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electronics fundamentals
circuits, devices, and applications
THOMAS L. FLOYD
DAVID M. BUCHLA
chapter 18
Electronics Fundamentals 8th edition
Floyd/Buchla
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 18
Summary
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.
Rin= ∞
Vin
AvVin
Rout= 0
Vout
+
Electronics Fundamentals 8th edition
Floyd/Buchla
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 18
Summary
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.
There are two inputs, labeled inverting and non inverting because of
the phase relation of the input and output signals.
inverting input
Rin
Vin
AvVin
Rout
non inverting input
Vout
+
Electronics Fundamentals 8th edition
Floyd/Buchla
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 18
Summary
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
+VCC
common-mode noise.
The signal at the collector
The signal at the collector of
Q1 is inverted.
of Q2 is not inverted.
RC1
Q1
The input is in
single-ended mode.
Electronics Fundamentals 8th edition
Floyd/Buchla
RC2
Q2
RE
-VEE
At the emitters, the
signal is ½ of the input.
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 18
Summary
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.
When the inputs are
out of phase, the
outputs are amplified
and larger than with
one input.
When
the inputs are in
phase, the outputs tend
to cancel and are near
zero.
Inputs
Inputsout
in phase
of phase
Electronics Fundamentals 8th edition
Floyd/Buchla
+VCC
RC1
Q1
RC2
Q2
RE
-VEE
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 18
Summary
Common-Mode Rejection Ratio (CMRR)
Many times, noise sources will induce an unwanted voltage
in a signal line. When the noise is induced in commonmode, the differential amplifier tends to cancel it. (The diffamp 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
Av ( d )
CMRR 
Acm
 Av ( d ) 

A
 cm 
CMRR can be expressed in decibels as CMRR  20log 
Electronics Fundamentals 8th edition
Floyd/Buchla
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 18
Summary
Common-Mode Rejection Ratio (CMRR)
A certain diff-amp has a differential voltage
gain of 500 and a common-mode gain of
0.1. What is the CMRR?
From the defining equation for CMRR:
Av ( d ) 500
CMRR 

 5000
Acm
0.1
Expressed in decibels, it is
 Av ( d ) 
CMRR  20log 
  20log  5000   74 dB
 Acm 
Electronics Fundamentals 8th edition
Floyd/Buchla
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 18
Summary
Common-Mode Rejection Ratio (CMRR)
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 differential signal is amplified by 100. Therefore,
the signal output is
Vout = Av(d) x Vin = 100 x 50 mV = 5.0 V
The common-mode gain can be found by
A
100
100
100
Acm  v ( d ) 
 4.5 
 0.0032
CMRR 90 dB 10
31,600
The noise is amplified by 0.0032. Therefore,
Vnoise = Acm x Vin = 0.0032 x 1.0 V = 3.2 mV
Electronics Fundamentals 8th edition
Floyd/Buchla
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 18
Summary
Op-amp parameters
Some important op-amp parameters are:
Input bias Average of input currents required to bias
current: the first stage of the amplifier:
I BIAS 
I1  I 2
2
Differential input Total resistance between the inverting and
resistance: non-inverting inputs
Common-mode Total resistance between each input and
input resistance: ground.
Input offset Absolute difference between the two bias
current: currents:
IOS  I1 - I 2
Electronics Fundamentals 8th edition
Floyd/Buchla
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 18
Summary
Op-amp parameters
Output The resistance when viewed from the output
resistance: terminal.
Common-mode Range of input voltages, which, when applied
input voltage to both inputs, will not cause clipping or other
range: distortion.
Common-mode Ratio of the differential gain to the commonrejection ratio mode gain. The differential gain for the op-amp
by itself is the same as its open loop gain.
CMRR 
Av ( d )
Acm

Aol
Acm
Slew rate: The maximum rate of change of the output in
response to a step input voltage.
Electronics Fundamentals 8th edition
Floyd/Buchla
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 18
Summary
Op-amp parameters
Vout (V)
What is the slew rate for
the output signal shown in
response to a step input?
12
10
0
-10
-12
25 ms
The output goes from -10 V to +10 V in 25 ms.
Slew rate =
Electronics Fundamentals 8th edition
Floyd/Buchla
Vout 20 V

 0.8 V/ms
t
25 ms
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 18
Summary
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.
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.
Electronics Fundamentals 8th edition
Floyd/Buchla
Vin
Vout
+
Vf
Feedback
network
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 18
Summary
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.
A basic configuration is a noninverting amplifier. The difference
between Vin and Vf is very small due to feedback. Therefore, Vin  V f .
+
Vin
Vout
Vf
Rf
Ri
Electronics Fundamentals 8th edition
Floyd/Buchla
Feedback
network
The closed-loop gain for the
noninverting amplifier can be
derived from this idea; it is
controlled by the feedback
Rf
resistors: A
 1
cl (NI)
Ri
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 18
Summary
Op amp circuits with negative feedback
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!
Rf
Virtual ground
Ri
Vin
+
Electronics Fundamentals 8th edition
Floyd/Buchla
Vout
The closed-loop gain for the
inverting amplifier can be
derived from this idea; again it
is controlled by the feedback
resistors: A  - R f
cl (I)
Ri
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 18
Summary
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 MW. 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.
Vin
+
-
Vout
Rf
Ri
Electronics Fundamentals 8th edition
Floyd/Buchla
Keep in mind that, although
Rin(NI) is extremely large, the opamp is a dc amplifier and still
requires a dc bias path for the
input.
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 18
Summary
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
+
-
Vout
Rf
Ri
Electronics Fundamentals 8th edition
Floyd/Buchla
Rout
Rout (NI) 
1  Aol 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).
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 18
Summary
What are the input and output resistances and the gain of
the noninverting amplifier? Assume the op amp has
Aol = 100,000, Rin = 2 MW, and Rout = 75 W.
The gain is
Vin
36 kW
Acl (I)  1 
 1
 25
Ri
1.5 kW
1
The feedback fraction is B 
 0.040
25
Rf
+
-
Vout
Rf
36 kW
Ri
1.5 kW
The input resistance is
Rin(NI)  1  Aol B  Rin  1+ 100,000 0.040  2 MW = 8 GW
Solution continued on next slide…
Electronics Fundamentals 8th edition
Floyd/Buchla
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 18
Summary
(continued)
Vin
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.
+
Vout
-
Rf
36 kW
Ri
1.5 kW
The output resistance is
Rout
75 W
Rout (NI) 

