Theory, worked examples and 35 previous-year questions with full solutions — IIT-JAM, GATE, JEST and CSIR NET-JRF, 2020 onwards.
Theory
An operational amplifier, or op-amp, is an integrated circuit designed to amplify weak electrical signals. It has two input terminals, the non-inverting input (marked \(+\)) and the inverting input (marked \(-\)), a single output terminal, and two supply terminals connected to a positive and a negative power supply. Its basic function is to amplify the voltage difference between its two inputs and to deliver the result at the output. The circuit symbol is shown below.
Op-amp circuits fall into two broad classes. In the open-loop configuration no part of the output is returned to the input; the op-amp then behaves as a very high-gain amplifier whose output is almost always saturated, and its chief use is as a comparator. In the closed-loop configuration part of the output is fed back to the input. With negative feedback the circuit becomes a stable, precisely controllable amplifier, while with positive feedback it is used for waveform generation. This classification is summarised below.
5.1 Characteristics of an op-amp
A practical op-amp departs from an ideal amplifier in a number of ways, each described by a parameter quoted on its data sheet. We define the most important of them here.
(a) Input offset voltage
Because the two input transistors of an op-amp are never perfectly matched, the output is not exactly zero when both inputs are at the same potential. The input offset voltage\(V_{io}\) is the voltage that must be applied between the two input terminals to make the output voltage zero. In Figure 5.1 the d.c. sources \(V_{dc1}\) and \(V_{dc2}\), each acting through a source resistance \(R_s\), are adjusted until \(V_o=0\); the input offset voltage is then
\[\boxed{V_{io}=\lvert V_{dc1}-V_{dc2}\rvert}.\]
Figure 5.1 Defining the input offset voltage \(V_{io}\).
The smaller the value of \(V_{io}\), the better matched are the input terminals of the op-amp.
(b) Input bias current
Each input terminal draws a small d.c. current, the base current of the input transistor. The input bias current\(I_B\) is defined as the average of the currents \(I_{B1}\) and \(I_{B2}\) that flow into the two input terminals:
\[\boxed{I_B=\frac{I_{B1}+I_{B2}}{2}}.\]
(c) Input offset current
The two bias currents are in general unequal. Their difference is the input offset current,
\[\boxed{I_{io}=\lvert I_{B1}-I_{B2}\rvert},\]
as indicated in Figure 5.2. The offset current is typically an order of magnitude smaller than the bias current.
Figure 5.2 Defining the input offset current \(I_{io}\).
(d) Differential input resistance
The differential input resistance\(R_i\), often called simply the input resistance, is the equivalent resistance measured at either the inverting or the non-inverting input terminal with the other input terminal connected to ground.
(e) Common-mode rejection ratio
When the same voltage \(v_{cm}\) is applied to both inputs, as in Figure 5.3, an ideal op-amp would produce no output, since the difference between its inputs is zero. A practical op-amp produces a small common-mode output \(v_{ocm}\), and we define the common-mode gain \(A_{cm}=v_{ocm}/v_{cm}\). The common-mode rejection ratio is the ratio of the differential voltage gain \(A_d\) to the common-mode voltage gain:
\[\boxed{\text{CMRR}=\frac{A_d}{A_{cm}}}.\]
Figure 5.3 The common-mode configuration.
Ideally \(\text{CMRR}=\infty\). Since unwanted noise and interference usually appear equally on both input lines, the CMRR measures the ability of the amplifier to reject noise: the larger the CMRR, the better the rejection. Because the ratio is very large it is normally quoted in decibels,
where in the second form \(v_{io}\) is the differential input voltage that would produce the same output as the common-mode input \(v_{cm}\) (so that \(A_dv_{io}=A_{cm}v_{cm}\)).
Example 5.1. An op-amp has \(\text{CMRR}=100\) dB and a differential gain \(A_d=10\). A common-mode voltage of 2 mV is applied to it. Find the common-mode output voltage.
