# Calculating power delivered by supply

• posted

I have a quick question that has been puzzling me. I am trying to calculate the internal power dissipated inside an op amp. One way of doing this is to determine the power dissipated in the load of the op- amp, then subtract that from the power delivered by a DC supply. The difference is the power dissipated in the output stage of the op-amp.

I have found several articles that touch on this subject, and they all state that the power delivered by the DC supply is the DC voltage multiplied by the average (not RMS) current.

My question is why is the average current being used rather than the RMS like the load calculations use?

• posted

"Kingcosmos"

** Because the power delivered by a supply of DC voltage is proportional to the average current draw.

The RMS value of a current is applicable only to its heating effect on a resistance through which that current is flowing.

A DC supply is not a resistor - as the voltage and current are not proportional.

....... Phil

• posted

The average power is the average of the instantaneous power which is the instantaneous product of voltage and current. However, if either of those two variables remains constant during the average, then the equivalent result will be had by multiplying the constant value of either current or voltage times the average of the other variable (either current or voltage). The average of a constant value times a varying value is the same as the constant times the average of the varying value. The constant can be pulled outside of the averaging process and multiplied after the average is done.

RMS has to do with the equivalent DC that produces the same heating effect on a resistor. So if you want the DC voltage that heats a resistor the same as a non DC waveform, you have to measure the RMS value of that waveform.

```--
Regards,

John Popelish```
• posted

Thanks for the replies. I have linked to my PDF calculating the internal power dissipation of a simple voltage follower. I believe the equations are correct. I have seen a Maxim App note that for some reason ignores the AC current portion from the supply.

If there are any errors please let me know.

• posted

Starting with the schematic at the bottom, I am not sure what the signal voltage is that is applied to the follower. After I understand this, I will plow through your integrals.

But, at the bottom, you seem to think that you can separate load power into a DC and an AC component, calculate those components separately, and then add those components together. I think that can be done only for linear processes, but power is not a linear process. You have to integrate the instantaneous power over a cycle and divide by the period to find the average resistor power.

```--
Regards,

John Popelish```
• posted

Hello John,

The input voltage to the follower is a peak voltage biased as Vcc/2. In the PDF I say 2Vp to give it an actual value, but I should have kept it generic for symbolic purposes.

Now that you mentioned it, you are right about power. Superposition cannot be applied in order to linearly sum powers. So would I have to find the total AVERAGE current AC+DC and intergrate over a cycle (or a

1/4 cycle works just as well).
• posted

For AC plus DC through a resistor, I think you need to integrate an integer number of half cycles (each having a quarter of a cycle of the negative side of the wave and a quarter of a cycle of the positive side of the wave. The math isn't any harder to just integrate a cycle.

As to the signal source, please label it the same way you include in in the integrals at the top. Thank you.

```--
Regards,

John Popelish```
• posted

The only item I can give you at the moment is Maxim's app note "One Resistor Takes the Heat" and trying to 'prove' their example. I am out of town and do not have access to a scanner.

I walked through the integral again using f(x)=3D Vdc + Vp*sin(x) from 0 to 2pi to end up with Vrms(total) =3D (Vdc^2 + Vprms^2)^(1/2).

This looks like an RSS of the RMS values. Also, this just so happens (probably not magically) to be equal to calculating the power from each component (AC and DC) separately and then summing them in a linear fashion.

So this cannot be superposition.

• posted

0

I get Vrms =3D sqrt(Vdc^2 + (1/2)Vp^2)

Makes sense, since if Vdc =3D 0 then=20 =20 Vrms =3D sqrt(0 + (1/2)Vp^2) =3D |Vp|/sqrt(2)

or

Vp =3D sqrt(2)*Vrms=20

as usual.

ElectronDepot website is not affiliated with any of the manufacturers or service providers discussed here. All logos and trade names are the property of their respective owners.