 level. This makes T3, T4, T9, and T10 ON. Suppose initially T1 is ON, T2 is OFF. The voltage across C1 is large and across C2 it is zero volt. During the refresh interval, VDD is applied through T10 and T4 to C1, since T2 is OFF. Therefore, current from VDD will flow through C1, allowing C1 to replenish any charge lost due to leakage. Since T1 is ON, hence C2 will not charge as rapidly as C1. Similarly VDD is applied to C2, which is in parallel to T1 when T1 is OFF and T2 is ON.

101

DE09
To other cells R efresh

DIGITALS ELECTRONICS
To other cells

V
DD

T A
X

9

T

10

To other cells ____ D ata line

D ata line

T

3

T

4

T W

1

T C
1

2

C

2

R

T D ata Input

5

T

6

T

7

A

T
Y

8

____ D ata Input

Fig.11(b) Logic Diagram of Dynamic MOS RAM Cell Q.23 Distinguish between ROM, PROM, EPROM, EEPROM. (4)

Ans: ROM: Read Only Memory is a Permanent Memory. In Permanent ROM, the data is permanently stored and cannot be changed. It can only be read from the memory. There cannot be a write operation because the specified data is programmed into the device by the manufacturer or the user. ROM is a Non-volatile memory. Some examples of ROM are conversion tables, pre-programmed instructions etc. PROM: Programmable Read Only Memory allows user to store the data. An instrument PROM programmer is used to store the required data. The process used is opening the links at bit locations using high current (this process is called burning in). Once this process has been done, the data is permanently stored and no change is possible. EPROM: EPROM means Erasable PROM. It can be reprogrammed by first erasing the existing program. EPROM uses N-MOSFET array with isolated gate structure. The isolated transistor gate has no electrical connection and can store an electrical charge indefinitely. The data bits in this memory array are represented by presence or absence of charge. Erasure is achieved by removing the gate charge. EPROM can be UV EPROM or EEPROM. UV EPROM means Ultra Violet Erasable PROM. Erasure is achieved by using ultra violet light. The light passes through a window in the IC package to the chip where there are stored charges. Thus the stored contents are erased. EEPROM: EEPROM means Electrically Erasable PROM. In this memory device, the erasure and programming is done by electrical pulses. Q.24 What is a universal gate? Give examples. Realize the basic gates with any one universal gate. (8)

102

DE09

DIGITALS ELECTRONICS Ans: Universal Gates: NAND and NOR are known as Universal gates The AND, OR, NOT gates can be realized using any of these two gates. The entire logic system can be implemented by using any of these two gates. These gates are easier to realize and consume less power than other gates. Realizations of NOT, AND and OR gates using NAND gates NOT GATE: Fig. 3(a) shows the realization of Inverter (NOT) gate using NAND gate. Both the inputs to the NAND gates are tied together so that the gate works as an inverter (NOT) gate.
_ A

A

Fig. 3(a) Realization of Inverter (NOT) gate using NAND gate AND GATE: Fig. 3(b) shows the realization of AND gate using two NAND gates. It has combination of two NAND gates gives AND operation. The first NAND gate has two inputs A and B. The two inputs to the second NAND gate are tied together and the output AB of the first gate is fed to this common terminal. The output is AB thus giving AND operation.
__ AB __ __ AB = AB

A B

. Fig. 3(b) Realization of AND gate using NAND gates
OR GATE: Fig. 3(c) shows the realization of OR gate using NAND gates. The two inputs from each of the first two NAND gates are tied together and fed by A and B as shown in the figure. The outputs are A and B . They are fed to as inputs to third NAND gate. The final output is A OR B thus giving OR operation.
_ A A ___ _ _ A B = A +B

_ B B

Fig. 3(c) Realization of OR gate using NAND gates Q.25 Give the circuit of a TTL NAND gate and explain its operation in brief. (6)

Ans: Operation of TTL NAND Gate: Fig.3(d) shows a TTL NAND gate with a totem pole output. The totem pole output means that transistor T4 sits atop T3 so as to give low output impedance. The low output impedance implies a short time constant RC so that the

