Test: Three Phase Converters - 2 - Electrical Engineering (EE) MCQ

Vo = 200 – 10×5 = 150V as the circuit is operating as an inverter Vo = -150V
Now, Vo = (3Vml/π) cos α
α = cos^-1(-150π/3√2×230) = 118.88°.

The RMS value of the supply current I_S over π radians would be
(I_S)² = (1/π) x (Im)² x (2π/3) = Im√2/√3.

 It uses 3 SCRs and 3 diodes, hence it is a semi-converter. Option (b) and (d) make no sense, because there can be no full controlled semi-converter.

 At any given time, one SCR and one diode is conducting, each conduct for 120° per cycle.

When one SCR conducts, a diode conducts along with it at a time to provide the path of current flow. . For example, if T2 starts conducting at 90° it will conduct till 90+120 = 210°. But while T2 is conducting, half of the time i.e. from 90 to 150 D1 is conducting and another half of the time D3 is conducting. T2 and D2 cannot conduct together as it will cause a short circuit. Hence, T2-D1 conduct for 60° and then T2-D3 conduct for another 60°.

For 3 – φ semi converter, for α = 125^o , thyristor and diode conducts for 55^o only. Freewheeling diode conducts for 105o – 55^o = 50^o .

For maximum output at a time two phases conducts and each set conducts for 360/6 and each diode 360/3 degree

 Ior = Vo/R = (3Vml/Rπ) cos α
P = 5 kW = Ior² x R = [(3Vml/π) cos α]² x 1/R] Therefore, Vs (line) = √50000 x (π/√2 x 3 x cos30) = 191.22 V
Vs (phase) = 191/√3 V.

Rectification efficiency = Pdc/Pac
Pdc = Vo x Io . . . (both average values of output current and voltage)
Pac = Vrms x Irms . . . (both rms values of input current and voltage)
For a semi-converter Irms = Io
Therefore, Rectification efficiency = 381/400 = 95.25 %.

 In case of a semi-converter operating with α < 60°, FD does not comes into play, as the voltage never falls to zero and gives no chance for the inductor to discharge.

PIV = √3 Vml in case of a semi-converter.

Average output voltage of 3 – Φ full converter

 It reduces the average value of the output voltage by introducing a overlap delay μ.

Half wave converter would require delta-star transformer, the secondary winding star connection requires a neutral point.

V_min = V_m sin(90 - 15) = 0.9659 V_m
V_max = V_m sin90 = V_m.

In case of a 3 phase bridge converter, the maximum value of voltage at the output terminal is always 1.5 Vm.

 In case of M-3 type of connection, the devices have to handle more peak inverse voltage (PIV) than the 3-phase full converter which has 6 SCRs.

= 3555.55 W
Total power = 3 × 3555.55
= 10.67 KW

Cos 90 = 0.

Semi-converters are single quadrant converters, because the voltage and current can only be both positive due to the diodes connected.