Power dissipation calculation uncertainty

R

Roque

Guest
If I have a power supply similar to a computer power supply where it's
rated at 300W, what is the proper way to figure out what my actual power
usage is when I have this system running under its maximum load?

Here's what I've done so far....

I have a power supply that provides 24.5 volts and 5.1 volts DC output,
I find I can run a total current of 9 amps continuous, or 20 amps peak
before tripping the internal protection.

When I put a current probe on the 24.5 and 5.1 volt lines,
using a digital scope with math functions, I get 8 amps RMS on the 24 volt
line and 1.15 amps RMS on the 5.1 volt line, in my worst case use of the
machine.

I'm trying to decide what power dissipation I have...

With my scope, I saved 30,000 points of voltage and current data against
the same time line, and I
imported this V, I data to a spreadsheet....and created a new calculation
where I multiplied V * I for each data point...for instantanious power per
sample. I then squared each V * I data value, summed it, divided by 30,000
samples, and took the square root...hoping to find RMS power.

With 8 amps at 24.5 volts and 1.15 amps at 5.1 volts, I found about 200W
power dissipation...

Does this make sense? I'm not experienced with "power" electronics to be
confident about it so I wanted to seek advice.

Thanks.
 
Roque wrote:

If I have a power supply similar to a computer power supply where it's
rated at 300W, what is the proper way to figure out what my actual power
usage is when I have this system running under its maximum load?

Here's what I've done so far....

I have a power supply that provides 24.5 volts and 5.1 volts DC output,
I find I can run a total current of 9 amps continuous, or 20 amps peak
before tripping the internal protection.

When I put a current probe on the 24.5 and 5.1 volt lines,
using a digital scope with math functions, I get 8 amps RMS on the 24 volt
line and 1.15 amps RMS on the 5.1 volt line, in my worst case use of the
machine.

I'm trying to decide what power dissipation I have...

With my scope, I saved 30,000 points of voltage and current data against
the same time line, and I
imported this V, I data to a spreadsheet....and created a new calculation
where I multiplied V * I for each data point...for instantanious power per
sample. I then squared each V * I data value, summed it, divided by 30,000
samples, and took the square root...hoping to find RMS power.

With 8 amps at 24.5 volts and 1.15 amps at 5.1 volts, I found about 200W
power dissipation...

Does this make sense? I'm not experienced with "power" electronics to be
confident about it so I wanted to seek advice.

Thanks.
Click to expand...
WOW, talk about *overkill*! :)
You don't need to calculate RMS when dealing with DC, you just multiply I by V.
So (24.5 x 8) + (5.1 x 1.15 ) = 201.865W.
 
On Thu, 07 Aug 2003 03:59:01 +0100, A E wrote:

WOW, talk about *overkill*! :)
You don't need to calculate RMS when dealing with DC, you just multiply
I by V. So (24.5 x 8) + (5.1 x 1.15 ) = 201.865W.
Click to expand...
What's a guy who couldn't see that doing with an oscilloscope capable of
doing what he did with it?

--
Then there's duct tape ...
(Garrison Keillor)
nofr@sbhevre.pbzchyvax.pb.hx
 
A E <aeisenhut@videotron.ca> wrote in message news:<3F31C075.BBC10E7B@videotron.ca>...
Roque wrote:

If I have a power supply similar to a computer power supply where it's
rated at 300W, what is the proper way to figure out what my actual power
usage is when I have this system running under its maximum load?

Here's what I've done so far....

I have a power supply that provides 24.5 volts and 5.1 volts DC output,
I find I can run a total current of 9 amps continuous, or 20 amps peak
before tripping the internal protection.

When I put a current probe on the 24.5 and 5.1 volt lines,
using a digital scope with math functions, I get 8 amps RMS on the 24 volt
line and 1.15 amps RMS on the 5.1 volt line, in my worst case use of the
machine.

I'm trying to decide what power dissipation I have...