103

DE09

DIGITALS ELECTRONICS

output can change quickly from one state to another. T1 is a multiple emitter transistor. This transistor can be thought of as a combination of many transistors with a common base and collector. Multiple emitter transistors with about 60 emitters have been developed. In the figure, T1 has 3 emitters so that there can be three inputs A, B, C. The transistor T2 acts as a phase splitter because the emitter voltage is out of phase with the collector voltage. The transistors T3 and T4 from the totem pole output, The capacitance CL represents the stray capacitance etc. The diode D is added to ensure that T4 is cut off when output is low. The voltage drop of diode D keeps the base-emitter junction of T4 reverse biased so that only T3 conducts when output is low. The operation can be explained briefly by three conditions as given below: Condition 1: At least one input is low (i.e., 0). Transistor T1 saturates. Therefore, the base voltage of T2 is almost zero. T2 is cut off and forces T3 to cut off. T4 acts like an emitter follower and couples a high voltage to load. Output is high (i.e. Y=1). Condition 2: All inputs are high. The emitter base junctions of T1 are reverse biased. The collector base junction of T1 is forward biased. Thus, T1 is in reverse active mode. The collector current of T1 flows in reverse direction. Since this current is flowing into the base of T2, the transistors T2 and T3 saturate and output Y is low. Condition 3: The circuit is operating under II when one of the input becomes low. The corresponding emitter base junction of T1 starts conducting and its base voltage drops to a low value. Therefore, T1 is in forward active mode. The high collector current of T1 removes the stored charge in T2 and T3 and therefore, T2 and T3 go to cutoff and T1 saturates and output Y returns to high.
V = + 5V cc 100 Ohms 4 KOhms 1.4 KOhms

T

4

T
A

1

T

2

Diode

C

B

T

3

. . . .

Y

C
1 KOhms . . . .

L

. Fig.3(d) Logic Diagram of TTL NAND Gate with Totem Pole Output
Q.26 With the help of a truth table explain the working of a half subtractor. Draw the logic diagram using gates. (8)

104

DE09

DIGITALS ELECTRONICS

Ans: Half Subtractor: A logic circuit for the subtraction of B (subtrahend) from A (minuend) where A and B are 1-bit numbers is referred to as a Half-Subtractor. The truth table for half subtractor is given in Table No.5.1. Here A and B are the two inputs and Di (difference) and Bo (borrow) are the two outputs. If B is larger than A (e.g., A=0 and B=1), a borrow is necessary, Inputs Outputs A B Di(Difference) Bo(Borrow)

0 0 1 1

0 1 0 1

0 1 1 0
Table 5.1

0 1 0 0

From the Truth Table, the logical expressions for Di and Bo are obtained as Di = A B + A B BO = A B
A B

D

i

A HALF SUBT RACT OR

D B

i

B

B o o

Fig.5(a) Logic Diagram of Half Subtractor

Fig.5(b) Block Diagram of Half Subtractor

In Table 5.1, input variable B is subtracted from A to give output Di(difference). If B is larger than A (e.g., A = 0 and B = 1), a borrow is necessary. In the Truth Table, inputs are A and B, Outputs are Di (difference) and BO (borrow). Hence, the Boolean expressions for the half subtractor from the Truth Table can be written as Di = A ⊕ B ---------(1) BO = A B --------(2) By combining Boolean Expressions (1) & (2), we get the logic circuit for Half Subtractor shown in fig.5(a) and its block diagram is shown in fig.5(b).
Q.27 Draw the logic diagram of a full subtractor using half subtractors and explain its working with the help of a truth table. (6) Ans:

105

DE09

DIGITALS ELECTRONICS Full Subtractor: A Full Subtractor has to take care of repeated borrow from the next higher bit. At any stage alongwith the two bits (one of which is to be subtracted from the other) is another input Bin, i.e., borrow bit from the Di and borrow Bo. Table shows the truth table.
B
in

