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## Q. 6.12

Apply Miller’s approximation to the above circuit if $C_L = 0$.

## Verified Solution

As illustrated in Fig. 6.25, the low-frequency gain from the gate to the source is equal to $(1/g_{mb})/[(1/g_m) + (1/g_{mb})] = g_m/(g_m+g_{mb})$. The Miller multiplication of $C_{GS}$ at the input is thus equal to $C_{GS}[1−g_m/(g_m+g_{mb})] = C_{GS} g_{mb}/(g_{m} + g_{mb}).$ 