Question 1 of 5: Bipolar Cascode Current Source — Output Resistance and Output Compliance

Question 1: Bipolar Cascode Current Source — Output Resistance and Output Compliance (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. (a) the small-signal resistance $R_O = v_x/i_x$ looking back into the output terminal, and (b) the lowest DC level that may be applied to $v_x$ before the current source stops behaving as one.

Approach. Fix the operating current from the mirror, convert it into the hybrid-$\pi$ parameters of $Q_2$ and $Q_3$, run a driving-point analysis at the collector of $Q_3$ (a common-base stage degenerated by the output resistance of $Q_2$), then read the compliance limit off the two saturation constraints that the fixed base bias imposes.

1. Part (a) — fix the bias current from the mirror. $Q_1$ is diode-connected and swallows the reference, so with matched devices the mirror copies it into $Q_2$. $Q_3$ is in series with $Q_2$, so the same current flows through the whole stack: $$I_{C1} = I_{C2} = I_{bias} = 10\ \text{mA}, \qquad I_{C3} \simeq I_{E3} = I_{C2} = 10\ \text{mA}.$$ Every small-signal parameter below is therefore evaluated at 10 mA.
2. Convert the operating point into hybrid-$\pi$ parameters. The transconductance, base input resistance and Early resistance follow directly from the operating current: $$g_{m3} = \frac{I_{C3}}{V_T} = \frac{10\ \text{mA}}{25\ \text{mV}} = 400\ \text{mA/V}, \qquad r_{\pi 3} = \frac{\beta}{g_{m3}} = \frac{100}{0.4} = 250\ \Omega,$$ $$r_{o2} = r_{o3} = \frac{V_A}{I_C} = \frac{100\ \text{V}}{10\ \text{mA}} = 10\ \text{k}\Omega.$$ The 250 Ω value of $r_{\pi 3}$ is the number that decides this question, so it is worth noting how small it is beside $r_{o2}$.
3. Set up the driving-point analysis at the output node. Drive the collector of $Q_3$ with $v_x$ and look for $i_x$. Seen from the emitter of $Q_3$, the mirror presents $r_{o2}$ to ground; the base of $Q_3$ is an AC ground, so $r_{\pi 3}$ appears in parallel with it. Writing KCL at the emitter node and eliminating $v_{\pi 3}$ gives the classic degenerated common-emitter result $$\boxed{\;R_O = r_{o3}\bigl[\,1 + g_{m3}\,(r_{\pi 3}\,\|\,r_{o2})\,\bigr] + (r_{\pi 3}\,\|\,r_{o2})\;}$$ in which the bracketed factor is the boost the cascode buys over a bare mirror.
4. Evaluate the degeneration resistance seen at the emitter of $Q_3$. Because $r_{\pi 3}$ is two orders of magnitude smaller than $r_{o2}$ it dominates the parallel pair, and that is exactly what limits the achievable boost: $$r_{\pi 3}\,\|\,r_{o2} = \frac{250 \times 10{,}000}{250 + 10{,}000} = 243.9\ \Omega .$$
5. Put the numbers into the boxed expression. The loop gain around the cascode device is $g_{m3}(r_{\pi 3}\|r_{o2}) = 0.4 \times 243.9 = 97.56$, so $$R_O = 10\ \text{k}\Omega \times (1 + 97.56) + 243.9\ \Omega = 985{,}610 + 244 \approx \boxed{\,986\ \text{k}\Omega\,}$$ i.e. very nearly 1 MΩ, against the 10 kΩ a plain mirror would offer — a hundred-fold improvement.
6. Check the answer against the ceiling that a bipolar cascode cannot cross. Since $r_{\pi 3}\|r_{o2} \le r_{\pi 3}$ and $g_{m3}r_{\pi 3} = \beta$, the boxed expression can never exceed $\beta r_{o3} = 100 \times 10\ \text{k}\Omega = 1.00\ \text{M}\Omega$. The computed 986 kΩ sits just under that ceiling, which is the signature of a correct answer. Dropping $r_{\pi 3}$ — the standard slip — would give $r_{o3}(1 + g_{m3}r_{o2}) = 40.0\ \text{M}\Omega$, forty times above a limit that no bipolar cascode can pass, and that impossibility is how the error is caught.
7. Part (b) — pin the internal node from the fixed base bias. The base of $Q_3$ is held at $V_{bias}$, so its emitter sits one diode drop below: $$V_E = V_{bias} - V_{BE(on)} = 3.0 - 0.7 = 2.3\ \text{V}.$$ That is also $V_{CE2}$, and since $2.3\ \text{V} \gg V_{CE(sat)} = 0.3\ \text{V}$ the mirror transistor $Q_2$ is comfortably active and never becomes the binding constraint.
8. Impose the saturation limit on the cascode device. $Q_3$ remains in the forward-active region only while its own collector–emitter voltage stays above $V_{CE(sat)}$, so the output terminal may not be pulled below $$v_{x,\min} = V_E + V_{CE(sat)} = V_{bias} - V_{BE(on)} + V_{CE(sat)} = 2.3 + 0.3 = \boxed{\,2.60\ \text{V}\,}$$ Below this the collector junction forward-biases, $R_O$ collapses and the branch current falls away from 10 mA. The usable compliance window is therefore $2.60\ \text{V} \le v_x \le V_{CC}$, i.e. 7.4 V of swing — the price paid for the hundred-fold gain in $R_O$, and the reason a designer keeps $V_{bias}$ only as high as the cascode actually needs.