INPUTS A
B

FULL SUBT RACT OR

D B

i

OUTPUTS o Fig.5(c) Block Diagram of Full Subtractor
B D D A i in

H ALF SU BTRAC TOR B o B

i

B

H ALF SU BTRAC TOR B o

o

Fig.5(d) Block Diagram of Full Subtractor as Combination of two Half Subtractors and OR Gate
HALF SUBTRACTOR

B in HALF SUBTRACTOR

Di D i D
A B

i

B o

B

OR GATE o

B

o

Fig.5(e) Logic Diagram of Full Subtractor

Fig.5(c) shows a block diagram for a full subtractor. It can be constructed from two Half Subtractors and an OR gate as shown in Fig.5(d). The logic diagram is shown in Fig.5(e). This logic diagram is as per the truth table of Table 5.1.

106

DE09 Inputs A B Bin Outputs Di

DIGITALS ELECTRONICS

BO

0 0 0 0 1 1 1 1

0 0 1 1 0 0 1 1

0 1 0 1 0 1 0 1

0 1 1 0 1 0 0 1

0 1 1 1 0 0 0 1

Table 5.1 Truth Table for Full Subtractor Q.28 Design a BCD to seven segment decoder that accepts a decimal digit in BCS and generates the appropriate output for segments in display indicator. (14) Ans: BCD-TO-7-Segment Decoder: A digital display that consists of seven LED segments is commonly used to display decimal numerals in digital systems. Most familiar examples are electronic calculators and watches where one 7-segment display device is used for displaying one numeral 0 through 9. For using this display device, the data has to be converted from some binary code to the code required for the display. Usually the binary code used is Natural BCD. Fig.6(a) shows the display device. Fig.6(b) shows the segments which must be illuminated for each of the numerals and Fig.6(c) gives the display system. a f e g d b c

Fig.6(a)
B C D

Fig.6(b)

A B BC D TO SEVEN SEGMEN T D ECODER

a b c d e f g

a f e g d b c

I N C P U D T S

Fig.6(c)

107

DE09

DIGITALS ELECTRONICS

Table 6.1 gives the truth table of BCD-to-7-segment Decoder. Here ABCD is the Natural BCD code for numerals 0 through 9. The K-maps for each of the outputs a through g are given in Fig.6(d), 6(f),6(h),6(j),6(l),6(n),6(p). The entries in the K-map corresponding to six binary combinations not used in the truth table are X –don’t care.
Decimal Displayed A B C D a b c d e f g Digit Inputs Outputs

0 1 2 3 4 5 6 7 8 9

0 0 0 0 0 0 0 0 1 1

0 0 0 0 1 1 1 1 0 0

0 0 1 1 0 0 1 1 0 0

0 1 0 1 0 1 0 1 0 1

1 0 1 1 0 1 0 1 1 1

1 1 1 1 1 0 0 1 1 1

1 1 0 1 1 1 1 1 1 1

1 0 1 1 0 1 1 0 1 0

1 1 0 0 1 0 0 0 0 1 0 1 1 1 0 0 1 1 0 1

0 0 1 1 1 1 1 0 1 1

Table 6.1 Truth Table of BCD-to-7 Segment Decoder (i) K-map and Logic Diagram for Digital Output ‘a’:

108

DE09

DIGITALS ELECTRONICS

The simplified expressions for the Fig.6(d) is given by a = B D + BD + CD + A and the logic diagram is given in Fig.6(e)
_ B _ D B D

a

C D _ A

Fig.6(e) Logic Diagram for Output ‘a’ (ii) K-map and Logic Diagram for Digital Output ‘b’:

The simplified expressions for the Fig.6(f) is given by b = B + C D + CD and the logic diagram is given in Fig.6(g)
_ C _ D C D B

b

Fig.6(g) Logic Diagram for Output ‘b’ (iii) K-map and Logic Diagram for Digital Output ‘c’:

