Question 10.5: Figure 10.21(a) shows an ideal voltage amplifier having a ga...

(a) For Z = 1 MΩ, employing Miller’s theorem results in the equivalent circuit in Fig. 10.21(b), where

Z_{1} = \frac{Z}{1  −  K} = \frac{1000  kΩ}{1  +  100} = 9.9  kΩ

Z_{2} = \frac{Z}{1  −  \frac{1}{K}} = \frac{1  MΩ}{1  +  \frac{1}{100}} = 0.99  MΩ

The voltage gain can be found as follows:

\frac{V_{o}}{V_{sig}} = \frac{V_{o}}{V_{i}}\frac{V_{i}}{V_{sig}} = −100 × \frac{Z_{1}}{Z_{1}  +  R_{sig}}

= −100 × \frac{9.9}{9.9  +  10} = −49.7  V/V

(b) For Z as a 1-pF capacitance—that is, Z = 1/sC = 1/s × 1 × 10⁻¹² —applying Miller’s theorem allows us to replace Z by Z₁ and Z₂, where

Z_{1} = \frac{Z}{1  −  K} = \frac{1/sC}{1  +  100} = 1/s (101  C)

Z_{2} = \frac{Z}{1  −  \frac{1}{K}} = \frac{1}{1.01} \frac{1}{sC} = \frac{1}{s (1.01  C)}

It follows that Z₁ is a capacitance 101 C = 101 pF and that Z₂ is a capacitance 1.01 C = 1.01 pF. The resulting equivalent circuit is shown in Fig. 10.21(c), from which the voltage gain can be found as follows:

\frac{V_{o}}{V_{sig}} = \frac{V_{o}}{V_{i}}\frac{V_{i}}{V_{sig}} = −100 \frac{1/sC_{1}}{1/(sC_{1})  +  R_{sig}}

= \frac{−100}{1  +  sC_{1} R_{sig}}

= \frac{−100}{1  +  s × 101 × 1 × 10^{−12} × 10 × 10^{3}}

= \frac{−100}{1  +  s × 1.01 × 10^{−6}}

This is the transfer function of a first-order low-pass network with a dc gain of –100 and a 3-dB frequency f_3dB of

f_{3dB} = \frac{1}{2π × 1.01 × 10^{−6}} = 157.6  kHz