Module: Electronics II
Module Number: 650321
th
Electronic Devices and Circuit Theory, 9 ed., Boylestad and Nashelsky
Philadelphia University
Faculty of Engineering
Communication and Electronics Engineering
Amplifier Circuits-III
Operational Amplifiers (Op-Amps):
An operational amplifier, or op-amp, is a very high gain differential amplifier
with high input impedance and low output impedance.
Typical uses of the operational amplifier are
 to provide voltage amplitude changes (amplitude and polarity),
 oscillators,
 filter circuits, and
 many types of instrumentation circuits.
An op-amp contains a number of differential amplifier stages to achieve a very
high voltage gain.
- Symbol and Terminals:
The following Figure shows a basic op-amp with two inputs and one output as would
result
using
a
differential
amplifier
input stage. Each input
results in either the
same or an opposite
polarity (or phase)
output, depending on
whether the signal is
applied to the plus (+)
or the minus (-) input.
To illustrate what an op-amp is, let's consider its ideal characteristics. A practical opamp, of course, falls short of these ideal standards, but it is much easier to understand
and analyze the device from an ideal point of view. These two considerations are
clearly shown in the Figure Below.
Firstly, the ideal op-amp has
- infinite voltage gain,
- infinite bandwidth,
- it has an infinite input impedance (open) so that it does not load the driving
source,
- a zero output impedance.
Lecturer: Dr. Omar Daoud
Part III
1
Module: Electronics II
Module Number: 650321
th
Electronic Devices and Circuit Theory, 9 ed., Boylestad and Nashelsky
The input voltage, V in appears between the two input terminals, and the output
voltage is A v V in as indicated by the internal voltage source symbol.
Secondly, characteristics of a practical op-amp are:
- very high voltage gain,
- very high input impedance,
- low output impedance, and
- wide bandwidth.
- Internal Block Diagram:
A typical op-amp is made up of three types of amplifier circuit: a differential
amplifier, a voltage amplifier, and a push-pull amplifier, as shown in the Figure
below.
A differential amplifier is the input stage for the op-amp; it provides amplification of
the difference voltage between the two inputs. The second stage is usually a class A
amplifier that provides additional gain. Some op-amps may have more than one
voltage amplifier stage. A push-pull class B amplifier is typically used for the output
stage.
Lecturer: Dr. Omar Daoud
Part III
2
Module: Electronics II
Module Number: 650321
th
Electronic Devices and Circuit Theory, 9 ed., Boylestad and Nashelsky
The Differential Amplifier Input Stage
The term differential comes from the amplifier's ability to amplify the difference of
two input signals applied to its inputs . Only the difference in the two signals is
amplified; if there is no difference, the output is zero .
A basic differential amplifier circuit and its symbol are shown in the Figure. The
transistors (Q 1 and Q 2 ) and the collector resistors (R c1 and R c2 ) are carefully matched
to have identical characteristics. Notice that the two transistors share a single emitter
resistor, R E .
The Differential Amplifier Operation:
-
-
Assume both bases are connected to ground,
The emitter voltage will be -0.7 V because the voltage drops across both baseemitter junctions are equa1.
The emitter currents are equal (I E1 = I E2 ) and each is one-half of the current
through R E .
The collector currents are both equal and are approximately equal to the emitter
currents. Because the collector currents are the same, the collector voltages are
also the same, which reflects the zero difference in the input voltages (both
bases are at 0V.
If the base of Q 1 is disconnected from ground and connected to a small positive
voltage,
- Q l will conduct more current because the positive voltage on its base
causes the emitter voltage to increase slightly.
- Although the emitter voltage is a little higher, the total current through R E
is nearly the same as before.
- The emitter current is now divided so that more of it is in Q l and less in
Q2.
- As a result, the collector voltage of Q 1 will decrease and the collector
voltage of Q 2 will increase, reflecting the difference in the input voltages
(one is 0 V and the other at a small positive value).
