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# Opamp Appication

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Submitted By lihaolee618
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EMT 212/4
ANALOG ELECTRONIC II
Chapter 2 – Op-amp Application

Content
1.

2.

Op-amp Application
 Introduction
 Inverting Amplifier
 Non-inverting Amplifier
 Voltage Follower / Buffer Amplifier
 Summing Amplifier
 Differencing Amplifier
 Integrator
 Differentiator
 Comparator
 Summary
Frequency Response

Op-amp Application

Introduction
Op-amps are used in many different applications. We will discuss the operation of the fundamental op-amp applications. Keep in mind that the basic operation and characteristics of the op-amps do not change — the only thing that changes is how we use them

Inverting Amplifier

Circuit consists of an op-amp and three resistors
The positive (+) input to the op-amp is grounded through
R2
The negative (-) input is connected to the input signal (via
R1) and also to the feedback signal from the output (via RF)

Inverting Amplifier
V

V+

Assume that amplifier operates in its linearly amplifying region. For an ideal op-amp, the difference between the input voltages V+ and V to the op-amp is very small, essentially zero; V   V   0
V  V 

Inverting Amplifier
V

V+

Hence;

Vin  V  Vin iin 

R1
R1

Inverting Amplifier

The op-amp input resistance is large, so the current into the +ve and –ve op-amp inputs terminal will be small, essentially zero
V   Vout  iin RF
Vin  V 
V  Vout 
RF
R1

V 0

Vin
 Vout 
RF
R1

Vout
RF
Av 

Vin
R1

Inverting Amplifier

Currents and voltages in the inverting op-amp

Inverting Amplifier - Example

Design an inverting amplifier with a specified voltage gain. Specification: Design the circuit such that the voltage gain is
Av = -5. Assume the op-amp is driven by an ideal sinusoidal source, vs = 0.1sin wt (V), that can supply a maximum current of 5 µA. Assume that frequency w is low so that any frequency effects can be neglected.

Inverting Amplifier – T-Network

R3 R3
R2
Av   (1 
 )
R1
R4 R2

Non-Inverting Amplifier

Circuit consists of an opamp and three resistors.
The negative (-) input to the op-amp is grounded through R1 and also to the feedback signal from the output (via RF).
The positive (+) input is connected to the input signal Non-Inverting Amplifier
Input current to op-amp is very small. No signal voltage is created across
R2 and hence V   vin
V   V so it follows that;

 V   vin

Non-Inverting Amplifier so RF and R1 carry the same current. Hence vout is related to V through a voltage-divider relationship  I  0

R1
V  vout R1  RF

R1 vin  vout R1  RF

vout
RF
Av 
 1 vin R1

Non-Inverting Amplifier

The output has the same polarity as the input

a positive input signal produces a positive output signal.

The ratio of R1 and RF determines the gain.

When a voltage is applied to the amplifier, the output voltage increases rapidly and will continue to rise until the voltage across R1 reaches the input voltage. Thus negligible input current will flow into the amplifier, and the gain depends only on R1 and RF

Non-Inverting Amplifier

The input resistance to the non-inverting amplifier is very high because the input current to the amplifier is also the input current to the op-amp, I+, which must be extremely small Non-Inverting Amplifier v1  v2 v1 i1  
R1
v1  vo i2 
R2
i1  i2 v1  vo v1 

R1
R2
vout
R2
Av 
 1 vin R1

Voltage Follower / Buffer Amplifier

This “buffer” is used to control impedance levels in the circuit
– it isolates part of the overall
(measurement) circuit from the output (driver).
The input impedance to the buffer is very high and its output impedance is low.
The output voltage from a source with high output impedance can, via the buffer, supply signal to one or more loads that have a low impedance.

Voltage Follower / Buffer Amplifier

High input impedance.
Low output impedance.
Voltage gain = 1

Vout  Vin
Vout
Av 
1
Vin

Summing Amplifier

The inverting amplifier can accept two or more inputs and produce a weighted sum. Using the same reasoning as with the inverting amplifier, that V ≈ 0.
The sum of the currents through R1, R2,…,Rn is:

Vn
V1 V2 iin  
 ... 
R1 R2
Rn

Summing Amplifier

The op-amp adjusts itself to draw iin through Rf (iin = if). if iin

Vout  iin R f
Rf
Rf 
 Rf
 V1
 R  V2 R  ...  VN R 

1
2
N 

The output will thus be the sum of V1,V2,…,Vn, weighted by the gain factors, Rf/R1 , Rf/R2 ….., Rf/Rn respectively.

