The ideal op-amp
The ideal op-amp is one with optimum characteristics,
which cannot be attained in the real world. Nevertheless,
actual op-amp circuits can often approach this ideal.
The ideal op amp has infinite voltage gain, infinite input resistance
(open), and zero output resistance.
Vin Rin= ∞ AvVin Vout
Rout= 0
+

The practical op-amp
Practical op-amps have limitations including power and
voltage limits. A practical op-amp has high voltage gain,
high input resistance, and low output resistance.
Vin
Rin AvVin Vout
Rout
There are two inputs, labeled inverting and non inverting because of
the phase relation of the input and output signals.
+

inverting input
non inverting
input
The differential amplifier
Most op-amps have a differential amplifier (“diff-amp”) as
the input stage. The differential amplifier has important
advantages over other amplifiers; for example it can reject
common-mode noise.
The input is in
single-ended mode.
RC1 RC2
RE
+VCC
VEE
Q1 Q2
At the emitters, the
signal is ½ of the input.
The signal at the collector of
Q1 is inverted.
The signal at the collector
of Q2 is not inverted.
Differential and common-mode signals
Signals can be applied to either or both inputs. If two input
signals are out of phase, they are in differential-mode. If the
signals are in phase, they are in common-mode.
RC1 RC2
RE
+VCC
VEE
Q1 Q2
When the inputs are
out of phase, the
outputs are amplified
and larger than with
one input.
Inputs out of phase
Inputs in phase
When the inputs are in
phase, the outputs tend
to cancel and are near
zero.
Common-Mode Rejection Ratio (CMRR)
Many times, noise sources will induce an unwanted voltage
in a signal line. When the noise is induced in common-
mode, the differential amplifier tends to cancel it. (The diff-
amp cannot reject any signal that is in differential mode.)
The ability to reject common-mode signals is measured with a
parameter called the common-mode rejection ratio (CMRR), which
is defined as ( )
CMRR
v d
cm
A
A

CMRR can be expressed in decibels as ( )
CMRR 20log v d
cm
A
A
 
  
 
Common-Mode Rejection Ratio (CMRR)
From the defining equation for CMRR:
( ) 500
CMRR
0.1
v d
cm
A
A
  
Expressed in decibels, it is
 
( )
CMRR 20log 20log 5000
v d
cm
A
A
 
  
 
 
A certain diff-amp has a differential voltage
gain of 500 and a common-mode gain of
0.1. What is the CMRR?
5000
74 dB
Common-Mode Rejection Ratio (CMRR)
The differential signal is amplified by 100. Therefore,
the signal output is
Vout = Av(d) x Vin = 100 x 50 mV =
( )
4.5
100 100 100
0.0032
CMRR 90 dB 10 31,600
v d
cm
A
A     
A certain diff-amp has Ad = 100 and a CMRR
of 90 dB. Describe the output if the input is a
50 mV differential signal and a common
mode noise of 1.0 V is present.
The common-mode gain can be found by
The noise is amplified by 0.0032. Therefore,
Vnoise = Acm x Vin = 0.0032 x 1.0 V = 3.2 mV
5.0 V
Op-amp parameters
Some important op-amp parameters are:
Input bias
current:
Differential input
resistance:
Common-mode
input resistance:
Input offset
current:
Average of input currents required to bias
the first stage of the amplifier:
Total resistance between the inverting and
non-inverting inputs
Total resistance between each input and
ground.
Absolute difference between the two bias
currents:
1 2
BIAS
2
I I
I


OS 1 2
I I I
 
Op-amp parameters
Output
resistance:
Common-mode
input voltage
range:
Common-mode
rejection ratio
Slew rate:
The resistance when viewed from the output
terminal.
Range of input voltages, which, when applied
to both inputs, will not cause clipping or other
distortion.
Ratio of the differential gain to the common-
mode gain. The differential gain for the op-amp
by itself is the same as its open loop gain.
( )
CMRR
v d ol
cm cm
A A
A A
 