109

DE09

DIGITALS ELECTRONICS

The simplified expressions for the Fig.6(h) is given by c = B + C + D and the logic diagram is given in Fig.6(i)
_ B C _ D

c

Fig.6(i) Logic Diagram for Output ‘ c ’ (iv) K-map and Logic Diagram for Digital Output‘d’:

110

DE09

DIGITALS ELECTRONICS

The simplified expressions for the Fig.6(j) is given by d = B D + C D + B C +B C D and the logic diagram is given in Fig.6(k)
_ B _ D C _ D _ B C B _ C D

d

Fig.6(k) Logic Diagram for Output ‘ d ’ (v) K-map and Logic Diagram for Digital Output ‘e’:

The simplified expressions for the Fig.6(l) is given by e = B D + C D and the logic diagram is given in Fig.6(m)
_ B _ D C _ D

e

Fig.6(m) Logic Diagram for Output ‘e’

111

DE09 (vi) K-map and Logic Diagram for Digital Output ‘f’:

DIGITALS ELECTRONICS

The simplified expressions for the Fig.6(n) is given by f = A + C D + B C +B D and the logic diagram is given in Fig.6(o)
_ C _ D B _ C B _ D _ A

f

Fig.6(o) Logic Diagram for Output ‘ f’

112

DE09

DIGITALS ELECTRONICS (vii)K-map and Logic Diagram for Digital Output ‘g’:

The simplified expressions for the Fig.6(p) is given by g = A + B C + B C+ C D and the logic diagram is given in fig.6(q).
B _ C _ B C C _ D _ A

g

Fig.6(q) Logic Diagram for Output ‘ g ’ Q.29 Explain the working of a demultiplexer with the help of an example. (6)

Ans: 1:4 Demultiplexer: Fig.7(a) shows the logic circuit of a 1:4 demultiplexer. It has two NOT gates, 4 AND gates, one data input line, 2 select lines (S0,S1) and four output lines (D0, D1,D2,D3). The data input line feeds all the AND gates. However, the two select lines enable only one gate at one time. If S1S0= 00 then the data goes to D0. if S1S0= 01, then the data goes toD1. If S1S0 = 10, then the data goes to D2 and if S1S0= 1, then the data goes to D3 .

113

DE09
DAT A INPUT

DIGITALS ELECTRONICS

D

0

S

0

D

1

S

1

D

2

O U T P U T S

D

3

Fig.7(a) Logic Circuit of 1:4 Demultiplexer Q.30 Give the truth table of S-R and D-flipflops. Convert the given S-R flipflop to a D-flipflop. (8)

Ans: The Truth Table of S-R Flip-Flop is shown in Fig.7(b) and truth table of D Flip-Flop is shown in Fig.7(c) Inputs Sn Rn Output Qn+1 Input Dn Output Qn+1

0 1 0 1

0 0 1 1

Qn 1 0 ? 0 1 0 1

Fig.7(b) Truth Table for S-R Flip-Flop

Fig.7(c) Truth Table for D-Flip-Flop

If we use only the middle two rows of the truth table of the S-R Flip-Flop shown in Fig.7(b) then we obtain a D-type Flip-Flop as shown in Fig.7(d) and 7(e). It has only one input referred to as D-input or Data Input. Its truth table is given in Fig.7(c) from which it is clear that the output Qn+1 at the end of the clock pulse equals the input Dn before the clock pulse. This is equivalent to saying that the input data appears at the output at the end of the clock pulse. Thus, the transfer of data from the input to the output is delayed and hence the name Delay (D) Flip-Flop. The D-type Flip-Flop is either used as a Delay Device or as a Latch to store 1-bit of binary information.

