Question 4 of 7

Question 4 (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Approach. (b) at midband every capacitor is a short, so $R_{out}$ is just what the $R_L$ branch sees looking back into the drain: $R_D\Vert r_o$. (c) the DC gate voltage from the $R_1$/$R_2$ divider must equal $V_{GS}+I_DR_S$, while $R_1\Vert R_2=R_{in}$. (d) with all capacitors shorted, $A_{vm}$ is the input-divider attenuation times $-g_m$ times the total drain-side parallel resistance. (e) find the dominant low-frequency pole among $\omega_{p1},\omega_{p2},\omega_{p3}$ (using the now-given $C_i$, $C_o$) and combine with the given $\omega_H$ to get the amplifier's overall passband.

1. Part (a) — role of $C_S$. $C_S$ is a bypass capacitor across the source resistor $R_S$. At DC, $R_S$ is left in the circuit to set the bias point (it sets $V_S=I_DR_S$, part of the gate-voltage design in part (c)). At signal frequencies, $C_S$ AC-shorts $R_S$ to ground, removing the source-degeneration gain loss that $R_S$ would otherwise cause — so the circuit gets DC bias stability from $R_S$ without sacrificing AC gain.
2. Part (b) — $R_{out}$. Looking back into the drain node from $R_L$ (figure above) with $v_{sig}=0$, the dependent source $g_mv_{gs}$ contributes zero current (its controlling $v_{gs}=0$), leaving only $R_D$ and $r_o$ in parallel: $$R_{out}=R_D\Vert r_o=\frac{(5\,\text{k})(500\,\text{k})}{5\,\text{k}+500\,\text{k}}=\boxed{4.95\ \text{k}\Omega}$$
3. Part (c) — $R_1$, $R_2$. The DC source voltage is $V_S=I_DR_S=(1\,\text{mA})(5\,\text{k}\Omega)=5\,\text{V}$, so the required gate voltage is $V_G=V_{GS}+V_S=1+5=6\,\text{V}$. With $V_G=V_{DD}\dfrac{R_2}{R_1+R_2}$ and $R_1\Vert R_2=420\,\text{k}\Omega$: $$\frac{R_2}{R_1+R_2}=\frac{6}{20}=0.30\qquad R_1\Vert R_2=0.30(0.70)(R_1+R_2)=420\,\text{k}\Omega\ \Rightarrow\ R_1+R_2=2\,\text{M}\Omega$$ $$R_1=0.70(2\,\text{M}\Omega)=\boxed{1.4\ \text{M}\Omega}\qquad R_2=0.30(2\,\text{M}\Omega)=\boxed{600\ \text{k}\Omega}$$ (Check: $R_1\Vert R_2=420\,\text{k}\Omega$; $V_G=20(600\text{k}/2\text{M})=6\,\text{V}$. ✓)
4. Part (d) — midband gain $A_{vm}=v_o/v_{sig}$. At midband all capacitors are shorts, so $v_{gs}$ equals the gate voltage and the drain sees $R_D\Vert R_L\Vert r_o$: $$R_D\Vert R_L\Vert r_o=\frac{1}{\frac{1}{5\text{k}}+\frac{1}{5\text{k}}+\frac{1}{500\text{k}}}=2487.6\ \Omega$$ The signal first divides down at the gate through $R_{sig}=1\,\text{k}\Omega$ and $R_{in}=420\,\text{k}\Omega$ (gate draws no current): $$A_{vm}=\frac{R_{in}}{R_{in}+R_{sig}}\times\big(-g_m(R_D\Vert R_L\Vert r_o)\big)=\frac{420}{421}\times\big(-4\times10^{-3}\times2487.6\big)=(0.99762)(-9.950)=\boxed{-9.93\ \text{V/V}}$$
5. Part (e) — bandwidth. With $R_1,R_2$ from part (c) and the given capacitor values, the three low-frequency break frequencies are $$\omega_{p1}=\frac{1}{(R_{sig}+R_{in})C_i}=\frac{1}{(421\,\text{k})(0.1\,\mu\text{F})}=23.8\ \text{rad/s}\qquad\omega_{p2}=\frac{1}{\left(R_S\Vert\frac{1}{g_m}\right)C_S}=\frac{1}{(238.1\,\Omega)(100\,\mu\text{F})}=42.0\ \text{rad/s}$$ $$\omega_{p3}=\frac{1}{(R_{out}+R_L)C_o}=\frac{1}{(9.95\,\text{k}\Omega)(0.01\,\mu\text{F})}=10{,}050\ \text{rad/s}$$ $\omega_{p3}$ is roughly 240–420× larger than $\omega_{p1}$ and $\omega_{p2}$, so it alone dominates and sets the amplifier's lower cutoff, $\omega_L\approx\omega_{p3}=10{,}050\,\text{rad/s}$. With the given upper cutoff $\omega_H=90{,}000\,\text{rad/s}$, the passband (bandwidth) is $$BW=\omega_H-\omega_L=90{,}000-10{,}050=\boxed{79{,}950\ \text{rad/s}\approx80\ \text{krad/s}}$$ or in Hz: $f_H=14.32\,\text{kHz}$, $f_L=1.60\,\text{kHz}$, $BW\approx\boxed{12.7\ \text{kHz}}$.