T HE O PERATIONAL A MPLIFIER
1 I NTRODUCTION TO THE O PERATIONAL A MPLIFIER
Early operational amplifiers (op-amps) were used primarily to perform mathematical
operations such as addition, subtraction, integration, and differentiation—hence the
term operational. These early devices were constructed with vacuum tubes and
worked with high voltages. Today’s op-amps are linear integrated circuits (ICs) that
use relatively low dc supply voltages and are reliable and inexpensive.
After completing this section, you should be able to
◆
Discuss the basic op-amp
◆
Recognize an op-amp symbol
◆
Identify terminals on op-amp packages
◆
Describe an ideal op-amp
◆
Describe a practical op-amp
Symbol and Terminals
The standard operational amplifier symbol is shown in Figure 1(a). It has two input
terminals, the inverting () input and the noninverting () input, and one output terminal.
The typical op-amp requires two dc supply voltages, one positive and the other negative, as
shown in Figure 1(b). Usually these dc voltage terminals are left off the schematic symbol
for simplicity, but they are always understood to be there. Typical IC packages are shown
in Figure 1(c).
FIGURE 1
+V
Op-amp symbols and packages.
Inverting
input
–
–
Output
Noninverting
input
+
+
–V
(a) Symbol
(b) Symbol with dc supply connections
8
1
DIP
4
8
1
DIP
4
8
1
SOIC
4
PLCC
(c) Typical packages. Looking from the top, pin 1 always is to the left of the
notch or dot on the DIP and SOIC packages. The dot indicates pin 1 on the
plastic-leaded chip carrier (PLCC) package.
The Ideal Op-Amp
To illustrate what an op-amp is, let’s consider its ideal characteristics. A practical op-amp,
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.
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T HE O PERATIONAL A MPLIFIER
The ideal op-amp has infinite voltage gain and an infinite input resistance (open), so that
it does not load the driving source. (Input resistance is sometimes called input impedance.)
Also, it has a zero output resistance. (Output resistance is sometimes called output impedance.) These characteristics are illustrated in Figure 2. The input voltage, Vin, appears
between the two input terminals, and the output voltage is AvVin, as indicated by the symbol
for internal voltage source. The concept of infinite input resistance is a particularly valuable
analysis tool for the various op-amp configurations, which will be discussed in Section 5.
FIGURE 2
Ideal op-amp representation.
AvVin
Rin = ⬁
Vin
Rout = 0
Vout
Av = ⬁
The Practical Op-Amp
Although modern IC op-amps approach parameter values that can be treated as ideal in many
cases, the ideal device can never be made. Any device has limitations, and the IC op-amp is no
exception. Op-amps have both voltage and current limitations. Peak-to-peak output voltage,
for example, is usually limited to slightly less than the two supply voltages. Output current is
also limited by internal restrictions such as power dissipation and component ratings.
Characteristics of a practical op-amp are high voltage gain, high input resistance, and
low output resistance, as illustrated in Figure 3.
FIGURE 3
Practical op-amp representation.
R in
Vin
AvVin
Vout
Rout
Internal Block Diagram of an Op-Amp
A typical op-amp is made up of three types of amplifier circuits: a differential amplifier, a
voltage amplifier, and a push-pull amplifier, as shown in Figure 4.
+
Vin
–
Differential
amplifier
input stage
Voltage
amplifier(s)
gain stage
Push-pull
amplifier
output
stage
Vout
FIGURE 4
Basic internal arrangement of an op-amp.
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T HE O PERATIONAL A MPLIFIER
A differential amplifier is the input stage for the op-amp; it has two inputs and provides
amplification of the difference voltage between the two inputs. The voltage amplifier is usually a class A amplifier that provides additional op-amp gain. Some op-amps may have more
than one voltage amplifier stage. A push-pull class B amplifier is used for the output stage.
SECTION 1
CHECKUP
Answers are at the end of the
chapter.
1. Sketch the symbol for an op-amp.
2. Describe the ideal op-amp.
3. Describe some of the characteristics of a practical op-amp.
4. List the amplifier stages in a typical op-amp.
2 T HE D IFFERENTIAL A MPLIFIER
The op-amp, in its basic form, uses differential amplifiers for its input stage. Because
the differential amplifier (diff-amp) is fundamental in the op-amp’s internal operation,
it is useful to have a basic understanding of this type of circuit.
