This set of Power Electronics Multiple Choice Questions & Answers (MCQs) focuses on “Single Phase Semi-Converters-2”.
1. A single-phase symmetrical semi-converter employs
a) one SCR and one diode in each leg
b) two SCRs and two diodes in each leg
c) two SCRs in each leg
d) two diodes in each leg
Explanation: A symmetrical semi-converter will have one SCR and one diode in each leg. Two legs connected in parallel with each other having a symmetrical configuration.
2. A single-phase asymmetrical semi-converter employs
a) one SCR and one diode in each leg
b) two SCRs in one leg and two diodes in the other
c) two SCRs in both the legs
d) two diodes in both the legs
Explanation: An asymmetrical semi-converter will two SCRs in one leg and two diodes in the other.
Explanation: At ωt = π, D1 is reversed biased but T1 is still ON due to the nature of the load. Hence, the load current or inductor current flows from T1 and D2 and short circuiting the load terminals. Hence, due to the diode D2 freewheeling action takes place without even having a FD diode.
Explanation: For the given circuit, ωt = π-α. tc = π – α/ω
α = π/6 and ω = 100 … (given)
tc = (π – π/6)/100.
5. For the circuit shown below, T1 and T2 (from top T1 and T2) are both fired at an angle α. Then from ωt = π to ωt = π+α
a) None of the devices conduct
b) T1 and one diode will conduct
c) T2 and one diode will conduct
d) Both the diodes will conduct
Explanation: From ωt = π to ωt = π+α, None of the SCRs are fired and both the diodes are forward biased due to the nature of the load. As such, they start to conduct and freewheel all the inductor energy.
6. In any AC-DC circuit, the freewheeling action
a) improves the power handling capabilities
b) increases the THD
c) improves CDF
d) all of the mentioned
Explanation: Freewheeling action reduces THD (total harmonic distortion), Improves CDF (current distortion factor and has no effect over the power handling capabilities as such.
7. A single-phase semi-converter is connected to a 230 V source and is feeding a load R = 10 Ω in series with a large inductance that makes the load current ripple free. Find the average output current for α = 45°.
a) 14 A
b) 17 A
c) 10 A
d) 0 A
Explanation: Vo =( Vm/π) (1 + cosα)
Io = Vo/R.
8. A single-phase semi-converter is connected to a 230 V source and is feeding a load R = 10 Ω in series with a large inductance that makes the load current ripple free. For α = 45°, find the rectification efficiency. The RMS value of output voltage is 219.3 V
a) 96.54 %
b) 75.25 %
c) 89.45 %
d) 80.58 %
Explanation: Vo =( Vm/π) (1 + cosα) = 176.72V
Io = Vo/R
Irms = Io = Vo/R = 17.67 A
Vrms = 219.3 V (given)
η = (Vo x Io) / (Vrms X Irms) = 0.8058 = 80.58 %.
9. A single-phase semi-converter circuit is supplying power to a motor load. The average value of load voltage is 176.72 V and the rms value is 219.3 V. Find the VRF (voltage ripple factor).
Explanation: First find the form factor (FF),
FF = Vrms/Vo = 1.241
Now, VRF = √(FF2-1) = 0.735.
10. For the same triggering angle and ratings
a) a semi-converter operates at lower output voltage than a full converter
b) a semi-converter operates at higher output voltage than a full converter
c) a semi-converter has lower values of input p.f as compared to a full converter
d) a semi-converter has more THD as compared to a full converter
Explanation: A semi-converter gives more output voltage than an equivalent full converter circuit. It also has less THD and high p.f (power factor).
Sanfoundry Global Education & Learning Series – Power Electronics.
To practice all areas of Power Electronics, here is complete set of 1000+ Multiple Choice Questions and Answers.