Question 3 of 5: Bipolar cascode (Razavi Example 9.9) — mid-band voltage gain

Question 3: Bipolar cascode (Razavi Example 9.9) — mid-band 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.

Find. The small-signal mid-band voltage gain v_OUT/v_IN of the cascode, loaded only by the ideal current source I₁.

Approach. Get the three small-signal parameters from the 1 mA bias, build the cascode output resistance by driving r_o2 from the degenerating resistance seen at Q₂'s emitter, then multiply by the transconductance of the input device.

1. Extract the small-signal parameters at 1 mA. Both transistors are in the same branch and carry $I_C \approx I_1 = 1$ mA (base currents are $I_C/\beta = 10\ \mu\text{A}$, a 1 % effect that the question invites us to ignore). Therefore $$g_m=\frac{I_C}{V_T}=\frac{1\ \text{mA}}{26\ \text{mV}}=38.46\ \text{mA/V},\qquad r_\pi=\frac{\beta}{g_m}=2.60\ \text{k}\Omega,\qquad r_o=\frac{V_A}{I_C}=5.00\ \text{k}\Omega.$$
2. Identify the topology. Q₁ is a common-emitter stage with its emitter grounded and v_IN at its base; Q₂ sits above it with its base held at the fixed d.c. potential V_b1 (a signal ground), so Q₂ is a common-base current buffer. The load is the ideal source I₁, which is an open circuit to the signal, so the whole gain is set by the resistance looking down into Q₂'s collector.
3. Find the resistance degenerating Q₂. Looking down from Q₂'s emitter, r_o1 appears in parallel with r_π2 (the latter runs to the a.c.-grounded base of Q₂): $$R_E = r_{o1}\parallel r_{\pi 2} = \frac{(5.00)(2.60)}{5.00+2.60} = 1.711\ \text{k}\Omega.$$ Because $V_A$ is only 5 V, r_o1 is small and r_π2 shunts a substantial part of the signal current — this is the term that separates a bipolar cascode from a MOS one.
4. Build the cascode output resistance. A transistor degenerated by R_E presents $$R_{out}=r_{o2}\bigl(1+g_{m2}R_E\bigr)+R_E = 5.00\bigl(1+38.46\times 1.711\bigr)+1.711\ \text{k}\Omega,$$ and with $g_{m2}R_E = 65.79$ this evaluates to $$R_{out} = 5.00(66.79)+1.71 = \boxed{336\ \text{k}\Omega}.$$
5. Check the result against the cascode ceiling. Since $R_E \le r_{\pi 2}$ and $g_{m2}r_{\pi 2}=\beta$, the output resistance of any bipolar cascode can never exceed $\beta r_o = 100 \times 5\ \text{k}\Omega = 500\ \text{k}\Omega$. The 336 kΩ obtained sits sensibly below that ceiling. (Dropping r_π2 and writing $r_{o2}(1+g_{m2}r_{o1}) = 966\ \text{k}\Omega$ — the standard error — would violate it outright.)
6. Convert to voltage gain. The transconductance of the composite is that of the input device, $G_m \approx g_{m1}$, because Q₂ passes essentially all of Q₁'s collector current to the output node. Hence $$A_v=\frac{v_{OUT}}{v_{IN}}=-G_m R_{out} = -(38.46\ \text{mA/V})(335.7\ \text{k}\Omega) = \boxed{-1.29\times 10^{4}}$$ i.e. a gain magnitude of about 12 900, or 82.2 dB, inverting.
7. Cross-check with an exact nodal solution. Solving the two-node small-signal network exactly (retaining r_π2 in the forward path as well as in R_E) gives −1.27×10⁴, 1.5 % below the $G_m R_{out}$ product. The small shortfall is precisely the fraction of Q₁'s collector current that leaks into r_π2 instead of reaching the output, which the $G_m \approx g_{m1}$ approximation ignores.

Final Results