Question 2 of 7: PMOS Current-Mirror-Biased CS Amplifier

Question 2: PMOS Current-Mirror-Biased CS Amplifier (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. $V_{DD}=5\text{V}$; $k_n'=k_p'=1\text{ mA/V}^2$ (matched process, only one $k'$ given); $W/L=10$ for all three transistors; $|V_t|=1\text{V}$; $V_A=100\text{V}$; $\lambda=0$ in the square-law DC equation; target $I_{D,Q3}=0.5\text{mA}$.

Find. (a) $R$; (b) small-signal model; (c) $A_v=v_o/v_i$.

Approach. $Q_3$ is diode-connected, so its own square-law equation plus $R$'s IV relation fixes $R$. $Q_2$ mirrors $Q_3$'s current 1:1 (same $V_{SG}$, same $W/L$) into $Q_1$'s drain, biasing $Q_1$ at the same $0.5\text{mA}$; the small-signal gain then follows from $Q_1$'s $g_{m1}$ driving the parallel combination of both transistors' output resistances.

1. Part (a) — solve for $R$. $Q_3$ diode-connected (assume $\lambda=0$): $I_{D3}=\tfrac12 k_p'(W/L)(V_{SG3}-|V_t|)^2$, so $$V_{OV3}=\sqrt{\frac{2I_{D3}}{k_p'(W/L)}}=\sqrt{\frac{2(0.5\text{mA})}{(1\text{mA/V}^2)(10)}}=\sqrt{0.1}=0.3162\text{V}$$ $$V_{SG3}=V_{OV3}+|V_t|=0.3162+1=1.3162\text{V}$$ The node between $Q_3$'s drain and $R$ sits at $V_{DD}-V_{SG3}=5-1.3162=3.6838\text{V}$ above ground, and carries the same $I_{D3}$ into $R$: $$R=\frac{V_{DD}-V_{SG3}}{I_{D3}}=\frac{3.6838}{0.5\text{mA}}=\boxed{7.37\text{ k}\Omega}$$
2. Part (b) — small-signal model. $Q_2$'s gate sits at the fixed DC bias set by the $Q_3$–$R$ reference (no signal reaches it, since $v_i$ only excites $Q_1$'s gate), so for small signals $Q_2$ collapses to its output resistance $r_{o2}$ from the AC-grounded $5\text{V}$ rail to $v_o$. $Q_1$ becomes the usual hybrid-$\pi$/T model: a dependent source $g_{m1}v_{gs1}$ ($v_{gs1}=v_i$, source grounded) in parallel with $r_{o1}$, both landing on node $v_o$, so $r_{o1}\parallel r_{o2}$ is the total load the current source drives.
3. Part (c) — AC gain. Both $Q_1$ and $Q_2$ carry $I_D=0.5\text{mA}$ (the mirror sets $I_{D1}=I_{D2}=I_{D3}$), so $$g_{m1}=\sqrt{2k_n'(W/L)I_{D1}}=\sqrt{2(1\text{mA/V}^2)(10)(0.5\text{mA})}=\sqrt{10}\text{ mA/V}=3.162\text{ mA/V}$$ $$r_{o1}=r_{o2}=\frac{V_A}{I_D}=\frac{100}{0.5\text{mA}}=200\text{ k}\Omega$$ $$A_v=\frac{v_o}{v_i}=-g_{m1}\big(r_{o1}\parallel r_{o2}\big)=-(3.162\text{mA/V})(100\text{k}\Omega)=\boxed{-316\text{ V/V}}$$