A common-source amplifier is fed from a signal source having a resistance $R_{\mathrm{sig}}=100 \mathrm{k} \Omega$ and has a load resistance $R_{L}=100 \mathrm{k} \Omega$. The MOSFET has $C_{g s}=0.1 \mathrm{pF}, C_{g d}=50 \mathrm{fF}, g_{m}=1 \mathrm{~mA} / \mathrm{V}$, and $r_{o}=$ $100 \mathrm{k} \Omega$. The total capacitance between the output node and ground is $C_{L}=0.1 \mathrm{pF}$.
(a) Find the midband gain.
(b) Use the method of open-circuit time constants to obtain an estimate of the upper 3-dB frequency, $f_{H}$.

Question

Quizplus · Accepted Answer

The Answer is Refer to Figure 10.1.1
$Refer to Figure 10.1.1 Figure 10.1.1 (a) At midband frequencies, \begin{aligned} \frac{V_{o}}{V_{ ext {sig }}} & =-g_{m}\left(r_{o} \| R_{L} ight) \ & =-1 imes(100 \| 100) \ & =-50 \mathrm{~V} / \mathrm{V} \end{aligned} (b) Capacitor C_{g s} sees a resistance R_{g_{s}}=R_{\mathrm{sig}}=100 \mathrm{k} \Omega Thus, \begin{aligned} au_{g s} & =C_{g s} R_{g s} \ & =0.1 imes 10^{-12} imes 100 imes 10^{3} \ & =10 \mathrm{~ns} \end{aligned} Capacitor C_{g d} sees a resistance R_{g d}=R_{L}^{\prime}+R_{ ext {sig }}\left(1+g_{m} R_{L}^{\prime} ight) where R_{L}^{\prime}=r_{o}\left\|R_{L}=100 ight\| 100=50 \mathrm{k} \Omega Thus, \begin{aligned} R_{g d} & =50+100 imes(1+1 imes 50) \ & =5150 \mathrm{k} \Omega \end{aligned} and \begin{aligned} au_{g d} & =C_{g d} R_{g d} \ & =50 imes 10^{-15} imes 5150 imes 10^{3} \ & =257.5 \mathrm{~ns} \end{aligned} Capacitor C_{L} sees a resistance R_{C L}=R_{L}\left\|r_{O}=100 ight\| 100=50 \mathrm{k} \Omega Thus, \begin{aligned} au_{C L} & =C_{L} R_{C L} \ & =0.1 imes 10^{-12} imes 50 imes 10^{3} \ & =5 \mathrm{~ns} \end{aligned} Finally, \begin{aligned} au_{H} & = au_{g s}+ au_{g d}+ au_{C L} \ & =10+257.5+5 \ & =272.5 \mathrm{~ns} \end{aligned} and \begin{aligned} f_{H} & =\frac{1}{2 \pi au_{H}}=\frac{1}{2 \pi imes 272.5 imes 10^{-9}} \ & =584 \mathrm{kHz} \end{aligned}$

Figure 10.1.1
(a) At midband frequencies,

\begin{aligned}\frac{V_{o}}{V_{ ext {sig }}} & =-g_{m}\left(r_{o} \| R_{L} ight) \& =-1 imes(100 \| 100) \& =-50 \mathrm{~V} / \mathrm{V}\end{aligned}

(b) Capacitor

C_{g s}

sees a resistance

R_{g_{s}}=R_{\mathrm{sig}}=100 \mathrm{k} \Omega

Thus,

\begin{aligned} au_{g s} & =C_{g s} R_{g s} \& =0.1 imes 10^{-12} imes 100 imes 10^{3} \& =10 \mathrm{~ns}\end{aligned}

Capacitor

C_{g d}

sees a resistance

R_{g d}=R_{L}^{\prime}+R_{ ext {sig }}\left(1+g_{m} R_{L}^{\prime} ight)

where

R_{L}^{\prime}=r_{o}\left\|R_{L}=100 ight\| 100=50 \mathrm{k} \Omega

Thus,

\begin{aligned}R_{g d} & =50+100 imes(1+1 imes 50) \& =5150 \mathrm{k} \Omega\end{aligned}

and

\begin{aligned} au_{g d} & =C_{g d} R_{g d} \& =50 imes 10^{-15} imes 5150 imes 10^{3} \& =257.5 \mathrm{~ns}\end{aligned}

Capacitor

C_{L}

sees a resistance

R_{C L}=R_{L}\left\|r_{O}=100 ight\| 100=50 \mathrm{k} \Omega

Thus,

\begin{aligned} au_{C L} & =C_{L} R_{C L} \& =0.1 imes 10^{-12} imes 50 imes 10^{3} \& =5 \mathrm{~ns}\end{aligned}

Finally,

\begin{aligned} au_{H} & = au_{g s}+ au_{g d}+ au_{C L} \& =10+257.5+5 \& =272.5 \mathrm{~ns}\end{aligned}

and

\begin{aligned}f_{H} & =\frac{1}{2 \pi au_{H}}=\frac{1}{2 \pi imes 272.5 imes 10^{-9}} \& =584 \mathrm{kHz}\end{aligned}

A Common-Source Amplifier Is Fed from a Signal Source Having Rsig=100kΩR_{\mathrm{sig}}=100 \mathrm{k} \OmegaRsig​=100kΩ

A Common-Source Amplifier Is Fed from a Signal Source Having $R_{\mathrm{sig}}=100 \mathrm{k} \Omega$