Firing angle in rectifiers and inverters.

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In summary, 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. 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.
  • #1
ramox3
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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
For inductive load

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?
 
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  • #2

Homework Statement



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.

Homework Equations


The Attempt at a Solution

This is my attempt ;

V out = 5400/ 20 = 270
For inductive load

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
Where is Fig 3?
 
  • #4
here's the figure :D

http://imageshack.us/photo/my-images/69/figure3t.png
 
  • #5
ramox3 said:

Homework Statement



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.

Homework Equations


The Attempt at a Solution

This is my attempt ;

V out = 5400/ 20 = 270
For inductive load

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?

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
uart said:
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.

but what is the relevant line voltage for the inverter? isn't it the same?
 
  • #7
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 textbooks 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):

[tex]P=V_{DC}I_{DC}[/tex]

where the voltage on the DC bus is given by:

[tex]V_{DC}=\frac{3\sqrt{2}}{\pi}V_{LL}\cos{\alpha}[/tex]

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

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

M.
 
  • #8
(Moderator's note -- two threads merged...)
 

1. What is a firing angle in rectifiers and inverters?

The firing angle in rectifiers and inverters is the angle at which the thyristors or other switching devices are turned on to control the flow of current through the circuit. It determines the amount of voltage or current that is supplied to the load.

2. How is the firing angle determined in a rectifier or inverter circuit?

The firing angle is determined by the control signal that is sent to the switching devices. This control signal is usually a pulse or a square wave that is synchronized with the AC input signal. The timing and duration of this signal can be adjusted to control the firing angle.

3. What is the significance of the firing angle in rectifiers and inverters?

The firing angle plays a crucial role in controlling the output voltage or current in a rectifier or inverter circuit. By adjusting the firing angle, the amount of power delivered to the load can be controlled, which can be useful for regulating the speed of motors or the brightness of lights.

4. What factors affect the firing angle in rectifiers and inverters?

The firing angle can be affected by a variety of factors, including the type and size of the switching devices, the input voltage and frequency, the load resistance, and the control signal. Changes in any of these factors can alter the firing angle and consequently, the output voltage or current.

5. How can the firing angle be optimized for efficient operation of a rectifier or inverter?

The firing angle can be optimized by carefully selecting the control signal and adjusting the timing and duration of the pulses. It is also important to consider the characteristics of the load and the input power source. Additionally, using more advanced control techniques such as pulse width modulation can help improve the efficiency of the circuit.

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