Current Transfer Ratio of a Three-Transistor Current Mirror: 25-Nav-B6 Ocean Engineering and Offshore Structures, Undated paper, Q1

Question 1: Current Transfer Ratio of a Three-Transistor Current Mirror (equal value)

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 identical npn transistors, common-emitter dc current gain $\beta$ (so $\alpha=\beta/(\beta+1)$), Early effect neglected. $I_1$ enters the node shared by $Q_2$’s collector and $Q_1$’s base; $I_2$ enters $Q_1$’s collector. $Q_1$’s emitter feeds $Q_3$’s collector and the common base line of the matched pair $Q_2,Q_3$, whose emitters are grounded.

Find. The current transfer ratio $I_2/I_1$ as a function of $\beta$ alone.

[Figure not reproduced: Figure 1 (redrawn for clarity) — the three-transistor network is the classical Wilson current mirror: $Q_2,Q_3$ are the matched pair (emitters grounded, bases tied at node M, $Q_3$ diode-connected), and $Q_1$ is the series transistor whose emitter returns the mirror pair’s base currents . See the official exam paper.]

Check — reading the figure. The printed caption for Figure 1 is internally inconsistent about which device is diode-connected and where the base lines run. The wiring reconstructed above is the only one consistent with a genuine three-transistor precision current mirror of the kind this question is built around, and it reduces to the textbook Wilson-mirror result derived below.

Approach. Write the collector/base/emitter currents of each transistor in terms of $\beta$, apply KCL at the input node and at $Q_1$’s emitter node, and eliminate the internal mirror current $I$.

Mirror pair. $Q_2$ and $Q_3$ share the same base-emitter voltage (bases tied, emitters grounded), so with matched devices their collector currents are equal: $I_{C2}=I_{C3}=I$.
KCL at node M ($Q_1$’s emitter). $Q_1$’s emitter current supplies $Q_3$’s collector current and both base currents of the mirror pair: $$I_{E1} = I_{C3} + I_{B2} + I_{B3} = I + \frac{I}{\beta} + \frac{I}{\beta} = I\left(1+\frac{2}{\beta}\right) = \frac{I(\beta+2)}{\beta}.$$
Relate $I_{E1}$ to $I_2$. $I_2 = I_{C1} = \alpha I_{E1} = \dfrac{\beta}{\beta+1}\cdot\dfrac{I(\beta+2)}{\beta} = \dfrac{I(\beta+2)}{\beta+1}.$
KCL at the input node (node X). $I_1$ supplies $I_{C2}$ and $Q_1$’s base current: $$I_1 = I_{C2} + I_{B1} = I + \frac{I_{E1}}{\beta+1} = I + \frac{I(\beta+2)}{\beta(\beta+1)} = \frac{I\left[\beta(\beta+1)+(\beta+2)\right]}{\beta(\beta+1)} = \frac{I(\beta^{2}+2\beta+2)}{\beta(\beta+1)}.$$
Form the ratio. The common factor $I$ and $(\beta+1)$ cancel: $$\frac{I_2}{I_1} = \frac{\dfrac{I(\beta+2)}{\beta+1}}{\dfrac{I(\beta^{2}+2\beta+2)}{\beta(\beta+1)}} = \boxed{\;\frac{I_2}{I_1} = \frac{\beta(\beta+2)}{\beta^{2}+2\beta+2} = 1-\frac{2}{\beta^{2}+2\beta+2}\;}$$ The error term falls as $2/\beta^{2}$, not the $2/\beta$ of a plain two-transistor mirror: at $\beta=100$, $I_2/I_1=0.99980$ (196 ppm error); even $\beta=50$ gives $0.99923$.

Quantity	Result
Current transfer ratio	$I_2/I_1 = \beta(\beta+2)/(\beta^{2}+2\beta+2)$
Numerical values	0.99923 ($\beta=50$); 0.99980 ($\beta=100$); 0.99995 ($\beta=200$)

Topic. The Wilson current mirror and second-order beta-error cancellation in bipolar current sources.

Key relations. Active-region BJT: $I_C=\beta I_B$, $I_E=(\beta+1)I_B=I_C/\alpha$, $\alpha=\beta/(\beta+1)$. Matched devices at equal $V_{BE}$ carry equal $I_C$. KCL at the input node and at the series transistor’s emitter node gives $I_2/I_1=\beta(\beta+2)/(\beta^{2}+2\beta+2)$.

Why this works. In a plain two-transistor mirror both base currents are stolen directly from the input node, so the error is first-order in $1/\beta$. Here the series transistor $Q_1$ closes a local feedback loop: if $I_2$ tends to rise, $Q_1$’s emitter current rises, raising the mirror pair’s current, which pulls the input node down and reduces $Q_1$’s own base drive. The loop pushes the residual error to second order, $2/\beta^{2}$, and simultaneously raises the output resistance — the reason the Wilson mirror is a standard high-compliance current source in bipolar analogue design.

Common pitfall. Writing $I_{E1}=I_{C3}$ and forgetting the two base currents drawn from node M collapses the answer to $I_2/I_1=1$ exactly, hiding the whole point of the question; a second common slip is using $I_C=\beta I_E$ in place of $I_C=\alpha I_E$ when splitting $I_{E1}$.

Source. Sedra & Smith, Microelectronic Circuits, 8th ed., §8.2, “The Wilson Current Mirror.”

Question 1 of 8: Current Transfer Ratio of a Three-Transistor Current Mirror

Question 1: Current Transfer Ratio of a Three-Transistor Current Mirror (equal value)