Slide 9.1CE164 – 2024 F. Sepulveda - CSEE - Essex University
Foundations of Electronics – II
CE164
Lecture 9 (Week 24)
Op-Amps:
• Differential amp
• Ideal vs. real
• Negative Feedback
Francisco Sepulveda
E-mail: f.sepulveda
moodle.essex.ac.uk/course/view.php?id=3644
Slide 9.2CE164 – 2024 F. Sepulveda - CSEE - Essex University
OpAmps
(operational amplifiers)
Reading:
Chapter 17, Floyd & Buchla (2014, 8th ed.)
Slide 9.3CE164 – 2024 F. Sepulveda - CSEE - Essex University
LM741 - www.ti.com
INSIDE AN OP-AMP
Slide 9.4CE164 – 2024 F. Sepulveda - CSEE - Essex University
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 ideal op-amp
Voltage gain Av = Vout / Vin
inverting input
non inverting
input
Slide 9.5CE164 – 2024 F. Sepulveda - CSEE - Essex University
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
Practical op-amps
Voltage gain Av = Vout / Vin
Slide 9.6CE164 – 2024 F. Sepulveda - CSEE - Essex University
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.
The differential amplifier
scenario 1: single–ended input
Slide 9.7CE164 – 2024 F. Sepulveda - CSEE - Essex University
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
Differential and common-mode signals
scenario 2: differential mode
Slide 9.8CE164 – 2024 F. Sepulveda - CSEE - Essex University
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
Inputs in phase
When the inputs are in
phase, the outputs
tend to cancel and are
near zero.
Differential and common-mode signals
scenario 3: common mode
noise tends to be in common mode
Slide 9.9CE164 – 2024 F. Sepulveda - CSEE - Essex University
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.
NB: 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)
Av(d): differential Av
Acm : common mode gain
Slide 9.10CE164 – 2024 F. Sepulveda - CSEE - Essex University
From the defining equation for
CMRR: ( ) 500
CMRR
0.1
v d
cm
A
A
  
Expressed in decibels, it is
 ( )CMRR 20log 20log 5000v 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)
Slide 9.11CE164 – 2024 F. Sepulveda - CSEE - Essex University
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
Common-Mode Rejection Ratio
(CMRR)
Slide 9.12CE164 – 2024 F. Sepulveda - CSEE - Essex University
Some other 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.
Absolute difference between the two bias currents:
1 2
BIAS
2
I I
I


OS 1 2I I I -
Op-amp parameters
Total resistance between each input and ground.
Rin
Rin+I1 = IB+
I2 = IB-
Slide 9.13CE164 – 2024 F. Sepulveda - CSEE - Essex University
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, i.e., the slope of the linear part of Vout.
Op-amp parameters (cont.)
Slide 9.14CE164 – 2024 F. Sepulveda - CSEE - Essex University
The output in the linear region goes from -10 V to +10 V in 25 ms.
What is the slew rate for the output
signal shown in response to a step
input?
20 V
Slew rate =
25 s
outV
t m

 

0.8 V/ms
Slew rate
Vin
Vout
Important: The slew rate can significantly distort higher frequency signals. We must choose
opAmps with a slew rate that is suitable for the intended frequencies. Faster is usually better,
but more expensive.
For sinusoidal signals: SlewRate
min
= 2 p f Vmax
Slide 9.15CE164 – 2024 F. Sepulveda - CSEE - Essex University
 Amplification on its own produces nonlinear distortions, unstable gain, and
noise amplification.
 In 1921, Harold S. Black realized that if he returned some of the output back
to the input in opposite phase, he had a means of canceling distortion.
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.
Negative feedback
The feedback in negative feedback is always on the
opAmp’s inverting input terminal, but the input
signal can be on either of the two input terminals
Slide 9.16CE164 – 2024 F. Sepulveda - CSEE - Essex University
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 non-inverting amplifier. The difference between Vin and
Vf is very small due to feedback and high open loop gain. Therefore,
(NI)
1
f
cl
i
R
A
R
 