The maximum rate of change of the output in
response to a step input voltage.
Op-amp parameters
The output goes from 10 V to +10 V in 25 s.
What is the slew rate for
the output signal shown in
response to a step input?
Vout (V)
12
10
10
12
0
25 s
20 V
Slew rate =
25 s
out
V
t 

 

0.8 V/s
Negative feedback
In 1921, Harold S. Black was working on the problem of
linearizing and stabilizing amplifiers. While traveling to work on
the ferry, he suddenly realized that if he returned some of the
output back to the input in opposite phase, he had a means of
canceling distortion. One of the most important ideas in
electronics was sketched out on his newspaper that morning.
+ Vout
Vin
Vf -
Feedback
network
The op-amp has a differential
amplifier as the input stage. When a
feedback network returns a fraction
of the output to the inverting input,
only the difference signal (Vin – Vf)
is amplified.
Op amp circuits with negative feedback
Negative feedback is used in almost all linear op-amp
circuits because it stabilizes the gain and reduces
distortion. It can also increase the input resistance.

+
Feedback
network
Vf
Vin
Rf
Ri
Vout
A basic configuration is a noninverting amplifier. The difference
between Vin and Vf is very small due to feedback. Therefore, .
in f
V V

(NI) 1 f
cl
i
R
A
R
 
The closed-loop gain for the
noninverting amplifier can
be derived from this idea; it
is controlled by the feedback
resistors:
Op amp circuits with negative feedback

+
Vin
Rf
Ri
Vout
The inverting amplifier is a basic configuration in which the
noninverting input is grounded (sometimes through a resistor to
balance the bias inputs). Again, the difference between Vin and Vf is
very small due to feedback; this implies that the inverting input is
nearly at ground. This is referred to as a virtual ground. The virtual
ground looks like ground to voltage, but not to current!
(I)
f
cl
i
R
A
R

The closed-loop gain for the
inverting amplifier can be
derived from this idea; again it
is controlled by the feedback
resistors:
Virtual ground
Input resistance for the noninverting amplifier
The input resistance of an op-amp without feedback is
Rin. For the 741C, the manufacturer’s specified value of
Rin is 2 M. Negative feedback increases this to
Rin(NI) = (1 + AolB)Rin. This is so large that for all
practical circuits it can be considered to be infinite.
Keep in mind that, although Rin(NI)
is extremely large, the op-amp is
a dc amplifier and still requires a
dc bias path for the input.

+
Vin
Rf
Ri
Vout
Output resistance for the noninverting amplifier
The output resistance of an op-amp without feedback
is Rout. Negative feedback decreases this by a factor of
(1 + AolB). This is so small that for all practical
circuits it can be considered to be zero.

+
Vin
Rf
Ri
Vout
(NI)
1
out
out
ol
R
R
A B


The low output resistance
implies that the output voltage
is independent of the load
resistance (as long as the
current limit is not exceeded).
    
 
in(NI) 1 1+ 100,000 0.040 2 M =
ol in
R A B R
   
The gain is
(I)
36 k
1 1
1.5 k
f
cl
i
R
A
R

    

25
The input resistance is
8 G
The feedback fraction is
1
0.040
25
B  

+
Vin
Rf
Ri
Vout
36 k
1.5 k
Solution continued on next slide…
What are the input and output resistances and the gain of
the noninverting amplifier? Assume the op amp has
Aol = 100,000, Rin = 2 M, and Rout = 75 
The last result illustrates why it is rarely
necessary to calculate an exact value for
the input resistance of a noninverting
amplifier. For practical circuits, you can
assume it is ideal.

+
Vin
Rf
Ri
Vout
36 k
1.5 k
(continued)
The output resistance is
  
(NI)
75
=
1 1+ 100,000 0.040
out
out
ol
R
R
A B

 

0.019 
This extremely small resistance is close to ideal. As in the
case of the input resistance, it is rarely necessary to calculate
an exact value for the noninverting amplifier.
Input resistance for the inverting amplifier
Recall that negative feedback forces the inverting input
to be near ac ground for the inverting amplifier. For
this reason, the input resistance of the inverting
amplifier is equal to just the input resistor, Ri. That is,
Rin(I) = Ri.
The low input resistance is
usually a disadvantage of this
circuit. However, because the
Rin(I) is equal to Ri, it can easily
be set by the user for those cases
where a specific value is needed.

