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SMPS components temperature?

Discussion in 'General Electronics Discussion' started by eem2am, Sep 5, 2013.

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  1. eem2am


    Aug 3, 2009

    Please can you evaluate this method of ensuring that the reading of infra red temperature monitors is correct? (it relates to accurately reading the temperature of components in an smps).

    Infrared thermometer:

    When testing an SMPS, it is essential to do thermal tests on the various components, eg the diode, the FET, the transformer (or inductor) ferrite, the current sense resistors, etc etc.

    The fastest way to test for temperature of any of these components is to use an infra-red temperature monitor, because it simply involves pointing the infra-red monitor at the component surface and reading off the temperature.

    However the problem with all of these devices is that the temperature reading is affected by the type and color of the component’s surface material. (eg shiny metal heatsink surfaces may often read lower temprature than they actually are)
    We cannot know whether the particular infra red monitor is calibrated to accurately read the temperature of the particular material that its being pointed at.

    Therefore, we must make up a “calibration SMPS”. This comprises an SMPS with thermocouples glued to each of its components.
    It should be ensured that this “calibration smps” comprises both shiny as well as anodised heat sinks etc etc. One simply then points the infra red monitor at the various components of this “calibration SMPS” and checks to see if the reading of the infra red monitor corresponds to the readings on the thermocouple reader for each type of component.

    Then one can confidently use this infra red monitor on the SMPS under test.

    Why do no companies do this?
  2. duke37


    Jan 9, 2011
    Gluing thermocouples to the components will give the wrong answer due to heat being conducted into the thermocouple. This can be reduced if the thermocouple is glued for say 10mm flat on the component.

    There are ways of measuring emissivity but it is complcated. The best way of getting the temperature would be to paint with a thin coat of matt black paint. You can compare the IR reading with a thermocouple embedded in a heat sink.
  3. eem2am


    Aug 3, 2009
    I thought that was the whole idea of gluing the thermocouple to the component surface?....certainly that's how we used to do it at Baldor, ....we didn't have 10mm of thermocouple glued to the surface, -but just the end where the two metals are soldered together.
  4. duke37


    Jan 9, 2011
    A thermocouple measures the temperature difference of the hot and cold junctions. Normally the hot junction is the one you are interested in. The junction will be heated by the component it is glued to and cooled by conduction along the wires. To get an accurate reading you should embed the junction in a drilled hole filled with thermal paste. This is obviously impossible with a small electronic component.

    I have used thermocouples with the two wires spot welded to the component. If the component is large, this gives a good result.

    Laying the wires along the component reduces the heat conducted from the junction. I presume that the components are small so the temperature will be affected by the heat conducted down the wires as well as the measurement error.

    Emissivity can be estimated by measuring reflectivity but it is easier to raise the emissivity by a suitable thin coating. You should find the efficacy of the coating by using a thermocouple embedded in a block of coated aluminium.
  5. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

    Jan 21, 2010
    Gluing a thermocouple to a component is fine, as long as the component is fairly massive compared to the thermocouple.

    A thermocouple attached to a TO-3 transistor is a very different thing to a thermocouple connected to a SOT-23 transistor.

    If you can model the heatsinking effect of the thermocouple then you can correct somewhat for it.

    One simple way is to operate a component dissipating 1W (say).

    Let's say that the component has a thermal resistance case to air of 20 degC per watt.

    You would expect the temperature of the component to be 20 degrees above ambient. But if you measure it with a thermocouple attached, you may find that it is only 18 degrees above ambient.

    I that temperature correct? No, the temperature may be higher without the thermocouple because the thermocouple itself is sinking away some heat.

    How can we figure out how much heat is being drawn off by the thermocouple? Simple. Connect another one. Once you do this, you'll find that the temperature read by the first thermocouple will drop slightly. Let's say it drops to 17.5 degrees.

    OK, now we can do some math to determine the heatsinking effect of the thermocouple.

    With one thermocouple, there was an 18 degC rise in temperature with 1W being dissipated. So the total heatsinking was 18 degC/W. It's not 20 because the board may be providing additional heatsinking, so may leads, and the thermocouple as well. And probably because the 20 degC/W figure is only an estimate.

    with the extra heatsinking, the total dissipation becomes 1/2 a degree cooler. Let's assume that the thermocouple is a heatsink in parallel with the existing heatsinks. And heatsinks are like resistors, so we can use the ressitors in parallel equations.

    1/17.5 = 1/18 + 1/x
    ==> 0.0571 = 0.0556 + 1/x
    ==> 0.0016 = 1/x
    ==> x = 630

    So the thermocouple is a 630 degreeC per watt heatsink.

    Now we can do some measurements with 1 and then 2 thermocouples, and calculate back to the temperature of the device.

    with one thermocouple we measure the temperature A. With 2 thermocouples we measure the temperature B.

    let's assume that X is the thermal resistance of the device (we want to solve for that) and Y is the calculates thermal resistance of the thermocouple.

    We know that A = P * 1/(1/X + 1/Y) and B = 1/(1/X + 2/Y)

    With a bit of rearranging, knowing that P is the same in both equations, we find that:

    (A - B)/X = (2B - A)/Y

    The only unknown is X, so X = Y*(A - B)/(2B - A)

    Let's assume that we see a 21.5 rise above ambient with one thermocouple and a 19.7 degree rise with two thermocouples.

    Using the formula above, we can calculate that the thermal resistance of the package is about 63.4 degC/W.

    Now we can calculate the Power being dissipated using one of the formula above

    A = P/(1/X + 1/Y)
    ==> P = A(1/X + 1/Y)
    = 21.5(1/63.4 + 1/630)
    = 373mA

    Now, knowing the thermal resistance and the power, we can determine the temperature

    T = 0.373 * 63.4
    = 23.6

    So the package has a temperature rise of 23.6 degrees above ambient with no thermocouple attached.

    The math for this can be done in a spreadsheet.

    As to whether or not you can do this in practice is another thing...

    Practically, as long as the thermal resistance of the package is less than a third of the thermal resistance of the probe, a straight line estimate (2*A - B) is going to be no more than 10% low.

    And if the thermal resistance of the package is less than 10% of the thermal resistance of the probe, the temperature A is going to be no more than 10% low.
  6. eem2am


    Aug 3, 2009
    @Steve & Duke37, I see your points........At Baldor they used to measure TO220 transistor body temperature with just one thermocouple....they didn't bother doing it the 2-thermocouple way........I suppose the most power components (except as you say, sot23's) have Rth's much lower than the thermocouple, and so maybe we can omit the 2 thermocouple way even though its more accurate.?
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