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I am working on estimating an electronic components junction temperature?

I am using a two-resistor model for modeling an electronic component thermally. Do any of you know of a good worked example that I could use as a reference?

A germanium diode has a junction temperature not to exceed 100C. How many watts will this diode dissipate at the maximum junction temperature?

Given:
Thermal resistance junction to case Trjc = 0.7C/watt
Thermal resistance case to ambient Trca = 40C/watt
Maximum ambient temperature = 30C

Solution:
Tj – Ta = q x (Trjc + Trca)
100 – 30 = q x (0.7 + 40)
70C = 40.7C/watt x q(watts)
q = 100/40.7 = 1.7 watts

If you know q and want to calculate Tj

Tj – Ta = q x (Trjc + Trca)
Tj – 30 = 1.7 x (0.7 + 40)
Tj = 1.7 x (0.7 + 40) + 30
Tj = 99C (with a little rounding error)

Heatsink and semiconductor manufacturers may have other examples in their application information.

This entry was posted on Monday, May 31st, 2010 at 2:53 am and is filed under electronic components. You can follow any responses to this entry through the RSS 2.0 feed. You can leave a response, or trackback from your own site.

3 Responses to “I am working on estimating an electronic components junction temperature?”

  1. mavis b Says:
    May 31st, 2010 at 8:43 am

    I don’t know any ‘reference’ other than the actual heat dissipated by a resister is the ‘true watts’ calculated by the constant 3.412 this gives the heat in Btu. It is basic physics.
    References :

  2. Chas EE Says:
    May 31st, 2010 at 8:56 am

    A germanium diode has a junction temperature not to exceed 100C. How many watts will this diode dissipate at the maximum junction temperature?

    Given:
    Thermal resistance junction to case Trjc = 0.7C/watt
    Thermal resistance case to ambient Trca = 40C/watt
    Maximum ambient temperature = 30C

    Solution:
    Tj – Ta = q x (Trjc + Trca)
    100 – 30 = q x (0.7 + 40)
    70C = 40.7C/watt x q(watts)
    q = 100/40.7 = 1.7 watts

    If you know q and want to calculate Tj

    Tj – Ta = q x (Trjc + Trca)
    Tj – 30 = 1.7 x (0.7 + 40)
    Tj = 1.7 x (0.7 + 40) + 30
    Tj = 99C (with a little rounding error)

    Heatsink and semiconductor manufacturers may have other examples in their application information.
    References :
    Aham Heat Sink Application Handbook 1974

  3. to_700million_idiots Says:
    May 31st, 2010 at 9:29 am

    Junction temperature is purely a function of the package

    You have the thermal resistance Junction to case
    and the thermal resistance Junction to ambient

    Just look at ANY data sheet

    if a component has 5 Deg/Watt to case and it’s max junction is 150 deg then you can have the Heatsink rise to 90 Degrees and still dissipate 12 Watts maximum
    References :

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