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Measuring Current From AC/DC Adapter

Discussion in 'General Electronics Discussion' started by XenonKilla, Nov 9, 2013.

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  1. XenonKilla

    XenonKilla

    2
    0
    Nov 9, 2013
    I've got an AC/DC Adapter that plugs into the wall that puts out 12v 3A and connects to a small DVR box. Well I wanted to know how much DC current the DVR box actually draws so I hooked up a multimeter on the DC side and the device was pulling about 1.5amps. I also have one of those "Kill A Watt EZ" meters that you plug into your wall and then plug your device into it and it will tell you how much current, voltage, watts etc that your device is pulling. When I used this device it gave me a 0.34amp reading while I was still getting 1.5amps from my multimeter.

    So my question is, is there a way to convert the AC amps into DC amps so that way if I wanted to test another device, I could simply plug that device into the Kill A Watt EZ and then convert the amp value into the DC amp value, rather than having to measure on the DC side using a multimeter.

    Here are the numbers I was getting...

    Multimeter on the DC side
    1.5amps

    Kill A Watt EZ on the AC side
    0.34amps
    122.5v
    20.5watts
    0.49pf

    Thanks in advance!
     
    Last edited: Nov 9, 2013
  2. duke37

    duke37

    5,364
    769
    Jan 9, 2011
    It looks promising. You are providing 18W and consuming 20.5W so the PSU is quite efficient.

    I would do the comparison using power rather than current because of the complication of the power factor.
     
  3. XenonKilla

    XenonKilla

    2
    0
    Nov 9, 2013
    I'm not wanting to compare the two, I'm just wondering if there is a way to calculate the DC amps by only knowing the AC amps value. Judging by your answer I'm assuming it's probably just best to measure AC and DC separate to get the correct values.....?
     
  4. duke37

    duke37

    5,364
    769
    Jan 9, 2011
    Your power supply seems to be 88% efficient.
    You measure the input power, multiply by 0.88 to get output power.
    W = V*I so I = W/12
    Putting it all together, I = inputP*0.88/12 or I = inputP*0.073

    Alternatively, the ratio of the currents is 4.4, so multiply the input current by 4.4.

    Try both methods with a variety of loads.
     
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