22-Elec-B5 Advanced Electronics · May 2015

Question 1 of 5: Bipolar Cascode with an Ideal Current-Source Load

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

Paper format. National Exams, May 2015 — 07-Elec-B5 Advanced Electronics. Three hours, CLOSED BOOK, any non-communicating calculator permitted. Five questions, 20 marks each; all five must be answered. The paper directs candidates to state any assumptions made, to treat ground and chassis as common, and to assume ideal op-amps on ±15 V rails unless a question says otherwise.

Reference texts. B. Razavi, Fundamentals of Microelectronics, 2nd ed. (Ch. 9, Cascode Stages and Current Mirrors — Question 1 cites Example 9.9 by name); A. S. Sedra & K. C. Smith, Microelectronic Circuits, 8th ed. (Ch. 8 Building Blocks of IC Amplifiers, Ch. 9 Differential and Multistage Amplifiers, Ch. 10 Frequency Response, Ch. 12 Output Stages and Power Amplifiers); P. R. Gray, P. J. Hurst, S. H. Lewis & R. G. Meyer, Analysis and Design of Analog Integrated Circuits, 5th ed. (Ch. 3–4 for the current-source-loaded pair).

How the figures were recovered. component placement, which terminal the input drives, and whether a load pair is a mirror or two independent sources. Where the printed data and the drawn topology cannot both be satisfied, the reading adopted is stated in a check note rather than silently chosen.

Question 1: Bipolar Cascode with an Ideal Current-Source 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. A two-transistor bipolar cascode: $Q_1$ is a common-emitter device driven at its base by $v_{IN}$ with its emitter grounded, and $Q_2$ is a common-base cascode device whose base sits at the DC potential $V_{b1}$. The load is the ideal current source $I_1$, which also sets the bias current of both devices.

Given data (Question 1)
Quantity	Symbol	Value
Bias current (both devices)	$I_1 = I_{C1} = I_{C2}$	$1\ \text{mA}$
Current gain	$\beta$	$100$
Early voltage	$V_A$	$5\ \text{V}$
Thermal voltage (Razavi convention)	$V_T$	$26\ \text{mV}$

Find. The small-signal voltage gain $A_v = v_{OUT}/v_{IN}$ of the stage.

Question 1 — bipolar cascode. The ideal current source I₁ is the load, so the gain is set entirely by the cascode's own output resistance.

Approach. Because the load is an ideal current source it contributes no shunt conductance, so the gain is simply the transconductance of the input device multiplied by the output resistance looking into the cascode collector; find the hybrid-π parameters first, then build the cascode output resistance, then multiply.

Small-signal parameters at the 1 mA operating point. Both devices carry the same collector current, so they share one set of parameters:$$\begin{aligned}g_m&=\frac{I_C}{V_T}=\frac{1\ \text{mA}}{26\ \text{mV}}=38.46\ \text{mA/V}\\r_\pi&=\frac{\beta}{g_m}=\frac{100}{38.46\ \text{mA/V}}=2.60\ \text{k}\Omega\\r_o&=\frac{V_A}{I_C}=\frac{5\ \text{V}}{1\ \text{mA}}=5.00\ \text{k}\Omega\end{aligned}$$The Early voltage here is unusually small, so $r_o$ is only twice $r_\pi$ — a detail that matters in the next step, because it means the parallel combination cannot be approximated by either one alone.
Resistance the cascode device sees at its emitter. Looking down from the emitter of $Q_2$ one sees the base resistance of $Q_2$ in parallel with the output resistance of $Q_1$:$$r_{\pi2}\parallel r_{o1}=\frac{2.60\times5.00}{2.60+5.00}\ \text{k}\Omega=1.711\ \text{k}\Omega$$The base branch is included because the base of $Q_2$ is an AC ground (it is held at the fixed potential $V_{b1}$), so $r_{\pi2}$ appears directly across the emitter node.
Output resistance of the cascode. Degenerating $Q_2$ with that resistance multiplies its own $r_{o2}$ by the usual cascoding factor:$$R_{out}=r_{o2}\bigl[1+g_{m2}\,(r_{\pi2}\parallel r_{o1})\bigr]+(r_{\pi2}\parallel r_{o1})$$Substituting the numbers, $g_{m2}(r_{\pi2}\parallel r_{o1}) =38.46\ \text{mA/V}\times1.711\ \text{k}\Omega=65.79$, so$$R_{out}=5.00\ \text{k}\Omega\times66.79+1.711\ \text{k}\Omega=\boxed{335.7\ \text{k}\Omega}$$
Voltage gain. An ideal current source has infinite output resistance, so nothing is placed in parallel with $R_{out}$ and the total load at the output node is $R_{out}$ itself. The transconductance driving that node is $g_{m1}$:$$A_v=\frac{v_{OUT}}{v_{IN}}=-g_{m1}R_{out}=-38.46\ \text{mA/V}\times335.7\ \text{k}\Omega$$$$\boxed{A_v=-1.291\times10^{4}\ \text{V/V}\;\;(82.2\ \text{dB})}$$The sign is negative because $Q_1$ inverts and the common-base device $Q_2$ passes the signal current through without a further inversion.
Sanity ceiling on the answer. A bipolar cascode can never beat $\beta r_o$, because the emitter-degeneration term is capped by $r_{\pi2}$ and $g_{m2}r_{\pi2}=\beta$:$$R_{out}\le\beta r_o=100\times5.00\ \text{k}\Omega=500\ \text{k}\Omega,\qquad |A_v|\le g_m\beta r_o=1.92\times10^{4}$$The computed $335.7\ \text{k}\Omega$ and $1.291\times10^{4}$ both sit below their ceilings, which confirms that $r_{\pi2}$ was not dropped by mistake. Omitting it would give $r_{o2}(1+g_mr_{o1})=965\ \text{k}\Omega$ — impossibly above the ceiling, and the single most common error on this question.

