Question 2 of 6: SCR turn-off time, and a half-wave controlled rectifier with an R-L load

Question 2: SCR turn-off time, and a half-wave controlled rectifier with an R-L load (20 points)

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 that lengthen an SCR's turn-off interval

The turn-off interval $t_q$ is the time that must elapse between the instant the anode current reaches zero and the instant forward voltage may safely be reapplied without the device latching on again. It is set by how quickly the excess carriers stored in the four layers are removed or recombine, and the following five factors govern that.

1. The on-state current immediately before commutation. Stored charge in the base regions is roughly proportional to the current that was flowing, so a device commutated from full rated current needs a substantially longer $t_q$ than the same device commutated from a light load. Manufacturers therefore quote $t_q$ at a stated on-state current, and a designer must derate the figure for higher currents.

2. Junction temperature. Carrier lifetime rises steeply with temperature, so recombination slows and $t_q$ lengthens — typically it roughly doubles between 25 °C and 125 °C. This is the single largest influence in practice, and it is why the circuit turn-off time must be checked at the maximum expected junction temperature rather than at ambient.

3. The magnitude and duration of the reverse voltage applied after commutation. Reverse bias actively sweeps stored charge out of the junctions through the reverse recovery current. A large reverse voltage held for a long interval shortens $t_q$; a device commutated with only a few volts of reverse bias, or one whose reverse voltage collapses quickly, recovers much more slowly.

4. The rate of fall of the anode current, $\mathrm{d}i/\mathrm{d}t$, at commutation. A rapid fall leaves less time for recombination during the decay itself and produces a larger reverse recovery current peak, so more charge must still be extracted after the current has crossed zero. A gentler decay does part of the recovery work for free.

5. The reapplied forward $\mathrm{d}v/\mathrm{d}t$ and its peak value. Even after the stored charge is gone, a fast-rising forward voltage injects capacitive displacement current into the gate-cathode junction and can re-trigger the device. The faster and higher the reapplied voltage, the longer the circuit must wait, so the effective $t_q$ demanded of the circuit rises with $\mathrm{d}v/\mathrm{d}t$. This is precisely why the RC snubber of Problem 1 also improves commutation margin.

Two secondary influences are worth naming: the device's own lifetime-control processing (gold or platinum doping, or electron irradiation, which trade forward drop for a fast $t_q$ — the difference between a phase-control and an inverter-grade thyristor), and the gate bias during recovery, since a negative gate bias helps extract charge from the cathode-side base.

Parts (b) and (c) — The half-wave controlled rectifier

Given.

Find. The delay angle at each of the two stated conduction angles; the load resistance implied by a 25 A mean output current at $\gamma = 147^\circ$; and the mean output current at $\gamma = 152^\circ$ with $R = 1.1\ \Omega$.

Approach. Get $\phi$ from the power factor, solve the extinction condition for $\alpha$ at each conduction angle, then use the fact that the mean inductor voltage is zero so that $I_{dc} = V_{dc}/R$ with $V_{dc}$ averaged over the full $2\pi$.

1. Part (b) — Load angle. The stated power factor is that of the series R-L load at supply frequency, so $\phi = \arccos(0.707) = 45.009^\circ$ and $\tan\phi = 1.00030$. The load angle is essentially 45°, which makes the exponential damping term unusually strong in this problem.
2. Solve the extinction condition for the delay angle. The current in a half-wave controlled rectifier with an R-L load obeys the same form as the controller of Problem 1, $i(\omega t) = \frac{V_m}{Z}\left[\sin(\omega t - \phi) - \sin(\alpha-\phi)e^{(\alpha-\omega t)/\tan\phi}\right]$, and setting $i(\beta) = 0$ with $\beta = \alpha + \gamma$ gives $$\sin(\alpha + \gamma - \phi) = \sin(\alpha - \phi)\,e^{-\gamma/\tan\phi}.$$ With $\gamma = 147^\circ$ the damping factor is $e^{-2.56563/1.00030} = 0.07690$, and solving numerically gives $$\boxed{\alpha = 75.755^\circ},\qquad \beta = 222.755^\circ .$$ Note that $\beta$ exceeds 180° by nearly 43°: the load inductance keeps the device conducting well into the negative half cycle.
3. Relate the mean output current to the load resistance. Over a complete cycle the mean voltage across an inductance is zero, so the whole of the mean output voltage appears across $R$, and $I_{dc} = V_{dc}/R$. For a half-wave converter the output is averaged over $2\pi$, giving $$V_{dc} = \frac{1}{2\pi}\int_{\alpha}^{\beta} V_m \sin\omega t\,\mathrm{d}(\omega t) = \frac{V_m}{2\pi}\left(\cos\alpha - \cos\beta\right).$$ Substituting, $V_{dc} = \frac{169.706}{2\pi}\left(0.24626 + 0.73402\right) = 26.479$ V.
4. Extract the load resistance. $$R = \frac{V_{dc}}{I_{dc}} = \frac{26.479}{25} = \boxed{1.0591\ \Omega}.$$ The reactance follows for completeness as $X = R\tan\phi = 1.0593\ \Omega$, so the load is very nearly a 1.06 $\Omega$ resistance in series with 1.06 $\Omega$ of reactance.
5. Part (c) — Re-solve the extinction condition at the new conduction angle. The load has not changed, so $\phi$ is still 45.009°; only $\gamma$ moves. With $\gamma = 152^\circ = 2.65290$ rad the damping factor becomes $e^{-2.65210} = 0.07047$, and the same equation returns $$\boxed{\alpha = 71.224^\circ},\qquad \beta = 223.224^\circ .$$ Firing 4.53° earlier buys 5° of extra conduction, and the extinction angle barely moves — the load, not the gate, decides where conduction ends.
6. Compute the new mean output current. With the wider window, $$V_{dc} = \frac{169.706}{2\pi}\left(\cos 71.224^\circ - \cos 223.224^\circ\right) = \frac{169.706}{2\pi}\left(0.32175 + 0.72872\right) = 28.375\ \text{V},$$ and with the stated $R = 1.1\ \Omega$, $$I_{dc} = \frac{28.375}{1.1} = \boxed{25.795\ \text{A}}.$$
7. Consistency check between the two parts. Part (b) deduced $R = 1.0591\ \Omega$ from the same physical load that part (c) then quotes as 1.1 $\Omega$ — a 3.9% difference, which is the rounding the examiner applied when handing the value back. That the two agree to within rounding confirms both the 2$\pi$ divisor and both firing angles; had the mean been averaged over $\pi$ instead, part (b) would have returned $R = 2.118\ \Omega$ and the two parts would be irreconcilable.

Final results.