Measuring the thermal impedance of heat sink - how to?

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I received a bunch of random heat sink material that a young EE pulled out of a dumpster at Carnegie Mellon. Most were for SCR's, a few could be deployed for amplifiers.

I attached a Caddock 100 Ohm 30W resistor to the plate and measured the temperature rise over time, applying 22.4V for a constant 5 watt dissipation. The Caddock, with thermal grease, dissipates most of its heat through the back so works fine.

Of course, it's in free air, so convection currents from any air movement affect the measurement.

I am open to suggestion. I was thinking to standardize to an Aavid Thermalloy sink with specified thermal resistance.

One paper suggested using several diodes to measure the temperatures across various parts of the heat sink.
You're doing it pretty much the same way I would. The catalog pages for heat sinks give rise in degrees C per watt after it stabilizes. It's going to vary with orientation and vary wildly with air flow. I don't think it's linear with power but can't say how much. If you do tests with forced air flow, ideally you should straighten the flow (stack of drinking straws?) and measure the velocity. I use one of the little Kestrel meters- Kestrel Meters Compare Chart - 1000-3500 Series Meters - KestrelMeters

An IR camera would be great, but way out of my budget. You also have to be very careful with emissivity and probably paint everything black to get a reliable reading. Another useful tool, though a bit pricey, for such work is the Thermoworks CFIR- ThermoWorks - Close Focus Infrared Thermometer The typical IR gun has way too large a measurement area, plus parallax. This thing measures a 0.1" area. Or, just probe the thing with a thermocouple or LM-series sensor. Naturally you also need to measure ambient.


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...measure the temperatures across various parts of the heat sink.

Why do you want to know that?

The main point is: how hot is the *transistor* !!

Since transistors are smaller than 100W resistors, you might possibly switch to a TO-220 or TO-3 device, monitored for Power.

If you read fine print in old datasheets they sometimes show a spec for drilling into the side of a TO-3 baseplate and inserting a thermocouple. Of couse you really want the die temp, not the base temp. But the sink can't do a lot to change the die-base resistance.

ThermalTrack devices have temp-diodes built in.

Heatsink performance typically "improves" at higher power because convection gets stronger. (Also radiation but it is rare for that to dominate before Silicon craps-out.)
For a highly conductive metal heatsink made of copper or aluminium the limiting factor is the metal to air interface, doesn't matter if the heat enters via one TO126 package or multiple TO3's, its going to perform pretty much the same.

I'd measure the temperature rise at a more realistic heat input, air-convection is a non-linear effect. Perhaps adjust the dissipation till the temperature rise is a particular value like 45C or whatever you consider a maximum operating temperature you'd like to see.

It also useful to measure the time-constant of the heatsink under full load (related to its thermal mass). Thermal mass by itself can be as useful as dissipation for stabilizing the bias point.
You can use the same principles as in this project:
A Thermal Ohmmeter
Of course, the project is vastly too complex, but comparable methods and strategies can be reused, in a much simpler way.
This instruments measures interface materials, but it could equally well measure heatsinks: the reference sensor would simply remain in the ambient air.
The rest is a matter of arithmetic, to present the data in °C/W
The complicator will be direct losses from the resistor to air and how similar or different that is from a power device mounted to the sink. While simple in construction this is a somewhat complex heat transfer problem. I would definitely mount the style of device you intend to use the sink with and measure temperature vs. dissipation...trying to use a quite different geometry as a surrogate will likely confound getting a close answer.
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