22-Mec-A5 Electrical and Electronics Engineering · December 2013

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

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

Paper format. PEO / Engineers Canada National Examinations, December 2013 — 07-Mec-A5, Electrical & Electronics Engineering (Mechanical Engineering discipline). Three hours, closed book, one approved Casio or Sharp calculator. Eight questions; any five constitute a complete paper and each question is of equal value, so each counted question carries 20 marks. All eight questions are solved below, because this set is a study resource rather than an exam script.

Reference texts. A. S. Sedra and K. C. Smith, Microelectronic Circuits, 8th ed. (BJT current mirrors, op-amp frequency response); M. M. Mano and M. D. Ciletti, Digital Design, 6th ed. (universal-gate synthesis, DeMorgan reduction); S. J. Chapman, Electric Machinery Fundamentals, 5th ed. (dc machines, induction-motor testing and slip); C. K. Alexander and M. N. O. Sadiku, Fundamentals of Electric Circuits, 7th ed. (phasor analysis, power factor, first-order transients); W. H. Hayt and J. A. Buck, Engineering Electromagnetics, 9th ed. (magnetic circuits and induced emf).

Check: front-page constant. Note [8] of the examination paper prints the permeability of free space as μ₀ = 4π × 10⁷ H·m⁻¹. The exponent has plainly lost its minus sign in printing; the physical constant is 4π × 10⁻⁷ H·m⁻¹, and that is the value used in Question 8. The other two constants, π = 3.14159 and 1 hp = 746 W, are used exactly as printed.

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 the same dc current gain $\beta$, wired as in Figure 1: the input current $I_1$ enters the node that joins the collector of $Q_2$ to the base of $Q_1$; the emitter of $Q_1$ feeds the node where $Q_3$ is diode-connected (collector tied to base); the bases of $Q_2$ and $Q_3$ are tied together and both emitters return to ground; the output current $I_2$ is the collector current of $Q_1$. All devices are assumed to be in the forward-active region with negligible Early effect, so matched base-emitter voltages imply matched collector currents.

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

[Figure not reproduced: Figure 1 redrawn. This is the Wilson current mirror: $Q_2$ and $Q_3$ are the matched pair, and $Q_1$ is the series-feedback device that returns the base-current error to the input node. See the official exam paper.]

Approach. Assign the common collector current of the matched pair as the single unknown, apply KCL at the emitter node of $Q_1$ and again at the input node, express both $I_1$ and $I_2$ in terms of that unknown, and divide so the unknown cancels.