Solution. From the definition of the CMRR in decibels,
The output voltage of an op-amp cannot change arbitrarily quickly, because the internal compensating capacitor can be charged only by a limited current. The slew rate is the maximum rate of change of the output voltage with respect to time,
usually expressed in V/\(\mu\)s. If a signal demands a faster change than the slew rate allows, the output is distorted.
For a sinusoidal output \(v_o=V_m\sin\omega t\) the rate of change is \(dv_o/dt=V_m\omega\cos\omega t\), whose maximum value, reached at the zero crossings, is \(V_m\omega\). Undistorted operation therefore requires
\[\boxed{V_m\omega=2\pi fV_m\le\text{SR}}.\]
Example 5.2. A voltage follower has the input \(v_{in}=v_m\sin\omega t\). Find the maximum angular frequency at which the output can follow the input without slew-rate distortion, for an op-amp of slew rate SR.
Solution. For a voltage follower \(v_o=v_{in}\), so
The highest frequency at which this does not exceed the slew rate is given by \(v_m\omega_{\max}=\text{SR}\), that is,
\[\omega_{\max}=\frac{\text{SR}}{v_m}.\]
Example 5.3. For a square-wave input of frequency 2 MHz and peak-to-peak amplitude 6 V, the output of an op-amp voltage follower is the triangular wave shown below. Find the slew rate of the op-amp.
Solution. The output rises linearly from \(-3\) V to \(+3\) V, a change of 6 V, and then falls linearly back. The period of the 2 MHz wave is \(0.5\,\mu\)s, so each rise occupies half a period, \(0.25\,\mu\)s. Since the output changes at the maximum possible rate throughout each ramp,
Example 5.4. An op-amp of slew rate \(0.5\ \text{V}/\mu\text{s}\) is used as an inverting amplifier with a gain of 50. Its voltage gain is flat up to 20 kHz. What is the maximum peak-to-peak input signal that can be applied without distorting the output?
Solution. At the highest frequency of interest, 20 kHz, the peak output \(V_m\) must satisfy \(2\pi fV_m\le\text{SR}\). Writing the slew rate as \(0.5\times10^{6}\) V/s,
The largest undistorted output is therefore \(2\times3.98=7.96\) V peak to peak. Since the gain is 50, the input must not exceed
\[\frac{7.96\ \text{V}}{50}=159\ \text{mV peak to peak}.\]
(g) Supply voltage rejection ratio
The input offset voltage of an op-amp changes slightly when the supply voltages change. If a change \(\Delta V\) in the supply voltages produces a change \(\Delta V_{io}\) in the input offset voltage, the supply voltage rejection ratio is defined as
It is usually quoted in microvolts per volt, and the lower its value, the better the op-amp. (Manufacturers often quote the reciprocal ratio, the power-supply rejection, as a positive number of decibels.)
(h) Large-signal voltage gain
Since the op-amp amplifies the difference between the voltages at its two inputs, its voltage gain is defined as
Because the output signal is very much larger than the input signal, this open-loop gain is commonly called the large-signal voltage gain.
(i) Output voltage swing
The output voltage swing is the range of output voltage that the op-amp can physically provide. The output never exceeds the limits \(\pm V_{o\max}\), which are a little smaller in magnitude than the supply voltages.
(j) Output resistance
The output resistance\(R_o\) is the equivalent resistance measured between the output terminal of the op-amp and ground.
(k) Gain–bandwidth product
The gain–bandwidth product is the bandwidth of the op-amp when its voltage gain is 1; we denote it by \(f_T\). Equivalent names are the unity-gain bandwidth (UGB), the closed-loop bandwidth and the small-signal bandwidth. For an op-amp whose gain falls off at \(-20\) dB/decade above its break frequency, the product of gain and bandwidth is the same for every gain setting, and is equal to \(f_T\), a result we derive in Section 5.5.
