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
- R/2R Ladder DAC
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: vout1
Rf
v in
1
R1
Differencing Amplifier
v in
Rf
R1 R f
v1
vout1
Rf
v in
1
R1
vout1
R f R f
1
R1 R1 R f
vout1
Rf
R
1
v1
v1
Differencing Amplifier
vout1
Rf
R
1
Thus the output is:
v1
vout 2
vout vout1 vout2
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
jC sC
Integrator
Vin
Vout
Vout
sCV out
The input current: I
Ri
Xc
1 /sC
Thus the output in time domain:
1
1
Vout
Vin
Vindt jCR 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 = 0.01F
Draw the waveform.
+2.5V
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 jC 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
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(2ft )
Signal maximum rate of change: 2πfK
To prevent distortion at the output, the rate of change must also be less than SR:
SR
f
2K
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
1.1106
K 0.48
Since the signal frequency, ω = 300 x 103 rad/s is less than the maximum value, no output distortion will result.
