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## Q. 6.22

Estimate the −3-dB bandwidth of a resistively-degenerated common-source stage. Assume λ = γ = 0.

## Verified Solution

Shown in Fig. 6.53(a), the small-signal model is of third order,$^8$ providing little intuition. The zero-value time constant method can give a rough estimate of the circuit’s bandwidth, thereby revealing the contribution of each capacitor.

We begin with the time constant associated with $C_{GS}$ and set $C_{G D } and C_L$ to zero. As depicted in Fig. 6.53(b), the resistance seen by $C_{GS} is V_X /I_X$ . We denote this resistance by $R_{CGS}$. Since $V_1 = V_X$ and the current flowing through $R_S$ is equal to $g_m V_1 − I_X = g_m V_X − I_X$ , we write a KVL as follows:

$I_X R_G = V_X + (g_m V_X − I_X )R_S$                                          (6.123)

obtaining

$R_{CGS} =\frac{ R_G + R_S} {1 + g_m R_S}$                                        (6.124)

For the resistance seen by $C_{G D}$, we have from Example 6.21

$R_{CGD} = R_D +\left(\frac{ g_m R_D}{1 + g_m R_S}+ 1\right)R_G$                                                                    (6.125)

Finally, the resistance seen by $C_L$ is simply equal to $R_D$. It follows that the −3-dB bandwidth is given by

$ω^{−1}_{−3d B} = \frac{R_G + R_S}{1 + g_m R_S}C_{GS} +\left[R_D +\left(\frac{g_m R_D}{1 + g_m R_S}+ 1\right) R_G \right] C_{G D} + R_DC_L$                                                  (6.126)

With no degeneration, this result reduces to Eq. (6.35). With a finite RS, the effect of $C_{GS} and R_G$ is reduced by a factor of $1 + g_m R_S,$ albeit at the cost of voltage gain.

$ω_{p1} = \frac{1}{R_S(1 + g_m R_D)C_{G D} + R_SC_{GS} + R_D(C_{G D} + C_{DB})}$                                        (6.35)

$^8$ This can be seen by observing that it is possible to impose three independent initial conditions across the three capacitors
without violating KVL.