Calculating power delivered by supply

On Feb 17, 7:46 pm, John Popelish <jpopel...@rica.net> wrote:
Kingcosmos wrote:
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.

http://nothingbutnode.freehostia.com/pdf/PowerSupply.pdf

If there are any errors please let me know.

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
Click to expand...
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).
 
Kingcosmos wrote:

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).
Click to expand...
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
 
Kingcosmos wrote:

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.

http://nothingbutnode.freehostia.com/pdf/PowerSupply.pdf

If there are any errors please let me know.
Click to expand...
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
 
On Feb 10, 5:23 pm, John Popelish <jpopel...@rica.net> wrote:
Kingcosmos wrote:
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?

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
Click to expand...
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.

http://nothingbutnode.freehostia.com/pdf/PowerSupply.pdf

If there are any errors please let me know.
 
K

Kingcosmos

Guest
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?
 
"Kingcosmos"
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?
Click to expand...

** 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
 
Kingcosmos wrote:
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?
Click to expand...
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
 
On Feb 19, 9:33 pm, John Popelish <jpopel...@rica.net> wrote:
Kingcosmos wrote:
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).

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
Click to expand...
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)= Vdc + Vp*sin(x) from 0
to 2pi to end up with Vrms(total) = (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.
 
"Kingcosmos" <distort10n@yahoo.com> wrote in message
news:0e665c3c-313f-4862-b47f-ff93abc96303@64g2000hsw.googlegroups.com
I walked through the integral again using f(x)= Vdc + Vp*sin(x) from 0
to 2pi to end up with Vrms(total) = (Vdc^2 + Vprms^2)^(1/2).
Click to expand...
I get Vrms = sqrt(Vdc^2 + (1/2)Vp^2)

Makes sense, since if Vdc = 0 then

Vrms = sqrt(0 + (1/2)Vp^2)
= |Vp|/sqrt(2)

or

Vp = sqrt(2)*Vrms

as usual.
 
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