Input current Calculation

Discussion in 'General Electronics Chat' started by Tahsina Hossain, Mar 11, 2015.

  1. Tahsina Hossain

    Thread Starter Member

    Jan 16, 2015
    40
    0
    Hi there kind people,

    I have developed an inverter of my own, now I want to measure the efficiency of my inverter, the formula I used for my simulations circuit is : η=(rms(I(R1))*rms(I(R1)*R1)/avg(V(Vdc:+)*abs(I(Vdc))),

    It might sound really naive but I am really confused on how to calculate my input current, I tried using a 1ohm 1W resistor to measure the input power and that only gave me around .635 amps.

    Can anyone please enlighten me how I can measure the input current properly. Thanks in advance!!

    P.s. the dc power supply I am using is 19v for now (will go upto 300 volt later) and it's showing me around .203 amps in the display of the dc power supply.
     
  2. crutschow

    Expert

    Mar 14, 2008
    12,993
    3,229
    Assuming the power supply meter is reasonably accurate then you input power would be .203A x 19V = 3.86W.
    To measure the current with a multimeter, you put the ammeter connections in series with the power.
     
    Roderick Young likes this.
  3. Roderick Young

    Member

    Feb 22, 2015
    408
    168
    In simulation, it looks like you're measuring the right things, although I'm a little confused as to why there is an avg() and abs() in the denominator of the fraction.

    On an actual physical circuit, putting a resistor in series with the positive supply and measuring the voltage drop across the resistor is a reasonable way to measure current, what I would call a sense resistor. It's best to use as low a value of sense resistor as practical, to avoid disturbing the circuit, of course. If you can find a 50 milliohm or 10 milliohm sense resistor, that would be better, if you have a voltmeter or scope that will measure millivolts. And my apologies if I'm stating the obvious, but you want the voltage difference across the resistor, so either must put your meter across the resistor, or measure the voltage to ground at each end, and take the difference of the two voltages for your calculation.
     
  4. uwed

    Member

    Mar 16, 2015
    64
    17
    If the converter efficiency is high (e.g. 95% and more), then measuring inaccurracies of current and voltage at input and output will start to distort the system efficiency significantly. If you simulate, make sure that switching losses are included, they make a large part of the converter losses. Some for transformer losses (if there is a transformer).

    efficiency_of_inverter= 1 - Pout/Pin

    Is R1 your output-load resistance? It looks like you forgot the "1-" in your equation (?)
     
  5. uwed

    Member

    Mar 16, 2015
    64
    17
    Sorry, I made a mistake. System efficiency is of course

    efficiency_of_inverter= Pout/Pin
     
  6. Tahsina Hossain

    Thread Starter Member

    Jan 16, 2015
    40
    0
    Dear All,
    I tried all of the procedures that you suggested still I am not confident about my calculations. I will post the full calculation later today then maybe you can tell me what silly mistake I am making.

    Thanks a lot again for all the help, I really appreciate how this forum people has always helped me in my dilemma time.

    Tahsina
     
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