Differential Pair with Current-Mirror Load: 98-Comp-A1, December 2015, Q4

Question 4: Differential Pair with Current-Mirror 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. Tail current $I=0.2\text{mA}$ ($I_{C1}=I_{C2}=I/2=0.1\text{mA}$); $\beta=100$; $V_A=100\text{V}$ (all four transistors); single-ended input at $Q_1$'s base, $Q_2$'s base grounded, ideal (infinite-resistance) tail source.

Given data
Quantity	Value
$I$ (tail)	$0.2\text{ mA}$
$\beta$	$100$
$V_A$	$100\text{ V}$
$V_T$ (assumed)	$25\text{ mV}$

Find. $R_i$, $R_o$, $G_m$, $A_{vo}=v_o/v_i$ (open circuit).

Approach. Because the tail current source is ideal, a single-ended input decomposes cleanly into a purely differential-mode signal at the pair (the common-mode component draws zero current, so it contributes nothing) — the input resistance is therefore the same $2r_\pi$ as for a fully differential drive. The current-mirror load converts the pair's differential output current into a SINGLE-ENDED current with full efficiency, giving $G_m=g_m$ (not $g_m/2$) and an output resistance $r_{o2}\|r_{o4}$.

Part (a) — input resistance. Each transistor carries $I_C=I/2=0.1\text{mA}$, so $$g_m=\frac{I_C}{V_T}=\frac{0.1\text{mA}}{25\text{mV}}=4\text{ mA/V},\qquad r_\pi=\frac{\beta}{g_m}=\frac{100}{4\text{mA/V}}=25\text{ k}\Omega$$ With an ideal (infinite) tail source, common-mode current is exactly zero, so all of the single-ended input current is differential-mode: $$R_i=2r_\pi=2(25\text{k}\Omega)=\boxed{50\text{ k}\Omega}$$
Part (b) — output resistance. The output node sees $Q_2$'s collector (NPN, in parallel) and $Q_4$'s collector (PNP mirror output), both driving into the high-impedance output: $$r_{o2}=r_{o4}=\frac{V_A}{I_C}=\frac{100\text{V}}{0.1\text{mA}}=1\text{ M}\Omega$$ $$R_o=r_{o2}\|r_{o4}=\boxed{500\text{ k}\Omega}$$
Part (c) — transconductance. The current mirror ($Q_3$ diode-connected, sensing $Q_1$'s signal current $g_mv_{id}/2$, and mirroring it 1:1 into $Q_4$'s branch) adds constructively to $Q_2$'s own signal current $g_mv_{id}/2$ at the output node, so the full $g_m$ appears at the output rather than half of it: $$G_m=\boxed{g_m=4\text{ mA/V}}$$

Fig. Q4 — output-referred small-signal model. The current-mirror load turns the pair's differential output current into a single dependent source $G_m v_{id}=g_mv_{id}$ driving $r_{o2}\|r_{o4}$ to ground.

Part (d) — open-circuit voltage gain. $$A_{vo}=\frac{v_o}{v_i}=-G_mR_o=-(4\text{mA/V})(500\text{k}\Omega)=\boxed{-2000\text{ V/V}}$$ (magnitude quoted; the sign follows the convention that $Q_1$'s base is the inverting input of this single-ended-output stage.)

Final Results — Question 4
Quantity	Value
$g_m$ (each transistor)	$4\text{ mA/V}$
$r_\pi$	$25\text{ k}\Omega$
$R_i$	$50\text{ k}\Omega$
$r_{o2}=r_{o4}$	$1\text{ M}\Omega$
$R_o$	$500\text{ k}\Omega$
$G_m$	$4\text{ mA/V}$
$A_{vo}$	$-2000\text{ V/V}$

Topic: BJT differential pair with a current-mirror active load.

Key relations: $g_m=I_C/V_T$, $r_\pi=\beta/g_m$, $r_o=V_A/I_C$; differential input resistance $R_{id}=2r_\pi$; current-mirror load gives $G_m=g_m$ and $R_o=r_{o,\text{NPN}}\|r_{o,\text{PNP}}$, so $A_{vo}=-g_m(r_{o2}\|r_{o4})$.

Why this works: With an ideal tail current source, the common-mode half-circuit sees infinite emitter degeneration and passes zero signal current, so a single-ended drive is entirely differential-mode at the transistor level — this is what lets $R_i=2r_\pi$ regardless of which input carries the signal. The mirror recovers what a plain resistive load would throw away (in a resistive-load diff-amp only $Q_2$'s own half of the differential current reaches a single-ended output, i.e. $G_m=g_m/2$); mirroring $Q_1$'s current into the same node doubles the effective transconductance.

Common pitfall: Using $G_m=g_m/2$ (correct for a resistive-load diff-amp with single-ended output) instead of $g_m$ — the whole point of the current-mirror load is to avoid that 2x loss.

Source: Sedra & Smith, Microelectronic Circuits, The Differential Pair; Current-Mirror-Loaded Differential Amplifier.

Question 4 of 7: Differential Pair with Current-Mirror Load

Question 4: Differential Pair with Current-Mirror Load (20 marks)