Question 10.11: Asource follower operated at gm = 2 mA/V and ro = 20 kΩ is f...
Asource follower operated at gm = 2 mA/V and ro = 20 kΩ is fed with a signal source for which Rsig = 10 kΩ and is loaded in a resistance RL = 20 kΩ. The MOSFET has Cgs = 20 fF, Cgd = 5 fF, and gmb = χ gm where χ = 0.2, and the total capacitance at the output CL = 15 fF. Determine AM, fT , fZ , Q, fP1, fP2, and f3dB.
R_{L}^{′} = R_{L} || r_{o} || \frac{1}{g_{mb}}
= 20 || 20 || \frac{1}{0.2 × 2} = 20 || 20 || 2.5 = 2 kΩ
A_{M} = \frac{R_{L}^{′}}{R_{L}^{′} + \frac{1}{g_{m}}} = \frac{2}{2 + \frac{1}{2}} = 0.8 V/V
f_{T} = \frac{g_{m}}{2π (C_{gs} + C_{gd})}
\frac{2 × 10^{−3}}{2π (20 + 5) × 10^{−15}}
= 12.7 GHz
f_{Z} = \frac{g_{m}}{2π C_{gs}} = \frac{2 × 10^{−3}}{2π × 20 × 10^{−15}}= 15.9 GHz
To evaluate Q we substitute the given component values into Eq. (10.128),
Q = \frac{\sqrt{b_{2}} }{b_{1}} = \frac{\sqrt{g_{m}R^{′}_{L} + 1}\sqrt{[(C_{gs} + C_{gd})C_{L} + C_{gs} C_{gd}] R_{sig}R^{′}_{L}} }{[C_{gs} + C_{gd} (g_{m}R^{′}_{L} + 1)] R_{sig} + (C_{gs} + C_{L})R^{′}_{L}} (10.128)
Q = 0.42
Thus the poles are real. Their frequencies can be obtained by finding the roots of the polynomial (1 + b1s + b2s²), where
b1 = τH = 104 ps
and
b2 = 1.9 × 10−21
Thus,
fP1 = 1.98 GHz
fP2 = 6.73 GHz
Since fP2/fP1 = 3.4 < 4, no dominant pole exists. An approximate value for fH can be obtained as
f_{H} = 1/\sqrt{\frac{1}{f_{p1}^{2}} + \frac{1}{f_{p2}^{2}} − \frac{2}{f_{Z}^{2}}} = 1.93 GHz
The exact value of fH can be found from the transfer function as 1.86 GHz, which is not much different from the approximate value.