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.

Answer: C

The maximum switching frequency is 5 kHz.

Each switching cycle must allow the SCR to turn on and then turn off, so the shortest possible period is the sum of the two times:

\[ \begin{aligned} T_{min} &= t_{on} + t_{off} \\ &= 50 + 150 = 200\ \mu s \\ f_{max} &= \frac{1}{200\ \mu s} = 5\ \text{kHz} \end{aligned} \]

20 kHz uses only the 50 µs turn-on time, about 6.7 kHz would use only the turn-off time, and 10 kHz and 2 kHz do not follow from either.

Answer: C

The gate current is a train of sharp pulses that decay exponentially, alternating in sign, waveform (C).

In this circuit the resistor R charges the capacitor C from the AC supply. When the capacitor voltage reaches the breakover voltage of the DIAC, the DIAC suddenly conducts and the capacitor discharges through it into the gate of the triac. The gate current therefore jumps up almost instantly and then falls as the capacitor discharges through the gate resistance, which is an exponential decay. The same thing happens on the negative half-cycle with the opposite sign, because the DIAC is symmetrical.

Waveform (A) is just the sine wave of the supply, with no triggering action. Waveform (B) is a smoothly rising curve with no sudden jump, which a breakover device cannot give. Waveform (D) shows a sudden rise but then a rounded fall that does not look like a capacitor discharge; C is the standard shape.

Answer: A

The control circuit is isolated by a pulse transformer.

A thyristor's gate and cathode sit at the potential of the power circuit, which may be hundreds of volts, while the firing circuit works at low voltage. A pulse transformer passes the short gate trigger pulse magnetically from the control side to the gate while keeping the two circuits electrically isolated. Opto-couplers are the other common method.

A high-voltage capacitor or an RC section would couple the circuits electrically rather than isolate them, and a relay is far too slow and bulky for gate triggering.

Answer: D

The anode current is \(I_A=\frac{\alpha_2I_G+I_{CBO1}+I_{CBO2}}{1-(\alpha_1+\alpha_2)}\).

In the two-transistor model, the thyristor is a p-n-p transistor Q1 and an n-p-n transistor Q2, with \(I_{K} = I_{A} + I_{G}\). The collector currents are \(I_{C1} = \alpha_{1}I_{A} + I_{CBO1}\) and \(I_{C2} = \alpha_{2}I_{K} + I_{CBO2}\), and the anode current is their sum:

\[ \begin{aligned} I_{A} &= \alpha_{1}I_{A} + \alpha_{2}(I_{A} + I_{G}) \\ &\quad + I_{CBO1} + I_{CBO2} \end{aligned} \]

Solving for \(I_{A}\) gives the keyed form. As \(\alpha_{1} + \alpha_{2}\) approaches 1 the denominator approaches zero and the anode current grows without limit: the device latches on. The other forms have the wrong sign for a leakage current, a plus in the denominator, \(\alpha_{1} - \alpha_{2}\), or \(\alpha_{1}\) on the gate current.

Answer: C

A good SCR will show high resistance in both directions.

Between anode and cathode an SCR is a four-layer PNPN device with three junctions. Whichever way the ohmmeter is connected, at least one junction is reverse biased, and without a gate trigger the SCR stays off, so the meter reads a high resistance both ways.

Low resistance in one or both directions would indicate a shorted device. Behaviour like a diode, low one way and high the other, would also mean the SCR is faulty.