+
Vin
Rf
Ri
Vout
Output resistance for the inverting amplifier
The equation for the output resistance of the inverting
amplifier is the essentially the same as the
noninverting amplifier:

+
Vin
Rf
Ri
Vout
(I)
1
out
out
ol
R
R
A B


Although Rout(I) is very small,
this does not imply that an op-
amp can drive any load. The
maximum current that the op-
amp can supply is limited; for
the 741C, it is typically 20 mA.

+
Vin
Rf
Ri
Vout
The gain is (I)
36 k
1.5 k
f
cl
i
R
A
R

  

36 k
1.5 k
24
The input resistance = Ri = 1.5 k
The output resistance is nearly identical to the noninverting
case, where it was shown to be negligible.
What is the input resistance and the gain
of the inverting amplifier?
Voltage-follower
The voltage-follower is a special case of the noninverting
amplifier in which Acl = 1. The input resistance is
increased by negative feedback and the output resistance
is decreased by negative feedback. This makes it an ideal
circuit for interfacing a high-resistance source with a low
resistance load.

+
Vin Vout
Operational
amplifier
Differential
amplifier
Common-mode
rejection ratio
(CMRR)
Selected Key Terms
An amplifier that produces an output
proportional to the difference of two inputs.
A measure of a diff-amp's or op-amp's
ability to reject signals that appear the same
on both inputs; the ratio of differential
voltage gain or open-loop gain (for op-amps)
to common-mode gain.
A special type of amplifier exhibiting very
high open-loop gain, very high input
resistance, very low output resistance, and
good rejection of common-mode signals.
Open-loop
voltage gain
Closed-loop
voltage gain
Noninverting
amplifier
Inverting
amplifier
Selected Key Terms
An op-amp closed-loop configuration in
which the input signal is applied to the
noninverting input.
The overall voltage gain of an op-amp with
negative feedback.
An op-amp closed-loop configuration in
which the input signal is applied to the
inverting input.
The internal voltage gain of an op-amp
without feedback.
Quiz
1. When two identical in-phase signals are applied to the
inputs of a differential amplifier, they are said to be
a. feedback signals.
b. noninverting signals.
c. differential-mode signals.
d. common-mode signals.
Quiz
2. Assume a differential amplifier has an input signal
applied to the base of Q1 as shown. An inverted replica
of this signal will appear at the
a. emitter terminals.
b. collector of Q1
c. collector of Q2
d. all of the above.
RC1 RC2
RE
VEE
Q1 Q2
Quiz
3. A differential amplifier will tend to reject
a. noise that is in differential-mode.
b. noise that is in common-mode.
c. only high frequency noise.
d. all noise.
Quiz
4. The average of two input currents required to bias the
first stage of an op-amp is called the
a. input offset current.
b. open-loop input current.
c. feedback current.
d. input bias current.
Quiz
5. The slew rate illustrated is
a. 0.5 V/s
b. 1.0 V/s
c. 2.0 V/s
d. 2.4 V/s
Vout (V)
12
10
10
12
0
10 s
Quiz
6. For the circuit shown, Vf is approximately equal to
a. Vin
b. Vout
c. ground.
d. none of the above.

