Regulated power supply Calculation

Discussion in 'The Projects Forum' started by Paras Bhanot, Mar 23, 2016.

  1. Paras Bhanot

    Thread Starter New Member

    Jun 22, 2015
    3
    0
    hello everyone ,

    i am working on home automation project in which Micro-controller (atmega328p ) is used . This micro-Controller is powered by Regulated power supply . I want to calculate Total power consumption in KWhr so that i can calculate my monthly electricity bill if i run my circuit 24/7.

    To simplify Calculations i replaced atmega32p with simple led. Then i measured input rms voltage and input rms current using multimeter on input side as shown in my attached circuit. i have got the following results ->

    V(rms) = 220 v and I(rms) = 6.6 mA

    So P(rms) =V(rms) X I(rms) = 220 X 6.6 mA = 1.45W

    in KWhr = (1.45W x 24 hrs) /1000 = 0.03484 which is obviously wrong

    Note : Cost of 1 unit = 8 bucks in my country

    What is the correct way to calculate Total power consumption.
     
  2. #12

    Expert

    Nov 30, 2010
    16,283
    6,795
    You forgot to multiply by days in a month.
     
  3. hp1729

    Well-Known Member

    Nov 23, 2015
    1,951
    219
    34.8 watt-hours per day. Sounds right. 0.034 kilo-watt-hours per day.
    What is your billing period? Monthly? Quarterly?
     
  4. Paras Bhanot

    Thread Starter New Member

    Jun 22, 2015
    3
    0
    it is Monthly .
     
  5. hp1729

    Well-Known Member

    Nov 23, 2015
    1,951
    219
    Then, as suggested, times 30 days for the month. About 1 kw/hr a month.
     
  6. Paras Bhanot

    Thread Starter New Member

    Jun 22, 2015
    3
    0
    :)
    thanks
     
  7. ronv

    AAC Fanatic!

    Nov 12, 2008
    3,289
    1,255
    Isn't there a sine or cosine involved here?
     
  8. crutschow

    Expert

    Mar 14, 2008
    13,006
    3,232
    For true power, yes.
    They actually calculated apparent power since the power factor wasn't included, but that's likely close enough to the real power for the purposes of the calculation. It is a pessimistic value so the real power will always be less.
     
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