Bipolar Cascode Voltage Gain: 22-Elec-B5 Advanced Electronics, December 2014, Q4

Question 4: 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.

Quantity	Symbol	Value
Bias current (both devices)	$I_1=I_{C1}=I_{C2}$	1 mA
Current gain	$\beta$	100
Early voltage	$V_A$	5 V
Thermal voltage	$V_T$	26 mV (see callout)
Load	$I_1$	ideal current source (infinite ac resistance)

Find. The small-signal voltage gain $v_{OUT}/v_{IN}$ of the cascode, with $Q_1$ as the input transconductor and $Q_2$ as the cascode device.

Figure 4.1 — Bipolar cascode: Q₁ is the common-emitter input device, Q₂ the common-base cascode biased at V_b1, and the load is the ideal current source I₁.

Approach. Compute the small-signal parameters at 1 mA, find the resistance $Q_2$ sees looking down into $Q_1$, use the cascode output-resistance formula to get the total resistance at the output node, and multiply by the input transconductance — the load contributes nothing because it is an ideal current source.

Evaluate the small-signal parameters at the operating point. Both transistors carry the same 1 mA, so they share one set of parameters: $$g_m=\frac{I_C}{V_T}=\frac{1\ \text{mA}}{26\ \text{mV}}=38.46\ \text{mA/V},\qquad r_\pi=\frac{\beta}{g_m}=\frac{100}{38.46\ \text{mA/V}}=2.60\ \text{k}\Omega,$$ $$r_o=\frac{V_A}{I_C}=\frac{5\ \text{V}}{1\ \text{mA}}=5.00\ \text{k}\Omega.$$ The very low Early voltage is what makes the cascode worth building here: a plain common-emitter stage would be limited to $-g_mr_o=-192$.
Find what the cascode device sees beneath it. Looking down from the emitter of $Q_2$ there are two paths to ac ground: the base–emitter resistance of $Q_2$ itself is not part of this, but the output resistance of $Q_1$ and the input resistance $r_{\pi2}$ appear in parallel at that node, $$R_{E2}=r_{\pi2}\,\|\,r_{o1}=\frac{(2.60)(5.00)}{2.60+5.00}=1.711\ \text{k}\Omega.$$ Unlike the MOS case, the bipolar cascode's own $r_\pi$ shunts the node, and it is this term that ultimately caps the achievable gain.
Compute the cascode output resistance. Degenerating $Q_2$ with $R_{E2}$ raises its output resistance in the usual way, $$R_{out}=r_{o2}+\big(1+g_{m2}r_{o2}\big)\big(r_{\pi2}\|r_{o1}\big),$$ $$R_{out}=5.00\ \text{k}\Omega+\big[1+(38.46\ \text{mA/V})(5.00\ \text{k}\Omega)\big](1.711\ \text{k}\Omega) =5.00+330.6=335.7\ \text{k}\Omega.$$ Written in Razavi's grouping, $r_{o2}\big[1+g_{m2}(r_{\pi2}\|r_{o1})\big]+(r_{\pi2}\|r_{o1})$, the same 335.7 kΩ results.
Establish the total resistance at the output node. The load is an ideal current source, so it presents infinite ac resistance and the only path from the output node to ac ground is through the cascode itself: $$R_{tot}=R_{out}\,\|\,\infty=R_{out}=335.7\ \text{k}\Omega.$$
Multiply by the input transconductance. The whole signal current $g_{m1}v_{IN}$ generated by $Q_1$ is delivered to the output node by $Q_2$ (the common-base device passes essentially all of its emitter current to its collector), so $$\boxed{\frac{v_{OUT}}{v_{IN}}=-g_{m1}R_{out} =-(38.46\ \text{mA/V})(335.7\ \text{k}\Omega)=-1.29\times10^{4}}$$ i.e. a magnitude of about 12 900, or 82.2 dB.
Check the result against the bipolar cascode's ceiling. No bipolar cascode can exceed $R_{out}=\beta r_o$, because $r_{\pi2}\|r_{o1}\le r_{\pi2}$ and $g_{m2}r_{\pi2}=\beta$; here that ceiling is $(100)(5\ \text{k}\Omega)=500$ kΩ, and the 335.7 kΩ obtained sits sensibly below it. Equivalently the gain can never exceed $g_m\beta r_o=1.92\times10^{4}$, so 12 900 is a plausible two-thirds of the theoretical maximum — a useful order-of-magnitude check before committing to a four-figure answer.

Check: the paper does not restate the thermal voltage in Question 4, and the question names Razavi's Example 9.9 as its source, so $V_T=26$ mV (that text's convention) is used above. Using instead the $V_T=25$ mV stated in Question 1 of this same paper gives $g_m=40$ mA/V, $r_\pi=2.50$ kΩ, $R_{out}=340$ kΩ and $v_{OUT}/v_{IN}=-1.36\times10^{4}$ — a 5 percent difference with an identical method, and either is defensible provided the assumption is stated, as the paper's own instruction 1 requires.

Final results.

Quantity	Result
$g_m$, $r_\pi$, $r_o$ (each device at 1 mA)	38.46 mA/V, 2.60 kΩ, 5.00 kΩ
Resistance below the cascode, $r_{\pi2}\\|r_{o1}$	1.711 kΩ
Cascode output resistance $R_{out}$	335.7 kΩ
Voltage gain $v_{OUT}/v_{IN}$	$-1.29\times10^{4}$ (82.2 dB, inverting)
Comparison: plain CE stage	$-192$ (67 times smaller)
Alternative at $V_T=25$ mV	$R_{out}=340$ kΩ, $A_v=-1.36\times10^{4}$

Topic. The bipolar cascode: raising output resistance, and hence voltage gain, without adding stages.

Key relations. $g_m=I_C/V_T$, $r_\pi=\beta/g_m$, $r_o=V_A/I_C$; degenerated output resistance $R_{out}=r_o+(1+g_mr_o)R_E$; cascode with $R_E=r_{\pi2}\|r_{o1}$; $A_v=-g_{m1}R_{tot}$; ceiling $R_{out}\le\beta r_o$.

Why this works. The gain of any transconductance stage is $g_m$ times whatever resistance the output current is forced into, so with a fixed bias current the only way to raise gain is to raise that resistance. A cascode does it by hiding the input device behind a common-base transistor: when the output node moves, the cascode's emitter barely moves, so $Q_1$ hardly notices and injects almost no extra current. The improvement factor is the cascode's own intrinsic gain $g_{m2}r_{o2}$, applied to the resistance beneath it. In bipolar circuits that resistance is limited by $r_{\pi2}$, since the base current of the cascode device is itself a leakage path, which is exactly why the answer stops at $\beta r_o$ rather than growing as $g_m^2r_o^2$ as it would in MOS.

Common pitfall. Forgetting $r_{\pi2}$ and writing $R_{out}=r_{o2}(1+g_mr_{o1})$ — that gives about 965 kΩ and a gain near 37 000, roughly three times too large and above the $\beta r_o$ ceiling, which is the tell that something has been dropped. The second trap is loading the output with a finite resistance that was not given: the current source here is explicitly ideal, and if it were a real transistor its own $r_o$ would dominate and cut the gain by more than an order of magnitude.

Source. Razavi, Fundamentals of Microelectronics, 2nd ed., §9.1 and Example 9.9 (p. 405), as cited on the paper; Sedra & Smith, Microelectronic Circuits, 8th ed., §8.5 (the BJT cascode).

Question 4 of 5: Bipolar Cascode Voltage Gain

Question 4: Bipolar Cascode Voltage Gain (20 marks)