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Chapter 6

Q. 6.10

Explain what happens to Eq. (6.47) if the circuit drives a large load capacitance.

\frac{V_X}{I_X}= \frac{1 + R_D(C_{G D} + C_{DB})s}{C_{G D}s(1 + g_m R_D + R_DC_{DB}s)}                                                               (6.47)

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Verified Solution

Merged with C_{DB}, the large load capacitance reduces the numerator to R_DC_{DB}s and the denominator to C_{G D}s(R_DC_{DB}s), yielding V_X /I_X ≈ 1/(C_{G D}s). In a manner similar to that in Example 6.7, the large load capacitance lowers the gain at high frequencies, suppressing Miller multiplication of C_{G D}.