114

DE09

DIGITALS ELECTRONICS
Pr
Pr S D S-R FLIP-FLOP CLK R _ Q Q

D

D FLIP-FLOP

Q

CLK _ Q

Cr

Cr

Fig.7 (d) S-R Flip-Flop converted into a D-Flip-Flop

Fig.7 (e) Logic Symbol of D Flip-Flop

Q.31 Define a register. Construct a shift register from S-R flip-flops. Explain its working.(8) Ans: Register: A register consists of a group of flip-flops and gates that effect their transition. The flip flops hold the binary information and the gates control when and how new information is transformed into the register. S-R Flip-Flop Shift Register: Shift registers can be built by using SR flip-flops.Fig.9(a) shows the 4-bit shift register, which uses RS flip-flops. It uses Four SR Flip-Flops in cascade and the inputs to the last three flip-flops in the chain receive complementary inputs, that is if S = 0, R = 1 and if S = 1, R = 0. The first flip-flop has complementary S and R inputs and, therefore, it behaves like a D-type flip-flop. Because of the Inverter in the clock line, data will be transferred to flip-flop outputs on the positive going edge of the clock pulse. There are two inputs A and B. Any one of the inputs can be used. Since a 1 input at A or B will be a 1 input at S of the first flip-flop, as a result of double complementation, a positive going clock pulse will produce an output of 1 at Q of the first flip-flop. Normally both A and B inputs of the NAND gate are connected together when data is being fed and the NAND is not required to serve as a gate.
SERIAL A INPUT B

S

Q
S-R FLIP-FLOP

S

Q
S-R FLIP-FLOP

S

Q
S-R FLIP-FLOP

S

Q
S-R FLIP-FLOP

SERIAL OUTPUTS

R

_ Q

R

_ Q

R

_ Q

R

_ Q

CLOCK

Fig.9(a)Logic Diagram of S-R Flip-Flop Shift Register Q.32 Explain how a shift register can be used as a ring counter giving the wave forms at the output of the flipflops. (6)

115

DE09

DIGITALS ELECTRONICS Ans: Shift Register as a Ring Counter: A Ring Counter is a Circular Shift Register with only one flip-flop being set at any particular time; all other are cleared. The single bit is shifted from one flip-flop to the other to produce the sequence of timing signals. Fig.9(b) shows a 4-bit shift register connected as a ring counter. The initial value of the register is 1000, which produces the variable T0. The single bit is shifted right with every clock pulse and circulates back from T3 to T0. Each flip-flop is in the 1 state once every four clock pulses and produces one of the four timing signals shown in Fig.9(c). Each output becomes 1 after the negative-edge transition of a clock pulse and remains 1 during the next clock pulse.

SH IFT R IGH T R EGISTER

T

0

T

T
1 2

T
3

Fig.9(b) 4-bit shift register connected as a ring counter.

C LOCK PU LSE

T

0

T

1

T T

2 3

Fig.9(c) Waveforms at the output of Flip-Flops Q.33 Differentiate between linear addressing and matrix addressing modes with examples. Which of them is the best method? (4) Ans: Linear Addressing: Addressing is the process of selecting one of the cells in a memory to be written into or to be read from. In order to facilitate selection, memories are generally arranged by placing cells in Linear form or Matrix form. Linear Addressing Mode: A single column that has n rows and 1 column (such as the 16X1 array of cells) shown in fig.11(a) is frequently called Linear Addressing. Selection of a cell simply means selection of the corresponding row and the column is used.

116

DE09
1 COLU MN

DIGITALS ELECTRONICS

1

1 2 3 16 R OWS

. . . . . .

16 X 1

16

Fig.11 (a) Linear Addressing Mode Matrix Addressing Mode: The arrangement that requires the fewest address lines is a square array of n rows and n columns for a total memory capacity of n x n = n2 cells. This arrangement of n rows and n columns is frequently referred to as Matrix Addressing which is shown in fig.11(b).
4 COLU MN S

1

2

3

4

1 4 R OWS 2 3 4

4x4

Fig.11(b) Matrix Addressing Mode Best Method: Matrix Addressing is the best method, because this configuration only requires 8 address lines (i.e.,4 rows and 4 columns), whereas Linear Addressing method requires a total of 17 address lines (i.e., 1 column and 16 rows).The square configuration is so widely used in industry. Q.34 Write short note on the following: Johnson counter. Ans: Johnson Counter: Johnson Counter is an synchronous counter, where all flip-flops are clocked simultaneously and the clock pulses drive the clock input of all the flip-flops together so that there is no propagation delay. Fig.11(e) shows the circuit of Johnson counter. In this case the D input of FF0 is driven by Q output of FF3, i.e., the complement of the output of the last flip flop is fed to the D of FF0. This feedback arrangement produces the sequence of states shown in Table 11.2. The 4 bit sequence (4)

