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Design >> Analog Design >> power calculation
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Message started by IanX on Apr 15^th, 2014, 1:40am

Title: power calculation
Post by IanX on Apr 15^th, 2014, 1:40am

Hello. everyone. I am doing the simulation in Cadence(IC6.1.6). I need to calculate the power provide by the source.

In the simulation, I use both the vdc and vpulse in the libary analogLib with conductive load.

I used two ways to get the power.

1.rms

With the tran analysis, I measure the voltage and the positive terminal of the vdc and vpulse to get the current waveform.

Then I use the calculator to get voltage and current with the 'rms' function. Then the power = voltage(rms)*current(rms).

2.Average

Still with same way above.

But I use the calculator to get V and I with 'average' function. Then the power = voltage(ave)*current(ave).

I can see there is a big difference in these two results, sometime up to 20% or even more.

I can guess there are some reasons related to the phenomena.

It's negative and positive current flowing into the vdc or vpulse source. Negative current means the current provided by the source. And this part of power is the part I need to calculate. And the positive part is the power that are not provided by the source. Then I should negelect. Though in my case, there is not much influence in my results. Because the positive loss is quite small.

Another reason is that the real source should contain resistor. When I use the ideal source, it didn't contain with the resistor. I can see a extremely large current peak, like 1KV.

I have searched so much material to verify which one is right. But it seems different people have different evident to support their ideas.

It really confuses me.

Title: Re: power calculation
Post by aaron_do on Apr 15^th, 2014, 2:27am

I think if you want to find the power leaving a voltage source, you should take the time average of (V*I)...

Aaron

Title: Re: power calculation
Post by IanX on Apr 15^th, 2014, 3:08am

why do you think it's right by average?

I know the definition of power is inter(v*I)/period. Then the results will be the same by 'average'. but why not rms? From the fundamental of circuit, I was taught to use the rms to calculate the power

Thanks

Title: Re: power calculation
Post by aaron_do on Apr 15^th, 2014, 10:27pm

Hi,

please note that I said average(V x I) not average(V) x average(I). If V and I are in phase, then average(V x I) will give the same result as rms(V) x rms(I). This would be the case of a sinusoid voltage across a resistor, for example. If you have a capacitor on the other hand, we know from theory that no net power will be transferred to the capacitor as the current and voltage will be 90 degrees out of phase. If you take rms(V) x rms(I) in this case, you would get the wrong result since you have removed the phase information from the V and I before multiplication.

From the opposite perspective, if you take a sinusoid voltage across a resistor, and you try and find the power as average(V) x average(I), you will get zero, which we know from theory is the wrong result.

If you want a deeper understanding of the definition of power, then you will need to revise your electrostatics classes...

Aaron

Title: Re: power calculation
Post by IanX on Apr 24^th, 2014, 7:49pm

Thanks for Aaron.

v,rms * I,rms is the apparent power. In some case, the active power equal to apparent power, we can just use this equation. However, if we just want to get the active power in most of the case of ac system, the equation is wrong since the power fact or is not included.

v,average * I,average is also wrong, since it doesn't have any physical meaning in ac system. However, this value is more closed to the active power valued. It could be roughly as a reference to estimated the active power.

Best regards.

Title: Re: power calculation
Post by aaron_do on Apr 24^th, 2014, 10:13pm

Hi IanX,

I think it depends entirely on your system. For some systems, Vav*Iav and Vrms*Irms might both be poor estimates of power. average(V*I) is correct. If you are dealing with sinusoids, you can use VI* (conjugate).

regards,
Aaron