how to measure efficiency of boost converter ?

[AdvaRes]

P3 could be used e.g. for the supply power of a logic gate. But it's not correct to apply an absolute value if the current sign is
changing, I3 = Avg(I(t)) would be correct.

Yeah sure FvM. I used your assumption I3 = Avg(I(t)) which is non other than Iavr (see above).
So now We have P1 = P2 mathematically correct but practically wrong. Which one is correct ?
Frankly, I tell you that I feel a little bit ashamed

I have to fill a table in which I have to put power consumption results and I dont know which one is the correct.

Please help me decide which power I have to put in my table.

 

[FvM]

P2 can't be right, because Irms is different from Iavr.

 

[AdvaRes]

P2 can't be right, because Irms is different from Iavr.

I also found that strange.

Generally not. It's true, if I and V have the same waveform.

Now I understand. Since Vdd and I(t) in my case have not the same variations. P=Irms*Vrms is wrong

This article confirm Indirectly your Idea but can mislead others like it does with me.

https://share-the-learning.blogspot.com/2007/05/difference-between-rms-power-and.html

Im now convinced that P1 is the real Power consumtion of my circuit.
Thanks you so much Professor !
Cheers,
Advares.

 

[erikl]

P1=∫I(t)*V(t)dt=∫I(t)dt * Vdd=Iavr*Vdd=5.9706507734989775E-6 Watt

Im now convinced that P1 is the real Power consumtion of my circuit.

Hi AdvaRes, just a reminder:
∫I(t)dt results in units of [As]. So don't forget to divide by the integration period!

 

[AdvaRes]

Hi Erikl,
Thanks for the note.
Actually there is no problem at all since I use the ADE calculator average fonction that calculate the average in the time interval of the plot window.