Question 5 of 5: Bipolar Cascode Amplifier — Voltage Gain

Question 5: Bipolar Cascode Amplifier — Voltage Gain (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. A two-transistor bipolar cascode: $v_{IN}$ drives the base of $Q_1$, whose emitter is grounded; the collector of $Q_1$ feeds the emitter of $Q_2$, whose base is held at the d.c. bias $V_{b1}$; the collector of $Q_2$ is the output node and is fed from $+V_{CC}$ by the ideal current source $I_1 = 1$ mA. Device data: $\beta = 100$, $V_A = 5$ V. The thermal voltage is taken as $V_T = 26$ mV, the convention of the text this question is drawn from.

Find. The small-signal voltage gain $v_{OUT}/v_{IN}$ of the cascode, driving the infinite resistance of the ideal current source.

Approach. The gain of any common-emitter-derived stage is $-g_{m1}R_{out}$; since the current-source load is ideal, $R_{out}$ is entirely the resistance looking into the collector of the cascode device, which is $r_{o2}$ boosted by the degeneration $Q_1$ provides.

1. Establish the small-signal parameters. Both transistors carry the same 1 mA, so they share one set of parameters: $$\begin{aligned}g_m&=\frac{I_C}{V_T}=\frac{1\times10^{-3}}{26\times10^{-3}} =38.46\ \text{mA/V}\\ r_\pi&=\frac{\beta}{g_m}=\frac{100}{38.46\times10^{-3}}=2.6\ \text{k}\Omega\\ r_o&=\frac{V_A}{I_C}=\frac{5}{1\times10^{-3}}=5\ \text{k}\Omega\end{aligned}$$ The Early voltage of 5 V is unusually low, which is what makes the cascode worth building here: a lone common-emitter stage would manage only $-g_mr_o=-192$.
2. Identify what degenerates the cascode device. Looking down from the emitter of $Q_2$ one sees two things in parallel: $Q_1$’s own output resistance $r_{o1}$, and $Q_2$’s base–emitter resistance $r_{\pi2}$, because $Q_2$’s base is a.c. grounded. Hence $$R_E=r_{\pi2}\parallel r_{o1}=\frac{(2.6)(5)}{2.6+5}=1.711\ \text{k}\Omega$$ This parallel pair is the whole reason a bipolar cascode behaves differently from a MOS one, where no $r_\pi$ exists to spoil the degeneration.
3. Compute the output resistance of the cascode. A transistor degenerated by $R_E$ presents at its collector $$R_{out}=r_{o2}\big[1+g_{m2}(r_{\pi2}\parallel r_{o1})\big]+(r_{\pi2}\parallel r_{o1})$$ $$=5000\big[1+(38.46\times10^{-3})(1711)\big]+1711=5000(66.79)+1711$$ which evaluates to $\boxed{R_{out} = 335.7\ \text{k}\Omega}$ — a 67-fold improvement on the 5 kΩ a single device offers.
4. Check $R_{out}$ against the hard ceiling. Because $r_{\pi2}\parallel r_{o1}\le r_{\pi2}$ and $g_{m2}r_{\pi2}=\beta$ exactly, no bipolar cascode can ever exceed $$R_{out}\le\beta\,r_o=(100)(5\ \text{k}\Omega)=500\ \text{k}\Omega$$ The computed 335.7 kΩ sits below it, as it must. This one-line test is the fastest way to catch the standard error of dropping $r_{\pi2}$ and writing $R_{out}=r_{o2}(1+g_{m2}r_{o1})=965$ kΩ, which is impossibly above the ceiling and should be rejected on sight.
5. Form the voltage gain. The ideal current source contributes infinite resistance, so the total load at the output node is $R_{out}$ alone, and $$\frac{v_{OUT}}{v_{IN}}=-g_{m1}R_{out}=-(38.46\times10^{-3})(335.7\times10^{3})$$ giving $\boxed{v_{OUT}/v_{IN} = -1.29\times10^{4}\ \ (82.2\ \text{dB, inverting})}$. The corresponding ceiling on the gain is $g_m\beta r_o=1.92\times10^{4}$, so the design realises about two thirds of the maximum a bipolar cascode can reach at this bias.
6. Record the alternative thermal-voltage convention. This question is Example 9.9 of Razavi’s Fundamentals of Microelectronics, which uses $V_T=26$ mV throughout; the boxed values follow that text. Taking $V_T=25$ mV instead would give $g_m=40$ mA/V, $R_{out}=340$ kΩ and $A_v=-1.36\times10^{4}$ — a 5.3 per cent spread on identical method. Both are defensible under the paper’s instruction to state assumptions; the cited text governs the boxed number.