Summing Amplifier

Special Cases for this Circuit:
1. If R1 = R2 =……= R then:
Vout  

if

iin

Rf
R

VIN 1  VIN 2  .....  VINn 

Summing Amplifier
2. If R1 = R2 = … = R and VIN1, VIN2, … are either 0V
(digital “0”) or 5V (digital “1”) then the output voltage is now proportional to the number of (digital)
1’s input. if iin

Summing Amplifier - Application

Digital to Analog Converter
- binary-weighted resistor DAC

Summing Amplifier - Application

Digital to Analog Converter

Differencing Amplifier

This circuit produces an output which is proportional to the difference between the two inputs

vout 

Rf
R1

v1  v 2 

Differencing Amplifier

The circuit is linear so we can look at the output due to each input individually and then add them
(superposition theorem)

Differencing Amplifier

Set v1 to zero. The output due to v2 is the same as the inverting amplifier, so

vout 2  

Rf
R1

v2

Differencing Amplifier

The signal to the non-inverting output, is reduced by the voltage divider:

v in 

Rf
R1  R f

v1

Differencing Amplifier

The output due to this is then that for a non-inverting amplifier: vout1

Rf 

v in
 1 

R1 

Differencing Amplifier

v in 

Rf
R1  R f

v1

vout1

Rf 

v in
 1 

R1 

vout1

R f  R f


 1 

R1  R1  R f



vout1

 Rf

R
 1

v1

v1

Differencing Amplifier

vout1

 Rf

R
 1

Thus the output is:

v1

vout 2  

vout  vout1  vout2 

Rf
R1

Rf
R1

v2

v1  v2 

Thus the amplifier subtracts the inputs and amplifies their difference.

Differencing Amplifier - Example

Determine voltage output, vo when vi1=+1 V, vi2=-1V,
R1=R3=10 kΩ, R2=20 kΩ and R4=21 kΩ
When vi1=vi2=+1V

Solution: vo=-4.0323
Solution: vo=0.0323

Integrator

The basic integrator is easily identified by the capacitor in the feedback loop.
A constant input voltage yields a ramp output. The input resistor and the capacitor form an RC circuit

Integrator

The slope of the ramp is determined by the RC time constant. The integrator can be used to change a square wave input into a triangular wave output

Integrator

The capacitive impedance:

1
1
Xc 

jC sC

Integrator

Vin
Vout
Vout


  sCV out
The input current: I 
Ri
Xc
1 /sC

Integrator

Vout
1

Vin
sCR i

Integrator

Thus the output in time domain:

Vout

1
1

Vin  
Vindt
jRi C
Ri C

Integrator
The input current:
Vin
Vout
Vout
I


 sCVout
Ri
Xc
1 / sC
Vout
1

Vin sCR The capacitive impedance:
1
1
Xc 

jC sC

Thus the output in time domain:
1
1
Vout  
Vin  
 Vindt jCR RC
Rate of change of the output:

Vout
Vin

t
RC

Integrator

The output voltage:

Vout is the same as the voltage on the negative side of the capacitor
When constant positive input voltage (step or pulse) is applied, the output ramp decreases negatively until the op-amp saturates at its maximum level

Usefulness:

Especially use in triangularwave oscillators

Integrator - Example

Determine the rate of change of the output voltage in response to the input square wave. The output voltage is initially zero. The pulse width is 100 s.
C = 0.01F
Draw the waveform.
+2.5V

-2.5V

_

Vin
R = 10 k

7

2

Vout

741
3

+

6
4

Integrator - Example
Solution:
(a) during capacitor charging
Vout
Vin
5V


 50 mV / s
t
RC
(10 k)(0.01 F )

during capacitor discharging
Vout
Vin
5V


 50 mV / s
t
RC
(10 k)(0.01 F )

Integrator - Example
(b)
+ 2.5V

Vin
- 2.5V

0

Vout
- 5V

100 s

100 s

100 s

Real Integrator
CF
RF

Vin

+V

R1

_

7

2
741
3

RP

+

6

Vout

4

-V

RP : added to the noninverting input to balance the effects of bias current (usually the value is similar to R1)
RF : provides the dc feedback and overcome the saturation problem, i.e. to reduce the error voltage at the output

Differentiator

V in
0

t0

t1

+

t2

V in

0

+

t0 - t1

V out
0

t0 t1 V out

t2

t

Differentiator

The differentiator does the opposite of the integrator in that it takes a sloping input and provides an output that is proportional to the rate of change of the input.
Note the capacitor is in the input circuit.
The output voltage can be determined by the formula below: Vout

dVin
R

Vin   RC
1
dt jC Differentiator
C

R

Vin

Vout iR iC

dVC d (Vin  0) dVin iC  C
C
C dt dt dt 0  Vout
Vout
iR 

R
R

------------> Eqn. 1

------------> Eqn. 2

dVin
Vout   RC dt Differentiator

When input is a positive-going ramp, the output is negative
(capacitor is charging)

When input is a negative-going ramp, the output is positive
(capacitor is discharging) – current is the opposite direction

Differentiator - Example

Determine the output voltage of the op-amp differentiator for the triangular-wave input shown
R
2.2 k

C

+ 5V

Vin

10 s

0
5 s

15 s

t

0.001 F

_

7

Vout

2
741

- 5V

3

+

6
4

Differentiator - Example

At t = 0, Vin is a positive-going ramp ranging from -5V to
+5V (i.e. +10V change) in 5 s
From t = 5 s to 10 s, Vin is a negative-going ramp ranging from +5V to -5V (i.e. -10V change) time constant, RC = (2.2 k)(0.001 ) = 2.2 s the slope, VC / t = 10 V/ 5 s = 2 V/ s

