Question 5 of 5: Bipolar Cascode — Voltage Gain

Question 5: Bipolar Cascode — 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 cascode: $Q_1$ is a grounded-emitter stage driven by $v_{IN}$, $Q_2$ is a common-base stage whose base is held at the d.c. level $V_{b1}$, and the pair is loaded by the ideal current source $I_1$:

Find. the small-signal voltage gain $v_{OUT}/v_{IN}$ of the cascode.

Approach. Compute the hybrid-$\pi$ parameters at 1 mA, recognise that the gain is the short-circuit transconductance multiplied by the total resistance at the output node, and evaluate that resistance from the cascode formula.

1. Find the small-signal parameters at the operating point. Both transistors carry the same 1 mA, so$$g_m = \frac{I_C}{V_T} = \frac{1\ \text{mA}}{26\ \text{mV}} = 38.5\ \text{mA/V}, \qquad r_{\pi} = \frac{\beta}{g_m} = 2.6\ \text{k}\Omega, \qquad r_o = \frac{V_A}{I_C} = 5\ \text{k}\Omega$$The Early voltage is unusually low, so $r_o$ is only 5 kΩ and the cascode's job of multiplying it is exactly what the question is testing.
2. Identify what loads the output node. The collector of $Q_2$ is tied to the ideal current source $I_1$, which by definition has infinite small-signal resistance. Nothing else is connected, so the total resistance at the output is the output resistance of the cascode itself, and the gain is $A_v = -G_m R_{out}$ with $G_m \approx g_{m1}$.
3. Find the resistance in the emitter of the upper device. Looking down from the emitter of $Q_2$ one sees the collector of $Q_1$, i.e. $r_{o1}$, in parallel with the base–emitter resistance of $Q_2$ itself:$$R_{E2} = r_{\pi 2} \parallel r_{o1} = \frac{(2.6)(5)}{2.6+5} = 1.711\ \text{k}\Omega$$Including $r_{\pi 2}$ is what separates the bipolar cascode from the MOS one, where the gate draws no current and the whole of $r_{o1}$ is available.
4. Evaluate the cascode output resistance. A common-base stage degenerated by $R_{E2}$ presents$$R_{out} = r_{o2}\left(1 + g_{m2}R_{E2}\right) + R_{E2} = 5\ \text{k}\Omega\,(1 + 65.8) + 1.71\ \text{k}\Omega = \boxed{336\ \text{k}\Omega}$$so the cascode has multiplied the 5 kΩ of a single transistor by a factor of 67.
5. Form the voltage gain. The input device converts $v_{IN}$ to a current $g_{m1}v_{IN}$, which is delivered essentially intact to the output node, so$$A_v = \frac{v_{OUT}}{v_{IN}} = -g_{m1}R_{out} = -(38.5\ \text{mA/V})(336\ \text{k}\Omega) = \boxed{-1.29 \times 10^{4}}$$that is, a magnitude of about 12 900 V/V or 82.2 dB, inverting.
6. Check the answer against the bipolar ceiling. The output resistance of a bipolar cascode can never exceed $\beta r_o$, because $R_{E2} \le r_{\pi 2}$ and $g_{m2}r_{\pi 2} = \beta$. Here that ceiling is $\beta r_o = 100 \times 5\ \text{k}\Omega = 500\ \text{k}\Omega$, and the 336 kΩ obtained sits comfortably below it. The corresponding gain ceiling is $g_m\beta r_o = 1.92\times 10^{4}$. An answer above either figure — for example the 965 kΩ that follows from dropping $r_{\pi 2}$ and writing $R_{out} = r_{o2}(1+g_{m2}r_{o1})$ — is wrong on inspection.
7. Refine the transconductance. Taking $G_m = g_{m1}$ is a slight over-estimate, because a little of the signal current is lost into $r_{o1}$ and $r_{\pi 2}$. Solving the two-node circuit exactly gives $A_v = -g_{m1}(r_{\pi 2}\parallel r_{o1})(1+g_{m2}r_{o2}) = -1.272\times 10^{4}$, i.e. $G_m = 0.985\,g_{m1}$ and a gain 1.5 per cent below the standard result. The textbook value $-1.29\times 10^{4}$ is the expected answer; the exact figure is quoted here as corroboration rather than as a correction.