Test: Firing Circuits
30 Questions MCQ Test Power Electronics | Test: Firing Circuits
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In a single pulse semi-converter using two SCRs, the triggering circuit must produce
A single phase semi-converter has only two SCRs & two diodes. Hence, only two pulses are required in each cycle, one in each half.
In a 3-phase full converter using six SCRs, gating circuit must provide
60° x 6(devices) = 360°.
In the complete firing circuit, the driver circuit consists of
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.
Find the average gate power dissipation (Pgav) when the maximum allowable gate power dissipation (Pgm) = 10 kW, with a duty cycle = 50 %.
(Pgm) = (Pgav)/Duty Cycle.
The magnitude of gate voltage and gate current for triggering an SCR is
Higher the temperature lesser will be the gate current required as the temperature must have already excited some of the atoms.
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.
(1+10 Ig).Ig = 5 Watts
Ig = 0.59 A.
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.
Es = Rs.Ig + Vg
Vg = 1+10 Ig
Therefore Rs = (15-1)/0.659.
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.
(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.
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.
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.
The R firing circuits cannot be used for alpha greater than 90 degrees.
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.
R = 100/2 = 50 Ohm.
The diode is placed between the resistances and gate which ensures that the current flows in one direction only.
In case of an RC half wave triggering circuit, the firing angle can be ideally varied between
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.
As R2 is the variable, R1 makes sure that the current does not exceed the maximum value when R2 is kept at zero position.
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
For turning on the device, the peak of gate voltage must be equal to the gate triggering voltage.
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.
The given circuit is a RC half-wave firing circuit.
The figure shown below is that of an RC firing circuit.
In case of negative cycle at Vs, the capacitor C
The current flows through Vs+ – C – D2 – Load – Vs.
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
R = (Vs-Vgt-Vd)/Igt.
For the following RC triggering circuit with R load and a firing angle of α, the voltage across the R load is zero for
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.
Circuit is that of a half-wave RC firing circuit where natural commutation takes place at π due to the sinusoidal voltage source.
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
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.
In case of an RC full wave firing circuit with R load, the voltage across the load is zero for____________
The SCR is not triggered(turned-on) until α, hence no current flows until α.
For an RC full wave firing circuit the empirical formula for calculating the value of RC is
RC = 157/ω, Where ω is the angular frequency.
Pulse triggering can be only used by the _____________ type of triggering circuit
R & RC produce prolonged pulses which increases the gate power dissipation. There is no RLC firing circuit.
The terminals are Emitter, Base1 & Base 2.
When the UJT is fired, C starts to discharge.
In the UJT firing circuit, the pulses are generated while the
When the C starts to discharge through the resistance with time constant RC the UJT turns on & hence pulses are generated.
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
T = 1/f = 1/2 kHz
R = T/C ln(1/1-η).
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