 VC
Vout  
 t
 VC
Vout  
 t

 RC  2V / s 2.2 s  4.4V

 RC   2V / s 2.2 s  4.4V

(+ve ramp)
(-ve ramp)

Differentiator - Example

Vin

0

t

+4.4 V
Vout

0
-4.4 V

5 s

10 s

15 s

t

A Practical Differentiator

A shunt capacitor and a series resistor are added – to reduce high frequency noise
CF
RF
Vin

R

C

_

7

2
741
3

RP

+

6
4

Vout

Comparator

The comparator is an op-amp circuit that compares two input voltages and produces an output indicating the relationship between them.
The inputs can be two signals (such as two sine waves) or a signal and a fixed dc reference voltage.
Comparators are most commonly used in digital applications. Comparator

Digital circuits respond to rectangular or square waves, rather than sine waves.
These waveforms are made up of alternating (high and low) dc levels and the transitions between them.

Transitions

"High" dc level
"Low" dc level

Comparator

Example:
Assume that the digital system is designed to perform a specific function when a sine wave input reaches a value of 10 V

Comparator

V ref

V

+

Variable
voltage source Comparator
Digital
system

V

Comparator

+V

With nonzero-level detection the voltage divider or zener diode sets the
VZ
reference voltage at which the op-amp turns goes to the maximum voltage level. R

V out
+
V in

(c) Zener diode sets reference voltage

Comparator
V REF
V in 0

t

+V out (max)
V out
0
V out (max)

t

Comparator Waveforms

Comparator

Remember that the comparator is configured in open-loop, making the gain very high. This is open-loop configuration.
This makes the comparator very susceptible to unwanted signals
(noise) that could cause the output to arbitrarily switch states.

Comparator

If the level of the pulse must be less than the output of a saturated op-amp, a zener-diode can be used to limit the output to a particular voltage. This is called output bounding.
Either positive, negative, or both halves of the output signal can be bounded by use of one or two zener diodes respectively

D1 D2
0

+V Z 2 + 0.7 V

V in R i
0

V Z 2  0.7 V

Comparator - Application

Over-Temperature Sensing circuit

Comparator - Application

Analog to Digital Converter

Summary

The summing amplifier’s output is the sum of the inputs.
An averaging amplifier yields an output that is the average of all the inputs.
The scaling adder has inputs of different weight with each contributing more or less to the input.
Integrators change a constant voltage input to a sloped output. Differentiators change a sloping input into a step voltage proportional to the rate of change.
The op-amp comparator’s output changes state when the input voltage exceeds the reference voltage.

Frequency Response of Op-amp

The “frequency response” of any circuit is the magnitude of the gain in decibels (dB) as a function of the frequency of the input signal.
The decibel is a common unit of measurement for the relative magnitude of two power levels. The expression for such a ratio of power is:

Power level in dB = 10log10(P1/P2)

Note: A decibel is one-tenth of a "Bel", a seldom-used unit named for Alexander Graham Bell, inventor of the telephone Frequency Response of Op-amp

The voltage or current gain of an amplifier expressed in dB is 20 log10|A|, where A = Vout/Vin.
The frequency response of an op-amp has a low-pass characteristic (passing low-frequency signals, attenuating high-frequency signals)

Frequency Response of Op-amp

The bandwidth is the frequency at which the power of the output signal is reduced to half that of the maximum output power.
This occurs when the power gain A drops by 3 dB. In
Figure shown before, the bandwidth is fc Hz.
For all op-amps, the Gain*Bandwidth product is a constant. Hence, if the gain of an op-amp is decreased, its operational bandwidth increases proportionally.

Frequency Response of Op-amp

Another parameter reflecting the op-amp’s ability to varying signals is the slew rate.
Vo
SR 
t
Slew rate provides a parameter specifying the maximum rate of change of the output voltage when driven by a large step-input signal.
If one tried to drive the output at a rate of voltage change greater than the slew rate:
 Output would not be able to change fast enough and would not vary over the full range expected
 Clipping or distortion

Frequency Response of Op-amp

The maximum frequency at which an op-amp may operate depends on both the bandwidth (BW) and SR parameters of the op-amp.
Sinusoidal signal: vo  K sin(2ft )
Signal maximum rate of change: 2πfK
To prevent distortion at the output, the rate of change must also be less than SR:
SR

f 

2K
SR

K

Frequency Response of Op-amp - Example
For the signal and circuit below, determine the maximum frequency that may be used. Op-amp slew rate is SR =
0.5 V/µs.

Frequency Response of Op-amp - Example

Solution

240k
A

 24
R1
10k
Rf

K  AVi  24(0.02)  0.48V
SR 0.5


 1.1106
K 0.48
Since the signal frequency, ω = 300 x 103 rad/s is less than the maximum value, no output distortion will result.