Question 6.21: A CG stage contains a resistance RG in series with the gate ...
A CG stage contains a resistance R_G in series with the gate [Fig. 6.50(a)]. If only C_{G D} is considered, determine the pole frequency.
As illustrated in Fig. 6.50(b), we remove C_{G D}, set V_{in} to zero, and apply a voltage (or current) source to measure the resistance seen by this capacitor. The voltage across R_S is equal to g_m V_1R_S, yielding
g_m V_1R_S + V_1 = −I_X R_G (6.116)
and hence V_1 = −I_X R_G/(1 + g_m R_S). Since the current flowing through R_D is equal to I_X − g_m V_1, we have
−I_X R_G + V_X = (I_X − g_m V_1)R_D (6.117)
Substituting for V_1, we obtain
\frac{V_X}{I_X}= R_D +\left(\frac{g_m R_D}{1 + g_m R_S}+ 1\right) R_G = R_{eq} (6.118)
The pole is given by 1/(R_{eq}C_{G D}). The reader is encouraged to determine the circuit’s transfer function directly and compare the mathematical labor.
Interestingly, as a result of R_G, the resistance seen by C_{G D} rises from R_D to R_D plus a multiple of R_G, the multiple given by the low-frequency gain of the C_G stage plus 1. It is also interesting to note that the circuit of Fig. 6.50(a) does not lend itself to Miller’s approximation (why?).