Regulated power supply Calculation - Page 1
Forum section: electronics / projects

author: parasbhanot
author_rank: Contributor
timestamp: March 23, 2016, 03:43:26 pm
content: 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 KW/hr 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 mASo P(rms) =V(rms) X I(rms) = 220 X 6.6 mA = 1.45Win KW/hr = (1.45W x 24 hrs) /1000 = 0.03484 which is obviously wrongNote : Cost of 1 unit = 8 bucks in my countryWhat is the correct way to calculate Total power consumption.

author: Ian.M
author_rank: Super Contributor
timestamp: March 23, 2016, 03:59:26 pm
content: There is no such thing as RMS power. V(rms) x I(rms) gives you VA not power, and you have no way of measuring the power factor.You cant do it with a simple multimeter. A bridge rectifier straight into a reservoir capacitor only pulls significant current at the crests of the AC voltage waveform. To get the true power you need test equipment that can average the product of the instantaneous current and voltage over a whole cycle.

author: Seekonk
author_rank: Super Contributor
timestamp: March 23, 2016, 04:24:42 pm
content: If you leave the center tap of your transformer connected to common, your power will be a lot higher.

author: suicidaleggroll
author_rank: Super Contributor
timestamp: March 23, 2016, 05:05:28 pm
content: Quote from: parasbhanot on March 23, 2016, 03:43:26 pmThat should be kWh, not kW/h. You did the math right, but the unit label is wrong. The number is correct, if your system is drawing 1.45W continuously (see the other posts for why this value is likely not correct), it will consume 34.8 Wh per day, or 12.7 kWh per year.

author: parasbhanot
author_rank: Contributor
timestamp: March 24, 2016, 02:04:47 am
content: After doing some research i have found the following results :1. True power(R)= Actually used by device2. Reactive power(Q)= unused / wasted power due to presence of inductive load (transformer in my case )3. Complex power(S)= combination of True power and Reactive powerS = R + j Q4. Apparent power|S|= (this is what i have measured with multimeter) Modulus of complex power .and Power factor=Cos(theta) = True Power / Apparent powerOne thing i don`t understand is that if Apparent power|S|is a modulus of Complex powerSwhich is nothing but combination of useful and wasted power then why should i calculate True power ?So Instead of calculating my electricity bill on True power (used power) i should calculate it on Apparent power(used power and wasted power ) as i have done previously but i am not sure .

author: pmbrunelle
author_rank: Regular Contributor
timestamp: March 24, 2016, 03:42:18 am
content: If you are billed for KWh consumed, this means that you are charged only for real power.While W and VA are dimensionally equivalent, in practice W is reserved for real power, and VA is reserved for reactive or apparent power.So your device is using 1.45 VA of apparent power. However, the real power is equal to or less than than the apparent power, so:Real power <= 1.45 WProbably you wouldn't mind paying for 1.45 W of real power (not expensive enough to worry about) all the time, so an upper bound estimate is all you need...Also, some of the formulas you have are applicable only for sinusoidal current at mains frequency, so it gets more complicated with something like a rectifier followed by a capacitor, which does not draw sinusoidal current.

author: parasbhanot
author_rank: Contributor
timestamp: March 24, 2016, 04:39:19 am
content: Quoteok thanks .So does that mean that i am on a right track to calculate upper bound of electricity bill ?QuoteDue to conservation of power P(in) = P(out). So output power(useful DC+ wasted Ac) should have no meaning in final calculations of upper bound of electricity bill.