Final results — Question 1
Quantity	Value
Transconductance $g_m$	$38.46\ \text{mA/V}$
Base resistance $r_\pi$	$2.60\ \text{k}\Omega$
Output resistance $r_o$	$5.00\ \text{k}\Omega$
Cascode output resistance $R_{out}$	$335.7\ \text{k}\Omega$
Voltage gain $v_{OUT}/v_{IN}$	$-1.291\times10^{4}\ \text{V/V}$ (82.2 dB)

Check: which thermal voltage. The question cites Razavi Example 9.9 by name but does not restate $V_T$; Razavi uses $26\ \text{mV}$ throughout, and that convention is adopted for the boxed answer. This paper does state $V_T = 25\ \text{mV}$ — but inside Question 2, for a different circuit. Carrying $25\ \text{mV}$ instead gives $g_m = 40.0\ \text{mA/V}$, $R_{out} = 340.0\ \text{k}\Omega$ and $A_v = -1.360\times10^{4}$, a 5 % shift with an identical method. Either is defensible under the paper’s own instruction to state assumptions; the cited text governs the boxed value.

Topic. Cascoding — raising output resistance without raising transconductance.

Key relations. $R_{out}=r_{o2}[1+g_{m2}(r_{\pi2}\parallel r_{o1})]+(r_{\pi2}\parallel r_{o1})$; ceiling $R_{out}\le\beta r_o$; $A_v=-g_{m1}R_{out}$ for an ideal current-source load; $g_m=I_C/V_T$, $r_\pi=\beta/g_m$, $r_o=V_A/I_C$.

Why this works. A common-emitter stage loaded by an ideal current source would give $-g_mr_o$, here only $-192$, because its own $r_o$ shunts the output node. Stacking a common-base device on top hides that node: a change in output voltage now has to fight its way through $Q_2$, and the fraction of it that reaches $Q_1$’s collector is reduced by roughly $g_{m2}$ times the emitter degeneration. The signal current, meanwhile, passes through the common-base device essentially unattenuated, so the transconductance is unchanged and the gain rises by the same factor the output resistance does. In a bipolar cascode that factor saturates: the base current of the cascode device provides a leakage path through $r_{\pi2}$, capping the whole structure at $\beta r_o$. A MOS cascode has no such path and is limited only by the second-order effects.

Common pitfall. Dropping $r_{\pi2}$ and writing $R_{out}=r_{o2}(1+g_{m2}r_{o1})$. That gives $965\ \text{k}\Omega$ here, which exceeds the $\beta r_o$ ceiling of $500\ \text{k}\Omega$ and is therefore visibly wrong; the ceiling check costs one line and catches the error every time.

Source. Razavi, Fundamentals of Microelectronics, 2nd ed., §9.1 and Example 9.9 (p. 405); Sedra & Smith, Microelectronic Circuits, 8th ed., §8.5 (The Cascode Amplifier).

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