Fundamentals-of-Microelectronics-Behzad-Razavi.pdf

85 design examples following our study of op amp

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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 grad- ually proceed to more challenging problems. Example 8.17 Design an inverting amplifier with a nominal gain of 4, a gain error of , 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 in Fig. 8.7(a), arriving at for a nominal gain of 4. Under these conditions, Eq. (8.21) demands that (8.100) and hence (8.101) 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 , gain error , closed-loop bandwidth 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 From Fig. 8.5 and Eq. (8.9), we have (8.102) The choice of and themselves depends on the “driving capability” (output resistance) of the op amp. For example, we may select k and k and check the gain error from (8.99) at the end. For a gain error of 1%, (8.103)
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BR Wiley/Razavi/ Fundamentals of Microelectronics [Razavi.cls v. 2006] June 30, 2007 at 13:42 407 (1) Sec. 8.5 Design Examples 407 and hence (8.104) Also, from (8.87), the open-loop bandwidth is given by (8.105) (8.106) (8.107) Thus, the op amp must provide an open-loop bandwidth of at least 500 kHz. Exercise Repeat the above example for a gain error of 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 (8.108) and, with , (8.109) (In discrete design, such a small capacitor value may prove impractical.) For an input given by , (8.110) with a maximum slope of (8.111) Equating this result to 0.1 V/ns gives (8.112)
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BR Wiley/Razavi/ Fundamentals of Microelectronics [Razavi.cls v. 2006] June 30, 2007 at 13:42 408 (1) 408 Chap. 8 Operational Amplifier As A Black Box 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 An op amp is a circuit that provides a high voltage gain and an output proportional to the difference between two inputs. Due to its high voltage gain, an op amp producing a moderate output swing requires only a very small input difference. The noninverting amplifier topology exhibits a nominal gain equal to one plus the ratio of two resistors. The circuit also suffers from a gain error that is inversely proportional to the gain of the op amp.
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