5.2 The ideal op-amp
(a) Characteristics of an ideal op-amp
An ideal op-amp would have the following electrical characteristics:
infinite voltage gain \(A\);
infinite input resistance \(R_i\), so that any signal source can drive it and there is no loading of the preceding stage;
zero output resistance \(R_o\), so that the output can drive any number of other devices;
zero output voltage when the input voltage is zero;
infinite bandwidth, so that a signal of any frequency from 0 to \(\infty\) Hz is amplified without attenuation;
infinite common-mode rejection ratio, so that the output common-mode noise voltage is zero;
infinite slew rate, so that changes in output voltage occur simultaneously with changes in input voltage.
No practical op-amp has these properties, but several of them can be closely approximated with a negative-feedback arrangement. In particular, the input resistance, the output resistance and the bandwidth can be brought close to their ideal values in this way, as we show in Sections 5.5 and 5.6.
(b) Equivalent circuit of an op-amp
Figure 5.4 shows the equivalent circuit of an op-amp. The input is represented by the resistance \(R_i\) between the two input terminals, and the output by a Thévenin equivalent: a voltage source \(Av_{id}\) in series with the resistance \(R_o\) seen looking back into the output terminal.
Figure 5.4 Equivalent circuit of an op-amp.
For this circuit the (open-circuit) output voltage is
\[\boxed{v_o=Av_{id}=A(v_1-v_2)},\]
where \(A\) is the large-signal voltage gain, \(v_{id}=v_1-v_2\) is the difference input voltage, \(v_1\) is the voltage at the non-inverting input terminal with respect to ground, and \(v_2\) is the voltage at the inverting input terminal with respect to ground. This equation shows that the output voltage is directly proportional to the algebraic difference between the two input voltages. In other words, the op-amp amplifies the difference between its input voltages, not the input voltages themselves, and the polarity of the output therefore depends on the polarity of the difference voltage.
(c) The ideal voltage transfer curve
The equation \(v_o=A(v_1-v_2)\) is the basic op-amp equation, in which the output offset voltage is taken to be zero. Plotting \(v_o\) against \(v_{id}\) for a fixed gain \(A\) gives the voltage transfer curve shown below. Near the origin the output is a straight line of slope \(A\); because \(A\) is so large, this linear region extends over only a tiny range of \(v_{id}\) (a few tens of microvolts). Outside it the output is clamped at the positive saturation voltage \(+V_{\text{sat}}\) or the negative saturation voltage \(-V_{\text{sat}}\), whose magnitudes are slightly smaller than the supply voltage \(V_{CC}\).
5.3 Open-loop op-amp configurations
In an amplifier the term open loop means that no part of the output signal is fed back to form part of the input; the loop that feedback would have formed is open. Connected in this way, the op-amp acts simply as a very high-gain amplifier. There are three open-loop configurations, classified according to the number of inputs used and the terminal to which the input is applied: the differential amplifier, the inverting amplifier and the non-inverting amplifier.
(a) The differential amplifier
Figure 5.5 shows the open-loop differential amplifier, in which the input signals \(v_{in1}\) and \(v_{in2}\) are applied to the non-inverting and inverting input terminals respectively. Since the op-amp amplifies the difference between the two signals, the configuration is called a differential amplifier. The source resistances \(R_{in1}\) and \(R_{in2}\) are normally negligible compared with the input resistance \(R_i\), so the voltage drops across them may be taken as zero; then \(v_1=v_{in1}\) and \(v_2=v_{in2}\), and
\[\boxed{v_o=A\left(v_{in1}-v_{in2}\right)}.\]
Figure 5.5 The open-loop differential amplifier.
Because no coupling capacitors are involved, the op-amp amplifies both a.c. and d.c. voltages, which makes it a very versatile device.
(b) The inverting amplifier
In the inverting amplifier only one input is used: the signal is applied to the inverting terminal and the non-inverting terminal is grounded, as shown below.
Since \(v_1=0\) V and \(v_2=v_{in}\),
\[\boxed{v_o=-Av_{in}}.\]
The negative sign indicates that the output is \(180^\circ\) out of phase with the input, or of opposite polarity. Thus, in the inverting amplifier, the input signal is amplified by the gain \(A\) and is also inverted at the output.
(c) The non-inverting amplifier
In the open-loop non-inverting amplifier, shown below, the input is applied to the non-inverting terminal and the inverting terminal is connected to ground.
