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Homework Help: Firing angle in rectifiers and inverters.

  1. Aug 16, 2011 #1
    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?
     
  2. jcsd
  3. Aug 17, 2011 #2
    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
    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?
     
  4. Aug 18, 2011 #3

    NascentOxygen

    User Avatar

    Staff: Mentor

    Where is Fig 3?
     
  5. Aug 18, 2011 #4
  6. Aug 18, 2011 #5

    uart

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    Science Advisor

    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.
     
  7. Aug 18, 2011 #6
    but what is the relevent line voltage for the inverter? isn't it the same?
     
  8. Aug 19, 2011 #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 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):

    [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.
     
  9. Aug 19, 2011 #8

    berkeman

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    Staff: Mentor

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