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Test: Firing Circuits


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30 Questions MCQ Test Power Electronics | Test: Firing Circuits

Test: Firing Circuits for Electrical Engineering (EE) 2023 is part of Power Electronics preparation. The Test: Firing Circuits questions and answers have been prepared according to the Electrical Engineering (EE) exam syllabus.The Test: Firing Circuits MCQs are made for Electrical Engineering (EE) 2023 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Firing Circuits below.
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Test: Firing Circuits - Question 1

 In a single pulse semi-converter using two SCRs, the triggering circuit must produce

Detailed Solution for Test: Firing Circuits - Question 1

A single phase semi-converter has only two SCRs & two diodes. Hence, only two pulses are required in each cycle, one in each half.

Test: Firing Circuits - Question 2

 In a 3-phase full converter using six SCRs, gating circuit must provide

Detailed Solution for Test: Firing Circuits - Question 2

60° x 6(devices) = 360°.

Test: Firing Circuits - Question 3

In the complete firing circuit, the driver circuit consists of

Detailed Solution for Test: Firing Circuits - Question 3

 The driver circuit consists of a pulse amplifier to increase the magnitude of the gate pulse to a sufficient value. The pulse transformer then provides pulses to individual SCRs.

Test: Firing Circuits - Question 4

Find the average gate power dissipation (Pgav) when the maximum allowable gate power dissipation (Pgm) = 10 kW, with a duty cycle = 50 %.

Detailed Solution for Test: Firing Circuits - Question 4

(Pgm) = (Pgav)/Duty Cycle.

Test: Firing Circuits - Question 5

 The magnitude of gate voltage and gate current for triggering an SCR is

Detailed Solution for Test: Firing Circuits - Question 5

Higher the temperature lesser will be the gate current required as the temperature must have already excited some of the atoms.

Test: Firing Circuits - Question 6

Find the amplitude of the gate current pulse, when the gate-cathode curve is given by the relation Vg = [(1+10) x Ig] The peak gate drive power is 5 Watts.

Detailed Solution for Test: Firing Circuits - Question 6

(1+10 Ig).Ig = 5 Watts
Ig = 0.59 A.

Test: Firing Circuits - Question 7

The gate-cathode curve for an SCR is given by the relation Vg = (1+10)Ig. The gate voltage source is a rectangular pulse of peak value 15 V and current = 0.659 A. Find the source resistance.

Detailed Solution for Test: Firing Circuits - Question 7

Es = Rs.Ig + Vg
Vg = 1+10 Ig
Therefore Rs = (15-1)/0.659.

Test: Firing Circuits - Question 8

Find the triggering frequency when the average gate power dissipation = 0.3 W and the peak gate drive power is 5 Watts. The gate source has a pulse width of 20 μsec duration.

Detailed Solution for Test: Firing Circuits - Question 8

(Pgm) = (Pgav)/Duty Cycle
Duty Cycle = f x T = (Pgav)/(Pgm)
Duty Cycle = 0.3/5
T = 20 μsec.
0.3/5 = f x T
f = (0.3)/(5 x 20 x 10-6) = 3000 Hz.

Test: Firing Circuits - Question 9

The duty cycle can be written as

Detailed Solution for Test: Firing Circuits - Question 9

The duty cycle is defined as the ratio of pulse-on period to periodic time of pulse.
The pulse on period is T, and the periodic time is 1/f.
It is to be noted that T = pulse width whereas f = (1/T1) = frequency of firing or pulse repetition rate.

Test: Firing Circuits - Question 10

The major function of the pulse transformer is to

Detailed Solution for Test: Firing Circuits - Question 10

Isolation of the two circuit is done by the transformer, as the transformer is a magnetically coupled device and any mishap at the load side will not damage the other side of the circuitry.

Test: Firing Circuits - Question 11

In a resistance firing circuit the firing angle

Detailed Solution for Test: Firing Circuits - Question 11

The R firing circuits cannot be used for alpha greater than 90 degrees.

Test: Firing Circuits - Question 12

For a R firing circuit, the maximum value of source voltage is 100 V. Find the resistance to be inserted to limit the gate current to 2 A.

Detailed Solution for Test: Firing Circuits - Question 12

R = 100/2 = 50 Ohm.

Test: Firing Circuits - Question 13

The diode in the R firing circuit

Detailed Solution for Test: Firing Circuits - Question 13

The diode is placed between the resistances and gate which ensures that the current flows in one direction only.