Here \(v_1=v_{in}\) and \(v_2=0\) V, so
\[\boxed{v_o=Av_{in}}.\]
The output voltage is therefore larger than the input voltage by the gain \(A\) and is in phase with the input signal.
Exam tip. When an op-amp is operated in open loop, its output is at either positive or negative saturation, or switches between the two saturation levels. For this reason open-loop configurations are not used in linear applications; they are used as comparators.
Example 5.5. Determine the output voltage of the open-loop differential amplifier in each of the following cases: (a) \(v_{in1}=5\,\mu\text{V}\) d.c., \(v_{in2}=-7\,\mu\text{V}\) d.c.; (b) \(v_{in1}=10\) mV r.m.s., \(v_{in2}=20\) mV r.m.s. The op-amp has \(A=200{,}000\), \(R_i=2\,\text{M}\Omega\), \(R_o=75\,\Omega\) and \(\pm V_{CC}=\pm15\) V.
This is only a theoretical value: the output cannot exceed the saturation voltage. The op-amp is therefore driven into saturation for almost the entire cycle, and the output is a clipped waveform that switches between \(+V_{\text{sat}}\) and \(-V_{\text{sat}}\approx\mp15\) V, in antiphase with \(v_{in1}\).
Example 5.6. In the op-amp shown below, \(V_1=2\,\mu\text{V}\) d.c. is applied to the inverting terminal and \(V_2=4\,\mu\text{V}\) d.c. to the non-inverting terminal. The open-loop gain is \(A_{OL}=2\times10^{5}\) and \(V_{CC}=\pm15\) V. (a) Find the output voltage. (b) Find the output voltage if \(V_1=-2\,\mu\text{V}\) and \(V_2=4\,\mu\text{V}\).
Solution. In the open-loop configuration the output is \(V_0=A_{OL}(V_2-V_1)\), where \(V_2\) is at the non-inverting terminal; the output varies only within \(\pm V_{\text{sat}}\).
(a) \(V_0=(4-2)\times10^{-6}\times2\times10^{5}=0.4\) V.
(b) \(V_0=\left[4-(-2)\right]\times10^{-6}\times2\times10^{5}=1.2\) V.
Both values lie within the saturation limits, so the op-amp operates linearly in each case.
5.4 Negative feedback
(a) The need for feedback
Since the open-loop gain of an op-amp is very large, only very small signals, of the order of microvolts or less and of very low frequency, can be amplified accurately without distortion. Such small signals are very susceptible to noise and are almost impossible to obtain cleanly in the laboratory. Moreover, the open-loop gain varies with temperature and from one device to another, which makes it unsuitable for most linear applications. Finally, the bandwidth of most open-loop op-amps, the band of frequencies over which the gain remains constant, is negligibly small, often only a few hertz. For these reasons the open-loop op-amp is impractical in a.c. applications.
The gain of an op-amp circuit can be both selected and controlled if the basic circuit is modified by the use of feedback: part of the output signal is returned to the input, either directly or through another network. If the signal fed back is of opposite polarity to the input signal, that is, out of phase with it by \(180^\circ\) (or an odd multiple of \(180^\circ\)), the feedback is called negative feedback. An amplifier with negative feedback has a self-correcting ability against any change in output voltage caused by changes in environmental conditions. If, on the other hand, the signal fed back is in phase with the input signal, so that it aids the input, the feedback is called positive feedback. Positive feedback is essential in oscillator circuits.
(b) Advantages and disadvantages of negative feedback
When used in amplifiers, negative feedback stabilises the gain and increases the bandwidth. It also reduces harmonic and non-linear distortion, reduces the effect of the input offset voltage on the output, and reduces the effect of variations in temperature and supply voltage on the output of the op-amp. Its principal disadvantage is that it reduces the voltage gain. In addition, it increases the output resistance in the case of current-series and current-shunt feedback amplifiers.