Lecturer: Dr. Omar Daoud
Part III
3
Module: Electronics II
Module Number: 650321
th
Electronic Devices and Circuit Theory, 9 ed., Boylestad and Nashelsky
-
-
If the base of Q l is placed back at ground and a small positive voltage is
connected to the base of Q 2 ,
- Q 2 will conduct more current,
- Q l will conduct less,
- The emitter current is now divided so that more of it is in Q 2 and less in
Q1,
- As a result, the collector voltage of Q 1 will increase and the collector
voltage of Q 2 will decrease.
The differential amplifier exhibits three modes of operation based on the type of
input signals. These modes are single-ended, differential, and common. Since
the differential amplifier is the input stage of the op-amp, the op-amp exhibits
the same modes.
Op-Amp Input Modes and Parameters
1) Input Signal Modes
 Single-Ended Mode: one input is grounded and a signal voltage is applied
only to the other input.
- In the case where the signal voltage is applied to the inverting input, an
inverted, amplified signal voltage appears at the output.
- In the case where the signal is applied to the noninverting input with
the inverting input grounded, a non inverted, amplified signal voltage
appears at the output.
 Differential Input Mode: Two opposite-polarity (out-of-phase) signals are
applied to the inputs. This type of operation is also referred to as doubleended. The amplified difference between the two inputs appears on the output.
Lecturer: Dr. Omar Daoud
Part III
4
Module: Electronics II
Module Number: 650321
th
Electronic Devices and Circuit Theory, 9 ed., Boylestad and Nashelsky
Common Mode Input: Two signal voltages of the same phase, frequency, and
amplitude are applied to the two inputs.
When equal input signals are applied to both
inputs, they cancel, resulting in a zero output
voltage. This action is called common-mode
rejection. Its importance lies in the situation
where an unwanted signal appears commonly
on both op-amp inputs.
2) Input Parameters
 Common Mode Rejection Ratio: Common-mode rejection means that this
unwanted signal will not appear on the output and distort the desired signal.
Common-mode signals (noise) generally are the result of the pick-up of
radiated energy on the input lines, from adjacent lines, the 60 Hz power line,
or other sources.
The measure of an amplifier's ability to reject common-mode signals is a
parameter called the CMRR (common-mode rejection ratio).
where Ad = differential gain of the amplifier
Ac = common-mode gain of the amplifier
Vd = difference voltage
Vc = common voltage
Ex. Calculate the CMRR for the circuit measurements shown in Fig.
Lecturer: Dr. Omar Daoud
Part III
5
Module: Electronics II
Module Number: 650321
th
Electronic Devices and Circuit Theory, 9 ed., Boylestad and Nashelsky
Solution:
Ex. Determine the output voltage of an op-amp for input voltages of Vi1 = 150 V,
Vi2 =140 V. The amplifier has a differential gain of Ad = 4000 and the value of
CMRR is:
(a) 100.
(b) 105.
Solution:
Lecturer: Dr. Omar Daoud
Part III
6
Module: Electronics II
Module Number: 650321
th
Electronic Devices and Circuit Theory, 9 ed., Boylestad and Nashelsky
 Common mode Input voltage range: It is the range of input voltages which,
when applied to both inputs, will not cause clipping or other output distortion.
Many op-amps have common-mode input voltage ranges of ±10 V with dc
supply voltages of ±15 V.
 Input Impedance: Two basic ways of specifying the input impedance of an
op-amp are the differential and the common mode.
 The differential input impedance is the total resistance between the
inverting and the noninverting inputs (measured by determining the
change in bias current for a given change in differential input
voltage).
 The common-mode input impedance is the resistance between each
input and ground (measured by determining the change in bias
current for a given change in common-mode input voltage).
 Input Offset Voltage: The ideal op-amp produces zero volts out for zero volts
in. In a practical op-amp, however, a small de voltage. V OUT(error) , appears at
the output when no differential input voltage is applied (causes a slight
mismatch of the base-emitter voltages of the differential amplifier input
stage of an op-amp) .
