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Inverter Circuit

Discussion in 'General Electronics Discussion' started by vinod chandran, Nov 14, 2012.

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  1. vinod chandran

    vinod chandran

    192
    2
    Jun 21, 2011
    Hi all,
    I built this circuit today(please look attached image) and is working superb. My battery 12v 7Ah. I want to build an inverter like this for my friend. But he need 150w output. This is only 60w. What changes should i made to obtain 150w with this circuit. Is there any formula to calculate output capacity and battery Ah?. I mean when we need 200w output, then calculate with a formula and get transformer ampere and battery Ah as result . What is the criteria for choosing right mosfets ?.
    -Vinod chandran.
     

    Attached Files:

  2. duke37

    duke37

    5,364
    769
    Jan 9, 2011
    200W output will need 17A or so input, the IRF540 fets should be able to handle this but may need a heat sink. Chose a fet with low on resistance and sufficient voltage capability.

    The transformer will need to be of sufficient size to handle the current.

    If you are taking 20A then a 20AHr battery will last less than 1Hr.
     
  3. vinod chandran

    vinod chandran

    192
    2
    Jun 21, 2011
    Hi duke37,
    Thank you for the reply. On resistance of IRF540 is 44 Mega ohms. What is the benefit from low on resistance ?. And how to calculate these parts?. I have a circuit with CD4047 which i successfully completed. All i want some formulas to calculate various parts for various output load. Is this possible?.
     
  4. gorgon

    gorgon

    603
    24
    Jun 6, 2011
    Hopefully it is 44 milliohm, not megaohm. To calculate the voltage drop over the IRF, multiply the resistance with the current through it. If you multiply again with the current you will get the power loss in the IRF. If it is activated only half the time, you can divide by 2.
     
    Last edited: Nov 15, 2012
  5. duke37

    duke37

    5,364
    769
    Jan 9, 2011
    I have a scrap of paper where I have jotted down the on resistance of a IRF540 as 77mohm. Perhaps I am wrong!

    Taking 17A as the current, the voltage drop = 17 * .077 = 1.3V this is lost voltage.
    The power in the fet = 17 * 1.3 = 22W, this is lost power. You can divide this by 2 as said but each fet will need to disipate more than 10W.

    The transformer winding is in series with the fet and will drop voltage and dissipate power in a similar way.

    Obviously, the lower the resistances, the less the power loss and the better the regulation.

    M = Mega
    m = milli
     
    Last edited: Nov 15, 2012
  6. vinod chandran

    vinod chandran

    192
    2
    Jun 21, 2011
    Hi, gorgon and duke37,
    That was my mistake. the resistance is 44 milliohm. Sorry for that. What i understand from your answers is - i need to replace the mosfet, transformer and battery for more output load. That means i don't need to change the basic circuit. Am i right ?.
     
    Last edited: Nov 15, 2012
  7. duke37

    duke37

    5,364
    769
    Jan 9, 2011
    You do not need to change the circuit or the fets. You could use a fet with a lower on resistance or place two in parallel to reduce the power loss. The transformer should be large enough to have a low resistance.

    A fet can dissipate about 0.5W without a heat sink but in this case you wiil need heatsinks.
     
  8. vinod chandran

    vinod chandran

    192
    2
    Jun 21, 2011
    Hi duke37,
    Thank you very much.
     
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