Test: Firing Circuits - Question 14

In case of an RC half wave triggering circuit, the firing angle can be ideally varied between

Detailed Solution for Test: Firing Circuits - Question 14

Unlike the R firing circuit, the RC firing circuits can be used to obtain firing angle greater than 180. Although practically 0 and 180 degree is improbable.

Test: Firing Circuits - Question 15

 In the figure given below, the resistance R1 is used to​

Detailed Solution for Test: Firing Circuits - Question 15

As R2 is the variable, R1 makes sure that the current does not exceed the maximum value when R2 is kept at zero position.

Test: Firing Circuits - Question 16

In case of a R firing with R2 as the variable resistance, Vgp (peak of gate voltage) and Vgt(gate triggering voltage) the value of R2 is so adjusted such that

Detailed Solution for Test: Firing Circuits - Question 16

For turning on the device, the peak of gate voltage must be equal to the gate triggering voltage.

Test: Firing Circuits - Question 17

In case of a R firing circuit with Vgp > Vgt

Detailed Solution for Test: Firing Circuits - Question 17

For the values of Vgp great than the gate triggering voltage the firing angle is less than 90°. And for Vgp = Vgt the firing angle is equal to 90°. Α cannot go beyond 90° in case of a R firing circuit.

Test: Firing Circuits - Question 18

The figure shown below is that of a

Detailed Solution for Test: Firing Circuits - Question 18

The given circuit is a RC half-wave firing circuit.

Test: Firing Circuits - Question 19

The figure shown below is that of an RC firing circuit.

In case of negative cycle at Vs, the capacitor C

Detailed Solution for Test: Firing Circuits - Question 19

The current flows through Vs+ – C – D2 – Load – Vs.

Test: Firing Circuits - Question 20

Find the value of R in case of an RC firing circuit which is to be turned on with a source voltage of 150 V and the following parameters.
Igt = 2A
Vd = 1.5V
Vgt = 125 V

Detailed Solution for Test: Firing Circuits - Question 20

R = (Vs-Vgt-Vd)/Igt.

Test: Firing Circuits - Question 21

 For the following RC triggering circuit with R load and a firing angle of α, the voltage across the R load is zero for

Detailed Solution for Test: Firing Circuits - Question 21

 The SCR is triggered at α therefore voltage appears across the load from ωt = α to π only as in case of a half wave firing circuit the SCR is naturally commuted at π.

Test: Firing Circuits - Question 22

For the following circuit, the SCR (thy2) is

Detailed Solution for Test: Firing Circuits - Question 22

Circuit is that of a half-wave RC firing circuit where natural commutation takes place at π due to the sinusoidal voltage source.

Test: Firing Circuits - Question 23

For the following RC triggering circuit with R load and a firing angle of α, the voltage across the SCR (thy2) will be zero for

Detailed Solution for Test: Firing Circuits - Question 23

 The SCR is triggered at α therefore voltage appears across the load from ωt = α to π only as in case of a half wave firing circuit the SCR is naturally commuted at π.

Test: Firing Circuits - Question 24

 In case of an RC full wave firing circuit with R load, the voltage across the load is zero for____________

Detailed Solution for Test: Firing Circuits - Question 24

 The SCR is not triggered(turned-on) until α, hence no current flows until α.

Test: Firing Circuits - Question 25

 For an RC full wave firing circuit the empirical formula for calculating the value of RC is

Detailed Solution for Test: Firing Circuits - Question 25

 RC = 157/ω, Where ω is the angular frequency.

Test: Firing Circuits - Question 26

Pulse triggering can be only used by the _____________ type of triggering circuit

Detailed Solution for Test: Firing Circuits - Question 26

R & RC produce prolonged pulses which increases the gate power dissipation. There is no RLC firing circuit.

Test: Firing Circuits - Question 27

The UJT terminals are

Detailed Solution for Test: Firing Circuits - Question 27

The terminals are Emitter, Base1 & Base 2.

Test: Firing Circuits - Question 28

In case of the UJT firing circuit, when the UJT turns on

Detailed Solution for Test: Firing Circuits - Question 28

When the UJT is fired, C starts to discharge.

Test: Firing Circuits - Question 29

In the UJT firing circuit, the pulses are generated while the

Detailed Solution for Test: Firing Circuits - Question 29

When the C starts to discharge through the resistance with time constant RC the UJT turns on & hence pulses are generated.

Test: Firing Circuits - Question 30

Find the value of the charging resistor in case of a UJT firing circuit with firing frequency of 2 kHz, C = 0.04 μF, η = 0.72

Detailed Solution for Test: Firing Circuits - Question 30

 T = 1/f = 1/2 kHz
R = T/C ln(1/1-η).

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