[SOLVED] Regulated power supply Calculation

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 mA

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

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

What is the correct way to calculate Total power consumption ?

This is circuit used :

https://obrazki.elektroda.pl/7398061300_1458794047.png

You are missing the big picture here. The power supply itself because of efficiency and type may be drawing from a few watts to 5 or 10 watts. The simplest way to measure power is with a "kill A watt" meter.

https://eshop.macsales.com/item/P3 ...dvwZzH2LkqLJkosch9ntDKN3rQLk3VDeQTBoCzMLw_wcB

If you have a 60 Hz transformer type supply then your idle current will be around 10% of the power supply wattage rating. Example 100 W power supply will draw 10 W just turned on. Switching power supply's can be much less, but depending how old may still draw a few watts just sitting there.

Also there is no attached circuit.

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 KW/hr = (1.45W x 24 hrs) /1000 = 0.03484 which is obviously wrong

What is the correct way to calculate Total power consumption ?

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That all seems correct.
But strictly speaking you are measuring volt amps (VAs) not watts.
But the difference should be minor.

You can buy watt meters fairly cheaply on e-bay, like the "kill a watt" mentioned above, and its hardly worth the trouble building your own.

Hi,

= 0.03484 which is obviously wrong

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Why do you think it is wrong?

From your schematic:
Let'say there is about 5mA load current.
An unloaded 12V AC transformer will easily give 20V at the capacitor.
The linear regulator needs to "dissipate" the difference of about 15V. This increases the total power by a factor of 4.
So I assume about 20V x 5mA = 100mW at transformer output.

The transformer input power is difficult to say, because of the unknown phase angle = cos(phi).
With 6.6mA x 220V you calculate the apparent power.
The effective power is less. And you have to pay for the effective power (x time) only.

I assume the effective power is somewhere inbetween 0.5W and 1.0W.
Lets say 0.8W. Multiplied with 24 h gives 19.2Wh = 0.019kWh per day
Multiplied with 360 it gives about 6.9kWh per year.

Klaus

Why do you think it is wrong?

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i did not expect it to be that low .

The effective power is less. And you have to pay for the effective power (x time) only.

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So we don`t pay for reactive power (wasted/unused power due to inductive transformer) ?

So we don`t pay for reactive power (wasted/unused power due to inductive transformer) ?

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It depends on the power usage meter. Some of the more inteligent ones take the V/I phase difference into consideration but older and mechanical meters do not. This is exactly the reason why electricity companies like power factor correction circuits in larger power supplies and it is now a legal requirement in many countries.

Brian.

betwixt said:

It depends on the power usage meter. Some of the more inteligent ones take the V/I phase difference into consideration but older and mechanical meters do not. ..............

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If you are saying, mechanical watt-hour meters do not take phase angle between voltage and current into consideration, that is incorrect.
Those meters do indeed only measure real power, not apparent power.

Those meters do indeed only measure real power, not apparent power.

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That is indeed correct.

So the conclusion is that i can only estimate apparent power consumption with simple multimeter. Since electricity bill is calculated at True power so i need special equipment like "kill A watt'' meter .

Am i correct ?

Hi,

Speaking for Germany:

Even the older mechanical meters (designed by Galileo Ferraris) are true active energy meters. I´ve never seen any other energy meter (in germany).
There are european regulations for energy meters (EN 50470), all pay attention to phase shift and are true active energy meters. Many decades before.

In former times the mains phase shift was more inductive, caused by motors, transformers, aerial cables and conventional ballast for neon and other flourescent lamps.
Nowadays it is more capacitive, caused by electronic motor control, electronic ballast, switched mode power supplies and the large capacitance of shielded underground cables.

Both causes increased current in the cables. And the loss in the cables is proportional to the square of the current (independent of the phase shift to the voltage).
This is why they want power factor correction.. simply lowering cable loss. And thus decreasing voltage drop in the cables.
The other point is (not directely related to phase shift) that nowadays regulations not only want simple power factor correction but additionally current waveform correction.
Usually a capacitor charged by a full bridge recitfier causes only current at the very top of a sine voltage. This was/is used in many SMPS. Nowadays power factor compensation circuits form the current more "sine shape".

Klaus


*****

So we don`t pay for reactive power (wasted/unused power due to inductive transformer) ?

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You don´t need to produce this power. It is not true energy. So don´t consider the kVAr is wasted. But it causes increased power loss in cables.
The reactive power causes only some % of true power (loss) in the cables.

And yes: We (households in Germany) don´t pay for reactive power. But industry with large ammount of reactive power need special energy meters that pay attention of reactive power.

Klaus

parasbhanot said:

So the conclusion is that i can only estimate apparent power consumption with simple multimeter. Since electricity bill is calculated at True power so i need special equipment like "kill A watt'' meter .

Am i correct ?

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If you want trouble and theoretical calculations (to see if theoretical calculations matches the measurements) you should do something like this.

Model your transformer:



Find its parameters by doing some experiments on the transformer called the open circuit test and the short circuit test. (not very hard but you would need some extra elements than a multimeter, like a wattmeter)

Then you can go on with calculations because now the power factor would be known.

So if you do not want this, then buy a wattmeter or "kill a watt" and with 2 measurements you have your active power .

2 measurements if you want to know your efficiency, 1 if only the consumption for your bill estimation.

parasbhanot said:

So the conclusion is that i can only estimate apparent power consumption with simple multimeter. Since electricity bill is calculated at True power so i need special equipment like "kill A watt'' meter .

Am i correct ?

Click to expand...

I am under the impression that the kill a watt meter does not measure true power either, but it is close enough for $20. The meter was meant to be inexpensive and easy to use.

I don't know about the "kill a watt" meter, but a couple of similar cheapie e-bay watt meters I have here, do display true power, and also display power factor as well, along with peak hold current. All of which can sometimes be very useful.

Its not too difficult with a microprocessor, once you have both a voltage waveform and a current waveform to calculate all the relevant values.

But with just a volt meter, and an amp meter, all you can do is multiply the two and get VA's.

i think i have got the basic idea.

Thank you everyone for taking time to join the discussion

FlapJack said:

I am under the impression that the kill a watt meter does not measure true power either, but it is close enough for $20. The meter was meant to be inexpensive and easy to use.

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It does measure true power, VA (apparent power), and the power factor, along with the line voltage and frequency.