Power electronics and drives miscellaneous Easy Questions and Answers | Page - 7

Power electronics and drives miscellaneous

A single - phase fully controlled thyristor bridge ac - dc converter is operating at a firing angle of 25° and an overlap angle 10° with constant dc output current of 20 A. The fundamental power factor (displacement factor) at input ac mains is

1. 0.78
1. 0.827
1. 0.866
1. 0.9

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I₀ = V_m [cosα - cos(α + μ)]
ωL_s

∴ 20 = 2 × √2 [cos25° - cos(25 + 10)]
2 × π × 100L_s

L_s = 0.0045 Henery
V_o = 2 × √2 × 230
π

cos25° - 2 × π × 50 × 0.0045 × 20 = 178.65
π

∴ Displacement factor = V_o I_o
V_s I_s

= 178.65 × 20 = 0.78
230 20

Correct Option: A

I₀ = V_m [cosα - cos(α + μ)]
ωL_s

∴ 20 = 2 × √2 [cos25° - cos(25 + 10)]
2 × π × 100L_s

L_s = 0.0045 Henery
V_o = 2 × √2 × 230
π

cos25° - 2 × π × 50 × 0.0045 × 20 = 178.65
π

∴ Displacement factor = V_o I_o
V_s I_s

= 178.65 × 20 = 0.78
230 20

A single phase full-wave half-controlled bridge converter feeds an inductive load. The two SCRs in the converter are connected to a common DC bus. The converter has to have a freewheeling diode

1. because the converter inherently does not provide for free-wheeling
1. because the converter does not provide for freewheeling for high values of triggering angles
1. or else the free-wheeling action of the converter will cause shorting of the AC supply
1. or else if a gate pulse to one of the SCRs is missed, it will subsequently cause a high load current in the other SCR

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Fig. Single-phase bridge type semi-converter: (a) circuit diagram, and (b) waveforms.
During the free-wheel period, π to (π + α), and 2π to (2π + α), energy stored in Inductance during (π + α) to π and α to π, gets dissipated partially in R and partially to the energy stored in load emf E. No energy is fed back to the source during free-wheeling period.
Therefore, in the absence of free-wheeling diode, T₁ and T₂ may conduct simultaneously which will came short-circuit of supply voltage V_S.

Correct Option: B

Fig. Single-phase bridge type semi-converter: (a) circuit diagram, and (b) waveforms.
During the free-wheel period, π to (π + α), and 2π to (2π + α), energy stored in Inductance during (π + α) to π and α to π, gets dissipated partially in R and partially to the energy stored in load emf E. No energy is fed back to the source during free-wheeling period.
Therefore, in the absence of free-wheeling diode, T₁ and T₂ may conduct simultaneously which will came short-circuit of supply voltage V_S.

A three - phase, fully controlled thyristor bridge converter is used as line commutated inverter to feed 50 kW power at 420 V dc to a three - phase, 415 V (line), 50 Hz ac mains. Consider dc link current to be constant. The rms current of the thyristor is

1. 119.05 A
1. 79.37 A
1. 68.73 A
1. 39.68 A

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P = V_d × I_d
∴ 50 × 1000 = 420 × I_d , I_d = 119.05
rms current of the thyristor = 119.05 = 39.68
3

Correct Option: D

P = V_d × I_d
∴ 50 × 1000 = 420 × I_d , I_d = 119.05
rms current of the thyristor = 119.05 = 39.68
3

“Six MOSFETs connected in a bridge configuration (having no other power device) MUST be operated as a Voltage Source Inverter (VSI)”. This statement is

1. True, because being majority carier devices, MOSFETs are voltage driven
1. True, because MOSFETs have inherently anti - parallel diodes
1. False, because it can be operated both as Current Source Inverter (CSI) or a VSI
1. False, because MOSFETs can be operated as excellent constant current sources in the saturation region

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NA

Correct Option: D

NA

In the circuit shown in the figure given below, the switch is operated at a duty cycle of 0.5. A large capacitor is connected across the load. The inductor current is assumed to be continuous.

The average voltage across the load and the average current through the diode will respectively be

1. 10 V, 2 A
1. 10 V, 8 A
1. 40 V, 2 A
1. 40 V, 8 A

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The current given is a step-up circuit and output voltage is given as,
V_o = V_s = 20 = 40 Volts
1 - α 1 - 0.5

and, average current through the diode

= I_L (1 – α) = 4 × (1 – 0.5) = 2 A

Correct Option: A

The current given is a step-up circuit and output voltage is given as,
V_o = V_s = 20 = 40 Volts
1 - α 1 - 0.5

and, average current through the diode

= I_L (1 – α) = 4 × (1 – 0.5) = 2 A