BJT Current-Mirror Bias Network: 17-Phys-A5, May 2018, Q1

Question 1: BJT Current-Mirror Bias Network (15 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. Three matched NPN transistors. $Q_3$: base grounded, collector fed from $+8\text{V}$ through $R_1$, emitter feeds directly into $Q_2$'s collector. $Q_1$ is diode-connected (collector shorted to base) and fed from $+8\text{V}$ through $R_2$; its base node also drives $Q_2$'s base. $Q_1$ and $Q_2$ emitters both return to $-5\text{V}$. $V_{BE}=0.7\text{V}$, $\beta=100$, $R_1=2\text{k}\Omega$, $R_2=3\text{k}\Omega$.

Find. The collector currents $I_{C1}$, $I_{C2}$, $I_{C3}$.

Fig. 1 — $Q_1$ is diode-connected and mirrors into $Q_2$ (both emitters at $-5\text{V}$); $Q_3$ (grounded base, collector via $R_1$) is a common-base (cascode) device stacked on $Q_2$: its emitter current is exactly $Q_2$'s collector current.

Approach (a). Use $Q_1$'s diode connection to fix the common base-node voltage, find the current $R_2$ delivers to that node, split it between the matched mirror pair including base currents, then follow $Q_3$'s emitter current down to its own collector current.

Fix the base-node voltage from $Q_1$'s diode connection. With $Q_1$'s base shorted to its collector, $V_{BE1}=0.7\text{V}$ across the emitter at $-5\text{V}$: $$V_C = V_{EE} + V_{BE} = -5 + 0.7 = -4.3\text{ V}.$$ This same node is $Q_2$'s base, so $V_{BE2}=0.7\text{V}$ too (matched pair, common base and emitter rails) — the textbook condition for $I_{C1}=I_{C2}$.
Current delivered by $R_2$ into the base node. $$I_{R2} = \frac{V_{CC}-V_C}{R_2} = \frac{8-(-4.3)}{3\text{k}\Omega} = 4.10\text{ mA}.$$ This current supplies $Q_1$'s collector current, $Q_1$'s own base current, and $Q_2$'s base current (all three terminals sit on the same node).
Split it between the matched pair, including both base currents. With $I_{C1}=I_{C2}$ (call it $I_C$) and $I_B=I_C/\beta$ for each transistor, $$I_{R2}=I_{C1}+\underbrace{I_{B1}}_{Q_1\text{ own base}}+\underbrace{I_{B2}}_{Q_2\text{ base}} = I_C\left(1+\frac{2}{\beta}\right).$$ Solving, $$\boxed{I_{C1}=I_{C2}=\frac{I_{R2}}{1+2/\beta}=\frac{4.10\text{ mA}}{1.02}=4.02\text{ mA}}.$$
Follow $Q_3$'s emitter current to $I_{C3}$. $Q_3$'s emitter is the ONLY thing feeding $Q_2$'s collector, so by KCL at that node $I_{E3}=I_{C2}=4.02\text{ mA}$. Since $I_{E3}=I_{C3}(1+1/\beta)$, $$\boxed{I_{C3}=I_{E3}\cdot\frac{\beta}{\beta+1}=4.02\text{ mA}\times\frac{100}{101}=3.98\text{ mA}}.$$

Final results
Quantity	Result
$I_{C1}$	$4.02\text{ mA}$
$I_{C2}$	$4.02\text{ mA}$
$I_{C3}$	$3.98\text{ mA}$

Check

Back-substituting gives $V_A$ (collector node of $Q_3$) $\approx 0.040\text{ V}$, only ~40 mV above $Q_3$'s grounded base — $Q_3$ is confirmed forward-active but with very little margin, so the given values put $Q_3$ right at the edge of saturation: any $R_1$ above about $2.01\text{ k}\Omega$ would drive it out of the active region.

Topic: BJT current mirror with a common-base (cascode) output transistor.

Key relations: Diode-connected BJT: $V_{BE}=V_T\ln(I_C/I_S)$, with $V_{BC}=0$ forcing $V_{BE}$ to set the mirror's reference current. Matched-pair mirror: $I_{C2}=I_{C1}$ when $V_{BE1}=V_{BE2}$. Beta error: $I_{R}=I_C(1+n/\beta)$ for $n$ base currents drawn from the reference node.

Why this works: $Q_1$ and $Q_2$ form a plain two-transistor mirror, so the reference current through $R_2$ must also supply both base currents ($n=2$ above), an error of order $2/\beta$. $Q_3$ is NOT a “beta helper”: its base goes to ground, not to the mirror’s base node, so it draws nothing from the reference. It is a common-base (cascode) device stacked on $Q_2$: $Q_2$ sets the current, $Q_3$ passes it up to $R_1$ (losing only its own base current, $I_{C3}=\alpha I_{C2}$) and holds $Q_2$’s collector at a fixed $-0.7\text{ V}$, which shields $Q_2$ from the voltage swing at $Q_3$’s collector.

Common pitfall: Forgetting that $Q_1$'s OWN base current is drawn from the same node as $Q_2$'s base current (a factor of $2/\beta$, not $1/\beta$) — a frequent slip is to write $I_{R2}=I_{C1}(1+1/\beta)$ and mirror $I_{C2}=I_{C1}$ separately, silently dropping one base-current term.

Source: Sedra-Smith §6.10, "Biasing Using a Diode-Connected Transistor" and current-mirror base-current compensation.

Question 1 of 6: BJT Current-Mirror Bias Network

Question 1: BJT Current-Mirror Bias Network (15 marks)