Power electronics and drives miscellaneous
- A three-phase current source inverter used for the speed control of an induction motor is to be realized using MOSFET switches as shown below. Switches S1 to S6 are identical switches.
The proper configuration for realizing switches S1 to S6 is
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NA
Correct Option: C
NA
- In the circuit shown, an ideal switch S is operated at 100 kHz with a duty ratio of 50%. Given that ∆ic is 1.6 A peak-to-peak and I0 is 5 A dc, the peak current in S is
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The size in inductor current during ON period.∆IL = Vs - V0 TON , TON = L∆L L Vs - V0
and during OFF - period-∆IL = - V0 TOFF , TOFF = L∆L T V0
Switching frequency,f = 1 = 1 . V0(Vs - V0) TON + TOFF L∆IL Vs ⇒ ∆IL = VsK(1 - K) .K = V0 fL Vs
The maximum source current is given by,Imax = Io + ∆L = 5 + 1.6 = 5.8 A 2 2
Correct Option: C
The size in inductor current during ON period.∆IL = Vs - V0 TON , TON = L∆L L Vs - V0
and during OFF - period-∆IL = - V0 TOFF , TOFF = L∆L T V0
Switching frequency,f = 1 = 1 . V0(Vs - V0) TON + TOFF L∆IL Vs ⇒ ∆IL = VsK(1 - K) .K = V0 fL Vs
The maximum source current is given by,Imax = Io + ∆L = 5 + 1.6 = 5.8 A 2 2
- The typcial ratio of latching current to holding current in a 20 A thyristor is
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NA
Correct Option: B
NA
- A half-controlled single-phase bridge rectifier is supplying an R-L load. It is operated at a firing angle α and the load current is continuous. The fr action of cycle that the freewheeling diode conducts is
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As seen in figure, FD works in the period of α radian, which is fraction of cycle = α π
Correct Option: D
As seen in figure, FD works in the period of α radian, which is fraction of cycle = α π
- The i-v characteristics of the diode in the circuit given below are
The current in the circuit is
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10 = (1000)i + v
= 1000(v - 0.7) + v 500
= 2(v – 0.7) + v = 2v – 1.4 + v = 3v – 1.4
⇒ 3v – 1.4 = 10
⇒ 3v = 10 + 1.4 = 11.4⇒ v = 11.4 = 3.8 3 ∴ i = v - 0.7 500 = 3.8 - 0.7 = 6.2 mA 500
Correct Option: D
10 = (1000)i + v
= 1000(v - 0.7) + v 500
= 2(v – 0.7) + v = 2v – 1.4 + v = 3v – 1.4
⇒ 3v – 1.4 = 10
⇒ 3v = 10 + 1.4 = 11.4⇒ v = 11.4 = 3.8 3 ∴ i = v - 0.7 500 = 3.8 - 0.7 = 6.2 mA 500