22-Mec-A5 Electrical and Electronics Engineering · December 2014
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Paper format. PEO / Engineers Canada National Examinations, December 2014 — 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, transformer magnetic circuits, 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).
Front-page constants. Note [8] of the examination paper supplies μ0 = 4π × 10−7 H·m−1, π = 3.14159 and 1 hp = 746 W. All three are used exactly as printed; μ0 enters in Question 4 and the horsepower conversion in Question 3.
Check: data-table exponent. The Question 4 specification table reads 3.77 × 10−2 m and 7.54 × 10−2 m, and only those values give a physically sensible core (a few centimetres of flux path rather than several hundred metres) and reluctances that land on the round figures 1.5 × 106 and 3.0 × 106 A/Wb. The negative exponents are used throughout.
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.
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.
| 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) |
