4
\$\begingroup\$

According to this chart of a diode's power dissipation over temperature, the power dissipation would decrease as temperature goes up.

But wouldn't it be the exact opposite and even more power be dissipated as heat, when the temperature goes up ?

enter image description here

\$\endgroup\$
8
\$\begingroup\$

But wouldn't it be the exact opposite and even more power be dissipated as heat, when the temperature goes up?

No. The ability of the diode to dissipate heat will be proportional to \$ \frac {\Delta T}{R_T} \$ where \$ \Delta T \$ is the difference in temperature between the chip and the ambient and \$ R_T \$ is the thermal resistance between the chip and ambient. The latter parameter covers the thermal resistance of the chip to case, case to heatsink (if provided) and heatsink to air or coolant.

Since the maximum temperature of the chip is limited to 175°C then as \$ \Delta T \$ decreases due to increasing ambient temperature the power to be dissipated must decrease to stay in the safe operating area (SOA).

Remember that temperature of the device rises until the power lost to cooling equals the electrical power input. \$ P_{IN} = P_{OUT} \$.


From the comments:

So the watts in the "Power dissipation" is actually current * voltage.

Yes, but remember it's just the voltage across the diode which, for silicon diodes, will be about 0.7 V. So at 1 A you will have \$ P = VI = 0.7 \times 1 = 0.7 \ \mathrm W \$ dissipated in the diode.

... and the more watts goes through the diode (if i can put it that way), ...

Probably not the best way. Watts that "go through" the diode won't affect it as they'll get to the load. Watts dissipated at the diode are the problem.

... the faster it heats ...

Yes.

... and so as the ambient temperature goes up the less watts are needed to increase it's temperature to 175C in which the diode would stop functioning?

Correct.

\$\endgroup\$
  • \$\begingroup\$ What is "The power to be dissipated" , isn't all power dissipation = heat dissipation ? \$\endgroup\$ – soundslikefiziks Dec 24 '17 at 19:00
  • \$\begingroup\$ I didn't use the expression, "The power to be dissipated" so I don't understand why you are quoting it (back) to me. You appear to have written two sentences but separated them by a comma so it's not clear what you are saying. "Isn't all power dissipation = heat dissipation?" Ultimately, maybe, but in the short term, no. For example, some power could be in the form of electromagnetic radiation such as radio waves or light. Only if that is absorbed by something will it be converted to and that at the point of absorption not at the source. \$\endgroup\$ – Transistor Dec 24 '17 at 19:06
  • \$\begingroup\$ You wrote "the power to be dissipated must decrease to stay in the safe operating area (SOA).", Are heat dissipation and Power dissipation interchangeable when we talk in electronic terms ? \$\endgroup\$ – soundslikefiziks Dec 24 '17 at 19:22
  • \$\begingroup\$ Got it, thanks. Yes, in this case all power dissipated will be in the form of heat. (In other cases it might not as explained above.) \$\endgroup\$ – Transistor Dec 24 '17 at 19:31
  • \$\begingroup\$ So the watts in the "Power dissipation" is actually current * voltage. and the more watts goes through the diode (if i can put it that way), the faster it heats and so as the ambient temperature goes up the less watts are needed to increase it's temperature to 175C in which the diode would stop functioning? \$\endgroup\$ – soundslikefiziks Dec 24 '17 at 20:24
12
\$\begingroup\$

The graph is not showing how much power the diode will dissipate as a function of temperature.

It shows permissible power dissipation as a function of ambient temperature. Think about it. If it's hot all around, less power will get something to a particular temperature than when it's cold all around.

In this case it looks like the diode stops functioning at 175 °C. If it's already at 174 °C, then you can't put much power into it before it won't function anymore. If it's at 25 °C, then you can put a lot more power into it before it gets to 175 °C and stops working.

\$\endgroup\$
  • \$\begingroup\$ so it shows "how much power would be needed to destroy the diode" at different temperatures ? \$\endgroup\$ – soundslikefiziks Dec 22 '17 at 23:07
  • 2
    \$\begingroup\$ No, it shows how much power the diode by itself can dissipate at different temperatures. This chart is means to aid in thermal design, to know where to place or not to place the diode on the board, and to figure out what size of heat sink you need. \$\endgroup\$ – Drunken Code Monkey Dec 23 '17 at 2:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.