1. For a string voltage of 3300 V, let there be six series connected SCRs each of voltage 600V. Then the string efficiency is
a) 99.36 %
b) 91.7 %
c) 98.54 %
d) 96 %
Explanation: String efficiency = 3300/(6 x 600) = 98.54.
2. The measure of reliability of string is given by the factor
a) DRF = 1 – String efficiency
b) DRF = 1 + String efficiency
c) DRF = String efficiency – 1
d) DRF = String efficiency x 2
Explanation: DRF is de-rating factor given by the above expression.
3. When an extra SCR is connected in series with a string
a) DRF decreases
b) DRF increases
c) DRF remains constant
d) None of the mentioned
Explanation:
DRF = 1 – String efficiency
String Efficiency = Rating of the whole string/(rating of one SCR x number of SCRs)
Extra SCR will reduce the string efficiency which in turn increase the DRF.
4. The most practical way of obtaining a uniform distribution of series connected SCRs is to
a) connect a resistor of value R in series with each of the series connected SCRs
b) connect a resistor of value R in parallel with each of the series connected SCRs
c) connect a resistor of value R in series with one of the series connected SCRs
d) connect a resistor of value R in parallel with one of the series connected SCRs
Explanation: For uniform distribution of voltage across series connected SCRs, a resistor of value R in parallel with each series connected SCR.
5. 3 SCRs are connected in series. The string efficiency is 91%. SCRs 1, 2 & 3 have leakage currents 4 mA, 8 mA & 12 mA. Which SCR will block more voltage?
a) SCR-1
b) SCR-2
c) SCR-3
d) All the three will block equal voltage
Explanation: The SCR with lower leakage current block more voltage.
6. Two parallel connect SCRs have same voltage drop (Vt) having rated current = 2I1. SCR-1 carries a current of I1=2.6 A whereas SCR-2 carries a current of I2=1.4 A. Find the string efficiency.
a) 45 %
b) 77 %
c) 92 %
d) 84 %
Explanation: The total current would be I1+I2 & rated current is 2I1
String efficiency = (I1+I2)/2I1.
7. SCRs with a rating of 1000 V & 200 A are available to be used in a string to handle 6 KV & 1 KV. Calculate the number of series & parallel units required in case the de-rating factor is 0.1. (Round off the fraction to the greatest & nearest integer)
a) Series = 7, Parallel = 6
b) Series = 6, Parallel = 7
c) Series = 6, Parallel = 6
d) Series = 7, Parallel = 7
Explanation: DRF = 1-S.E
Therefore
0.1 = (1-6000/1000Ns) = (1-1000/200Np)
Ns = 6.6 = 7(say)
Np = 5.5 = 6(say)
8. _________ device from the thyristor family has its gate terminal connected to the n-type material near the anode.
a) SCR
b) RCT
c) PUT
d) SUT
Explanation: PUT is Programmable Unijunction Transistor which is a p-n-p-n device just like the SCR with its gate connected to the n-type material.
9. The Programmable Unijunction Transistor (PUT) turns on & starts conducting when the
a) gate voltage exceeds anode voltage by a certain value
b) anode voltage exceeds gate voltage by a certain value
c) gate voltage equals the anode voltage
d) gate is given negative pulse w.r.t to cathode
Explanation: The device only starts to conduct when the forward anode to cathode voltage exceeds the applied gate to cathode voltage.
10. The equivalent circuit of SUS (Silicon Unilateral Switch) consists of
a) a diode in series with a PUT
b) a diode in parallel with a PUT
c) a diode in anti-parallel with a PUT
d) two diodes
Explanation: It is a diode connected in anti-parallel with a PUT.