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This board has been very helpful to me, I am stuck with a small problem again. The datasheet is here.

Part name: SMDJ series.

I have a pulse train coming from an ignition system, the power I calculate is about 4 Watts of Pavg. When I refer to the datasheet above, we can see it can withstand about 6.5W of Pavg. But when I calculate the Tj, it exceeds, I get \$75C\times4 = 300C\$ above ambient. Which spec should I rely on: Pavg as stated by manufacturer, or Tj as calculated?

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    \$\begingroup\$ It should be implicit that you have to meet all of the maximum ratings simultaneously. In other words, the device can handle 6.5W of power only if you keep it below its maximum temperature (by heatsinking it or something). \$\endgroup\$
    – Dave Tweed
    May 16, 2014 at 17:44
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    \$\begingroup\$ In any case you have to extrapolate VBR (clamping)for other than 25degC junction temperatures, using formula like ΔVBR=αT(VBR)*(Tj-25)*VBR, as well as the new peak power. \$\endgroup\$
    – GR Tech
    May 17, 2014 at 3:22
  • \$\begingroup\$ Thank all for the valuable comments. I find other Leaded part showing dissipation spec TL @75C is 6.5W datsheet here ; littelfuse.com/~/media/electronics/datasheets/tvs_diodes/…. In figure 6, does this means it can dissipate with deration for leads lengths being atleast 10mm each side. \$\endgroup\$
    – Adi
    May 17, 2014 at 8:45
  • \$\begingroup\$ I get 4W*15C=70C aabove ambient, is it ambient ?. \$\endgroup\$
    – Adi
    May 17, 2014 at 8:51
  • \$\begingroup\$ @GR TEch , how do I find αT , its not given in datasheet. \$\endgroup\$
    – Adi
    May 17, 2014 at 10:14

2 Answers 2

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The data sheet shows that the device can dissipate 6.5 watts on an Infinite Heatsink.

The value you are using (75 degrees C per watt) is for dissipation to ambient (No Heatsink). So for dissipating to ambient, the device can't really dissipate 6.5 watts.

Assuming 20 degrees ambient temperature, and you doing the math, you could only do 1.73 watts dissipation to ambient.

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As it says right in the datasheet, it can withstand 6.5W dissipation on an infinite heatsink. An unheatsinked chip will increase 75C per watt, limiting it to 2W before it reaches its maximum operating temperature. The point of a heatsink is to reduce its junction thermal resistance to some value less than 75 so that it can dissipate more power without hitting 150C. The 6.5W figure is saying that even if you kept it cold, that's the limit for the internal workings of the chip.

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