(c) Block-diagram representation of feedback configurations
An op-amp circuit that uses feedback is called a feedback amplifier. It is also called a closed-loop amplifier, because the feedback forms a closed loop between the input and the output. A feedback amplifier consists essentially of two parts, the op-amp and the feedback circuit, which may be represented by two blocks. There are four ways of connecting these two blocks, shown in Figure 5.6.
In the voltage-series and voltage-shunt arrangements of Figure 5.6(a) and (b), the voltage across the load resistor \(R_L\) is the input to the feedback circuit, and the feedback quantity, either a voltage or a current, is the output of the feedback circuit and is proportional to the output voltage. In the current-series and current-shunt arrangements of Figure 5.6(c) and (d), the load current \(i_L\) flows into the feedback circuit, whose output (either a voltage or a current) is proportional to the load current. The word series indicates that the feedback signal is applied in series with the input signal as a voltage; shunt indicates that it is applied in parallel with the input as a current.
5.5 Voltage-series feedback: the non-inverting amplifier
(a) The circuit and its gains
The voltage-series feedback amplifier is shown in Figure 5.7. The input \(v_{in}\) is applied to the non-inverting terminal, and the potential divider formed by \(R_F\) and \(R_1\) forms the feedback circuit, which returns the fraction \(v_f\) of the output to the inverting terminal. We define three gains:
\[\begin{aligned}\text{open-loop gain}\quad &A=\frac{v_o}{v_{id}},\\ \text{closed-loop gain}\quad &A_F=\frac{v_o}{v_{in}},\\ \text{gain of the feedback circuit}\quad &B=\frac{v_f}{v_o}.\end{aligned}\]
Figure 5.7 The voltage-series feedback amplifier.
Kirchhoff’s voltage law for the input loop gives
\[v_{id}=v_{in}-v_f,\]
where \(v_{in}\) is the input voltage, \(v_f\) the feedback voltage and \(v_{id}\) the difference input voltage. The feedback voltage always opposes the input voltage (it is out of phase by \(180^\circ\) with respect to the input, in the sense that it is subtracted from it), so the feedback is negative.
(b) Closed-loop voltage gain
The output voltage is \(v_o=A(v_1-v_2)\), with \(v_1=v_{in}\). Since the input current of the op-amp is negligible, \(R_F\) and \(R_1\) carry the same current and form a simple potential divider, so
the ideal closed-loop gain is the reciprocal of the gain of the feedback circuit. Dividing the numerator and denominator of the exact gain by \(R_1+R_F\) gives the general form
\[\boxed{A_F=\frac{A}{1+AB}},\]
where the product \(AB\) is called the loop gain. When \(AB\gg1\) this again reduces to \(1/B\).
(d) Difference input voltage
Since \(v_{id}=v_o/A\) and \(v_o\) is finite while \(A\) is very large (\(A\to\infty\) ideally),
With negative feedback the op-amp adjusts its output so that the voltage at the inverting terminal follows the voltage at the non-inverting terminal. This virtual short between the inputs, together with the negligible input current, is the key to analysing every negative-feedback op-amp circuit.
(e) Input resistance with feedback
Let \(R_i\) be the open-loop input resistance of the op-amp and \(R_{iF}\) the input resistance of the amplifier with feedback. The input current is \(i_{in}=v_{id}/R_i\), and from the input loop
The input resistance of the op-amp with voltage-series feedback is \((1+AB)\) times that without feedback.
(f) Output resistance with feedback
The output resistance is the resistance seen looking back into the feedback amplifier from the output terminal. To find it we set \(v_{in}=0\) and apply a test voltage \(v_o\) at the output. Then \(v_{id}=-v_f=-Bv_o\), and, neglecting the small current taken by the feedback divider, the current into the output terminal is
The output resistance of the voltage-series feedback amplifier is \(1/(1+AB)\) times the output resistance \(R_o\) of the op-amp.
