# Measuring Power dissipation emperically

Discussion in 'Electronic Design' started by Paul E. Schoen, Apr 13, 2008.

1. ### Paul E. SchoenGuest

Device packages and heat sinks are rated in Degrees C per watt. So you can
get a good idea of the power by reading the temperature rise above ambient
after it has stabilized.

TO-220 packages in free air can handle about 1 or 2 watts. A reasonable
"rule of thumb" (or index finger) is to touch the package with a wet
finger. If it sizzles it's too hot. If you can hold your finger on it you
are well within safe limits. Something like 50 degrees C above ambient, or
about 75 C, is OK, and that's about like a fresh cup of coffee. You can
also use a thermistor or thermocouple to get a more accurate reading.

If you have a good LTSpice model, you can read device power by computing
the instantaneous voltage times current, and taking the average over a
period of time after the circuit has stabilized.

Paul

2. ### Jon SlaughterGuest

How does one go about measuring the power dissipation of a device(low
voltage) emperically? Stick it in water? Use a large heat sink? (talking
about the long term average and not instantanous obviously)

3. ### PaulGuest

I agree with the other poster about
using the thermal resistance of the case-to-
ambient (aka, "Theta-CA"). Not Theta-JA, because
you won't have much chance to measure the junction
temperature. Use the themocouple to measure the
case temp, and you should have a good idea of the
dissipated power.

Getting fancy, with big bucks, you would
measure the case temp with an Infra-red camera,
after spraying everything with flat-black paint,
to get the black body emissivity.

4. ### Phil AllisonGuest

"Jon Slaughter"

** Depends on the item and the level of power involved - dickhead.

Usually , one just measures the power INPUT, subtracts any power OUTPUT
and what is left over MUST be the power dissipation.

It's called using logic.

....... Phil

5. ### Jon SlaughterGuest

Yeah, thats what I was thinking. Although this isn't necessarily the
internal temperature I suppose I can calulate that from the datasheet?
hehe, I was hoping something a bit more accurate
How accurate would this be?

6. ### DaveNGuest

But Phil,

The OP asked for an empirical method and what you describe, whilst more
probably usual, is not empirical!

7. ### Phil AllisonGuest

"Dave Nutcase Fuckwit "

** Of course it is - FUCKWIT .

It involves physical measurements on REAL things.

Piss off - you asinine ASD fucked pedant.

...... Phil

8. ### DaveNGuest

To be empirical the measurement has to be of the power dissipation. Your
method measures the other things and then calculates the power dissipation.

9. ### Phil AllisonGuest

"Dave Nutcase Fucking **** "

** BOLLOCKS !!

It measures power dissipation by use of DIRECT MEASUREMENTS and the of the
conservation of energy principle.

It therefore is EMPIRICAL !!

YOU have no idea of the definition of the term.

So, **** the HELL OFF -

you ASININE , ASD fucked, pedantic TWAT !!

...... Phil

10. ### DaveNGuest

Hi Phil,

Whilst your right that it is based on experiment and therefore the data that
is calculated can be considered to be empirically based it is not in itself
emperical as it was not observed.

The problem with this method is that in complex systems it may be difficult
or even impossible to measure all the useful power outputs and just leave
the power loss(dissipation) as a calculated result.

I have many times made measurements of the power loss on a system as it can
be relatively straight forward; put the thing in a box, run it and measure
steady state temperatures. The box will have been previously characterised
to determine its thermal characteristics so then the system power loss can
be calculated using the same characterisation formula.

I still think your mother would be disappointed.

11. ### Phil AllisonGuest

"Dave Nutcase Fucking **** "

** Go drop dead you -

STUPID Pile of AUTISTIC DUNG !!!

** BOLLOCKS !!!!!!!!!!!!!!!!!!!!!!!!!!!!!

** IRRELEVANT to the issue YOU raised !!!!!!!!!!!

Y O U F U C K W I T !!

** Errr - so THAT does not involve making * logical deductions * from
the laws of nature ????

OKeeeee.............

Leeme tell ya somethin here - sunshine.

YOU have no damn idea who your father was, cos you mother has no idea
either.

Cos she was a pox ridden, ASD fucked whore.

And you are an abortion the crawled out of the bucket and got away.

....... Phil

12. ### DaveNGuest

Again Phil, thanks for your quick response.

It's not irrelevant as the OP was looking for methods of making the
measurement. Note I have changed the subject back to the original from the
whatever you feel appropriate.
I never suggested that logical deductions weren't required. If that is how
you read it then too bad. The issue is over what measurements are used to
make the deductions/calculations and whether these are considered to be
empirical measurements or empirically based data. Sometimes the differences
are not significant, sometimes they are and that has to be down to the
judgement and justification from the engineer and the specific application.

I am not sure how this is relevant to either measuring empirically or by
calculation the power dissipation in an electronic circuit. Please
elaborate so that I can improve my understanding.

