Fundamentals of Microelectronics

The system is still of first order and the pole of the closed-loop transfer function is given by

As depicted in Fig. 8.37, the bandwidth of the closed-loop circuit is substantially higher than that of the op amp itself. This improvement, of course, accrues at the cost of a proportional reduction in gain—from A0 to 1 + R2A0=(R1 + R2).

Figure 8.37 Frequency response of (a) open-loop op amp and (b) closed-loop circuit.

Example 8.14

A noninverting amplifier incorporates an op amp having an open-loop gain of 100 and bandwidth of 1 MHz. If the circuit is designed for a closed-loop gain of 16, determine the resulting bandwidth and time constant.

Solution

For a closed-loop gain of 16, we require that 1 + R1=R2 = 16 and hence

Given by j!p;closedj,1, the time constant of the circuit is equal to 2.51 ns.

Exercise

Repeat the above example if the op amp gain is 500.

The above analysis can be repeated for the inverting amplifier as well. The reader can prove that the result is similar to (8.87).

The finite bandwidth of the op amp may considerably degrade the performance of integrators. The analysis is beyond the scope of this book, but it is outlined in Problem 57 for the interested reader.

Another critical issue in the use of op amps is stability; if placed in the topologies seen above, some op amps may oscillate. Arising from the internal circuitry of the op amp, this phenomenon often requires internal or external stabilization, also called “frequency compensation.” These concepts are studied in Chapter 12.

Slew Rate In addition to bandwidth and stability problems, another interesting effect is observed in op amps that relates to their response to large signals. Consider the noninverting configuration shown in Fig. 8.38(a), where the closed-loop transfer function is given by (8.86). A small step of V at the input thus results in an amplified output waveform having a time constant equal to j!p;closedj,1 [Fig. 8.38(b)]. If the input step is raised to 2 V , each point on the output waveform also rises by a factor of two.7 In other words, doubling the input amplitude doubles both the output amplitude and the output slope.

Figure 8.38 (a) Noninverting amplifier, (b) input and output waveforms in linear regime, (c) input and output waveforms in slewing regime.

In reality, op amps do not exhibit the above behavior if the signal amplitudes are large. As illustrated in Fig. 8.38(c), the output first rises with a constant slope (i.e., as a ramp) and eventually settles as in the linear case of Fig. 8.38(b). The ramp section of the waveform arises because, with a large input step, the internal circuitry of the op amp reduces to a constant current source charging a capacitor. We say the op amp “slews” during this time. The slope of the ramp is called the “slew rate” (SR).

Slewing further limits the speed of op amps. While for small-signal steps, the output response is determined by the closed-loop time constant, large-signal steps must face slewing prior to linear settling. Figure 8.39 compares the response of a non-slewing circuit with that of a slewing op amp, revealing the longer settling time in the latter case.

It is important to understand that slewing is a nonlinear phenomenon. As suggested by the waveforms in Fig. 8.38(c), the points on the ramp section do not follow linear scaling (if x ! y, then 2x 6!2y). The nonlinearity can also be observed by applying a large-signal sine wave to the circuit of Fig. 8.38(a) and gradually increasing the frequency (Fig. 8.40). At low frequencies, the op amp output “tracks” the sine wave because the maximum slope of the sine wave remains less than the op amp slew rate [Fig. 8.40(a)]. Writing Vin(t) = V0 sin !t and Vout(t) = V0(1 + R1=R2) sin !t, we observe that

7Recall that in a linear system, if x(t) ! y(t), then 2x(t) ! 2y(t).

Vin

t

Without

Slewing

Vout

With

Slewing

t

Figure 8.39 Output settling speed with and without slewing.

(c)

Figure 8.40 (a) Simple noninverting amplifier, (b) input and output waveforms without slewing, (c) input and output waveforms with slewing.

The output therefore exhibits a maximum slope of V0!(1 + R1=R2) (at its zero crossing points), and the op amp slew rate must exceed this value to avoid slewing.

What happens if the op amp slew rate is insufficient? The output then fails to follow the sinusoidal shape while passing through zero, exhibiting the distorted behavior shown in Fig. 8.40(b). Note that the output tracks the input so long as the slope of the waveform does not exceed the op amp slew rate, e.g., between t1 and t2.