With my scope, I saved 30,000 points of voltage and current data against
the same time line, and I
imported this V, I data to a spreadsheet....and created a new calculation
where I multiplied V * I for each data point...for instantanious power per
sample. I then squared each V * I data value, summed it, divided by 30,000
samples, and took the square root...hoping to find RMS power.

With 8 amps at 24.5 volts and 1.15 amps at 5.1 volts, I found about 200W
power dissipation...

Does this make sense? I'm not experienced with "power" electronics to be
confident about it so I wanted to seek advice.

Thanks.

WOW, talk about *overkill*! :)
You don't need to calculate RMS when dealing with DC, you just multiply I by V.
So (24.5 x 8) + (5.1 x 1.15 ) = 201.865W.
Click to expand...
although the unit is rated at a total of 300W, don't you think that
the wattage on each tap may be different? The 24V line may be rated up
to 300W (12A) but I would be very suprised if the 5V line could
deliver anything like the 60A that would be required to get 300W @ 5V.

I would suppose that the unit is designed to run some light
electronics off the 5V line and some kind of heavier electric
actuators off the 24V line. With these assumptions I would expect the
5V line to only be rated at a few Amps, while the 24V line would be
rated much higher. You would only need to exceed the limit on the
lowest rated line to trip the protection circuit (say, by drawing 2A
on the 5V line that may only be rated for 1.5A)

- Jeff Dutky
 
Jeffrey Dutky wrote:

A E <aeisenhut@videotron.ca> wrote in message news:<3F31C075.BBC10E7B@videotron.ca>...
Roque wrote:

If I have a power supply similar to a computer power supply where it's
rated at 300W, what is the proper way to figure out what my actual power
usage is when I have this system running under its maximum load?

Here's what I've done so far....

I have a power supply that provides 24.5 volts and 5.1 volts DC output,
I find I can run a total current of 9 amps continuous, or 20 amps peak
before tripping the internal protection.

When I put a current probe on the 24.5 and 5.1 volt lines,
using a digital scope with math functions, I get 8 amps RMS on the 24 volt
line and 1.15 amps RMS on the 5.1 volt line, in my worst case use of the
machine.

I'm trying to decide what power dissipation I have...

With my scope, I saved 30,000 points of voltage and current data against
the same time line, and I
imported this V, I data to a spreadsheet....and created a new calculation
where I multiplied V * I for each data point...for instantanious power per
sample. I then squared each V * I data value, summed it, divided by 30,000
samples, and took the square root...hoping to find RMS power.

With 8 amps at 24.5 volts and 1.15 amps at 5.1 volts, I found about 200W
power dissipation...

Does this make sense? I'm not experienced with "power" electronics to be
confident about it so I wanted to seek advice.

Thanks.

WOW, talk about *overkill*! :)
You don't need to calculate RMS when dealing with DC, you just multiply I by V.
So (24.5 x 8) + (5.1 x 1.15 ) = 201.865W.

although the unit is rated at a total of 300W, don't you think that
the wattage on each tap may be different? The 24V line may be rated up
to 300W (12A) but I would be very suprised if the 5V line could
deliver anything like the 60A that would be required to get 300W @ 5V.
Click to expand...
Yeah, so? The OP clearly stated the current on each rail, the numbers didn't trigger any
alarms for me.
 
Fred Abse wrote:

On Thu, 07 Aug 2003 03:59:01 +0100, A E wrote:

WOW, talk about *overkill*! :)
You don't need to calculate RMS when dealing with DC, you just multiply
I by V. So (24.5 x 8) + (5.1 x 1.15 ) = 201.865W.

What's a guy who couldn't see that doing with an oscilloscope capable of
doing what he did with it?
Click to expand...
Bah, have you been in a university lately? The kids there have equipment that
took a building full of engineers to design, and is used for the most trivial
tasks. That's life, electronics is a commodity now.
 
If it's reality, then does it have to make sense?

"The difference between fiction and reality is that fiction has to make
sense." Tom Clancey.

The 300W is probably the input power to the supply at the rated line
voltage. Some goes to heat etc. If the supply is 75% efficient, then 225W
will get to the loads.
Zoramy
 
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