This set of Power Electronics online test focuses on “Thyristors-4”.
1. A tangent drawn from the Y-axis to the Pavg on the gate characteristics gives the value of the
a) maximum value of gate-source resistance
b) minimum value of gate-source resistance
c) maximum value of gate-source power
d) minimum value of gate-source power
View Answer
Explanation: It gives the min gate to source resistance.
2. Higher the magnitude of the gate pulse
a) lesser is the time required to inject the charges
b) greater is the time required to inject the charges
c) greater is the value of anode current
d) lesser is the value of anode current
View Answer
Explanation: Lesser time is required to inject the charges & turn on the device with higher gate pulse magnitude.
3.The average gate power dissipation for an SCR is 0.5 Watts the voltage applied to the gate is Vg = 10 V. What is the maximum value of current Ig for safe operation?
a) 0.25 A
b) 10 A
c) 0.05 A
d) 0.1 A
View Answer
Explanation: Vg.Ig = 0.5 W, the power dissipation mustn’t exceed the average power dissipation.
4. For an SCR, the gate-cathode characteristic has a slop of 130. The gate power dissipation is 0.5 watts. Find Ig
a) 0.62 A
b) 620 mA
c) 62 mA
d) 6.2 mA
View Answer
Explanation: Vg/Ig = 130 .. (given)
Vg.Ig = 0.5 watts .. (given)
use both the given data & find the gate current.
5. The two transistor model of the SCR can obtained by
a) bisecting the SCR vertically
b) bisecting the SCR horizontally
c) bisecting the SCRs top two & bottom two layers
d) bisecting the SCRs middle two layers
View Answer
Explanation: The two transistor model consists of p-n-p and n-p-n transistors, of which the middle n-p layer is common in both the transistors.
6. Latching current for an SCR is 100 mA, DC source of 200 V is also connected from the SCR to the L load. Compute the minimum width of the gate pulse required to turn on the device. Take L = 0.2 H.
a) 50 μsec
b) 100 μsec
c) 150 μsec
d) 200 μsec
View Answer
Explanation: For L load, E = L di/dt
I = E/L t
Therefore, 0.100 = 200t/0.2
T = 100 μsec.
7. The gate-source voltage is Es = 16 V and the load line has a slope of 128 V/A. Calculate the gate current for an average gate power dissipation of 0.5 W.
a) 62.5 mA
b) 100.25 mA
c) 56.4 mA
d) 80.65 mA
View Answer
Explanation: Load line is nothing but Rs
Es = 16V
Vg.Ig = 0.5
Rs = 128
We have Es = Ig x Rs + Vg.
8. From the two transistor (T1 & T2) analogy of SCR, the total anode current of SCR is ___________ in the equivalent circuit.
a) the sum of both the base currents
b) the sum of both the collector current
c) the sum of base current of T1 & collector current of T2
d) the sum of base current of T2 & collector current of T1
View Answer
Explanation: The sum of both the collector currents of T1 and T2 forms the total anode current of SCR. Refer the model.
9. Consider the two transistor analogy of SCR, if α1 & if α2 are the common-base current gains of both the transistors then to turn-on the device
a) α1 + α2 should approach zero
b) α1 x α2 should approach unity
c) α1 – α2 should approach zero
d) α1 + α2 should approach unity
View Answer
Explanation: To turn on the device sum of both the current gains should approach unity value.
10. Latching current for an SCR is 100 mA, a dc source of 200 V is also connected to the SCR which is supplying an R-L load. Compute the minimum width of the gate pulse required to turn on the device. Take L = 0.2 H & R = 20 ohm both in series.
a) 62.7 μsec
b) 100.5 μsec
c) 56.9 μsec
d) 81 μsec
View Answer
Explanation: E = Ri + L di/dt
Solve the above D.E for I & substitute the above values.
t = 100.503 μsec.
Sanfoundry Global Education & Learning Series – Power Electronics.
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