 The input offset voltage, V os , is the differential dc voltage required between
the inputs to force the output to zero volts (Typical values of in the range of
2 mV or less, while it is 0V in ideal cases).
 Input Offset Current: Ideally, the two
input bias currents are equal, and thus
their difference is zero. In a practical
op-amp, however, the bias currents are
not exactly equal.
The input offset current, I os , is the
difference of the input bias currents,
expressed as an absolute value
(Actual magnitudes of offset current are usually at least an order of
magnitude (ten times) less than the bias current).
Vos  I 1 Rin  I 2 Rin  I 1  I 2 Rin
Vos  I os Rin
VOUT ( error )  Av I os Rin
Lecturer: Dr. Omar Daoud
Part III
7
Module: Electronics II
Module Number: 650321
th
Electronic Devices and Circuit Theory, 9 ed., Boylestad and Nashelsky
 Slew Rate: The maximum rate of change of the output voltage in response to a
step input voltage is the slew rate of an op-amp.
 The slew rate is dependent upon the high-frequency response of the
amplifier stages within the op-amp.
 A pulse is applied to the input and the resulting ideal output voltage is
indicated as below.
 The width of the input pulse must be sufficient to allow the output to
"slew" from its lower limit to its upper limit.
 A certain time interval Δt is required for the output voltage to go from
its lower limit – V max to its upper limit +V max once the input step is
applied.
Vout
t
where ΔV out =+V max - (- V max ), The unit of slew rate is volts per microsecond (V/s).
Slew Rate=
Ex. The output voltage of a certain op-amp
appears as shown in the given Figure in response
to a step input. Determine the slew rate.
Solution:
Slew Rate=
9  (9)
 18V/μs
1  10 6
 Input Bias Current: The input terminals of a bipolar differential amplifier are
the transistor bases and, therefore, the input currents are the base currents. The
input bias current is the dc current required by the inputs of the amplifier to
properly operate the first stage (It is the average of both input currents).
Lecturer: Dr. Omar Daoud
Part III
8
Module: Electronics II
Module Number: 650321
th
Electronic Devices and Circuit Theory, 9 ed., Boylestad and Nashelsky
Comparison between Op-amps with and without negative feedback:
Practical Op-Amps Circuits:
1) Inverting Op-Amp
The signal input is applied to the inverting (–) input. The non-inverting input (+) is
grounded. The resistor R f is the feedback resistor. It is connected from the output to the
negative (inverting) input. This is negative feedback.
Apply kvl to the input loop
V1  I i  R1      (1)
But, I i   I f      (2)
Apply kvl to the output loop
Vo  I f  R f      (3)
substitute 1 and 2 in 3
Rf
Rf
Vo  
Vi  Av  
Ri
Ri
2) Noninverting Op-Amps
Lecturer: Dr. Omar Daoud
Part III
9
Module: Electronics II
Module Number: 650321
th
Electronic Devices and Circuit Theory, 9 ed., Boylestad and Nashelsky
3) Summing Op-Amps
Because the op-amp has high input impedance, the multiple inputs are treated as separate
inputs.
4) Integrator Op-Amps
The output is the integral of the input. Integration is the operation of summing the area
under a waveform or curve over a period of time. This circuit is useful in low-pass filter
circuits and sensor conditioning circuits.
Lecturer: Dr. Omar Daoud
Part III
10
Module: Electronics II
Module Number: 650321
th
Electronic Devices and Circuit Theory, 9 ed., Boylestad and Nashelsky
5) Differentiator Op-Amps
The differentiator takes the derivative of the input. This circuit is useful in high-pass filter
circuits.
Ex. Calculate the output voltages V2 and V3 in the given circuit below.
