# Firing angle in rectifiers and inverters.

1. Aug 16, 2011

### ramox3

Hello everyone I am a newbie to electrical engineering, I have this problem I've been strugling with, no textbook seems to cover this problem..

A three-phase, 440-V, generator delivers 5400W of active power through two recti-fies, A and B, into a three-phase, 230-V, line as shown in Figure 3.
The dc current through the inductor is 20A. If power loss in both rectifiers are negligible, calculate:

(a). Firing angle of the rectifier A;
(b). Firing angle of the rectifier (inverter) B.

This is my attempt ;

V out = 5400/ 20 = 270

VOUT = 1.35 × VLINE × cos α

Where ‘α’ is the firing angle of the rectifier.

Therefore,

cos α = V / 1.35 x Vline

α = 29

how does this seem? and how is the inverter's firing angle different?

2. Aug 17, 2011

### ramox3

1. The problem statement, all variables and given/known data

Hello everyone I am a newbie to electrical engineering, I have this problem I've been strugling with, no textbook seems to cover this problem..

A three-phase, 440-V, generator delivers 5400W of active power through two recti-fies, A and B, into a three-phase, 230-V, line as shown in Figure 3.
The dc current through the inductor is 20A. If power loss in both rectifiers are negligible, calculate:

(a). Firing angle of the rectifier A;
(b). Firing angle of the rectifier (inverter) B.

2. Relevant equations

3. The attempt at a solutionThis is my attempt ;

V out = 5400/ 20 = 270

VOUT = 1.35 × VLINE × cos α

Where ‘α’ is the firing angle of the rectifier.

Therefore,

cos α = V / 1.35 x Vline

α = 29

how does this seem? and how is the inverter's firing angle different?

3. Aug 18, 2011

### Staff: Mentor

Where is Fig 3?

4. Aug 18, 2011

### ramox3

5. Aug 18, 2011

### uart

Seems ok. And you can use the same formula (with the relevant line voltage) for the inverter side.

BTW The constant 1.35 comes from 3 sqrt(2) / pi.

6. Aug 18, 2011

### ramox3

but what is the relevent line voltage for the inverter? isn't it the same?

7. Aug 19, 2011

### Mbert

Not sure this is the right place to post this (perhaps this is why no one answers, but I don't know I'm new to PF). Here's some help:

In fact, there are a lot of text books that cover this. For instance see "Power electronics", from Mohan. I can't see your Figure 3, but I assume you have a perfect voltage source connected to 2 full-bridge thyristor converters (1 rectifier, 1 inverter) and some load. There should also be an inductor on the DC bridge. To find the solution for rectifier A (source):

$$P=V_{DC}I_{DC}$$

where the voltage on the DC bus is given by:

$$V_{DC}=\frac{3\sqrt{2}}{\pi}V_{LL}\cos{\alpha}$$

where $V_{LL}$ is the line-line voltage of the AC side. The DC bus current is given in the problem, which is $I_{DC}$. Just isolate these for $\alpha$...

For rectifier (inverter) B, same procedure, but by using a negative power and use the 230V line-line voltage. Isolate again for $\alpha$... This angle should be between 90 and 180 degrees since this converter is operating in inverter mode.

M.

8. Aug 19, 2011

### Staff: Mentor

(Moderator's note -- two threads merged...)