BJT Cascode Differential Pair with Wilson Current-Mirror Load: 98-Comp-A1, May 2016, Q4

Question 4: BJT Cascode Differential Pair with Wilson 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 $I=0.2\text{mA}$; $\beta=100$; $V_A=100\text{V}$ (every transistor); $V_T=25\text{mV}$; differential input between the $Q_1,Q_2$ bases; single-ended output $V_o$ at the joined $Q_4/Q_5$ collectors.

Find. $R_i$, $R_o$, $G_m$, $A_{vo}=G_mR_o$.

Approach. Identify the blocks from the figure. $Q_1$–$Q_4$ are a cascode differential pair: $Q_4$ sits on top of $Q_2$, so looking down from $V_o$ you see a cascode. $Q_5$, $Q_6$, $Q_7$ are a Wilson current mirror: the input current from $Q_3$ enters node $X$, which drives $Q_5$'s base, and $Q_5$'s emitter returns through $Q_7$. Looking up from $V_o$ you therefore see the Wilson mirror's output resistance. Then use the standard results: differential $R_i=2r_\pi$, cascode $R_o\approx\beta r_o$, Wilson $R_o\approx\beta r_o/2$, and $G_m=g_m$ for a mirror-loaded pair.

Check — drawn base tie. As printed, the $Q_6/Q_7$ base tie lands on node $X$ ($Q_6$'s own collector), the same node that drives $Q_5$'s base. Taken literally for DC, that leaves $V_{EC7}=V_{EB6}-V_{EB5}\approx0$, so $Q_7$ would be saturated. The standard Wilson connection takes the bases from $Q_7$'s collector (the $Q_5$ emitter) instead, and that is the reading used below. The small-signal answers barely depend on the choice: the exact nodal solution gives $R_o=33.1\text{M}\Omega$, $A_{vo}=1.311\times10^5$ for the Wilson tie and $R_o=33.6\text{M}\Omega$, $A_{vo}=1.303\times10^5$ for the tie exactly as drawn, both within $1.5\%$ of the hand results below.

Bias point (every transistor). Each half of the pair carries $I/2$, and the cascodes and the 1:1 mirror carry the same current: $$I_{C}=\frac{I}{2}=0.1\text{mA},\quad g_m=\frac{I_C}{V_T}=\frac{0.1\text{mA}}{25\text{mV}}=4\text{ mA/V},\quad r_\pi=\frac{\beta}{g_m}=25\text{k}\Omega,\quad r_o=\frac{V_A}{I_C}=1\text{M}\Omega$$
Part (a) — input resistance. A differential signal sees the two base–emitter resistances in series (the ideal tail source is an open circuit): $$R_i=2r_\pi=2(25\text{k}\Omega)=\boxed{50\text{ k}\Omega}$$ (The exact nodal solve gives $50.4\text{k}\Omega$; the small extra comes from $r_o$.)
Part (b) — output resistance. Looking down into $Q_4$, whose emitter is loaded by $r_{o2}$: $$R_{\text{down}}=r_{o4}+(1+g_mr_{o4})\big(r_{o2}\parallel r_{\pi4}\big)=1\text{M}+(4001)(24.39\text{k})=98.6\text{ M}\Omega\approx\beta r_o$$ Looking up into the Wilson mirror output ($Q_5$): $$R_{\text{up}}\approx\frac{\beta r_o}{2}=\frac{100\times1\text{M}\Omega}{2}=50\text{ M}\Omega$$ This is the standard Wilson-mirror result: the $Q_6$–$Q_7$ feedback loop around $Q_5$ gives only half the boost of a plain cascode. The two resistances are in parallel at $V_o$: $$R_o=R_{\text{down}}\parallel R_{\text{up}}=98.6\text{M}\parallel50\text{M}=33.2\text{M}\Omega\approx\frac{\beta r_o}{3}=\boxed{33.3\text{ M}\Omega}$$ (The exact nodal solve gives $33.1\text{M}\Omega$.)
Part (c) — transconductance. With $V_o$ shorted, $Q_4$ delivers $g_mv_{id}/2$ to the output. The Wilson mirror copies $Q_3$'s $-g_mv_{id}/2$ and inverts it, adding another $g_mv_{id}/2$. The two halves add: $$G_m=g_m=\boxed{4.00\text{ mA/V}}$$ (The exact nodal solve gives $3.96\text{mA/V}$; the difference is base-current loss in the cascodes and the mirror.)
Part (d) — open-circuit voltage gain. $$A_{vo}=G_m\,R_o=g_m\frac{\beta r_o}{3}=\frac{\beta V_A}{3V_T}=(4\text{mA/V})(33.3\text{M}\Omega)=\boxed{1.33\times10^{5}\text{ V/V}}\ (102.5\text{ dB})$$ (The exact nodal solve gives $1.31\times10^5$.)

Final Results — Question 4
Quantity	Value
$R_i$	$2r_\pi=50\text{ k}\Omega$
$R_o$	$\beta r_o\parallel\beta r_o/2\approx33.3\text{ M}\Omega$ (exact $33.1\text{ M}\Omega$)
$G_m$	$4.00\text{ mA/V}$
$A_{vo}$	$\approx1.33\times10^5$ V/V (exact $1.31\times10^5$)

Topic: BJT cascode differential amplifier with a Wilson current-mirror load.

Key relations: $g_m=I_C/V_T$, $r_\pi=\beta/g_m$, $r_o=V_A/I_C$; cascode $R_{out}=r_o+(1+g_mr_o)(R_E\parallel r_\pi)\approx\beta r_o$ when $R_E\gg r_\pi$; Wilson mirror $R_{out}\approx\beta r_o/2$; mirror-loaded differential pair $G_m=g_m$, so $A_{vo}=g_m(\beta r_o\parallel\beta r_o/2)=\beta V_A/(3V_T)$.

Why this works: Both sides of the output node are high-impedance. The cascode lifts the input side to about $\beta r_o$, and the Wilson mirror lifts the load side to about $\beta r_o/2$, while the mirror still returns the other half of the differential current. That is how one gain stage biased at $0.1\text{mA}$ reaches a gain above $10^5$.

Common pitfall: Misreading which device cascodes which. $Q_5$ sits between $Q_7$ and the output, not on top of $Q_4$. If you treat the load side as a bare $r_o$, $R_o$ collapses to about $1\text{M}\Omega$ and the gain comes out roughly $30\times$ too low. Also check $(1+g_mr_o)$ carefully: with $g_mr_o=4000$, not $400$, the cascode reaches about $\beta r_o$.

Source: Sedra & Smith, Microelectronic Circuits, Ch.8 (cascode amplifiers, Wilson current mirror, differential pairs with active loads).

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

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