+
Feedback
network
Vf
Vin
Rf
Ri
Vout
Quiz
7. For the inverting amplifier shown, the input resistance is
closest to
a. zero
b. 10 k
c. 2 M
d. 8 G

+
Vin
Rf
Ri
Vout
150 k
10 k
Quiz
8. For the inverting amplifier shown, the output resistance
is closest to
a. zero
b. 10 k
c. 150 k
d. 8 G

+
Vin
Rf
Ri
Vout
150 k
10 k
Quiz
9. The gain of the inverting amplifier shown is
a. 1
b. 10
c. 15
d. 16

+
Vin
Rf
Ri
Vout
150 k
10 k
Quiz
10. A voltage follower has
a. current gain.
b. voltage gain.
c. both of the above.
d. none of the above.
Quiz
Answers:
1. d
2. b
3. b
4. d
5. c
6. a
7. b
8. a
9. c
10. a

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operational amplifier and it's applications

  • 1. The ideal op-amp The ideal op-amp is one with optimum characteristics, which cannot be attained in the real world. Nevertheless, actual op-amp circuits can often approach this ideal. The ideal op amp has infinite voltage gain, infinite input resistance (open), and zero output resistance. Vin Rin= ∞ AvVin Vout Rout= 0 + 
  • 2. The practical op-amp Practical op-amps have limitations including power and voltage limits. A practical op-amp has high voltage gain, high input resistance, and low output resistance. Vin Rin AvVin Vout Rout There are two inputs, labeled inverting and non inverting because of the phase relation of the input and output signals. +  inverting input non inverting input
  • 3. The differential amplifier Most op-amps have a differential amplifier (“diff-amp”) as the input stage. The differential amplifier has important advantages over other amplifiers; for example it can reject common-mode noise. The input is in single-ended mode. RC1 RC2 RE +VCC VEE Q1 Q2 At the emitters, the signal is ½ of the input. The signal at the collector of Q1 is inverted. The signal at the collector of Q2 is not inverted.
  • 4. Differential and common-mode signals Signals can be applied to either or both inputs. If two input signals are out of phase, they are in differential-mode. If the signals are in phase, they are in common-mode. RC1 RC2 RE +VCC VEE Q1 Q2 When the inputs are out of phase, the outputs are amplified and larger than with one input. Inputs out of phase Inputs in phase When the inputs are in phase, the outputs tend to cancel and are near zero.
  • 5. Common-Mode Rejection Ratio (CMRR) Many times, noise sources will induce an unwanted voltage in a signal line. When the noise is induced in common- mode, the differential amplifier tends to cancel it. (The diff- amp cannot reject any signal that is in differential mode.) The ability to reject common-mode signals is measured with a parameter called the common-mode rejection ratio (CMRR), which is defined as ( ) CMRR v d cm A A  CMRR can be expressed in decibels as ( ) CMRR 20log v d cm A A       
  • 6. Common-Mode Rejection Ratio (CMRR) From the defining equation for CMRR: ( ) 500 CMRR 0.1 v d cm A A    Expressed in decibels, it is   ( ) CMRR 20log 20log 5000 v d cm A A          A certain diff-amp has a differential voltage gain of 500 and a common-mode gain of 0.1. What is the CMRR? 5000 74 dB
  • 7. Common-Mode Rejection Ratio (CMRR) The differential signal is amplified by 100. Therefore, the signal output is Vout = Av(d) x Vin = 100 x 50 mV = ( ) 4.5 100 100 100 0.0032 CMRR 90 dB 10 31,600 v d cm A A      A certain diff-amp has Ad = 100 and a CMRR of 90 dB. Describe the output if the input is a 50 mV differential signal and a common mode noise of 1.0 V is present. The common-mode gain can be found by The noise is amplified by 0.0032. Therefore, Vnoise = Acm x Vin = 0.0032 x 1.0 V = 3.2 mV 5.0 V
  • 8. Op-amp parameters Some important op-amp parameters are: Input bias current: Differential input resistance: Common-mode input resistance: Input offset current: Average of input currents required to bias the first stage of the amplifier: Total resistance between the inverting and non-inverting inputs Total resistance between each input and ground. Absolute difference between the two bias currents: 1 2 BIAS 2 I I I   OS 1 2 I I I  