After completing this section, you should be able to
◆
Explain the basic operation of a differential amplifier
◆
Describe single-ended input operation
◆
Describe differential input operation
◆
Describe common-mode input operation
◆
Define common-mode rejection ratio (CMRR)
◆
Describe how diff-amps are used within an op-amp
A differential amplifier is an amplifier that produces an output proportional to the difference of two inputs. A basic differential amplifier circuit and its symbol are shown in
Figure 5. The diff-amp stages that make up part of the op-amp provide high voltage gain
and common-mode rejection (defined later in this section). Notice that the differential
amplifier has two outputs where the op-amp has only one output. Also, there is a positive
and a negative supply voltage (VCC and VEE). The differential amplifiers in this text use
BJTs. FETs can be used where very high input resistance is required.
FIGURE 5
+VCC
Basic differential amplifier.
RC2
RC1
Output
1
Output
2
1
1
Input
1
Input
2
Inputs
Outputs
2
2
Q2
Q1
RE
–VEE
(a) Circuit
838
(b) Symbol
T HE O PERATIONAL A MPLIFIER
Basic Operation
Although an op-amp typically has more than one diff-amp stage, we will use a single diffamp to illustrate the basic operation. The following discussion is in relation to Figure 6 and
consists of a basic dc analysis of the diff-amp’s operation.
First, when both inputs are grounded (0 V), the emitters are at 0.7 V, as indicated in
Figure 6(a). It is assumed that the transistors are identically matched by careful process
control during manufacturing so that their dc emitter currents are the same when there is no
input signal. Thus,
IE1 = IE2
Since both emitter currents combine through RE,
IRE
IE1 = IE2 =
2
where
IRE =
VE - VEE
RE
+VCC
IC1
–
VC1
+VCC
IC2
RC2
RC1
+
IC1
2
1
Q1
1
IE1
Q2
– 0.7 V
+
VC2
–
–
VC1
IC2
RC2
RC1
+
+VB
2
2
1
Q1
1
VB – 0.7 V
IE1
IE2
Q2
RE
+
VC2
–
2
IE2
RE
–VEE
–VEE
(a) Both inputs grounded
(b) Bias voltage on input 1 with input 2 grounded
+VCC
IC1
–
VC1
IC2
RC2
RC1
+
+
2
1
Q1
1
IE1
Q2
VB – 0.7 V
2
VC2
–
+VB
IE2
RE
–VEE
(c) Bias voltage on input 2 with input 1 grounded
FIGURE 6
Basic operation of a differential amplifier (ground is 0 V) showing relative changes in the collector voltages.
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T HE O PERATIONAL A MPLIFIER
Based on the approximation that IC ⬵ IE, it can be stated that
IC1 = IC2 ⬵
IRE
2
Since both collector currents and both collector resistors are equal (when the input voltage
is zero),
VC1 = VC2 = VCC - IC1RC1
This condition is illustrated in Figure 6(a).
Next, input 2 remains grounded and a positive bias voltage is applied to input 1, as
shown in Figure 6(b). The positive voltage on the base of Q1 increases IC1 and raises the
emitter voltage. This action reduces the forward bias (VBE) of Q2 because its base is held at
0 V (ground), thus causing IC2 to decrease. The net result is that the increase in IC1 causes
a decrease in VC1, and the decrease in IC2 causes an increase in VC2, as shown.
Finally, input 1 is grounded and a positive bias voltage is applied to input 2, as shown in
Figure 6(c). The positive bias voltage causes Q2 to conduct more, thus increasing IC2. Also,
the emitter voltage is raised. This reduces the forward bias of Q1 because its base is held at
ground and causes IC1 to decrease. The result is that the increase in IC2 produces a decrease
in VC2, and the decrease in IC1 causes VC1 to increase as shown.
Modes of Signal Operation
Single-Ended Input When a diff-amp is operated in the single-ended mode, one input
is grounded and the signal voltage is applied only to the other input, as shown in Figure 7.
In the case where the signal voltage is applied to input 1, as in part (a), an inverted,
amplified signal voltage appears at output 1 as shown. Also, a signal voltage appears in
FIGURE 7
+VCC
Single-ended input operation of a
differential amplifier.
RC2
RC1
2
1
Vout1
1
Q1
Q2
Vout 2
2
Vin1
RE
Ve
–VEE
(a)
+VCC
RC2
RC1
1
Vout1
1
2
Q1
Q2
Vout 2
2
Vin2
RE
Ve
(b)
840
–VEE
T HE O PERATIONAL A MPLIFIER
phase at the emitter of Q1. Since the emitters of Q1 and Q2 are common, this emitter signal
becomes an input to Q2, which functions as a common-base amplifier. The signal is amplified by Q2 and appears, noninverted, at output 2. This action is illustrated in part (a).