Example 8.15

The internal circuitry of an op amp can be simplified to a 1-mA current source charging a 5-pF capacitor during large-signal operation. If an amplifier using this op amp produces a sinusoid with a peak amplitude of 0.5 V, determine the maximum frequency of operation that avoids slewing.

404 Chap. 8 Operational Amplifier As A Black Box

Solution

That is, for frequencies above 63.7 MHz, the zero crossings of the output experience slewing.

Exercise

Plot the output waveform if the input frequency is 200 MHz.

Equation (8.91) indicates that the onset of slewing depends on the closed-loop gain, 1 + R1=R2. To define the maximum sinusoidal frequency that remains free from slewing, it is common to assume the worst case, namely, when the op amp produces its maximum allowable voltage swing without saturation. As exemplified by Fig. 8.41, the largest sinusoid permitted at the output is given by

where Vmax and Vmin denote the bounds on the output level without saturation. If the op amp provides a slew rate of SR, then the maximum frequency of the above sinusoid can be obtained by writing

Called the “full-power bandwidth,” !FP serves as a measure of the useful large-signal speed of the op amp.

8.4.4 Finite Input and Output Impedances

Actual op amps do not provide an infinite input impedance 8 or a zero output impedance—the latter often creating limitations in the design. We analyze the effect of this nonideality on one circuit here.

Consider the inverting amplifier shown in Fig. 8.42(a), assuming the op amp suffers from an output resistance, Rout. How should the circuit be analyzed? We return to the model in Fig. 8.1 and place Rout in series with the output voltage source [Fig. 8.42(b)]. We must solve the circuit in the presence of Rout. Recognizing that the current flowing through Rout is equal to (,A0vX , vout)=Rout, we write a KVL from vin to vout through R2 and R1:

Figure 8.42 (a) Inverting amplifier, (b) effect of finite output resistance of op amp.

To construct another equation for vX, we view R1 and R2 as a voltage divider:

Substituting for vX in (8.97) thus yields

The additional terms ,Rout=R1 in the numerator and Rout=R2 in the denominator increase the gain error of the circuit.

Example 8.16

An electrical engineering student purchases an op amp with A0 = 10; 000 and Rout = 1 and constructs the amplifier of Fig. 8.42(a) using R1 = 50 and R2 = 10 . Unfortunately, the circuit fails to provide large voltage swings at the output even though Rout=R1 and Rout=R2 remain much less than A0 in (8.99). Explain why.

Solution

For an output swing of, say, 2 V, the op amp may need to deliver a current as high as 40 mA to R1 (why?). Many op amps can provide only a small output current even though their small-signal output impedance is very low.

8Op amps employing MOS transistors at their input exhibit a very high input impedance at low frequencies.

Exercise

If the op amp can deliver a current of 5 mA, what value of R1 is acceptable for output voltages as high as 1 V?

8.5 Design Examples

Following our study of op amp applications in the previous sections, we now consider several examples of the design procedure for op amp circuits. We begin with simple examples and gradually proceed to more challenging problems.

Example 8.17

Design an inverting amplifier with a nominal gain of 4, a gain error of 0:1%, and an input impedance of at least 10 k . Determine the minimum op amp gain required here.

Solution

For an input impedance of 10 k , we choose the same value of R2 in Fig. 8.7(a), arriving at R1 = 40 k for a nominal gain of 4. Under these conditions, Eq. (8.21) demands that

Exercise

Repeat the above example for a nominal gain of 8 and compare the results.

Example 8.18

Design a noninverting amplifier for the following specifications: closed-loop gain = 5, gain error = 1%, closed-loop bandwidth = 50 MHz. Determine the required open-loop gain and bandwidth of the op amp. Assume the op amp has an input bias current of 0.2 A.

Solution

The choice of R1 and R2 themselves depends on the “driving capability” (output resistance) of the op amp. For example, we may select R1 = 4 k and R2 = 1 k and check the gain error from (8.99) at the end. For a gain error of 1%,

Exercise

Repeat the above example for a gain error of 2% and compare the results.

