Question 2 of 6: SCR Turn-Off Time, and a Half-Wave Controlled Rectifier with an Inductive Load

Question 2: SCR Turn-Off Time, and a Half-Wave Controlled Rectifier with an Inductive 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 that lengthen the turn-off interval of an SCR

The turn-off interval of a thyristor — the circuit-commutated turn-off time $t_q$ — is the minimum time for which the device must be held at reverse or zero voltage after its anode current has fallen to zero before forward voltage may safely be re-applied. It is not a fixed catalogue number but a strong function of how the device is being operated, and a converter design that ignores that dependence commutation-fails in service even though the data-sheet number looked adequate.

1. The magnitude of the forward current before commutation. The stored charge in the lightly doped base regions grows with the on-state current, and every one of those excess carriers must recombine or be swept out before the blocking junction can support forward voltage. A device commutated from full rated current has a measurably longer $t_q$ than the same device commutated from a light load, which is why $t_q$ is always quoted with the test current.

2. The rate of fall of anode current at commutation. A steep $-\mathrm{d}i/\mathrm{d}t$ forces a large reverse-recovery current and leaves the remaining charge distributed deep in the base, where the reverse field reaches it poorly. A gentler current slope allows more of the charge to recombine in place and shortens the recovery. Commutating-circuit inductance is therefore a design variable, not an incidental parasitic.

3. The magnitude of the reverse voltage applied during the recovery interval. Reverse bias actively sweeps carriers out of the junction region. A generous reverse voltage shortens turn-off; a circuit that applies only a volt or two of reverse bias — or none at all, merely holding the device at zero — leaves recombination as the only removal mechanism and can double the effective time required.

4. Junction temperature. Carrier lifetime in silicon increases markedly with temperature, so recombination slows and $t_q$ rises — typically by a factor of roughly two between 25 °C and the 125 °C rated junction temperature. Turn-off time must therefore be evaluated at the hottest credible operating point, not at ambient, and adequate heat-sinking is part of the commutation design.

5. The rate of re-applied forward voltage, and the gate condition. Even after the charge has cleared, a forward $\mathrm{d}v/\mathrm{d}t$ above the device rating injects enough displacement current through the junction capacitance to re-trigger conduction, so the practical turn-off interval is whatever time the circuit needs to keep the device inside its $\mathrm{d}v/\mathrm{d}t$ limit as well as clear of charge. A negative gate bias during recovery extracts carriers from the gate region and shortens the interval; a floating or positive gate lengthens it. Device construction sits behind all five factors: lifetime control by gold doping or electron irradiation produces fast inverter-grade thyristors with $t_q$ of 10–50 µs at the price of higher on-state voltage drop, whereas converter-grade devices intended for 50/60 Hz line commutation may need 100–200 µs.

Parts (b) and (c) — Delay angle, load resistance, and mean output current

Given. A single-thyristor (half-wave) controlled rectifier is supplied from 120 V rms and feeds a series R–L load whose power factor is 0.707; two operating points are specified, one by conduction angle and mean current, the other by conduction angle and resistance.

Find. The delay angle at each conduction angle, the resistance that makes the mean current 25 A in case (b), and the mean current that results from $R=1.1\ \Omega$ in case (c).

Approach. The extinction condition of Question 1 applies unchanged, because the conduction interval of a half-wave rectifier is governed by the same differential equation; only the averaging interval differs. Solve it for $\alpha$ at each $\gamma$, then integrate the normalised current over one full cycle and use the power factor to convert between $Z$ and $R$.

1. Load angle from the power factor. $$\varphi=\cos^{-1}(0.707)=45.01^\circ,\qquad \tan\varphi=1.0004 ,$$ so this load is very nearly the classic $X=R$ case; the reactance and the resistance will come out equal to three figures.
2. Delay angle for $\gamma=147^\circ$. Substituting into $$\sin(\alpha+\gamma-\varphi)=\sin(\alpha-\varphi)e^{-\gamma/\tan\varphi}$$ and solving numerically, $$\boxed{\alpha_b=75.75^\circ},\qquad \beta_b=\alpha_b+\gamma_b=222.75^\circ .$$ The extinction angle exceeds 180° by nearly 43°, confirming that the inductance is doing substantial work.
3. Average the current over the whole cycle. A half-wave rectifier delivers one pulse per supply period, so $$I_{\text{dc}}=\frac{1}{2\pi}\int_{\alpha}^{\beta}i\,\mathrm{d}(\omega t) =\frac{V_m}{2\pi Z}\int_{\alpha}^{\beta}u\,\mathrm{d}(\omega t),$$ with $u$ the same normalised waveform as before. Numerically $\int u=1.38662$ for this window.
4. Solve for the resistance. Substituting $Z=R/\cos\varphi$ and rearranging for $R$, $$R=\frac{V_m\cos\varphi}{2\pi I_{\text{dc}}}\int_{\alpha}^{\beta}u\,\mathrm{d}(\omega t) =\frac{169.71\times0.707}{2\pi\times25}\times1.38662,$$ $$\boxed{R=1.059\ \Omega},\qquad Z=1.498\ \Omega,\qquad X_L=1.059\ \Omega .$$
5. Delay angle for $\gamma=152^\circ$. Re-solving the extinction condition with the new conduction angle, $$\boxed{\alpha_c=71.22^\circ},\qquad \beta_c=223.22^\circ .$$ Firing 4.5° earlier buys 5° of extra conduction, and the extinction angle barely moves: once the load is strongly inductive, $\beta$ is far more stable than $\alpha$.
6. Mean current at the new operating point. The window integral becomes $\int u=1.48593$, and with the stated $R=1.1\ \Omega$, $$I_{\text{dc}}=\frac{V_m\cos\varphi}{2\pi R}\int_{\alpha}^{\beta}u\,\mathrm{d}(\omega t) =\frac{169.71\times0.707}{2\pi\times1.1}\times1.48593,$$ $$\boxed{I_{\text{dc}}=25.80\ \text{A}} .$$
7. Check both answers against the mean output voltage. The inductor can support no average voltage, so the mean load voltage must equal $I_{\text{dc}}R$ exactly. Evaluating the rectifier expression directly, $$V_{\text{dc}}=\frac{V_m}{2\pi}\left(\cos\alpha-\cos\beta\right):\qquad 26.479\ \text{V}=25\times1.0591,\quad 28.375\ \text{V}=25.795\times1.1 .$$ Both close to six figures, which validates the delay angles, the integrals and the impedance split simultaneously.

The one structural difference from Question 1 deserves emphasis, because it is the commonest way to lose the whole of this problem. There the divisor was $2\pi$ per device in a full-wave arrangement that used two devices; here the divisor is $2\pi$ for the output, because a half-wave circuit produces only one pulse per period. Averaging over $\pi$, as one would for a full-wave bridge, would halve the required resistance in part (b) and double the current in part (c).