Re: Measuring Power dissipation emperically


"Jon Slaughter" <Jon_Slaughter@xxxxxxxxxxx> wrote in message
news:KmtMj.1575$I55.1322@xxxxxxxxxxxxxxxxxxxxxxxxxxxxx

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."

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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