13. ### RFI-EMI-GUYGuest

Power consumed less work done = power dissipated

--
Joe Leikhim K4SAT

"Treason doth never prosper: what's the reason?
For if it prosper, none dare call it treason."

14. ### Jon SlaughterGuest

Um, if, say, you put the device in water, the power dissipation will change
the temperature of the water. This is how they measure the power dissipation
of resistors(or used too) and also other things. Don't you ever remember
doing those experiments in physics?
Same page:
"In a second sense "empirical" in science may be synonymous with
"experimental.""
I'm looking something more precise than 2 or 3 orders of magnitude.

For example, I know that I could use water. Put the device in the water and
run it for a while. The water will heat up. Since one knows the specific
heat capacity of water,

http://en.wikipedia.org/wiki/Specific_heat_capacity

Its not difficult to measure the initial and final temperature of the water
to get the power dissipation... actually its quite easy(much easier than
measuring the temperature of the device itself IMO... although more work to
setup). The formula is a simpel calculation.

BTW, empirical doesn't mean no calculation even experiments require
calculations(you might have to add 1 + 1 to get 2.

"...such methods are opposed to theoretical ab initio methods which are
purely deductive and based on first principles."

15. ### Paul E. SchoenGuest

You may be confusing energy and power. Water can be used to determine the
amount of energy that was used from time A to time B, by the temperature
rise, assuming the water is well insulated so there is no heat (energy)
loss. This will be in joules, watt-hours, btus, or calories.

The power dissipation (watts, btu/sec, etc) will result in a temperature
rise that will stabilize according to the thermal conductivity of the
environment. For better accuracy, this should be adjusted so you get a
temperature rise large enough to measure accurately, and the measurement
should be done close to the device, taking into account the Deg C / Watt of
the heat sink or package to free air.

You could figure out power by measuring the energy after a period of time,
and it will be P = E/t, or dE/dt. But if you wait until the temperature
stabilizes, you will not get a correct reading. And you must make sure that
the water is continually stirred.

You can also measure power by comparing a known device (like a TO220
resistor) with the unknown (MOSFET with arbitrary waveform), and adjust the
power in the known device with a DC source so that the temperature
differential is zero. You just need to be sure the packages and any heat
sinking are identical.

Paul

16. ### Phil AllisonGuest

"Jeff Liebermann" <>

** ( snip piles of absurd drivel )

You are nothing but a TOTAL FUCKING IDIOT

PISS OFFFFFFFF !!!

....... Phil

17. ### Phil AllisonGuest

"Jeff Liebermann" <>

** ( snip piles of this wanker's anencephalic drivel )

You are nothing but a

TOTAL FUCKING ILLITERATE IDIOT

PISS OFFFFFFFF !!!

....... Phil

18. ### Phil AllisonGuest

"Jeff Liebermann" <>

** ( snip piles of this wanker's anencephalic drivel )

You are nothing but a

TOTAL FUCKING ILLITERATE IDIOT

& CHARLATAN

( It's a big word - look it up ... )

SO PISS OFFFFFF !!!

....... Phil

19. ### Jon SlaughterGuest

No you don't realize that the only way one can reach equillibrium is if the
water is loosing exactly the energy that the device is giving it.

In a truely insulated system there can be no equillibrium so one just has to
calculate the temperature and time and can determine P.

P = c*m*dT/dt

If the system is not insulated then the loss of heat is exactly balanced by
the gain in heat at equillibrium. Knowing and knowing the how the container
looses heat(its thermal resistance) then one knows the power dissipation of
the device.

e.g., suppose I have a container of water and I put a resistor in it. When
the resistor is dissipating P watts and the water is in equillibrium then I
know the thermal resistance of the water. (its dT/P)

I could farther just make a graph using the resistor at diffrent P's and
then use that to determine unknown devices power just by measuring the
temperature(yes, its that easy) when at equillibrium.

e.g., if the resistor raised the temperature 10 C at 10W and another device
raised the temperature 10 C then it must too be dissipating 10W. I'd imagine
the curve would be fairly linear for low enough power dissipations(nothing
that will make a phase change of course).

Theoretically its simple... I'm not so sure in practice. Not sure how
deionized the water much be, if the thermal inertia of the water is a +
or -(should help with an average but might mean one has to wait an aweful
long time for proper readings) or what. This method definitely would make it
easy to automate but one would have to make assumptions about how long
equillibrium would take(which isn't a real problem I imagine cause it should
happen pretty quick unless one has a lot of water).

In any case its no big deal. I might try it and see as I think it could be a
pretty good method. (probably pretty accurate too if the setup is decent)

20. ### Arlet OttensGuest

You could stick in a box, and watch the rise of the air temperature in
the box.

Then repeat the experiment, but replace your device with a resistor.
Adjust the current to get the same rise in temperature.