Question 2 of 7: Current-Mirror-Biased CMOS Common-Source Amplifier
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
98-Comp-A1, Electronics — National Exams, December 2016. Open-book, 3 hours; seven 20-mark questions, five to be marked (all seven answered below as a complete study resource, per the standing "answer all M" rule). Unless stated otherwise, diode drops $V_D=0.7\text{V}$.
Reference texts: Sedra & Smith, Microelectronic Circuits (diode limiters, MOSFET current-mirror biasing and CS amplifiers, active-RC matched-feedback filters, BJT current-source-biased CE amplifiers, RC-ladder (Wien-bridge type) sinusoidal oscillators, CMOS static logic synthesis and sizing, flash-ADC comparator/encoder networks) — the single reference text covering every question on this paper.
Given. $I_{\text{ref}}=100\,\mu\text{A}$; $k_n'=100\,\mu\text{A/V}^2$, $k_p'=50\,\mu\text{A/V}^2$; $(W/L)=5$ for $Q_1,Q_2,Q_3$; $|V_t|=1\text{V}$; $V_A=50\text{V}$ (both devices); $\chi=0.2$ (body-effect factor, unused below — see the check note).
Given data
Quantity
Value
$I_{\text{ref}}$
$100\,\mu\text{A}$
$k_n'$
$100\,\mu\text{A/V}^2$
$k_p'$
$50\,\mu\text{A/V}^2$
$(W/L)$ all transistors
$5$
$V_A$
$50$ V
Find. Small-signal model; input resistance $R_i$; AC voltage gain $v_o/v_i$.
Approach. $Q_3$–$Q_2$ is a $1{:}1$ current mirror (equal $(W/L)$), so $Q_2$'s drain current equals $I_{\text{ref}}$, and since $Q_2$ stacks directly on $Q_1$ at the shared drain node $V_o$, $Q_1$ also carries $I_D=I_{\text{ref}}=100\,\mu\text{A}$. $Q_3$'s gate/drain node is fed by the DC current source $I_{\text{ref}}$ only (no AC component), so $Q_2$'s gate is AC ground and $Q_2$ behaves as a simple output resistance $r_{o2}$ — the classic current-mirror active-load common-source stage.
Part (b) — input resistance. $v_i$ drives $Q_1$'s gate directly; an ideal MOSFET gate draws no DC (or low-frequency AC) current, so
$$R_i=\boxed{\infty}$$
(this is the standard advantage of a MOSFET gate input over a BJT base, whose finite $r_\pi$ was needed in Question 4).
Part (c) — small-signal gain. $Q_1$'s transconductance, from $g_m=\sqrt{2k_n'(W/L)I_D}$:
$$g_{m1}=\sqrt{2(100\,\mu\text{A/V}^2)(5)(100\,\mu\text{A})}=\sqrt{1.0\times10^{-7}}=\boxed{316\,\mu\text{A/V}}$$
Both $Q_1$ and $Q_2$ carry the same $100\,\mu\text{A}$ and share $V_A=50\text{V}$, so
$$r_{o1}=r_{o2}=\frac{V_A}{I_D}=\frac{50}{100\,\mu\text{A}}=\boxed{500\text{ k}\Omega}$$
The output node sees $r_{o1}\parallel r_{o2}=250\text{ k}\Omega$, giving
$$A_v=-g_{m1}(r_{o1}\parallel r_{o2})=-(316\,\mu\text{A/V})(250\text{ k}\Omega)=\boxed{-79.1\text{ V/V}}$$
Fig. Q2(a) — small-signal model: $Q_1$ is the dependent source $g_{m1}v_{gs1}$ (driven directly by $v_i$ since the source is grounded), loaded by $r_{o1}\parallel r_{o2}$ ($Q_2$'s gate is AC ground via the DC-only mirror reference).
Final Results — Question 2
Quantity
Value
$I_{D1}=I_{D2}$
$100\,\mu\text{A}$
$g_{m1}$
$316\,\mu\text{A/V}$
$r_{o1}=r_{o2}$
$500$ k$\Omega$
$R_i$
$\infty$
$A_v$
$-79.1$ V/V
Check: the body-effect factor $\chi=0.2$ is given but not needed here — $Q_1$'s source is tied directly to (AC) ground, so $v_{sb1}=0$ and no body-transconductance term enters the gain; $\chi$ would matter only for a stage whose source itself carries a signal (e.g. a source follower or a degenerated stage). The paper's printed "$W/L{=}50$" is read as a stray repeat of the nearby "$k_p'{=}50$" figure and superseded by the paper's own explicit "$(W/L)=5$ for all transistors," which is used throughout.