117

DE09

DIGITALS ELECTRONICS

has a total of 8 states ( n bit sequence will have 2n states). Thus an n bit Johnson counter will have a modulus of 2n. The Q output of each stage feeds the D input of next stage. But the Q output of the last stage feeds the D input of first stage. The counter fills up with 1’s from left to right and then fills up 0s again as shown in Table 11.2. Fig. 11(f) shows the waveshapes/timing diagram of 4 bit Johnson counter.
Q

Q D Q

0

Q D

1

2

Q Q
3

3

0

0

1

Q
1

D

2

Q
2

D3

FF

0

FF

FF

1

2

FF

3 _ Q3

CLOCK

Fig.11(e) Logic Diagram of Johnson counter

Clock Pulse 0 1 2 3 4 5 6 7

Q0 0 1 1 1 1 0 0 0

Q1 0 0 1 1 1 1 0 0

Q2 0 0 0 1 1 1 1 0

Q3 0 0 0 0 1 1 1 1

Table 11.2 Sequence of states of 4 bit Johnson Counter
1 C LK 3 4 6 7

2

5

8

Q

0

Q
1

Q

2

Q

3

Fig.11(f) Timing Diagram of 4-bit Johnson Counter

118

DE09

DIGITALS ELECTRONICS

Q.35 The voltage waveforms shown in Fig.1 are applied at the inputs of 2-input AND and OR gates. Determine the output waveforms. (3)

Ans: The Output waveforms for AND and OR gates are shown in fig.3(a)
1

A
0 1
0 1 2 3 4 5

time (ms)

B
0 1 AN D OF A & B i.e., A X B 0 1 OR OF A & B i.e., A + B 0 time (ms)

Fig.3(a) Output Waveforms Q.36 What are the advantages of CMOS logic and explain CMOS Inverter with the help of a neat (7) circuit diagram. Ans: Advantages of CMOS Logic: (i) The power dissipation is minimum of all the logic families (ii) LSI & VLSI are possible CMOS Inverter: The basic CMOS logic circuit is an inverter shown in Fig.5(a). For this circuit the logic levels are 0 V (logic 0) and VCC (logic 1). When Vi = Vcc, T1 turns ON and T2 turns OFF. Therefore VO ≈ 0 V and since the transistors are connected in series, the current ID is very small. On the other hand, when Vi = 0 V, T1 turns OFF and T2 turns ON giving an output voltage VO ≈ Vcc and ID is again very small. In either logic state, T1 or T2 is OFF and the quiescent power dissipation which is the product of the OFF leakage current and Vcc is very low. More complex functions can be realized by combinations of inverters.

119

DE09
+V

DIGITALS ELECTRONICS

CC

S

2

T (p-channel)
2

D V i 2

I

D V o G

D D
1 1

T (n-channel) S
1

Fig.5(a) Logic Diagram of CMOS Inverter Q.37 What is Tri-state logic and explain Tri-state logic inverter with the help of a circuit diagram. Give its Truth Table. (7) Ans: Tri-state Logic: In normal logic circuits, there are two states of the output, LOW and HIGH. If the output is not in the LOW state, it is definitely in the other state (HIGH). Similarly, if the output is not in the HIGH state, it is definitely in the LOW state. In complex digital systems like microcomputers and microprocessors, a number of gate outputs may be required to be connected to a common line which is referred to as a bus which in turn may be required to drive a number of gate inputs. When a number of gate outputs are connected to the bus, Totem pole TTL outputs leads to heating of the ICs which may get damaged and Open-collector TTL outputs causes the problems of loading and speed of operation. To overcome these difficulties, in addition to low impedance outputs 0 & 1, there is a third state known as the High-impedance state. Such logic circuits in which the output can have three states is called tri-state logic. In the Tri-state Logic, in addition to low impedance outputs 0 & 1, there is a third state known as the High-impedance state. When the gate is disabled, it is in the third state. Tri-state Logic Inverter: The functional diagram of Tri-state Logic Inverter is shown in fig.5(b) and its logic diagram is shown in fig. 5(c). When the control input is LOW, the drive is removed from T3 & T4. Hence both T3 & T4 are cut-off and the output is in the third state. When the control input is HIGH, the output Y is Logic 1 or 0 depending on the data input. Truth table of Tri-state Logic Inverter is given in Table No.5.1

