(A) an NMOS Common-Source Amplifier with a Simple PMOS Current-Source$C_{L}=10 \mathrm{pF}$

Question

Examlex · Accepted Answer

The Answer is (a)
$(a) Figure 10.3.1 shows the amplifier together with its output equivalent circuit. The dc gain is given by \begin{aligned} \frac{V_{o}}{V_{ ext {sig }}} & =-g_{m}\left(r_{o 1} \| r_{o 2} ight) \ & =-g_{m}\left(r_{o} \| r_{o} ight)=-\frac{1}{2} g_{m} r_{o} \ & =-\frac{1}{2} imes 1 imes 20=-10 \mathrm{~V} / \mathrm{V} \end{aligned} The 3-dB frequency is determined from the equivalent circuit as \begin{aligned} f_{3 d B} & =\frac{1}{2 \pi C_{L}\left(r_{O 1} \| r_{O 2} ight)} \ & =\frac{1}{2 \pi C_{L} imes \frac{1}{2} imes r_{O}} \ & =\frac{1}{\pi imes 10 imes 10^{-12} imes 20 imes 10^{3}} \ & =1.6 \mathrm{MHz} \ \mathrm{GB} & =10 imes 1.6=16 \mathrm{MHz} \end{aligned} (b) Figure 10.3.2 shows the cascode amplifier together with its output equivalent circuit. The dc gain is given by \frac{V_{o}}{V_{ ext {sig }}}=-g_{m 1}\left(R_{o 2} \| R_{o 3} ight) where \begin{aligned} & g_{m 1}=g_{m}=1 \mathrm{~mA} / \mathrm{V} \ & R_{o 2}=R_{o 3}=\left(g_{m} r_{o} ight) r_{o}=(1 imes 20) imes 20=400 \mathrm{k} \Omega \end{aligned} Thus, \frac{V_{o}}{V_{ ext {sig }}}=-1 imes(400 \| 400)=-200 \mathrm{~V} / \mathrm{V} The 3-dB frequency is determined from the equivalent circuit as \begin{aligned} f_{3 d B} & =\frac{1}{2 \pi C_{L}\left(R_{O 2} \|\left(R_{O 3} ight) ight.} \ & =\frac{1}{2 \pi imes 10 imes 10^{-12} imes(400 \| 400) imes 10^{3}} \ & =80 \mathrm{kHz} \ \mathrm{GB} & =200 imes 80 imes 10^{3}=16 \mathrm{MHz} \end{aligned}$

Figure 10.3.1 shows the amplifier together with its output equivalent circuit. The dc gain is given by

\begin{aligned}\frac{V_{o}}{V_{ ext {sig }}} & =-g_{m}\left(r_{o 1} \| r_{o 2} ight) \& =-g_{m}\left(r_{o} \| r_{o} ight)=-\frac{1}{2} g_{m} r_{o} \& =-\frac{1}{2} imes 1 imes 20=-10 \mathrm{~V} / \mathrm{V}\end{aligned}

The 3-dB frequency is determined from the equivalent circuit as

\begin{aligned}f_{3 d B} & =\frac{1}{2 \pi C_{L}\left(r_{O 1} \| r_{O 2} ight)} \& =\frac{1}{2 \pi C_{L} imes \frac{1}{2} imes r_{O}} \& =\frac{1}{\pi imes 10 imes 10^{-12} imes 20 imes 10^{3}} \& =1.6 \mathrm{MHz} \\mathrm{GB} & =10 imes 1.6=16 \mathrm{MHz}\end{aligned}

(b)
Figure 10.3.2 shows the cascode amplifier together with its output equivalent circuit. The dc gain is given by

\frac{V_{o}}{V_{ ext {sig }}}=-g_{m 1}\left(R_{o 2} \| R_{o 3} ight)

where

\begin{aligned}& g_{m 1}=g_{m}=1 \mathrm{~mA} / \mathrm{V} \& R_{o 2}=R_{o 3}=\left(g_{m} r_{o} ight) r_{o}=(1 imes 20) imes 20=400 \mathrm{k} \Omega\end{aligned}

Thus,

\frac{V_{o}}{V_{ ext {sig }}}=-1 imes(400 \| 400)=-200 \mathrm{~V} / \mathrm{V}

The 3-dB frequency is determined from the equivalent circuit as

\begin{aligned}f_{3 d B} & =\frac{1}{2 \pi C_{L}\left(R_{O 2} \|\left(R_{O 3} ight) ight.} \& =\frac{1}{2 \pi imes 10 imes 10^{-12} imes(400 \| 400) imes 10^{3}} \& =80 \mathrm{kHz} \\mathrm{GB} & =200 imes 80 imes 10^{3}=16 \mathrm{MHz}\end{aligned}

(A) an NMOS Common-Source Amplifier with a Simple PMOS Current-Source $C_{L}=10 \mathrm{pF}$

Essay

Correct Answer:
Verified

(A) an NMOS Common-Source Amplifier with a Simple PMOS Current-Source CL=10pFC_{L}=10 \mathrm{pF}CL​=10pF

Essay

Correct Answer:VerifiedShow Answer

(A) an NMOS Common-Source Amplifier with a Simple PMOS Current-Source $C_{L}=10 \mathrm{pF}$

Correct Answer:
Verified