NMOS Common-Source Amplifier with Diode-Connected Load: 98-Comp-A1, December 2015, Q2

Question 2: NMOS Common-Source Amplifier with Diode-Connected Load (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.

Given. $k_n=k_n'(W/L)=10\text{ mA/V}^2$ for $Q_1,Q_2$ (identical devices); $V_{tn}=1\text{V}$, $|V_A|=100\text{V}$, $V_{DD}=5\text{V}$, $V_i=2\text{V}$.

Given data
Quantity	Value
$k_n'$	$1\text{ mA/V}^2$
$W/L$ ($Q_1,Q_2$)	$10$
$V_{tn}$	$1\text{V}$
$\|V_A\|$	$100\text{V}$
$V_{DD}$	$5\text{V}$

Find. $I_{D1}$ and $V_o$ at $V_i=2\text{V}$; small-signal equivalent circuit; small-signal AC gain $v_o/v_i$.

Approach. $Q_1$ (gate driven directly by $V_i$, source grounded) sets the stack current from its own saturation equation; $Q_2$ (diode-connected, so always saturated) must carry that same current since the two devices are in series with nothing else at the shared node, which pins $V_o$. Linearise both around this point for the small-signal gain, recognising $Q_2$'s diode connection turns it into a resistive load of value $1/g_{m2}\|r_{o2}$.

Part (a) — $I_{D1}$. $Q_1$: $V_{GS1}=V_i=2\text{V}$, so $V_{ov1}=V_{GS1}-V_{tn}=1\text{V}$ (saturation, confirmed once $V_o$ is found below). $$I_{D1}=\tfrac12 k_n V_{ov1}^2=\tfrac12(10\text{mA/V}^2)(1)^2=\boxed{5\text{ mA}}$$
Part (b) — $V_o$. $Q_2$ is diode-connected ($V_{GS2}=V_{DD}-V_o$, $V_{DS2}=V_{GS2}$, always saturated) and carries the same $5\text{mA}$ (single series current path): $$5\text{mA}=\tfrac12(10)(V_{DD}-V_o-1)^2\ \Rightarrow\ (V_{DD}-V_o-1)^2=1\ \Rightarrow\ V_{DD}-V_o-1=+1$$ $$V_o=V_{DD}-2=\boxed{3\text{ V}}$$ Check: $V_{DS1}=V_o=3\text{V}\ge V_{ov1}=1\text{V}$ — $Q_1$ saturated as assumed.

Fig. Q2(c) — small-signal model. $Q_1$'s dependent source $g_{m1}v_i$ drives the parallel combination of both output resistances and $Q_2$'s diode-connected resistance $1/g_{m2}$ (since $Q_2$'s own drain-gate tie forces $v_{gs2}=-v_o$, making it look purely resistive).

Part (d) — small-signal gain. $g_{m1}=g_{m2}=k_nV_{ov}=(10\text{mA/V}^2)(1\text{V})=10\text{ mA/V}$ (identical devices, identical $5\text{mA}$, identical $1\text{V}$ overdrive — $Q_2$'s overdrive is $V_{DD}-V_o-V_{tn}=1\text{V}$, matching $Q_1$'s). $r_{o1}=r_{o2}=|V_A|/I_D=100/5\text{mA}=20\text{k}\Omega$. The load seen at $v_o$ is $r_{o1}\|r_{o2}\|(1/g_{m2})$: $$R_L=\left(\frac{1}{20\text{k}}+\frac{1}{20\text{k}}+g_{m2}\right)^{-1}=99.0\,\Omega$$ $$A_v=\frac{v_o}{v_i}=-g_{m1}R_L=-(10\text{mA/V})(99.0\,\Omega)=\boxed{-0.990\text{ V/V}}$$ Close to the simple $-g_{m1}/g_{m2}=-1$ estimate (identical devices), reduced slightly by the two $r_o$'s loading the diode-connected resistance.

Final Results — Question 2
Quantity	Value
$I_{D1}$	$5\text{ mA}$
$V_o$	$3\text{ V}$
$g_{m1}=g_{m2}$	$10\text{ mA/V}$
$r_{o1}=r_{o2}$	$20\text{ k}\Omega$
$A_v=v_o/v_i$	$-0.990\text{ V/V}$

Topic: NMOS common-source stage with a diode-connected NMOS active load.

Key relations: Saturation $I_D=\tfrac12 k_n V_{ov}^2$; diode connection forces $V_{DS}=V_{GS}$ (always saturated) and gives a small-signal resistance $1/g_m\|r_o\approx1/g_m$; $g_m=k_nV_{ov}=\sqrt{2k_nI_D}$; $r_o=|V_A|/I_D$.

Why this works: With no other branch at the shared node, $Q_1$ and $Q_2$ must carry the identical current — this single constraint (not two independent bias equations) is what pins $V_o$. Because $Q_2$'s gate is tied to its own drain, any change in the shared node's small-signal voltage appears directly as $-v_{gs2}$, so it presents itself as a two-terminal resistor rather than a true dependent source to the rest of the circuit.

Common pitfall: Forgetting to include $r_{o1}$ in parallel with the diode-connected load when it is comparable in size to $1/g_{m2}$ — here $r_{o1}=20\text{k}\Omega\gg1/g_{m2}=100\,\Omega$, so it barely matters (0.99 vs 1.00), but for lower-$V_A$ processes it would.

Source: Sedra & Smith, Microelectronic Circuits, MOSFET Biasing and Small-Signal Operation; Diode-Connected Loads.

Question 2 of 7: NMOS Common-Source Amplifier with Diode-Connected Load

Question 2: NMOS Common-Source Amplifier with Diode-Connected Load (20 marks)