In the case where the signal is applied to input 2 with input 1 grounded, as in Figure 7(b),
the emitter signal becomes an input to Q1, and an inverted, amplified signal voltage appears
at output 2. In this situation, Q1 acts as a common-base amplifier, and a noninverted, amplified signal appears at output 1.
Differential Input In the differential mode, two signals of opposite polarity (out of
phase) are applied to the inputs, as shown in Figure 8(a). This type of operation is also
referred to as double-ended. Each input affects the outputs, as you will see in the following
discussion.
Vp
Vin1
Vin2
Vin1
1
2
–
+
1
1
–
Vout1
1
Vp
2
2
+
2
Vout2
(b) Outputs due to Vin1
(a) Differential inputs
2Vp
Vp
1
–
Vout1
1
Vin1
1
–
1
Vp
Vin2
2
+
2
(c) Outputs due to Vin2
Vout2
Vin2
2
+
2Vp
2
(d) Total outputs due to differential inputs
FIGURE 8
Differential operation of a differential amplifier.
Figure 8 (b) shows the output signals due to the signal on input 1 acting alone as a
single-ended input. Figure 8(c) shows the output signals due to the signal on input 2 acting
alone as a single-ended input. In parts (b) and (c), notice that the signals on output 1 are of the
same polarity. The same is also true for output 2. By superimposing both output 1 signals and
both output 2 signals, you get the total differential operation, as pictured in Figure 8(d).
Common-Mode Input One of the most important aspects of the operation of a diff-amp
can be seen by considering the common-mode condition. Any ac or dc voltage that is
identical on both inputs is a common-mode signal. Consider the two signals shown in
Figure 9(a), which have the same amplitude, frequency, and phase relationship. Again, by
considering each input as acting alone, you can understand the basic operation.
Figure 9 (b) shows the output signals due to the signal only on input 1, and part (c)
shows the output signals due to the signal on only input 2. Notice in parts (b) and (c) that
the signals on output 1 are of the opposite polarity, and so are those on output 2. When the
input signals are applied to both inputs, the outputs are superimposed and they cancel,
resulting in a near zero output voltage, as shown in Figure 9(d).
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T HE O PERATIONAL A MPLIFIER
Vin1
Vin2
1
2
–
+
Vin1
1
2
2
(a) Common-mode inputs
1
Vin2
2
–
+
1
1
–
+
Vout1
Vout2
2
(b) Outputs due to Vin1
1
2
Vin1
Vout1
Vout2
(c) Outputs due to Vin2
1
1
–
0V
Output signals of
equal amplitude but
0 V opposite phase
2
2
cancel, producing
Vin2
0 V on each output.
(d) Ouputs cancel when common-mode signals are applied
+
FIGURE 9
Common-mode operation of a differential amplifier.
This action is called common-mode rejection. Its importance can be seen in the situation where an unwanted signal appears commonly on both diff-amp inputs. Commonmode rejection means that this unwanted signal will not appear on the outputs to 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, or the 60 Hz power line, or other
sources. Differential amplifiers are very efficient at rejecting low-frequency commonmode signals like power line interference, but they usually cannot reject very high interfering signals such as those caused by high-frequency switching noise. In cases where
high-frequency noise is present, shielding the signal lines may be necessary.
Common-Mode Rejection Ratio, CMRR
Desired signals appear on only one input or with opposite polarities on both input lines.
These desired signals are amplified and appear on the outputs as previously discussed.
Unwanted signals (noise) appearing with the same polarity on both input lines are essentially canceled by the diff-amp and do not appear on the outputs. The measure of an amplifier’s ability to reject common-mode signals is a parameter called the common-mode
rejection ratio (CMRR).
Ideally, a diff-amp provides a very high gain for desired signals (single-ended or differential), and zero gain for common-mode signals. Practical diff-amps, however, do exhibit a
very small common-mode gain (usually much less than 1), while providing a high differential voltage gain (usually several thousand). The higher the differential gain with respect
to the common-mode gain, the better the performance of the diff-amp in terms of rejection
of common-mode signals. This suggests that a good measure of the diff-amp’s performance in rejecting unwanted common-mode signals is the ratio of the differential gain, Av(d),
to the common-mode gain, Acm. This ratio is the common-mode rejection ratio, CMRR.
