# Measuring Power dissipation emperically

• posted

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

• posted

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)

• posted

On Apr 13, 2:49=EF=BF=BDam, "Jon Slaughter" wrot= e:

snipped-for-privacy@news.coretel.net...

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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.

• posted

"Jon Slaughter"

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

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

• posted

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?

• posted

But Phil,

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

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DaveN```
• posted

You're measuring the wrong thing. What you want is the temperature of the device, not the power dissipation. Power dissipated and converted to heat is nothing more than AC/DC power input, minus whatever goes out. You can measure those all you want, but it won't tell you the final temperature of the device. Over a long term, you can use stick on temperature indicators for measuring the upper extreme, or whatever data logger you can borrow, for continuous measurement.

However, you wanted an empirical measurement:

which implies that one cannot use test equipment or measuring devices, and must use your senses (touch, observation, smell, hearing, etc). Been there, done that. Put your finger on the heat source. If you can hold it there, the cooling system is working. If you have to remove your finger after about 15 seconds, it's at the upper border of acceptable. If you burn the hell out of your finger, you need a bigger heat sink.

You can also detect if something is over temperature by smell. If the device smells like it's melting or burning, it's too hot. Similarly, you can also test for overtemperature by sight. If there's smoke or flames coming from the device, you've exceeded the manufacturers suggested operating conditions.

I'm not sure what you mean by "long term average". I would be rather uncomfortable leaving my finger on the test device for several days or weeks. Perhaps you had something more specific in mind? Maybe empirical isn't the correct word?

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Jeff Liebermann     jeffl@cruzio.com
150 Felker St #D    http://www.LearnByDestroying.com```
• posted

Power consumed less work done = power dissipated

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Joe Leikhim K4SAT
• posted

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,

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."

• posted

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

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I sure do. I also remember putting it in an insulated (adiabatic) environment, so that none of the heat generated leaks out to the room. However, doing the test in an Coleman ice chest caused a problem. Since the system was insulated, the heat would just accumulate. The more power we pumped in, the hotter it became. It never reached equilibrium. If we kept applying power to the resistor, it would eventually get hot enough to melt the ice chest.

However, running the same test in an uninsulated container yields a different result. The thermal gradient (degrees-C/watt) between the resistor and the accompanying heat sink, and that between the heat sink and the air are easily calculated or measured. Applying power over a long period of time results in equilibrium, as the air removes as much of the heat as possible. Move the air past the heat sink, and even more heat is removed.

However, you do have a point. Your original two sentence question managed to misuse the definitions of power, dissipation, energy, and temperature. I have no clue as to what you are trying to actually measure, empirically or otherwise. That's why I suggested that perhaps you might want to measure the temperature (empirically with your finger).

The real problem is that you haven't supplied sufficient information with which to answer your question. What are you measuring (units of measure)? What devices are involved? What resources do you have? What are you really trying to accomplish? We'll be happy to help, if possible (unless this is a student exam question).

I've seen it misused both ways. If you simply dropped the term "empirically", it would mean exactly what you're suggesting. Measurements are done "experimentally" as in standard procedures and methods. However, adding the term "empirically" implies that the standard methods of measurement are somehow inadequate, and that it is "dependent on evidence or consequences that are observable by the senses". Note the word senses. If you want to emulate a thermometer with your "senses", I supplied the method to do so.

Actually, I know doctors and nurses that can measure temperature with the back of their hand to within a few degrees. Granted, the range is limited, but the accuracy is amazing. Incidentally, 2 orders of magnitude is roughly the difference between freezing and boiling (in degrees Celsius). With 3 orders of magnitude, your system will probably ignite.

Sure. Please re-read my comments at the beginning of this rant. If you insulate the system, the heat will accumulate. If you expose your water bucket to the environment, the environment becomes part of your thermal calculations. For example, stupid things like someone opening the door, and creating a draft, will affect your results.

Hmmm.... I smell high skool or college fizzix homework. Waco, Texas?

OF course, you're correct. However, you have apparently also missed my point. You haven't defined (or controlled) what you're measuring. By supplying a rather useless and possibly humorous answer, I had hoped to kick you into some altered reality where people that ask question also do their homework first, and supply sufficient information to answer your question. Incidentally, you're asking in the wrong newsgroup. This is a fizzix problem, not an electronic design problem.

Yep. Just one problem. You haven't supplied any additional information sufficient to answer your question or do your homework. Keep trying but realize that you're not going to learn anything have people in a newsgroup do your homework.

```--
Jeff Liebermann     jeffl@cruzio.com
150 Felker St #D    http://www.LearnByDestroying.com```
• posted

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)

• posted

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.

• posted

Yes, this is what I'm talking about with water. Air has less thermal inertia so the measurements will be more unstable. (unless you can control it precisely)

Also, with water its easier to use the specific heat capacity to get another measurement. (easier to measure water's mass and it has extremely high thermal inertia)

• posted

ny power OUTPUT

This is probably the best way.

Measure the DC current, and multiply with the DC voltage to get the total DC power draw.

Then subtract the RF power IN, from the RF power Out (like in a P.A.E. calculation). Then take this, and subtract from the DC power.

The remaining power has to be the dissipated power as heat.

• posted

I wouldn't recommend water. Even if you use distilled water, you risk that residues on the PCB dissolve in water, and create free ions. This ruins the board, and may even create significant leakage currents.

Some folks have put their PC motherboard in regular cooking oil, and it works fine. Of course, it's really messy, so you'll end up sacrificing the board. Here's an example:

Using a insulated box with air is certainly the simplest. You can reduce the errors by measuring inside and outside temperature, and averaging the delta over a long period. Do the same with a resistor, possibly mounted on the same (empty) PCB, or same heatsink, and match the current until you get the same delta.

• posted

The water was only a example... It has a much larger specific heat than almost any other substance.

I saw that site you mentioned several years ago: If you go a few pages back:

"the hardware was placed in a container in which five gallons of de-ionized water were poured. To everybody's amazement, the system ran solidly for a period of five minutes before crashing. We repeated the assembly numerous times after the hardware was dried. The expensive components had suffered no damage. Accordingly, this solution was deemed unviable."

So water might work for low voltage with no problem for small devices. I'm not talking about sticking a whole pcb in there but just an IC.

I imagine water might be able to work for a cpu cooler too(I was plan on trying the oil idea when I brought a new computer). If one could remove the ions periodically then it might not be a problem. (no ions should be able to get in if sealed properly)

There are probably some better liquids out there but water was just an example. (and probably would work just for simple cases)

• posted

Measure the RMS voltage across it and the RMS current through it for as long as you want. Multiplying those figures gives you the power dissipation.

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• posted

No, it doesn't.

Best regards, Spehro Pefhany

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• posted

RMS phase measrements :)

martin

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