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Chapter 6

Q. 6.22

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

Step-by-Step

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.

6.53