Current Transfer Ratio of a Three-Transistor Mirror: 22-Mec-A5 Electrical and Electronics Engineering, May 2013, Q1

Question 1: Current Transfer Ratio of a Three-Transistor Mirror (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. Three identical npn transistors, each with common-emitter dc current gain $\beta$ (so $\alpha = \beta/(\beta+1)$). The reference current $I_1$ enters the collector of $Q_2$ and the base of $Q_1$; the output current $I_2$ enters the collector of $Q_1$. $Q_1$’s emitter feeds the collector of $Q_3$ and the common base line of $Q_2$ and $Q_3$, whose emitters return to ground. Transistors are assumed matched and in the active region, and the Early effect is neglected.

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

[Figure not reproduced: Figure 1 — the circuit of the question paper redrawn: $Q_2$ and $Q_3$ form a mirror pair whose common base is driven from $Q_1$’s emitter. This is the Wilson current mirror. See the official exam paper.]

Approach. Recognise the topology as a Wilson mirror, set the matched pair’s collector current as the unknown $I$, then write KCL at the two nodes that carry $I_1$ and at $Q_1$’s emitter, expressing every base current as (collector current)/$\beta$.

Fix the matched pair. $Q_2$ and $Q_3$ share the same base node and the same grounded emitter node, so they see an identical base–emitter voltage. Being identical devices, they carry identical collector currents: $$I_{C2} = I_{C3} = I, \qquad I_{B2} = I_{B3} = \frac{I}{\beta}.$$ The value of $I$ never has to be evaluated — it cancels in the ratio.
KCL at $Q_1$’s emitter node. That node collects $Q_3$’s collector current and supplies both base currents of the mirror pair: $$I_{E1} = I_{C3} + I_{B2} + I_{B3} = I + \frac{2I}{\beta} = I\left(\frac{\beta+2}{\beta}\right).$$ This is the term that makes the Wilson mirror better than a simple two-transistor mirror: the base currents are drawn from $Q_1$’s emitter, not from the input node.
Split $Q_1$’s emitter current. For any active-region transistor $I_C = \alpha I_E$ and $I_B = I_E/(\beta+1)$, with $\alpha = \beta/(\beta+1)$. Hence $$I_2 = I_{C1} = \frac{\beta}{\beta+1}\,I_{E1} = I\,\frac{\beta+2}{\beta+1}, \qquad I_{B1} = \frac{I_{E1}}{\beta+1} = I\,\frac{\beta+2}{\beta(\beta+1)}.$$
KCL at the input node. The input current splits between $Q_2$’s collector and $Q_1$’s base: $$I_1 = I_{C2} + I_{B1} = I + I\,\frac{\beta+2}{\beta(\beta+1)} = I\,\frac{\beta(\beta+1) + \beta + 2}{\beta(\beta+1)} = I\,\frac{\beta^{2}+2\beta+2}{\beta(\beta+1)}.$$
Form the ratio. Dividing, the unknown $I$ and the factor $(\beta+1)$ both cancel: $$\frac{I_2}{I_1} = \frac{\dfrac{\beta+2}{\beta+1}}{\dfrac{\beta^{2}+2\beta+2}{\beta(\beta+1)}} = \boxed{\;\frac{I_2}{I_1} = \frac{\beta(\beta+2)}{\beta^{2}+2\beta+2} = 1 - \frac{2}{\beta^{2}+2\beta+2}\;}$$
Interpret the error term. The second form shows the mirror error falls as $2/\beta^{2}$ rather than the $2/\beta$ of a simple mirror. With $\beta = 100$ the ratio is $0.99980$, an error of $196$ ppm; even a modest $\beta = 50$ gives $0.99923$. That $\beta^{2}$ dependence is the whole point of the Wilson configuration.

Quantity	Result
Mirror-pair collector currents	$I_{C2}=I_{C3}=I$ (cancels)
$Q_1$ emitter current	$I_{E1}=I(\beta+2)/\beta$
Output current	$I_2 = I(\beta+2)/(\beta+1)$
Input current	$I_1 = I(\beta^{2}+2\beta+2)\,/\,\beta(\beta+1)$
Current transfer ratio	$\mathbf{I_2/I_1 = \beta(\beta+2)/(\beta^{2}+2\beta+2)}$
Numerical values	0.99923 ($\beta=50$); 0.99980 ($\beta=100$); 0.99995 ($\beta=200$)

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

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

Why this works. In a plain two-transistor mirror the two base currents are stolen from the input node, so $I_2/I_1 = 1/(1+2/\beta)$ and the error is first order in $1/\beta$. The Wilson topology adds a series transistor whose emitter, not the input node, supplies those base currents, and closes a negative-feedback loop: if $I_2$ tries to rise, $Q_1$’s emitter current rises, which raises the mirror pair’s current, which pulls the input node down and reduces $Q_1$’s base drive. The loop forces the residual error to appear only at second order, $2/\beta^2$. The same feedback raises the output resistance to roughly $\beta r_o/2$, which is why the circuit is the standard high-compliance current source in bipolar analogue ICs.

Common pitfall. Writing $I_{E1} = I_{C3}$ and forgetting the two base currents of the mirror pair — that error produces $I_2/I_1 = 1$ exactly and hides the whole point of the question. A second frequent slip is using $I_C = \beta I_E$ instead of $I_C = \alpha I_E$ when splitting $Q_1$’s emitter current.

Source. Sedra & Smith, Microelectronic Circuits, 8th ed., §8.2 (“The Wilson current mirror”).

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

Question 1: Current Transfer Ratio of a Three-Transistor Mirror (20 marks)