# Calculate power value from a input current peak

#### Chatanas

Nov 28, 2020
3
Hello folks,

I have a really simple problem.
I'm probing the input current peak of a system and I'd like to calculate the equivalent power consumption.

This peak current is about 3.3A for 2ms approximately and the system is powered by a 12V power-supply. The signal form is not exactly this one but I simplified it for the sake of the exercise. I attached the screenshot of the real one to this post (and before you ask, the unit is actually in A, but I forgot to change the unit in the oscilloscope).

If the peak lasted 1s it would be simple and I would only have to write the traditional equation P = I x V = 3.3A x 12V = 39.6W. But as this peak last only 2ms, then it cannot be that high.

Then, I would have written something like (3.3A x 2ms) x 12V... However, it would not define a power anymore, but an energy, unless I say it is a normalization to 1s like in the following formula:

Did I write this equation right?

I acquired the value samples from an oscilloscope to calculate the power by transcripted the equation in the discrete-domain. Here is how I computed it:

With N the number of sample during the peak and M the total number of sample during 1s.
I considered Vn and Vm as constants equals to 12V since the voltage value barely changed during the current peak.

However, summing the first part of the equation ( ) already gives me a result too high to be realistic.
Indeed, in my example I have M = 10^5 samples, N = 127 sample and sum(VnxIn) = 1700W.
Then it gives a result of 2.16W, for only 2ms (while it has to be 39.6W for a full second).

Thanks a lot for your help!

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#### Kabelsalat

Jul 5, 2011
176
Well - this doesn't look like an ordinary RC discharge curve, so it's tricky if you cannot describe this area with some ordinary math function.
Have you considering using Simpsons formula ?

#### Harald Kapp

##### Moderator
Moderator
Nov 17, 2011
12,774
Here is how I computed it:
I think that is where you err. The equation should look like this:

Reason:
1. count starts at 1 if you have M samples. If count started at 0, the total number of samples is M+1
2. Assume Vn and In are constant, not time variable. Then sum(Vn × In)1N = N × Vn × In. In your equation you have N². That's where your too large value comes from.

#### Chatanas

Nov 28, 2020
3
Hello Kabelsalat and Harald Kapp and thanks for your replies.

Kabelsalat said:
Have you considering using Simpsons formula ?
I didn't, but after reading some more about it, it is a good method, indeed
As I was looking for more information (mainly to know why the equation takes this form) I found this website, which performs the mathematical demonstration... Very interesting!
I won't necessarily use it for this application, but I'll keep it in mind if ever I need extra accuracy in my calculation for the future.
Thanks a lot for your help!

Harald Kapp said:
1. count starts at 1 if you have M samples. If count started at 0, the total number of samples is M+1
2. Assume Vn and In are constant, not time variable. Then sum(Vn × In)1N = N × Vn × In. In your equation you have N². That's where your too large value comes from.
1- Indeed, I didn't write properly the formula. Sorry about it and thanks for the correction!
2- You're absolutely right! I knew there was something wrong in this equation, but I did not found what it was. I correct it in my equation and it worked! Now I'm looking at the equation this way, it's just a basic equation to find the average and I feel a bit stupid to not have thought about it... I should not have been properly awake
In any case, it's exactly what I was looking for! Thanks again!

Beside these methods, I slept on my issue and came with this equation, which seemed to work properly, too:

(edit: replace the 2 under the epsilon by 1)

With Tn the time when the sample In has been measured, V = 12V and T = 1s.

This way, the equation sums the successive rectangles with a length of In and a width of (Tn - Tn-1) composing the total area covers by the curve.
Then, to remove the part in excess (or missing) of each rectangle when the curve is falling (or rising), I would write this equation:

(edit: replace the 2 under the epsilon by 1)

It is not as accurate as the Simpson's method, but I think it is a bit more accurate than the average equation.
What do you think about it?

#### Chatanas

Nov 28, 2020
3
And my bad: it is of course not an epsilon but a sigma
Shame on me

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