Fundamentals of Microelectronics

Due to the negative gain, the circuit is called the “inverting amplifier.” As with its noninverting counterpart, the gain of this circuit is given by the ratio of the two resistors, thereby experiencing only small variations with temperature and process.

It is important to understand the role of the virtual ground in this circuit. If the inverting input of the op amp were not near zero potential, then neither Vin=R2 nor Vout=R1 would accurately represent the currents flowing through R2 and R1, respectively. This behavior is similar to a seesaw [Fig. 8.7(c)], where the point between the two arms is “pinned” (e.g., does not move), allowing displacement of point A to be “amplified” (and “inverted”) at point B.

The above development also reveals why the virtual ground cannot be shorted to the actual ground. Such a short in Fig. 8.7(b) would force to ground all of the current flowing through R2, yielding Vout = 0. It is interesting to note that the inverting amplifier can also be drawn as shown in Fig. 8.8, displaying a similarity with the noninverting circuit but with the input applied at a different point.

Vout

Vin

Figure 8.8 Inverting amplifier.

In contrast to the noninverting amplifier, the topology of Fig. 8.7(a) exhibits an input impedance equal to R2—as can be seen from the input current, Vin=R2, in Fig. 8.7(b). That is, a lower R2 results in a greater gain but a smaller input impedance. This trade-off sometimes makes this amplifier less attractive than its noninverting counterpart.

Let us now compute the closed-loop gain of the inverting amplifier with a finite op amp gain. We note from Fig. 8.7(a) that the currents flowing through R2 and R1 are given by (Vin,VX)=R2 and (VX , Vout)=R1, respectively. Moreover,

As expected, a higher closed-loop gain ( ,R1=R2) is accompanied with a greater gain error. Note that the gain error expression is the same for noninverting and inverting amplifiers.

Example 8.4

Design the inverting amplifier of Fig. 8.7(a) for a nominal gain of 4, a gain error of 0:1%, and an input impedance of at least 10 k .

Solution

Since both the nominal gain and the gain error are given, we must first determine the minimum op amp gain. We have

Exercise

Repeat the above example for a gain error of 1% and compare the results.

In the above example, we assumed the input impedance is approximately equal to R2. How accurate is this assumption? With A0 = 5000, the virtual ground experiences a voltage of ,Vout=5000 ,4Vin=5000, yielding an input current of (Vin + 4Vin=5000)=R1. That is, our assumption leads to an error of about 0:08%—an acceptable value in most applications.

8.2.3 Integrator and Differentiator

Our study of the inverting topology in previous sections has assumed a resistive network around the op amp. In general, it is possible to employ complex impedances instead (Fig. 8.9). In analogy with (8.16), we can write

Z 1

Z 2

Vout

Vin

Figure 8.9 Circuit with general impedances around the op amp.

where the gain of the op amp is assumed large. If Z1 or Z2 is a capacitor, two interesting functions result.

Integrator Suppose in Fig. 8.9, Z1 is a capacitor and Z2 a resistor (Fig. 8.10). That is, Z1 = (C1s),1 and Z2 = R1. With an ideal op amp, we have

C1

R1 X

Vout

Vin

Figure 8.10 Integrator.

Providing a pole at the origin,4 the circuit operates as an integrator (and a low-pass filter). Figure 8.11 plots the magnitude of Vout=Vin as a function of frequency. This can also be seen in the time domain. Equating the currents flowing through R1 and C1 gives

Equation (8.29) indicates that Vout=Vin approaches infinity as the input frequency goes to zero. This is to be expected: the capacitor impedance becomes very large at low frequencies, approaching an open circuit and reducing the circuit to the open-loop op amp.

As mentioned at the beginning of the chapter, integrators originally appeared in analog computers to simulate differential equations. Today, electronic integrators find usage in analog filters, control systems, and many other applications.

Example 8.5

Plot the output waveform of the circuit shown in Fig. 8.12(a). Assume a zero initial condition across C1 and an ideal op amp.

4Pole frequencies are obtained by setting the denominator of the transfer function to zero.

Solution

When the input jumps from 0 to V1, a constant current equal to V1=R1 begins to flow through the resistor and hence the capacitor, forcing the right plate voltage of C1 to fall linearly with time while its left plate is pinned at zero [Fig. 8.12(b)]:

(Note that the output waveform becomes “sharper” as R1C1 decreases.) When Vin returns to zero, so do the currents through R1 and C1. Thus, the voltage across the capacitor and hence Vout remain equal to ,V1Tb=(R1C1) (proportional to the area under the input pulse) thereafter.

Exercise

Repeat the above example if V1 is negative.

The above example demonstrates the role of the virtual ground in the integrator. The ideal integration expressed by (8.32) occurs because the left plate of C1 is pinned at zero. To gain more insight, let us compare the integrator with a first-order RC filter in terms of their step response.

As illustrated in Fig. 8.13, the integrator forces a constant current (equal to V1=R1) through the capacitor. On the other hand, the RC filter creates a current equal to (Vin , Vout)=R1, which decreases as Vout rises, leading to an increasingly slower voltage variation across C1. We may therefore consider the RC filter as a “passive” approximation of the integrator. In fact, for a large R1C1 product, the exponential response of Fig. 8.13(b) becomes slow enough to be approximated as a ramp.