  • 9. Op-amp parameters Output resistance: Common-mode input voltage range: Common-mode rejection ratio Slew rate: The resistance when viewed from the output terminal. Range of input voltages, which, when applied to both inputs, will not cause clipping or other distortion. Ratio of the differential gain to the common- mode gain. The differential gain for the op-amp by itself is the same as its open loop gain. ( ) CMRR v d ol cm cm A A A A   The maximum rate of change of the output in response to a step input voltage.
  • 10. Op-amp parameters The output goes from 10 V to +10 V in 25 s. What is the slew rate for the output signal shown in response to a step input? Vout (V) 12 10 10 12 0 25 s 20 V Slew rate = 25 s out V t      0.8 V/s
  • 11. Negative feedback In 1921, Harold S. Black was working on the problem of linearizing and stabilizing amplifiers. While traveling to work on the ferry, he suddenly realized that if he returned some of the output back to the input in opposite phase, he had a means of canceling distortion. One of the most important ideas in electronics was sketched out on his newspaper that morning. + Vout Vin Vf - Feedback network The op-amp has a differential amplifier as the input stage. When a feedback network returns a fraction of the output to the inverting input, only the difference signal (Vin – Vf) is amplified.
  • 12. Op amp circuits with negative feedback Negative feedback is used in almost all linear op-amp circuits because it stabilizes the gain and reduces distortion. It can also increase the input resistance.  + Feedback network Vf Vin Rf Ri Vout A basic configuration is a noninverting amplifier. The difference between Vin and Vf is very small due to feedback. Therefore, . in f V V  (NI) 1 f cl i R A R   The closed-loop gain for the noninverting amplifier can be derived from this idea; it is controlled by the feedback resistors:
  • 13. Op amp circuits with negative feedback  + Vin Rf Ri Vout The inverting amplifier is a basic configuration in which the noninverting input is grounded (sometimes through a resistor to balance the bias inputs). Again, the difference between Vin and Vf is very small due to feedback; this implies that the inverting input is nearly at ground. This is referred to as a virtual ground. The virtual ground looks like ground to voltage, but not to current! (I) f cl i R A R  The closed-loop gain for the inverting amplifier can be derived from this idea; again it is controlled by the feedback resistors: Virtual ground
  • 14. Input resistance for the noninverting amplifier The input resistance of an op-amp without feedback is Rin. For the 741C, the manufacturer’s specified value of Rin is 2 M. Negative feedback increases this to Rin(NI) = (1 + AolB)Rin. This is so large that for all practical circuits it can be considered to be infinite. Keep in mind that, although Rin(NI) is extremely large, the op-amp is a dc amplifier and still requires a dc bias path for the input.  + Vin Rf Ri Vout
  • 15. Output resistance for the noninverting amplifier The output resistance of an op-amp without feedback is Rout. Negative feedback decreases this by a factor of (1 + AolB). This is so small that for all practical circuits it can be considered to be zero.  + Vin Rf Ri Vout (NI) 1 out out ol R R A B   The low output resistance implies that the output voltage is independent of the load resistance (as long as the current limit is not exceeded).
  • 16.        in(NI) 1 1+ 100,000 0.040 2 M = ol in R A B R     The gain is (I) 36 k 1 1 1.5 k f cl i R A R        25 The input resistance is 8 G The feedback fraction is 1 0.040 25 B    + Vin Rf Ri Vout 36 k 1.5 k Solution continued on next slide… What are the input and output resistances and the gain of the noninverting amplifier? Assume the op amp has Aol = 100,000, Rin = 2 M, and Rout = 75 
  • 17. The last result illustrates why it is rarely necessary to calculate an exact value for the input resistance of a noninverting amplifier. For practical circuits, you can assume it is ideal.  + Vin Rf Ri Vout 36 k 1.5 k (continued) The output resistance is    (NI) 75 = 1 1+ 100,000 0.040 out out ol R R A B     0.019  This extremely small resistance is close to ideal. As in the case of the input resistance, it is rarely necessary to calculate an exact value for the noninverting amplifier.
