Question 2 of 6: SCR Turn-Off and a Half-Wave Controlled Rectifier

Question 2: SCR Turn-Off and a Half-Wave Controlled Rectifier (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. A half-wave phase-controlled rectifier is supplied from a 120 V (rms) single-phase source and feeds a series R–L load of power factor 0.707, i.e. load angle $\phi = 45^\circ$. Two operating points are specified by their conduction angles.

Find. Five factors that set the SCR turn-off time; then the delay angle and load resistance at $\gamma = 145^\circ$, and the delay angle and mean current at $\gamma = 150^\circ$ with that same resistance.

Approach. Part (a) is descriptive. Parts (b) and (c) use the same extinction condition as Question 1 to convert a stated conduction angle into a delay angle, after which the fact that an inductor supports no average voltage collapses the whole problem to one algebraic relation between $V_m$, $\alpha$, $\beta$, $R$ and $I_{dc}$.

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

The turn-off (circuit-commutated) 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. It is set by how quickly the excess stored charge in the four-layer structure is removed or recombines, and the following five factors dominate it.

1. The magnitude of the forward current before commutation. The stored charge in the base regions is roughly proportional to the on-state current, so a device carrying its full rated current has far more charge to sweep out than one lightly loaded. A thyristor rated $t_q = 40\ \mu$s at rated current may recover in half that at ten per cent of rated current.

2. The junction temperature. Minority-carrier lifetime rises strongly with temperature, so recombination slows. Between a 25 °C and a 125 °C junction the turn-off time of a typical converter-grade device roughly doubles. Data sheets always quote $t_q$ at the maximum rated junction temperature for exactly this reason.

3. The rate of decay of the anode current, $di/dt$, at commutation. A rapid current fall leaves more of the stored charge trapped in the base rather than carried out by the conduction current, increasing the reverse-recovery charge and lengthening the interval over which the device can still be re-triggered.

4. The magnitude and duration of the reverse voltage applied after commutation. A large reverse anode voltage actively sweeps carriers out of the junction and shortens $t_q$; a device commutated with only a diode drop of reverse bias recovers much more slowly. This is why forced-commutation circuits are designed to hold reverse voltage for a stated minimum time, and it is a design variable rather than a device property.

5. The rate of reapplication of forward voltage, $dv/dt$. Displacement current $C_j\,dv/dt$ through the junction capacitance can re-trigger a device whose charge has not fully recombined. A high reapplied $dv/dt$ therefore demands a longer effective turn-off time; snubber design and $dv/dt$ rating are two faces of the same constraint.

Two further influences are worth noting for completeness: a reverse gate bias applied during recovery shortens $t_q$ by extracting base charge, and lifetime-control processing in manufacture (gold doping or electron irradiation) trades a shorter $t_q$ for a higher on-state voltage — the distinction between converter-grade and inverter-grade thyristors.

Parts (b) and (c) — Quantitative solution

1. Part (b) — convert the conduction angle to a delay angle. The load angle follows from the stated power factor, $\phi = \cos^{-1}(0.707) = 45^\circ$, so $\tan\phi = 1$ and the extinction condition of Question 1 becomes $$\sin(\alpha+\gamma-45^\circ)=\sin(\alpha-45^\circ)\,e^{-\gamma}\quad(\gamma\ \text{in rad}).$$ With $\gamma_b = 145^\circ = 2.53073$ rad the damping factor is $e^{-2.53073}=0.07958$, and solving numerically gives $$\boxed{\alpha_b = 77.55^\circ},\qquad \beta_b=\alpha_b+\gamma_b=222.55^\circ .$$ The current therefore extinguishes $42.5^\circ$ after the supply zero — the load, not the gate, fixes $\beta$.
2. Use the zero-average-voltage property of the inductor to find $R$. Over a complete period the mean voltage across $L$ is zero, so the mean load voltage appears entirely across $R$. For a half-wave circuit there is one pulse per period, so the average is taken over $2\pi$: $$V_{dc}=\frac{1}{2\pi}\int_{\alpha}^{\beta}V_m\sin\theta\,d\theta =\frac{V_m(\cos\alpha-\cos\beta)}{2\pi}=I_{dc}R .$$ Substituting $\cos 77.55^\circ = 0.21566$, $\cos 222.55^\circ = -0.73674$ and $V_m = 169.71$ V: $$\begin{aligned} V_{dc}&=\frac{169.71\times 0.95240}{2\pi}=25.72\ \text{V},\ R&=\frac{25.72}{25}=\boxed{1.029\ \Omega}. \end{aligned}$$ The $2\pi$ divisor is the whole trap in this family: averaging over $\pi$, as one would for a full-wave bridge, halves $R$ and poisons part (c).
3. Part (c) — repeat the extinction solution at the wider conduction angle. With $\gamma_c = 150^\circ = 2.61799$ rad the damping factor is $e^{-2.61799}=0.07293$ and the same equation returns $$\boxed{\alpha_c = 73.04^\circ},\qquad \beta_c = 223.04^\circ .$$ Advancing the gate by $4.51^\circ$ has moved the extinction angle by only $0.49^\circ$: at strong inductance $\beta$ is far more stable than $\alpha$, which is the physical point the question is built around.
4. Evaluate the new mean current at the same load resistance. $$I_{dc,c}=\frac{V_m(\cos\alpha_c-\cos\beta_c)}{2\pi R} =\frac{169.71\times 1.02272}{2\pi\times 1.02895}=\boxed{26.85\ \text{A}}.$$ Five degrees of extra conduction have raised the mean current by 7.4 per cent, from 25.0 A to 26.85 A.
5. Independent check on both operating points. The mean-voltage identity used above is itself the check, because it was derived from the inductor rather than from the current waveform: at case (b), $V_{dc}=25.72$ V against $I_{dc}R = 25\times 1.02895 = 25.72$ V; at case (c), $V_{dc}=27.62$ V against $26.85\times 1.02895 = 27.62$ V. Both close to six figures, which simultaneously validates $\alpha$, $\beta$ and $R$.

It is worth recording how badly the familiar closed-form estimate $\alpha \approx 180^\circ+\phi-\gamma$ performs at this power factor. It predicts $80.0^\circ$ against the true $77.55^\circ$ and $75.0^\circ$ against $73.04^\circ$ — errors of $2.45^\circ$ and $1.96^\circ$, tracking the damping factor $e^{-\gamma/\tan\phi}$ exactly as theory says it should. At $\cos\phi = 0.707$ that estimate is a sanity check only.