(g) Bandwidth with feedback
The bandwidth of an amplifier is the range of frequencies over which its gain remains constant. The open-loop gain of a compensated op-amp falls off above a break frequency \(f_0\) as
which has the same form with the low-frequency gain \(A/(1+AB)\) and the break frequency
\[\boxed{f_F=f_0(1+AB)}.\]
The bandwidth of the non-inverting amplifier with feedback is thus its bandwidth without feedback multiplied by \((1+AB)\). Since the gain is divided by the same factor, the gain–bandwidth product \(A_Ff_F=Af_0\) is unchanged; it equals the unity-gain bandwidth \(f_T\), so that \(f_F=f_T/A_F\).
(h) Total output offset voltage with feedback
Without feedback, even a tiny input offset voltage drives the output into saturation, so the total output offset voltage is \(\pm V_{\text{sat}}\). Since feedback reduces the gain of the non-inverting amplifier from \(A\) to \(A/(1+AB)\), it reduces the total output offset voltage by the same factor:
\[\boxed{V_{ooT}=\frac{\text{total output offset voltage without feedback}}{1+AB}=\frac{\pm V_{\text{sat}}}{1+AB}}.\]
This analysis shows that the non-inverting amplifier with feedback has the characteristics of a nearly perfect voltage amplifier: a very high input resistance, a very low output resistance, a stable voltage gain, a large bandwidth and very little (ideally zero) output offset voltage.
(i) The voltage follower
The lowest gain that a non-inverting amplifier with feedback can provide is 1. It is obtained by opening \(R_1\) (\(R_1\to\infty\)) and shorting \(R_F\) (\(R_F=0\)), which gives the circuit of Figure 5.8. The whole of the output voltage is fed back to the inverting terminal, so the gain of the feedback circuit is \(B=1\) and \(A_F=1\), that is, \(v_o=v_{in}\).
Figure 5.8 The voltage follower.
Putting \(B=1\) in the general results, and using \(1+A\simeq A\),
The voltage follower is also called a non-inverting buffer, because when placed between two networks it removes the loading of the second on the first.
Example 5.7. An op-amp with \(A=200{,}000\), \(R_i=2\,\text{M}\Omega\), \(R_o=75\,\Omega\), \(f_0=5\) Hz and \(\pm V_{CC}=\pm15\) V is connected as a non-inverting amplifier with \(R_1=1\,\text{k}\Omega\) and \(R_F=10\,\text{k}\Omega\). Compute \(A_F\), \(R_{iF}\), \(R_{oF}\), \(f_F\) and \(V_{ooT}\).
Example 5.8. For the ideal op-amp circuit shown below, \(R_1=10\,\text{k}\Omega\), \(R_2=2\,\text{k}\Omega\) and \(R_L=50\,\text{k}\Omega\). For an applied input voltage \(V=10\) mV, find the current through \(R_2\).
Solution. The circuit is a non-inverting amplifier with \(V\) at the non-inverting input and the inverting input connected to the junction of \(R_1\) and \(R_2\). By the virtual short the voltage across \(R_2\), which is connected from that junction to ground, is \(V\) itself. Hence
5.6 Voltage-shunt feedback: the inverting amplifier
(a) The circuit
Figure 5.9 shows the voltage-shunt feedback amplifier. The non-inverting terminal is grounded, and the feedback circuit consists of the single resistor \(R_F\) connected from the output to the inverting terminal; an extra resistor \(R_1\) is connected in series with the input signal source \(v_{in}\). The feedback current is applied in parallel (shunt) with the input current at the inverting terminal.
Figure 5.9 The voltage-shunt feedback amplifier, and the same circuit redrawn.
(b) Closed-loop voltage gain
We apply Kirchhoff’s current law at the input node \(v_2\). The input current \(i_{in}\) divides into the feedback current \(i_F\) and the bias current \(I_B\) drawn by the op-amp; since \(I_B\to0\) as \(R_i\to\infty\),
The negative sign shows that the output and input are \(180^\circ\) out of phase, or of opposite polarities, which is why the circuit is called an inverting amplifier with feedback.
Dividing the numerator and denominator of the exact gain by \(R_1+R_F\),
is a voltage attenuation factor and \(B=R_1/(R_1+R_F)\) is, as before, the gain of the feedback circuit. Note that \(K+B=1\).