D ATA IN PUT C ON TROL

D ATA OU TPU T

Fig.5(b) Functional Diagram of Tri-state Logic Inverter

120

DE09

DIGITALS ELECTRONICS
+V
CC

C ontrol T
5

T

4

T D ata input

1

T

2

Y D at a out put T
3

Fig.5(c) Logic Diagram of Tri-state Logic Inverter

Data Input 0 1 0 1

Control 0 0 1 1

Data Output High - Z High - Z 1 0

Table 5.1 Truth Table of Tri-state Logic Inverter Q.38 What is a digital comparator. Explain the working of a 2-bit digital comparator with the help of (6) Truth Table. Ans: Digital Comparator: The comparison of two numbers is an operation that determines if one number is greater than, less than, or equal to the other number. A Digital comparator is a combinational circuit that compares two numbers, A and B, and determines their relative magnitudes. The outcome of the comparison is specified by three binary variables that indicate whether A > B, A = B, or A < B. Comparators can be designed for comparing multibit numbers. Figure 6(e) shows the block diagram of an n-bit comparator. It receives two n-bit numbers A and B as inputs and the outputs are A > B, A = B, and A < B. Depending upon the relative magnitude of the two numbers, one of the outputs will be HIGH. Table 6.2 gives the truth table of a 2-bit comparator. (I) If the magnitude of the inputs A and B are equal (i.e., A = B):Consider two numbers, A and B as inputs with two digits each i.e., A1, A0 and B1, B0. The two numbers are equal if all pairs of significant digits are equal i.e., if A1 = 0, A0 = 0, B1 = 0, B0 = 0, then A1 = B1 and A0 = B0. For example if A1 = 0, A0 = 0, B1 = 0, B0 = 0, then pairs of significant digits i.e., A1 = B1 = 0 and A0 = B0 =0. Output for this combination becomes 1 for A = B and 0 for A < B and A > B. This is given in the Truth Table.

121

DE09

DIGITALS ELECTRONICS (II) If the magnitude of the input A is greater than or less than B (i.e., A > B or A < B):To determine if A is greater than or less than B, we inspect the relative magnitude of pairs of significant digits starting from the most significant position. If the two digits are equal, we compare the next lower significant pair of digits. This comparison continues until a pair of unequal digits is reached. (i)If the input A is greater than B (i.e., A > B): If the corresponding digit of A is 1 and that of B is 0, we conclude that A > B. For example if A1 = 0, A0 = 1, B1 = 0, B0 = 0, then pairs of significant digits are A1 = B1 =0, and A0 (i.e., digit 1) > B0(i.e., digit 0). This is shown in the Truth Table. (ii) If the input A is less than B (i.e., A < B):If the corresponding digit of A is 0 and that of B is 1, we conclude that A < B. For example if A1 = 0, A0 = 0, B1 = 0, B0 = 1, then pairs of significant digits are A1 = B1 =0, and A0 ( i.e., digit 0) < B0 (i.e., digit 1). This is shown in the Truth Table.

Inputs A1 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 A0 0 0 0 0 1 1 1 1 0 0 0 0 1 1 1 1 B1 0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1 B0 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 A>B 0 0 0 0 1 0 0 0 1 1 0 0 1 1 1 0

Outputs A=B 1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1 A