Lecturer: Dr. Omar Daoud
Part III
11
Module: Electronics II
Module Number: 650321
th
Electronic Devices and Circuit Theory, 9 ed., Boylestad and Nashelsky
Solution:
Ex. What range of output voltage is developed in the given circuit below.
Solution:
Lecturer: Dr. Omar Daoud
Part III
12
Module: Electronics II
Module Number: 650321
th
Electronic Devices and Circuit Theory, 9 ed., Boylestad and Nashelsky
Ex. The differential amplifier shown below has a differential gain of 2500 and a
CMRR of 30000.
 In part (a), a single ended input of 500V rms is applied and at the same time a
100mV, 60Hz common-mode interference signal appears on both inputs as a
result of radiated pick-up from ac power system.
 In part (b), differential input signals of 500V rms each are applied to the
inputs. The common-mode interference is the same as in part (a).
1) Determine the common-mode gain,
2) Express the CMRR in dB,
3) Determine the rms output signal for both parts a and b,
4) Determine the rms interference voltage on the output.
Solution:
Ad
Ad
 Ac 
 0.083
Ac
CMRR
2) CMRR dB =20log(30000)=89.5dB
3) (a) vo1  Ad vid  2500  500  1.25Vrms
1) CMRR=
(b) vo1  Ad vid  2500  500   500    2.5Vrms
4) voc  Ac vic  0.083  100m   8.3mVrms
Ex. In the given op-amp, R i =10k, R f =220k, and
from the datasheet Z in =2M, Z out =75, and A d
(A ol )=200000.
a) Determine the input and output impedances,
b) A c (A cl ).
Solution:
The attenuation B 
Ri
 0.0435
Ri  R f
Z in ( NI )  Z in (1  Ad B )  17.4G
Z out ( NI ) 
Ac 
Z out
 8.6 
(1  Ad B )
1
 23
B
Lecturer: Dr. Omar Daoud
Part III
13
Module: Electronics II
Module Number: 650321
th
Electronic Devices and Circuit Theory, 9 ed., Boylestad and Nashelsky
Ex. In the given op-amp, R i =1k,
R f =100k, and from the datasheet
Z in =4M, Z out =50, and A d (A ol )=200000.
c) Determine the input and output
impedances,
d) A c (A cl ).
Solution:
The attenuation B 
Ri
 0.009901
Ri  R f
Z in ( NI )  Ri  1.0k
Z out ( NI ) 
Ac  
Rf
Ri
Z out
 101m
(1  Ad B)
 100
Ex. Determine the output for the circuit of the given Fig. with components Rf =1M, R1=100k,
R2 =50k, and R3 =500k.
Solution:
Lecturer: Dr. Omar Daoud
Part III
14
Module: Electronics II
Module Number: 650321
th
Electronic Devices and Circuit Theory, 9 ed., Boylestad and Nashelsky
Op-Amps Applications (Active Filters):
A filter circuit can be constructed using passive components: resistors and
capacitors. An active filter additionally uses an amplifier to provide voltage
amplification and signal isolation or buffering (The high input impedance of
the op-amp prevents excessive loading of the driving source. and the low
output impedance of the op-amp prevents the filter from being affected by
the load that it is driving. Active filters are also easy to adjust over a wide
frequency range without altering the desired response).
 A filter that provides a constant output from dc up to a cutoff frequency
fOH and then passes no signal above that frequency is called an ideal lowpass filter.
 A filter that provides or passes signals above a cutoff frequency fOL is a
high-pass filter.
 When the filter circuit passes signals that are above one ideal cutoff
frequency and below a second cutoff frequency, it is called a bandpass
filter.