Exploit the matched pair. Transistors $Q_2$ and $Q_3$ share a base node and both emitters are grounded, so they see identical base-emitter voltages. Identical devices at identical $V_{BE}$ carry identical currents: $$I_{C2}=I_{C3}\equiv I_C,\qquad I_{B2}=I_{B3}=\frac{I_C}{\beta}$$ Everything below is written in terms of this one current $I_C$, which will cancel at the end.
Apply KCL at the emitter node of $Q_1$. That node supplies the collector of the diode-connected $Q_3$ plus the base currents of both members of the pair: $$I_{E1}=I_{C3}+I_{B3}+I_{B2}=I_C+\frac{2I_C}{\beta}=I_C\,\frac{\beta+2}{\beta}$$ The factor $\beta+2$ rather than $\beta$ is the whole story of this circuit: the two base currents are the only imperfection in the mirror.
Split the emitter current of $Q_1$ into its collector and base parts. For a transistor in the active region, $I_C=\alpha I_E$ with $\alpha=\beta/(\beta+1)$, and $I_B=I_E/(\beta+1)$. Therefore the output current is $$I_2=I_{C1}=\frac{\beta}{\beta+1}\,I_{E1}=\frac{\beta}{\beta+1}\cdot I_C\,\frac{\beta+2}{\beta}=I_C\,\frac{\beta+2}{\beta+1}$$ and the base current drawn by $Q_1$ from the input node is $$I_{B1}=\frac{I_{E1}}{\beta+1}=\frac{I_C(\beta+2)}{\beta(\beta+1)}$$
Apply KCL at the input node. The source current $I_1$ divides between the collector of $Q_2$ and the base of $Q_1$: $$I_1=I_{C2}+I_{B1}=I_C+\frac{I_C(\beta+2)}{\beta(\beta+1)}=I_C\,\frac{\beta(\beta+1)+\beta+2}{\beta(\beta+1)}=I_C\,\frac{\beta^{2}+2\beta+2}{\beta^{2}+\beta}$$ Both $I_1$ and $I_2$ are now proportional to $I_C$, so the ratio is independent of the operating current.
Form the ratio. Dividing and cancelling $I_C$ and the common factor $(\beta+1)$, $$\frac{I_2}{I_1}=\frac{\beta+2}{\beta+1}\cdot\frac{\beta^{2}+\beta}{\beta^{2}+2\beta+2}=\frac{\beta(\beta+2)}{\beta^{2}+2\beta+2}$$ $$\boxed{\;\frac{I_2}{I_1}=\frac{\beta^{2}+2\beta}{\beta^{2}+2\beta+2}=1-\frac{2}{\beta^{2}+2\beta+2}\;}$$
Read the engineering meaning off the error term. The mirror error is $2/(\beta^{2}+2\beta+2)$, which falls off as $1/\beta^{2}$ rather than the $1/\beta$ of a simple two-transistor mirror. At $\beta=100$ the error is $2/10\,202 = 1.96\times10^{-4}$, that is 0.0196 %.

The reason the transfer ratio is so close to unity is worth stating explicitly: $Q_1$ is not merely an output device, it is a feedback element. If the output current tries to rise, the emitter current of $Q_1$ rises, which raises the current in the diode-connected $Q_3$, which raises the base voltage of $Q_2$ and pulls more of $I_1$ into the collector of $Q_2$ — leaving less to drive the base of $Q_1$. The loop opposes the original disturbance, and the residual base-current error appears only in second order.

Final results — Question 1
Quantity	Result
Current transfer ratio	$I_2/I_1=\dfrac{\beta^{2}+2\beta}{\beta^{2}+2\beta+2}$
Equivalent error form	$I_2/I_1 = 1-\dfrac{2}{\beta^{2}+2\beta+2}$
Ratio at $\beta=10$	0.98361 (1.64 % low)
Ratio at $\beta=50$	0.99923 (0.077 % low)
Ratio at $\beta=100$	0.99980 (0.0196 % low)

Topic. The Wilson current mirror and second-order suppression of base-current error.

Key relations. Matched devices at equal $V_{BE}$ carry equal $I_C$; $I_C=\alpha I_E$ with $\alpha=\beta/(\beta+1)$; $I_B=I_E/(\beta+1)$; KCL at each internal node.

Why this works. A basic two-transistor mirror loses the base currents of both devices out of the input node, giving a transfer ratio of $1/(1+2/\beta)$ and hence an error of order $2/\beta$ — about 2 % at $\beta=100$. The Wilson topology adds a third transistor in series with the output and closes a negative-feedback loop through the diode-connected device, so the error that survives is of order $2/\beta^{2}$. Numerically, a device with $\beta=100$ improves from roughly 2 % error to roughly 0.02 %, a hundredfold gain in accuracy for one extra transistor and no extra quiescent current. This is why the Wilson mirror is the standard choice for bias chains in analog integrated circuits where $\beta$ is neither large nor well controlled.

Common pitfall. Forgetting that the emitter node of the output transistor must supply two base currents, not one, which gives $(\beta+1)$ where $(\beta+2)$ belongs and destroys the $\beta^{2}$ behaviour. The second frequent slip is to write $I_2=I_{E1}$ instead of $I_2=\alpha I_{E1}$; the collector current is always the emitter current less the base current.

Source. Sedra & Smith, Microelectronic Circuits, 8th ed., Ch. 8 (current mirrors and the Wilson configuration).

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