= 0.019 W
1  Aol B 1+ 100,000  0.040 
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.
Electronics Fundamentals 8th edition
Floyd/Buchla
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 18
Summary
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.
Rf
Vin
Ri
+
Electronics Fundamentals 8th edition
Floyd/Buchla
Vout
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.
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 18
Summary
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:
Rout
Rout (I) 
Rf
Vin
Ri
+
Electronics Fundamentals 8th edition
Floyd/Buchla
Vout
1  Aol B
Although Rout(I) is very small,
this does not imply that an opamp can drive any load. The
maximum current that the opamp can supply is limited; for
the 741C, it is typically 20 mA.
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 18
Summary
Rf
What is the input resistance and the gain
of the inverting amplifier?
Ri
Vin
1.5 kW
The gain is Acl (I)  -
Rf
Ri
-
36 kW
 -24
1.5 kW
36 kW
-
Vout
+
The input resistance = Ri = 1.5 kW
The output resistance is nearly identical to the noninverting
case, where it was shown to be negligible.
Electronics Fundamentals 8th edition
Floyd/Buchla
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 18
Summary
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
-
Electronics Fundamentals 8th edition
Floyd/Buchla
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 18
Selected Key Terms
Operational
amplifier
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.
Differential
amplifier
An amplifier that produces an output
proportional to the difference of two inputs.
Common-mode
rejection ratio
(CMRR)
Electronics Fundamentals 8th edition
Floyd/Buchla
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 commonmode gain.
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 18
Selected Key Terms
Open-loop The internal voltage gain of an op-amp
voltage gain without feedback.
Closed-loop The overall voltage gain of an op-amp with
voltage gain negative feedback.
Noninverting An op-amp closed-loop configuration in
amplifier which the input signal is applied to the
noninverting input.
Inverting An op-amp closed-loop configuration in
amplifier which the input signal is applied to the
inverting input.
Electronics Fundamentals 8th edition
Floyd/Buchla
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 18
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.
Electronics Fundamentals 8th edition
Floyd/Buchla
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 18
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.
RC1
RC2
b. collector of Q1
c. collector of Q2
d. all of the above.
Q1
Q2
RE
-VEE
Electronics Fundamentals 8th edition
Floyd/Buchla
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 18
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.
Electronics Fundamentals 8th edition
Floyd/Buchla
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 18
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.
Electronics Fundamentals 8th edition
Floyd/Buchla
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 18
Quiz
5. The slew rate illustrated is
a. 0.5 V/ms
Vout (V)
12
10
b. 1.0 V/ms
0
c. 2.0 V/ms
d. 2.4 V/ms
Electronics Fundamentals 8th edition
Floyd/Buchla
-10
-12
10 ms
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 18
Quiz
6. For the circuit shown, Vf is approximately equal to
a. Vin
+
Vin
-
b. Vout
c. ground.
d. none of the above.
Electronics Fundamentals 8th edition
Floyd/Buchla
Vout
Vf
Rf
Ri
Feedback
network
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 18
Quiz
7. For the inverting amplifier shown, the input resistance is
closest to
Rf
a. zero
b. 10 kW
c. 2 MW
Vin
Ri
10 kW
150 kW
-
Vout
+
d. 8 GW
Electronics Fundamentals 8th edition
Floyd/Buchla
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 18
Quiz
8. For the inverting amplifier shown, the output resistance
is closest to
Rf
a. zero
b. 10 kW
c. 150 kW
Vin
Ri
10 kW
150 kW
-
Vout
+
d. 8 GW
Electronics Fundamentals 8th edition
Floyd/Buchla
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 18
Quiz
9. The gain of the inverting amplifier shown is
a. -1
Rf
b. -10
c. -15
d. -16
Electronics Fundamentals 8th edition
Floyd/Buchla
Vin
Ri
10 kW
150 kW
-
Vout
+
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 18
Quiz
10. A voltage follower has
a. current gain.
b. voltage gain.
c. both of the above.
d. none of the above.
Electronics Fundamentals 8th edition
Floyd/Buchla
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 18
Quiz
Answers:
Electronics Fundamentals 8th edition
Floyd/Buchla
1. d
6. a
2. b
7. b
3. b
8. a
4. d
9. c
5. c
10. a
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.