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98-Comp-A1 · May 2014

Question 2 of 7: MOS Amplifier with Current-Mirror Active Load

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

98-Comp-A1, Electronics — National Exams, May 2014. Open-book, 3 hours; seven 20-mark questions, five to be marked (all seven answered below as a complete study resource). Unless stated otherwise, diode drops $V_D=0.7\text{V}$.

Reference texts: Sedra & Smith, Microelectronic Circuits (diode limiters/rectifiers, MOSFET/BJT small-signal amplifiers, active loads and current mirrors, active-RC filters, BJT switches, RTL logic, CMOS inverter design, RC oscillators, ring oscillators, ADC/DAC) — the single reference text covering every question on this paper.

Question 2: MOS Amplifier with Current-Mirror Active 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. $Q_1$ (NMOS, common-source, gate$=V_i$, drain$=V_o$) with active load $Q_2$ (PMOS, drain$=V_o$), mirrored 1:1 off $Q_3$ (PMOS, diode-connected) by reference current $I_{REF}$.

Given data
QuantityValue
$k_n=k_n'(W/L)$$1\text{mA/V}^2\times10=10\text{ mA/V}^2$
$k_p=k_p'(W/L)$$40\mu\text{A/V}^2\times10=0.4\text{ mA/V}^2$
$V_{tn}=|V_{tp}|$$1\text{V}$
$|V_A|$ (all devices)$100\text{V}$
$R_L$ (part d)$100\text{k}\Omega$
Target gain (part c)$200\text{ V/V}$

Find. Small-signal AC circuit, general gain expression, $I_{REF}$ for $|A_v|=200$, loaded gain with $R_L=100\text{k}\Omega$.

v_i gate ($R_i=\infty$) g_m1 v_gs1 v_gs1=v_i r_o1 r_o2 ($Q_2$, $v_{gs2}=0$) v_o
Fig. Q2-a — Small-signal AC equivalent. $Q_1$'s gate is an ideal open circuit; its drain drives $v_o$ through $g_{m1}v_{gs1}$ in parallel with $r_{o1}$. $Q_2$'s gate sits at AC ground (fed only through the diode-connected mirror off an ideal current source), so $v_{gs2}=0$ and $Q_2$ contributes only $r_{o2}$ at the output node.

Approach. Find the shared DC bias current from the 1:1 mirror, compute $g_m,r_o$ for $Q_1$, note $Q_2$ degenerates to $r_{o2}$ only (its gate carries no AC signal), then read gain/resistances off the single-node output circuit.

  1. Part (a)/(b) — bias current and gain expression. $Q_3$ is diode-connected and forced to $I_{D3}=I_{REF}$; the 1:1 mirror sets $I_{D2}=I_{REF}$, and KCL at $V_o$ (only DC path) forces $I_{D1}=I_{D2}=I_{REF}$ too. $$g_{m1}=\sqrt{2k_nI_{REF}}\qquad r_{o1}=r_{o2}=\frac{V_A}{I_{REF}}$$ $$A_v=\frac{v_o}{v_i}=-g_{m1}(r_{o1}\|r_{o2})=-g_{m1}\cdot\frac{r_{o1}}{2}=-\sqrt{2k_nI_{REF}}\cdot\frac{V_A}{2I_{REF}}$$
  2. Part (c) — solve for $I_{REF}$ at $|A_v|=200$. Substituting $k_n=10\text{mA/V}^2$, $V_A=100\text{V}$ and solving $|A_v(I_{REF})|=200$ numerically (bisection; monotonic in $I_{REF}$): $$I_{REF}=\boxed{1.25\text{ mA}}$$ Check: $g_{m1}=\sqrt{2(10)(1.25)}=5.00\text{ mA/V}$, $r_{o1}=r_{o2}=100/1.25=80\text{k}\Omega$, $A_v=-5.00\times(80/2)=\boxed{-200\text{ V/V}}$ ✓.
  3. Part (d) — loaded gain, $R_L=100\text{k}\Omega$. $R_L$ adds in parallel with $r_{o1}\|r_{o2}$: $$R_{out}'=r_{o1}\|r_{o2}\|R_L=\left(\frac{1}{80}+\frac{1}{80}+\frac{1}{100}\right)^{-1}\text{k}\Omega=28.57\text{k}\Omega$$ $$A_{v,L}=-g_{m1}R_{out}'=-(5.00\text{mA/V})(28.57\text{k}\Omega)=\boxed{-142.9\text{ V/V}}$$
Final Results — Question 2
QuantityValue
$I_{REF}$ for $|A_v|=200$$1.25\text{ mA}$
$g_{m1}$$5.00\text{ mA/V}$
$r_{o1}=r_{o2}$$80\text{ k}\Omega$ each
$A_v$ (open circuit)$-200\text{ V/V}$
$A_v$ with $R_L=100\text{k}\Omega$$-142.9\text{ V/V}$