We now examine the performance of the integrator for A0 < 1. Denoting the potential of the virtual ground node in Fig. 8.10 with VX, we have

C1s

Exercise

What value of RX is necessary if CX = C1?

Differentiator If in the general topology of Fig. 8.9, Z1 is a resistor and Z2 a capacitor (Fig.

Exhibiting a zero at the origin, the circuit acts as a differentiator (and a high-pass filter). Figure 8.16 plots the magnitude of Vout=Vin as a function of frequency. From a time-domain perspective, we can equate the currents flowing through C1 and R1:

Example 8.7

Plot the output waveform of the circuit shown in Fig. 8.17(a) assuming an ideal op amp.

Solution

At t = 0,, Vin = 0 and Vout = 0 (why?). When Vin jumps to V1, an impulse of current flows through C1 because the op amp maintains VX constant:

Figure 8.17(b) depicts the result. At t = Tb, Vin returns to zero, again creating an impulse of current in C1:

We can therefore say that the circuit generates an impulse of current [ C1V1 (t)] and “amplifies” it by R1 to produce Vout. In reality, of course, the output exhibits neither an infinite height (limited by the supply voltage) nor a zero width (limited by the op amp nonidealities).

Exercise

Plot the output if V1 is negative.

It is instructive to compare the operation of the differentiator with that of its “passive” counterpart (Fig. 8.18). In the ideal differentiator, the virtual ground node permits the input to change the voltage across C1 instantaneously. In the RC filter, on the other hand, node X is not “pinned,” thereby following the input change at t = 0 and limiting the initial current in the circuit to V1=R1.

388 Chap. 8 Operational Amplifier As A Black Box

If the decay time constant, R1C1, is sufficiently small, the passive circuit can be viewed as an approximation of the ideal differentiator.

Figure 8.18 Comparison of differentiator and RC circuit.

Let us now study the differentiator with a finite op amp gain. Equating the capacitor and resistor currents in Fig. 8.15 gives

Example 8.8

Determine the transfer function of the high-pass filter shown in Fig. 8.19 and choose RX and CX such that the pole of this circuit coincides with (8.55).

CX

VinVout

RX

Figure 8.19 Simple high-pass filter.

Solution

The capacitor and resistor operate as a voltage divider:

The circuit therefore exhibits a zero at the origin (s = 0) and a pole at ,1=(RX CX). For this pole to be equal to (8.55), we require

Exercise

What is the necessary value of CX if RX = R1?

An important drawback of differentiators stems from the amplification of high-frequency noise. As suggested by (8.42) and Fig. 8.16, the increasingly larger gain of the circuit at high frequencies tends to boost noise in the circuit.

The general topology of Fig. 8.9 and its integrator and differentiator descendants operate as inverting circuits. The reader may wonder if it is possible to employ a configuration similar to the noninverting amplifier of Fig. 8.5 to avoid the sign reversal. Shown in Fig. 8.20, such a circuit provides the following transfer function

Vout

Vin

Z 1

Z 2

Figure 8.20 Op amp with general network.

if the op amp is ideal. Unfortunately, this function does not translate to ideal integration or differentiation. For example, Z1 = R1 and Z2 = 1=(C2s) yield a nonideal differentiator (why?).

8.2.4 Voltage Adder

The need for adding voltages arises in many applications. In audio recording, for example, a number of microphones may convert the sounds of various musical instruments to voltages, and these voltages must then be added to create the overall musical piece. This operation is called “mixing” in the audio industry. 5 For example, in“noise cancelling” headphones, the environmen-

5The term “mixing” bears a completely different meaning in the RF and wireless industry.

tal noise is applied to an inverting amplifier and subsequently added to the signal so as to cancel itself.

Figure 8.21 depicts a voltage adder (“summer”) incorporating an op amp. With an ideal op amp, VX = 0, and R1 and R2 carry currents proportional to V1 and V2, respectively. The two currents add at the virtual ground node and flow through RF :

This circuit can therefore add and amplify voltages. Extension to more than two voltages is straightforward.

Equation (8.63) indicates that V1 and V2 can be added with different weightings: RF =R1 and RF =R2, respectively. This property also proves useful in many applications. For example, in audio recording it may be necessary to lower the “volume” of one musical instrument for part of the piece, a task possible by varying R1 and R2.

The behavior of the circuit in the presence of a finite op amp gain is studied in Problem 31.

8.3 Nonlinear Functions

It is possible to implement useful nonlinear functions through the use of op amps and nonlinear devices such as transistors. The virtual ground property plays an essential role here as well.

8.3.1 Precision Rectifier

The rectifier circuits described in Chapter 3 suffer from a “dead zone” due to the finite voltage required to turn on the diodes. That is, if the input signal amplitude is less than approximately 0.7 V, the diodes remain off and the output voltage remains at zero. This drawback prohibits the use of the circuit in high-precision applications, e.g., if a small signal received by a cellphone must be rectified to determine its amplitude.

It is possible to place a diode around an op amp to form a “precision rectifier,” i.e., a circuit that rectifies even very small signals. Let us begin with a unity-gain buffer tied to a resistive load [Fig. 8.22(a)]. We note that the high gain of the op amp ensures that node X tracks Vin (for both positive and negative cycles). Now suppose we wish to hold X at zero during negative cycles, i.e., “break” the connection between the output of the op amp and its inverting input. This can be