(c) The inverting input at virtual ground
The difference input voltage is ideally zero, so the voltage \(v_2\) at the inverting terminal is approximately equal to the voltage \(v_1\) at the non-inverting terminal. Since the non-inverting terminal is grounded, \(v_2\) is approximately at ground potential, and the inverting terminal is said to be at virtual ground: it is held at zero potential, yet no current flows from it to ground. With \(v_1=v_2=0\) V the gain follows at once:
Looking into the circuit from the source, we see \(R_1\) in series with the resistance from the inverting node to ground. That node is connected to ground through \(R_i\) and, through \(R_F\), to an output that moves by \(-A\) times its voltage. A change \(v_2\) therefore drives the current \(v_2(1+A)/R_F\) through \(R_F\), so \(R_F\) appears at the input as \(R_F/(1+A)\) (the Miller effect), in parallel with \(R_i\). Hence
Since \(v_1=0\) V and \(v_1=v_2\), the input current is \(v_{in}/R_1=i_{in}\), and therefore
\[\boxed{v_o=-i_{in}R_F}.\]
If the combination of \(v_{in}\) and \(R_1\) is replaced by a current source \(i_{in}\), as in the figure below, the output voltage becomes proportional to the input current. In other words, the circuit converts an input current into a proportional output voltage. Because the input is at virtual ground, the source sees zero load resistance, which is exactly what an ideal current meter requires.
Example 5.9. An op-amp with \(A=200{,}000\), \(R_i=2\,\text{M}\Omega\), \(R_o=75\,\Omega\), \(f_0=5\) Hz and \(\pm V_{CC}=\pm15\) V is connected as an inverting amplifier with \(R_1=470\,\Omega\) and \(R_F=4.7\,\text{k}\Omega\). Compute \(A_F\), \(R_{iF}\), \(R_{oF}\), \(f_F\) and \(V_{ooT}\).
Solution. The feedback and attenuation factors are
The output resistance, bandwidth and offset are the same as for the non-inverting amplifier of the earlier example, because the loop gain \(AB\) is the same; only the gain and the input resistance differ.
Example 5.10. In the op-amp circuit shown in the figure, find the load current \(i_L\).
Solution. Let \(v\) be the common voltage of the two input terminals. Kirchhoff’s current law at the inverting terminal, where \(v_i\) acts through \(R_1\) and the feedback resistor is also \(R_1\), gives
The load current is set by \(v_i\) and \(R_2\) alone and is independent of \(R_L\), so the circuit acts as a voltage-controlled current source.
Example 5.11. Assuming the op-amp to be ideal, find the voltage gain of the amplifier shown below.
Solution. The non-inverting terminal is grounded, so the inverting terminal A is at virtual ground, \(V_A=0\). The resistor \(R_2\) runs from A to the output line, and \(R_3\) runs from the output line to ground. \(R_3\) is therefore simply a load across the output; the ideal op-amp, having zero output resistance, maintains \(V_0\) whatever current \(R_3\) draws. The current \(V_i/R_1\) flowing into A through \(R_1\) must all flow on through \(R_2\), so
5.7 Analysing op-amp circuits with negative feedback
The examples of this section show how circuits built from inverting and non-inverting stages are analysed. Two rules, both consequences of very large gain with negative feedback, suffice for an ideal op-amp: the two input terminals are at the same potential (the virtual short, or virtual ground when one input is grounded), and no current flows into either input terminal. With these rules we write Kirchhoff’s current law at the inverting node, or use the standard results \(-R_F/R_1\) and \(1+R_F/R_1\) together with superposition. We give ideal values throughout; a practical op-amp cannot deliver an output beyond its saturation voltage.
