22-Elec-B8 Power Electronics and Drives · December 2014

Question 2 of 6: Controlled Half-Wave Rectifier with a Back-EMF Load

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

Question 2: Controlled Half-Wave Rectifier with a Back-EMF Load (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.

Part (a) — Five factors governing the SCR turn-off interval

The turn-off interval $t_q$ of a thyristor is the minimum time for which the device must be held at or below zero anode voltage after the anode current has ceased, before forward voltage may be safely reapplied. It is not a fixed catalogue number but a strong function of the operating conditions, and a converter designed against the datasheet minimum without allowing for the following five effects will suffer commutation failure.

1. The anode current immediately before commutation. The excess-carrier charge stored in the wide, lightly doped base regions is roughly proportional to the forward current, and all of it must recombine or be swept out before the device regains its blocking capability. Doubling the load current can lengthen $t_q$ by tens of per cent, so the worst case for commutation is full load, not light load.

2. The rate of fall of the anode current, $-di/dt$. A commutating circuit that drives the current to zero very rapidly does not give the stored charge time to recombine in place; instead a large reverse recovery current flows, the recovered charge $Q_{rr}$ rises, and the effective turn-off time increases. Softening the current fall with a small series inductance is a standard remedy.

3. Junction temperature. Carrier lifetime rises steeply with temperature, so $t_q$ typically doubles between a junction at $25^\circ$C and one at $125^\circ$C. Turn-off time must always be quoted, and designed for, at the maximum rated junction temperature.

4. The magnitude and duration of the reverse voltage applied after current zero. A substantial reverse bias actively sweeps carriers out of the junction and shortens the recovery; a circuit that merely holds the device near zero volts relies on recombination alone and needs a much longer interval. This is why forced-commutation circuits are sized to hold a definite reverse voltage across the outgoing device.

5. The rate of reapplication of forward voltage, $dv/dt$. Charging current flowing into the depletion capacitance of the blocking junction, $i = C\,dv/dt$, acts like gate current; if forward voltage is restored too quickly the device retriggers without a gate pulse even though its recombination is complete. Snubber networks limit the reapplied $dv/dt$ precisely so that the circuit turn-off time can be kept close to the device $t_q$.

Two further influences are worth naming because they are the designer's levers rather than nuisances: the gate condition during recovery — a reverse-biased or resistively shorted gate removes carriers from the gate region and measurably shortens $t_q$ — and the device construction itself, since lifetime-controlled 'inverter-grade' thyristors reach $5$–$50\ \mu\text{s}$ against $100$–$200\ \mu\text{s}$ for a phase-control device, at the cost of a higher on-state voltage.

Parts (b) and (c) — Back-EMF load calculations

Given. A controlled half-wave rectifier fed from a 120 V (rms) supply drives a resistance in series with the internal EMF of a dc motor; at 20 A average output current the measured conduction angle is $140^\circ$ and the minimum possible firing angle is $15^\circ$.

Find. The back EMF $E_c$, the firing angle $\alpha$ and the load resistance $R$; then, with the firing angle re-set to $30^\circ$, the average power absorbed by $E_c$ and the corresponding mechanical output in horsepower.

Approach. Recognise that $\alpha_{min}$ is the angle at which the supply first reaches $E_c$; that fixes $E_c$ and, by symmetry, the extinction angle. The conduction angle then gives $\alpha$, the average-current integral gives $R$, and repeating the integral at the new firing angle gives the power delivered to the EMF.

1. Part (b) — recover the back EMF from the minimum firing angle. The thyristor can only be turned on while the supply exceeds the opposing EMF, so the earliest possible firing instant is where $V_m\sin\omega t = E_c$: $$E_c = V_m \sin\alpha_{min} = 169.71 \times \sin 15^\circ = \boxed{43.92\ \text{V}}$$ Everything else in this question follows from that one observation.
2. Part (b) — the extinction angle is set by the load. Once fired, the device conducts only while $v_s > E_c$; there is no inductance to carry current beyond that point, so conduction ends where the supply falls back through $E_c$ on the descending side: $$\beta = 180^\circ - \alpha_{min} = 165^\circ$$ Because $\beta$ is fixed, specifying the conduction angle is equivalent to specifying the firing angle: $$\alpha = \beta - \gamma = 165^\circ - 140^\circ = \boxed{25^\circ}$$
3. Part (b) — average current gives the resistance. During conduction the loop equation is $V_m\sin\omega t = E_c + iR$, and the rectifier is half-wave so the average is taken over the full period $2\pi$: $$I_{avg} = \frac{1}{2\pi R}\int_{\alpha}^{\beta}\left(V_m\sin\omega t - E_c\right) d(\omega t) = \frac{V_m(\cos\alpha - \cos\beta) - E_c(\beta - \alpha)}{2\pi R}$$
4. Substitute and solve for $R$. With $\cos 25^\circ - \cos 165^\circ = 1.8722$ and $\beta - \alpha = 2.4435\ \text{rad}$, $$V_m(\cos\alpha - \cos\beta) = 317.71\ \text{V}, \qquad E_c(\beta-\alpha) = 107.31\ \text{V}$$ so the numerator is 210.40 V and $$R = \frac{210.40}{2\pi \times 20} = \boxed{1.674\ \Omega}$$
5. Part (c) — re-fire at $30^\circ$ and recompute the current. The extinction angle does not move — it is set by $E_c$, which the motor's speed and field fix, not by the firing circuit — so $\beta$ stays at $165^\circ$ and the conduction angle shrinks to $135^\circ$. Repeating the integral with $\cos 30^\circ - \cos 165^\circ = 1.8320$ and $\beta-\alpha = 2.3562\ \text{rad}$ gives a numerator of 207.38 V and $$I_{avg} = \frac{207.38}{2\pi \times 1.674} = \boxed{19.71\ \text{A}}$$ Delaying the firing by five degrees has cost only 1.4 per cent of current, because the extra $5^\circ$ is removed from the shoulder of the sine wave where $v_s - E_c$ is smallest.
6. Part (c) — power absorbed by the EMF. $E_c$ is constant, so the average power it absorbs is simply the product of the EMF and the average current: $$P_{E_c} = E_c\,I_{avg} = 43.92 \times 19.71 = \boxed{865.9\ \text{W}}$$
7. Part (c) — convert to mechanical output. The internal EMF of a dc machine is the seat of electromechanical conversion: the power absorbed there is the developed mechanical power, the copper loss having already been accounted for in $R$. Hence $$P_{mech} = \frac{865.9}{745.7} = \boxed{1.16\ \text{hp}}$$ The remaining $I_{rms}^2 R$ appears as heat in the series resistance and armature circuit.