Example 8.19

Design an integrator for a unity-gain frequency of 10 MHz and an input impedance of 20 k . If the op amp provides a slew rate of 0.1 V/ns, what is the largest peak-to-peak sinusoidal swing at the input at 1 MHz that produces an output free from slewing?

Solution

From (8.29), we have

In other words, the input peak-to-peak swing at 1 MHz must remain below 3.18 V for the output to be free from slewing.

Exercise

How do the above results change if the op amp provides a slew rate of 0.5 V/ns?

8.6 Chapter Summary

The inverting amplifier configuration provides a nominal gain equal to the ratio of two resistors. Its gain error is the same as that of the noninverting configuration. With the noninverting input of the op amp tied to ground, the inverting input also remains close to the ground potential ans is thus called a “virtual ground.”

If the feedback resistor in an inverting configuration is replaced with a capacitor, the circuit operates as an integrator. Integrator find wide application in analog filters and analog-to- digital converters.

Problems

1.Actual op amps exhibit “nonlinear” characteristics. For example, the voltage gain may be equal to 1000 for ,1 V < Vout < +1 V, 500 for 1 V < jVoutj < 2 V, and close to zero for

jVoutj > 2 V.

(a)Plot the input/output characteristic of this op amp.

(b)What is the largest input swing that the op amp can sense without producing “distortion” (i.e., nonlinearity)?

2. An op amp exhibits the following nonlinear characteristic:

Sketch this characteristic and determine the small-signal gain of the op amp in the vicinity

of Vin1 , Vin2 0.

3.A noninverting amplifier employs an op amp having a nominal gain of 2000 to achieve a nominal closed-loop gain of 8. Determine the gain error.

4.A noninverting amplifier must provide a nominal gain of 4 with a gain error of 0:1%. Compute the minimum required op amp gain.

5.Looking at Equation (8.11), an adventurous student decides that it is possible to achieve a zero gain error with a finite A0 if R2=(R1 + R2) is slightly adjusted from its nominal value.

(a)Suppose a nominal closed-loop gain of 1 is required. How should R2=(R1 + R2) be chosen?

(b)With the value obtained in (a), determine the gain error if A0 drops to 0:6A0.

6.A noninverting amplifier incorporates an op amp having an input impedance of Rin. Modeling the op amp as shown in Fig. 8.43, determine the closed-loop gain and input impedance.

Vin2

Figure 8.43

What happens if A0 ! 1?

7. A noninverting amplifier employs an op amp with a finite output impedance, Rout. Representing the op amp as depicted in Fig. 8.44, compute the closed-loop gain and output

Vin2

Figure 8.44

impedance. What happens if A0 ! 1?

8. In the noninverting amplifier shown in Fig. 8.45, resistor R2 deviates from its nominal value

R2

Figure 8.45

9.The input/output characteristic of an op amp can be approximated by the piecewise-linear behavior illustrated in Fig. 8.46, where the gain drops from A0 to 0:8A0 and eventually to

Vout

Figure 8.46

zero as jVin1 ,Vin2j increases. Suppose this op amp is used in a noninverting amplifier with a nominal gain of 5. Plot the closed-loop input/output characteristic of the circuit. (Note that the closed-loop gain experiences much less variation; i.e., the closed-loop circuit is much more linear.)

10.A truck weighing station incorporates a sensor whose resistance varies linearly with the weight: RS = R0 + W . Here R0 is a constant value, a proportionality factor, and W the weight of each truck. Suppose RS plays the role of R2 in the noninverting amplifier (Fig. 8.47). Also, Vin = 1 V. Determine the gain of the system, defined as the change in Vout

R1

RS

Figure 8.47

divided by the change in W .

11.Calculate the closed-loop gain of the noninverting amplifier shown in Fig. 8.48 if A0 = 1. Verify that the result reduces to expected values if R1 ! 0 or R3 ! 0.

Figure 8.48

12.An inverting amplifier must provide a nominal gain of 8 with a gain error of 0:2%. Determine the minimum required op amp gain.

13. The op amp used in an inverting amplifier exhibits a finite input impedance, Rin. Modeling the op amp as shown in Fig. 8.43, determine the closed-loop gain and input impedance.