The closed-loop gain for the non-
inverting amplifier can be derived
from this idea; it is controlled by the
feedback resistors:
Op-amp with negative feedback:
non-inverting (input) case
Acl = closed loop gain
Slide 9.17CE164 – 2024 F. Sepulveda - CSEE - Essex University
 The inverting amplifier: the non-inverting input is grounded (sometimes through
a resistor to balance the bias inputs).
 As the difference between Vin and Vf is very small due to feedback, 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:
Op-amp with negative feedback:
inverting input case
Vf
Vf
If = Iin
Slide 9.18CE164 – 2024 F. Sepulveda - CSEE - Essex University
 The input resistance of an op-amp without feedback is Rin. For the 741C, the
manufacturer’s specified value of Rin is 2 MW.
 Negative feedback increases this to Rin(NI) = (1 + Aol B)Rin. This is so large
that for all practical circuits it can be considered to be infinite.
-
+Vin
Rf
Ri
Vout
(effective) Input resistance for the
non-inverting amplifier
Aol = open loop gain
B = feedback fraction =
=

( + )
=
1


Slide 9.19CE164 – 2024 F. Sepulveda - CSEE - Essex University
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).
Output resistance for the
non-inverting amplifier
Slide 9.20CE164 – 2024 F. Sepulveda - CSEE - Essex University
      in(NI) 1 1+ 100,000 0.040 2 M = ol inR A B R   W
The closed-loop gain is:
(I)
36 k
1 1
1.5 k
f
cl
i
R
A
R
W
    
W
25
The input resistance is
8 GW
The feedback fraction is
1
0.040
25
B  
-
+
Vin
Rf
Ri
Vout
36 kW
1.5 kW
Solution continues on next slide…
What are the input and output resistances and the gain of
the non-inverting amplifier below? Assume the opamp has
Aol = 100,000, Rin = 2 MW, and Rout = 75 W.
(NI)
f
cl
i
A
Slide 9.21CE164 – 2024 F. Sepulveda - CSEE - Essex University
The last result illustrates why it is rarely
necessary to calculate an exact value for
the input resistance of a non-inverting
amplifier. For practical circuits, you can
assume it is ideal.
-
+
Vin
Rf
Ri
Vout
36 kW
1.5 kW
(continued)
The output resistance is
  (NI)
75
=
1 1+ 100,000 0.040
out
out
ol
R
R
A B
W
 

0.019 W
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 non-
inverting amplifier.
Slide 9.22CE164 – 2024 F. Sepulveda - CSEE - Essex University
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 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
Input resistance for the
inverting amplifier
Slide 9.23CE164 – 2024 F. Sepulveda - CSEE - Essex University
The equation for the output resistance of the inverting amplifier
is essentially the same as for the non-inverting 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.
Output resistance for the
inverting amplifier
Slide 9.24CE164 – 2024 F. Sepulveda - CSEE - Essex University
-
+
Vin
Rf
Ri
Vout
The gain is (I)
36 k
1.5 k
f
cl
i
R
A
R
W
 -  - 
W
36 kW
1.5 kW
-24
The input resistance = Ri = 1.5 kW
The output resistance is nearly identical to the
non-inverting case, where it was shown to be
negligible (0.019 W).
What is the input resistance and
the closed-loop gain of this
inverting amplifier?
Slide 9.25CE164 – 2024 F. Sepulveda - CSEE - Essex University
The voltage-follower is a special case of the non-inverting
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
Voltage Follower
Slide 9.26CE164 – 2024 F. Sepulveda - CSEE - Essex University
Negative feedback summary
Purpose:
 Desensitize the output gain to variations in other circuit components.
 Reduce non-linear gain distortions (equivalent to keeping the gain
constant)
 Reduce the effect of noise
 Control the input and output impedances
 Extend the bandwidth of the amplifier.
Main disadvantage:
 Reduced gain compared to other configurations.
-
+
Vin
Rf
Ri
Vout
-
+Vin
Rf
Ri
Vout
inverting
non-Inverting