Thyristor Characteristics and Triggering Questions and Answers

Answer: A

The minimum gate pulse width is \(8 \times 10^{-3}\) s.

With a purely inductive load, the anode current after turn-on rises linearly, \(i = \frac{V}{L}t\). The gate pulse must last until this current reaches the latching current, or the thyristor turns off when the pulse is removed.

\[ \begin{aligned} t &= \frac{L\,I_{L}}{V} = \frac{10 \times 0.08}{100} \\ &= 8 \times 10^{-3}\ \text{s} = 8\ \text{ms} \end{aligned} \]

80 ms comes from leaving out the factor 10 in the conversion of 80 mA to amperes, and 1 ms or 10 ms do not let the current reach 80 mA, since at 1 ms it is only 10 mA.

Answer: A

The device can be operated up to about 230 V.

The voltage safety factor is the ratio of the PIV rating to the peak working voltage. With a factor of 2, the peak working voltage is half the rating, and for a sinusoidal supply the rms value is the peak divided by \(\sqrt{2}\).

\[ \begin{aligned} V_{m} &= \frac{650}{2} = 325\ \text{V} \\ V_{rms} &= \frac{325}{\sqrt{2}} \approx 229.8 \approx 230\ \text{V} \end{aligned} \]

325 V is the permissible peak, not the supply voltage rating, and 459 V is \(650/\sqrt{2}\), which ignores the safety factor. 650 V would leave no margin at all.

Answer: A

A snubber is used for \(\frac{dv}{dt}\) protection.

A snubber is a series RC circuit connected across the thyristor. A fast-rising forward voltage can charge the junction capacitance enough to turn the thyristor on without a gate signal. The snubber capacitor slows the rate of rise of voltage across the device, and the resistor limits the discharge current when it turns on.

Protection against a high \(\frac{di}{dt}\) is given by a small inductor in series with the thyristor, not by the snubber. Phase shifting and triggering are jobs of the gate firing circuit.

Answer: D

The official key is d only (decreasing the circuit inductance L).

The turn-off time of an SCR is the time the device needs, after its current has fallen to zero, to regain forward blocking. In a series RL circuit the current decays with time constant L/R, so a smaller inductance lets the current fall away faster and shortens the time before the device can start to recover. That supports (d).

What could not be confirmed: the same time-constant argument would also favour a larger R (option (a), giving 'a and d'), and no textbook or reference was found that states which of the two the question intends. The key accepts only (d), so (c), increasing L, and (b), decreasing R, are both the wrong direction and are rejected, and the official key is kept.

Answer: C

The ideal pulse has a short rise time with pulse width = 6 μs.

The gate pulse must last longer than the SCR's turn-on time, so that the anode current has risen above the latching current before the gate drive is removed; otherwise the SCR turns off again when the pulse ends. With a turn-on time of 5 μs, a 6 μs pulse is long enough while a 3 μs pulse is not.

A short (fast) rise time makes turn-on quick and spreads conduction across the junction rapidly, reducing local heating and di/dt stress. A long rise time slows turn-on, so both 'long rise time' options are worse.