Lecturer: Dr. Omar Daoud
Part III
15
Module: Electronics II
Module Number: 650321
th
Electronic Devices and Circuit Theory, 9 ed., Boylestad and Nashelsky
 Active Low Pass Filters (LPF):
First Order LPF:
Av  1 
f OH 
RF
RG
1
2R1C1
Second Order LPF:
f OH 
Lecturer: Dr. Omar Daoud
1
2 R1C1 R2 C 2
Part III
16
Module: Electronics II
Module Number: 650321
th
Electronic Devices and Circuit Theory, 9 ed., Boylestad and Nashelsky
 Active High Pass Filters (HPF):
 Active Band Pass Filters (BPF):
Lecturer: Dr. Omar Daoud
Part III
17
Module: Electronics II
Module Number: 650321
th
Electronic Devices and Circuit Theory, 9 ed., Boylestad and Nashelsky
Ex. For the given Active BPF circuit, if R 1 =R 3 =30k, R 2 =R 4 =51k,
R A2 =R B2 =10k, R A1 = 15k, R B1 =30k, C 1 = C 2 = C 3 =0.01F and C 2 = 0.02
F then find
a) The circuit BW,
b) The geometric center frequency, and
c) The Q value (Quality factor is defined as the peak energy stored in the
circuit divided by the average energy dissipated in it per cycle at resonance).
Solution:
f c1 ( HPF ) 
f c 2 ( LPF ) 
1
2 R A1 RB1C A1C B1
1
2 R A 2 RB 2 C A2 C B 2
 750 Hz
 1130 Hz
BW  f c 2  f c1  380 Hz
fo 
Q
f c1 f c 2  921Hz
fo
 2.42
BW
Lecturer: Dr. Omar Daoud
Part III
18
Module: Electronics II
Module Number: 650321
th
Electronic Devices and Circuit Theory, 9 ed., Boylestad and Nashelsky
Multiple-Feedback Band-Pass Filter

Another
type
of
filter
configuration, shown in the
Figure below, it is a multiplefeedback bandpass filter. The
two feedback paths are through
R 2 and C 1 ). Components R 1
and C 1 provide the low-pass
response, and R 2 and C 2
provide the high-pass response.
The maximum gain Ao occurs
at the center frequency.
 Assume that C 1 >C 2  f C1 <f C2
1) C 2 is Open circuit as long as C 1 is open circuit V out =0V.
2) f C1 <f in <f C2  C 1 is short circuit and C 2 is still open circuit
Thus, it acts as inverting amplifier  V out exists.
3) f in ≥f C2  Both C 1 and C 2 are short circuit
R f is shorted  V out =0V.
1
fo 

2
R1 // R3 R2 C1C 2
1
2C
R1 // R3 R2

, making C1  C 2  C
1
2C
R1  R3 
R1 R3 R2
R1 
Q
Q
Q
, R2 
, R3 
2f o CAo
f o C
2f o C 2Q 2  Ao
Ao 
R2
2 R1


Ao  2Q 2 , This a limitation on the gain to give a positive value for R3
Ex.
Determine the
center
frequency,
maximum gain, and
bandwidth for the
filter in the given
Figure.
Solution:
Lecturer: Dr. Omar Daoud
Part III
19
Module: Electronics II
Module Number: 650321
th
Electronic Devices and Circuit Theory, 9 ed., Boylestad and Nashelsky
R1  R3 
fo 
1
2C
Ao 
R2
180k

 1.32
2 R1 2(68k)
R1 R3 R2

1
2 (0.01F )
68k  2.7k   736Hz
68k 180k 2.7k 
Q  f o CR2   736Hz 0.01F 180k   4.16
BW 
fo 736Hz

 177 Hz
4.16
Q
 Active Notch/Band-Stop Filters (NF):
1) f in <f C2  C 1 and C 2 is open circuit V out =0V from the inverting
stage but there is a noninverting one.
2) f C1 <f in <f C2  C 1 is short circuit  V i is shown to be common
mode signal.
3) f in ≥f C2  Both C 1 and C 2 are short circuit
 C 2 is short out the signal at inverting input  V out comes from
the noninverting input (amplify the difference between the two
inputs).
1
fo 
, Ao  1
2 R1 R2 C1C 2
Lecturer: Dr. Omar Daoud
Part III
20