Equation 1
CMRR ⴝ
Av(d )
Acm
The higher the CMRR, the better. A very high value of CMRR means that the differential
gain Av(d) is high and the common-mode gain Acm is low.
The CMRR is often expressed in decibels (dB) as
Equation 2
842
CMRR ⴝ 20 log a
Av(d )
Acm
b
T HE O PERATIONAL A MPLIFIER
EXAMPLE 1
A certain diff-amp has a differential voltage gain of 2000 and a common-mode gain of
0.2. Determine the CMRR and express it in decibels.
Solution
Av(d) 2000 and Acm 0.2. Therefore,
CMRR =
Av(d)
Acm
=
2000
= 10,000
0.2
Expressed in decibels,
CMRR = 20 log(10,000) = 80 dB
Related Problem*
Determine the CMRR and express it in dB for an amplifier with a differential voltage
gain of 8500 and a common-mode gain of 0.25.
*Answers are at the end of the chapter.
A CMRR of 10,000, for example, means that the desired input signal (differential) is
amplified 10,000 times more than the unwanted noise (common-mode). So, as an example,
if the amplitudes of the differential input signal and the common-mode noise are equal, the
desired signal will appear on the output 10,000 times greater in amplitude than the noise.
Thus, the noise or interference has been essentially eliminated.
Example 2 illustrates further the idea of common-mode rejection and the general signal
operation of the diff-amp.
EXAMPLE 2
The diff-amp shown in Figure 10 has a differential voltage gain of 2500 and a CMRR
of 30,000. In part (a), a single-ended input signal of 500 mV rms is applied. At the
same time a 100 mV, 60 Hz common-mode interference signal appears on both inputs
as a result of radiated pick-up from the ac power system. In part (b), differential input
signals of 500 mV rms each are applied to the inputs. The common-mode interference
is the same as in part (a).
(a) Determine the common-mode gain.
(b) Express the CMRR in dB.
(c) Determine the rms output signal for parts (a) and (b).
(d) Determine the rms interference voltage on the output.
Vin1
Vin1
1
–
1
Vout1
–
Vout1
+
Vout2
Vcm
Vcm
2
+
2
Vout2
Vcm
Vcm
(a)
(b)
Vin2
FIGURE 10
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T HE O PERATIONAL A MPLIFIER
Solution
(a) CMRR Av(d)>Acm. Therefore,
Acm =
Av(d)
CMRR
=
2500
= 0.083
30,000
(b) CMRR 20 log(30,000) 89.5 dB
(c) In Figure 10(a), the differential input voltage, Vin(d), is the difference between the
voltage on input 1 and that on input 2. Since input 2 is grounded, the voltage is
zero. Therefore,
Vin(d) = Vin1 - Vin2 = 500 mV - 0 V = 500 mV
The output signal voltage in this case is taken at output 1.
Vout1 = Av(d)Vin(d) = (2500)(500 mV) = 1.25 V rms
In Figure 10(b), the differential input voltage is the difference between the two
opposite-polarity, 500 mV signals.
Vin(d) = Vin1 - Vin2 = 500 mV - ( -500 mV) = 1000 mV = 1 mV
The output signal voltage is
Vout1 = Av(d)Vin(d) = (2500)(1 mV) = 2.5 V rms
This shows that a differential input (two opposite-polarity signals) results in a gain
that is double that for a single-ended input.
(d) The common-mode input is 100 mV rms. The common-mode gain Acm is 0.083.
The interference voltage on the output, therefore, is
Acm =
Vout(cm)
Vin(cm)
Vout(cm) = AcmVin(cm) = (0.083)(100 mV) = 8.3 mV
Related Problem
SECTION 2
CHECKUP
The amplifier in Figure 10 has a differential voltage gain of 4200 and a CMRR of
25,000. For the same single-ended and differential input signals as described in the
example: (a) Find Acm. (b) Express the CMRR in dB. (c) Determine the rms output
signal for parts (a) and (b) of the figure. (d) Determine the rms interference (commonmode) voltage appearing on the output.
1. Distinguish between differential and single-ended inputs.
2. Define common-mode rejection.
3. For a given value of differential gain, does a higher CMRR result in a higher or lower
common-mode gain?
3 O P -A MP PARAMETERS
In this section, several important op-amp parameters are defined. (These are listed in
the objectives that follow.) Also, several popular IC op-amps are compared in terms of
these parameters.
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