Test: Three Phase Converters - 1 - Electrical Engineering (EE) MCQ

It uses three SCRs with a three-phase transformer. M-3 type 3-pulse converters are not practically used.

 In a three phase controller, the actually conduction starts at 30° + α. Hence, ωt = 30+15 = 45°.

In three-phase three pulse converter the conduction can start only after 30°. As each SCR conducts from 120°, T2 would conduct on 30+120+n° = 150+n° and so on.

Each conduct for 120° per cycle is the firing angle is less than 30°. 120 x 3 = 360°.

The circuit is that of a three-pulse M-3 connection. The firing angle is less than 30°. Therefore, each device conducts for an angle of 120°.
Vo = 3 x [ 1/2π ∫ Vmp sin⁡ωt d(ωt) ] Where, the integration runs from α+π/6 to α+5π/6.
Vo = (3√3/2π) x Vmp x cosα.

When firing angle is more than 30°, T1 would conduct from 30 + α to 180°. Irrespective of the firing angle, T1 will be turned on at 180° because it conducts first which means it is connected to the R phase and the phase sequence is R-Y-B. As R starts at 0° its value is 0 at 180° which reverse biases the SCR T1.

Vo = 3 x [ 1/2π ∫ Vmp sin⁡ωt d(ωt) ] Where, the integration runs from α+π/6 to π. Because conduction takes place from 30 + α to 180° for T1 and than the waveform is symmetrical for all other SCRs.
Vo = (3√3/2π) x Vmp x [1+cos(30+α)].

The output voltage is proportional to cosine of the firing angle α. If α goes above 90° then the output voltage is negative, meaning that it is operating as an inverter.

We need , Vo = 0.5 Vom.
α>30° hence we use the equation Vo = (3√3/2π) x Vmp x [1+cos(30+α)] √3Vmp = Vml = √2×230
Therefore, Vo = (3/2π) x √3 Vml x [1+cos(30+α)] = 0.5Vom
(1/√3) x [1+cos(30+α)] = Vo x 2π/3Vml = Vo/Vom = 1/2
α = 67.7°.

Vo = (3√3/2π) x Vmp x cosα = 233.874 V.
Ia = Io/3 = 12 A.

 Three legs having two SCRs each, six in total.

When α is less than 90°, the SCRs conduct for 120° and the current and voltage are positive on an average hence, the power flows from AC source to DC load.

Assuming the phase sequence is R-Y-B. T1 would start conducting at 30+60 = 90°, T2 at 90+210/2 = 150°. This is because after T1, T3 would conduct from the upper group, as T2 belongs to the lower group it will start to conduct exactly between T1 and T3 i.e. between 90 and 210(90+120) which is 150°.

 Let’s say T1, T3 and T5 belong to the positive group and T2, T4 and T6 to the negative group. At any given time one SCR from each group conducts. e.g. T1 and T6 or T1 and T2.

Every SCR conducts for 120°. This means that the SCRs from the positive group are fired 120° among themselves, same is true for SCRs from negative group. For example, if T1 starts conducting at 90° it will conduct till 90+120 = 210°. But while T1 is conducting, half of the time i.e. from 90 to 150, T6 is conducting and another half of the time T2 is conducting. Hence, commutation (change in the SCR which is conducting) takes place every 60 degrees irrespective of the firing angle. Construct the firing sequence table for better understanding.

The negative group of SCRs has T2, T4 and T6. The conduct as T2-T4-T6, as T2 is connected to the B phase, T4 to the R phase and like-wise.

For α = 90 degrees, the voltage waveform is equally symmetrical about the ωt axis, hence the average value is zero. This can also be found by using the formula for average output voltage,
Vo = (3Vml/π) cos α,
For α = 90°, cosα = 0, Vo = zero.

Vo = 210 V
Vo = E + Io x R
210 = 200 + 0.5 x Io
Io (Current through the battery) = 20 A.

Vo = (3Vml/π) cos α
α = cos^-1(210π/3√2×230) = 47.453°.

 Iavg = Irms = 20 A
P = E x Iavg + Irms² x R = 4200 W.