When the same op-amp receives signals at both inputs, superposition gives a useful short cut. If a voltage \(V_-\) is applied through \(R_1\) to the inverting terminal, with feedback resistor \(R_F\), and a voltage \(V_+\) is applied through \(R_2\) to the non-inverting terminal, which is returned to ground through \(R_3\), then
If the resistors are chosen so that \(R_F/R_1=R_3/R_2\), the factor multiplying \(V_+\) becomes \(\dfrac{R_1+R_F}{R_1}\cdot\dfrac{R_3}{R_2+R_3}=\dfrac{R_F}{R_1}\), and
Exam tip. Check the ratio condition before using the short cut \(V_0=(R_F/R_1)(V_+-V_-)\); when it does not hold, add the inverting and non-inverting contributions separately.
Example 5.13. For the op-amp network shown below, find the current \(I_0\).
Solution. The left-hand op-amp is an inverting amplifier with \(R_1=2\,\text{k}\Omega\) and \(R_F=10\,\text{k}\Omega\) driven by 2 V, so its output, the node \(b\) at the left of the \(5\,\text{k}\Omega\) resistor, is
The right-hand op-amp is a non-inverting amplifier with 8 V at its non-inverting input, \(1\,\text{k}\Omega\) from its inverting input to ground and \(10\,\text{k}\Omega\) feedback, so its output, the node \(a\) at the right of the \(5\,\text{k}\Omega\) resistor, is
Each ideal op-amp holds its output voltage regardless of the current it delivers, so the current through the \(5\,\text{k}\Omega\) resistor, in the direction of the arrow from \(a\) to \(b\), is
Example 5.14. For the circuit shown below, find the voltage gain \(V_0/V_{in}\), taking the input end of the upper resistor \(R\) to be at ground potential.
Solution. The upper op-amp receives \(V_{in}\) at its non-inverting terminal, with \(R\) to ground and \(5R\) as feedback; it is a non-inverting amplifier, so
The lower op-amp receives \(V_{in}\) through \(R\) at its inverting terminal, with \(2R\) as feedback and its non-inverting terminal grounded; it is an inverting amplifier, so
The resistor \(R_2\) from the output to ground is merely a load; since the ideal op-amp has zero output resistance, it does not affect \(V_0\).
Example 5.16. In the op-amp circuit shown below, find the value of \(V_0\).
Solution. We use superposition. With the 5 V source set to zero, the non-inverting terminal is at zero potential (no current flows in the \(10\,\text{k}\Omega\) and \(20\,\text{k}\Omega\) resistors), and the circuit is an inverting amplifier for the 3 V input:
With the 3 V source set to zero, the 5 V input reaches the non-inverting terminal through the divider of \(10\,\text{k}\Omega\) and \(20\,\text{k}\Omega\):
In this circuit \(R_F/R_1=10\) but \(R_3/R_2=9\), so the short cut does not apply.
Example 5.18. For the op-amp circuits shown below, find \(V_{01}/V_{02}\).
Solution. Op-amp 1 is a non-inverting amplifier with \(R_1=1\,\text{k}\Omega\) and \(R_F=10\,\text{k}\Omega\), so
\[V_A=\left(1+\frac{10}{1}\right)V_i=11V_i.\]
Op-amp 2, whose output is connected directly to its inverting input, is a voltage follower (buffer), so \(V_{01}=V_A=11V_i\). Op-amp 3 is an inverting amplifier with equal input and feedback resistors \(R\), so
Example 5.19. In the op-amp circuit shown below, the feedback element is a diode whose current is \(I=I_0e^{v/\eta V_T}\) for a forward voltage \(v\). Find the output voltage.
Solution. The inverting terminal is at virtual ground, so the input current is \(I_1=v_{in}/R\), and since no current enters the op-amp it all flows through the diode: \(I_1=I_2\). The diode’s anode is at the virtual ground and its cathode at the output, so its forward voltage is \(v=0-v_0=-v_0\). Hence
The output is proportional to the logarithm of the input, so the circuit is a logarithmic amplifier (it requires \(v_{in}>0\) to forward-bias the diode).
Example 5.20. In the op-amp circuit shown below, find the output voltage.
Solution. In the first stage the non-inverting terminal B is grounded, so by the virtual-ground principle \(V_A=V_B=0\). Kirchhoff’s current law at A, with currents in milliamperes and resistances in kilohms, gives
