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Designing a DC link Capacitor

Discussion in 'Electronics Homework Help' started by diya0908, Feb 27, 2013.

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  1. diya0908

    diya0908

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    Feb 27, 2013
    hi, I would appreciate is some one could help me with this question by giving me clues about what formula i could use and how to proceed.

    Ques:

    The DC supply for the voltage source inverter of the VFD is derived using six pulse diode converter.
    Supply voltage is 600V, 3phase, 50Hz. Xdā€™ā€™=12.5 pu ,X/R ratio as 20. Model the load seen by the DC link capacitor as pure resistor. Take rated load current as 900A.
    (i) Design DC link capacitor to get DC voltage with 10% ripple at rated load. (Hint : use simplified formula).
    (ii) Simulate the above system. (Assume suitably any missing data)
    (iii) Draw the circuit diagram.


    P.S: PLEASE HELP ME WITH CLUES TO SOLVE THE PROBLEM
     
  2. Miguel Lopez

    Miguel Lopez

    252
    64
    Jan 25, 2012
    Do you mean a three phase bridge rectifier supplying a pure resistive load? 600V - 900A? Do you need to calculate the smoothing capacitor for 10% ripple?
     
  3. diya0908

    diya0908

    7
    0
    Feb 27, 2013
    yes sir, i myself am struggling to understand the question, anyhow I feel it is a 3phase six pulse diode rectifier supplying a resistive load and having a supply voltage of 600V.

    the first part of the quest asks to design a capacitor, but where will it be placed?? and how do we calculate it?
     
  4. diya0908

    diya0908

    7
    0
    Feb 27, 2013
    [​IMG]


    Ok so now i get it, the dc link capacitor is placed after the rectifier. But the "load seen by the DC link capacitor is a pur resistor"-- could you explain me what does this mean.
     
  5. Miguel Lopez

    Miguel Lopez

    252
    64
    Jan 25, 2012
    I guess we're talking about something like this. The value of the capacitor is not correct. I just wrote any number. Anyway in this case it means that you will supply a load of only 1,55 ohms. I think the 900A value is too high.

    Althought the 3 phase bridge rectifier gives a decent DC level at the output, to have 10% of ripple it needs a huge amount of microfarads. You should check this:

    http://en.wikipedia.org/wiki/Rectifier#Three-phase_bridge_rectifier
    http://en.wikipedia.org/wiki/Ripple_(electrical)
    http://www.google.com/url?sa=t&rct=...mYD4Bg&usg=AFQjCNFcx4c_dSXE3OkBoNM1ioBD9bM-eg

    It means that voltage and current will be in phase. No reactive effect (inductive or capacitive). The ripple can cause the reactances to behave different to a pure resistor.
    http://en.wikipedia.org/wiki/Electrical_reactance
     

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    Last edited: Feb 27, 2013
  6. diya0908

    diya0908

    7
    0
    Feb 27, 2013
    yes sir, you are correct, the value of current is too high and the value of resistance is therefore low. I have gone through the links and i believe now i know how to find the value of the DC link Capacitor (C=I/Vf)

    now my question is what does Xd''=12.5 and X/R ratio mean??
     
  7. diya0908

    diya0908

    7
    0
    Feb 27, 2013
    Also i would like to know how did you calculate the value of resistance in your diagram
     
  8. Miguel Lopez

    Miguel Lopez

    252
    64
    Jan 25, 2012
    You can use Ohm's Law, but to be honest, in the diagram I used "trial and error" in the simulation software.

    R = V / I

    http://en.wikipedia.org/wiki/Ohm's_law
     
  9. diya0908

    diya0908

    7
    0
    Feb 27, 2013
    Could you tell me which program you have used to simulate the circuit.
     
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