  • 18. Input resistance for the inverting amplifier Recall that negative feedback forces the inverting input to be near ac ground for the inverting amplifier. For this reason, the input resistance of the inverting amplifier is equal to just the input resistor, Ri. That is, Rin(I) = Ri. The low input resistance is usually a disadvantage of this circuit. However, because the Rin(I) is equal to Ri, it can easily be set by the user for those cases where a specific value is needed.  + Vin Rf Ri Vout
  • 19. Output resistance for the inverting amplifier The equation for the output resistance of the inverting amplifier is the essentially the same as the noninverting amplifier:  + Vin Rf Ri Vout (I) 1 out out ol R R A B   Although Rout(I) is very small, this does not imply that an op- amp can drive any load. The maximum current that the op- amp can supply is limited; for the 741C, it is typically 20 mA.
  • 20.  + Vin Rf Ri Vout The gain is (I) 36 k 1.5 k f cl i R A R      36 k 1.5 k 24 The input resistance = Ri = 1.5 k The output resistance is nearly identical to the noninverting case, where it was shown to be negligible. What is the input resistance and the gain of the inverting amplifier?
  • 21. Voltage-follower The voltage-follower is a special case of the noninverting amplifier in which Acl = 1. The input resistance is increased by negative feedback and the output resistance is decreased by negative feedback. This makes it an ideal circuit for interfacing a high-resistance source with a low resistance load.  + Vin Vout
  • 22. Operational amplifier Differential amplifier Common-mode rejection ratio (CMRR) Selected Key Terms An amplifier that produces an output proportional to the difference of two inputs. A measure of a diff-amp's or op-amp's ability to reject signals that appear the same on both inputs; the ratio of differential voltage gain or open-loop gain (for op-amps) to common-mode gain. A special type of amplifier exhibiting very high open-loop gain, very high input resistance, very low output resistance, and good rejection of common-mode signals.
  • 23. Open-loop voltage gain Closed-loop voltage gain Noninverting amplifier Inverting amplifier Selected Key Terms An op-amp closed-loop configuration in which the input signal is applied to the noninverting input. The overall voltage gain of an op-amp with negative feedback. An op-amp closed-loop configuration in which the input signal is applied to the inverting input. The internal voltage gain of an op-amp without feedback.
  • 24. Quiz 1. When two identical in-phase signals are applied to the inputs of a differential amplifier, they are said to be a. feedback signals. b. noninverting signals. c. differential-mode signals. d. common-mode signals.
  • 25. Quiz 2. Assume a differential amplifier has an input signal applied to the base of Q1 as shown. An inverted replica of this signal will appear at the a. emitter terminals. b. collector of Q1 c. collector of Q2 d. all of the above. RC1 RC2 RE VEE Q1 Q2
  • 26. Quiz 3. A differential amplifier will tend to reject a. noise that is in differential-mode. b. noise that is in common-mode. c. only high frequency noise. d. all noise.
  • 27. Quiz 4. The average of two input currents required to bias the first stage of an op-amp is called the a. input offset current. b. open-loop input current. c. feedback current. d. input bias current.
  • 28. Quiz 5. The slew rate illustrated is a. 0.5 V/s b. 1.0 V/s c. 2.0 V/s d. 2.4 V/s Vout (V) 12 10 10 12 0 10 s
  • 29. Quiz 6. For the circuit shown, Vf is approximately equal to a. Vin b. Vout c. ground. d. none of the above.  + Feedback network Vf Vin Rf Ri Vout
  • 30. Quiz 7. For the inverting amplifier shown, the input resistance is closest to a. zero b. 10 k c. 2 M d. 8 G  + Vin Rf Ri Vout 150 k 10 k
  • 31. Quiz 8. For the inverting amplifier shown, the output resistance is closest to a. zero b. 10 k c. 150 k d. 8 G  + Vin Rf Ri Vout 150 k 10 k
  • 32. Quiz 9. The gain of the inverting amplifier shown is a. 1 b. 10 c. 15 d. 16  + Vin Rf Ri Vout 150 k 10 k
  • 33. Quiz 10. A voltage follower has a. current gain. b. voltage gain. c. both of the above. d. none of the above.
  • 34. Quiz Answers: 1. d 2. b 3. b 4. d 5. c 6. a 7. b 8. a 9. c 10. a