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I'm using the OPA551PA amplifier to charge a capacitor to various voltages. After thermal calculations, the worst case scenario is depicted here. TA = 40C, Vin = 36V, Vout = 18V, Iload = 50mA. Under these parameters, this chip should achieve a junction temperature of 125C, which is obviously too hot. There's a caveat however - this is a transient condition. We are charging a capacitor, so the charging current dramatically decreases as a function of time. My graph plots the charging current of the capacitor and the resulting thermal heating vs time. You can see that we are only in worrisome territory for some ~200ms. In the next 500ms it quickly stabilizes to the resting junction temperature that's calculated at the quiescent current. This transient also only happens once every 5 seconds. Again, this is worst case! The typical condition is the calculated junction temperature is greater that 80C for only 50ms, thereafter returning to 40C after 200ms - this process happening once every 4 seconds.

My question is, how long does it actually take a chip to heat up to the calculated junction temperature? If this was a sustained load, obviously this would be a problem. But does the chip in real life actually reach those temperatures in a <150ms timeframe?

Its like putting a pan on the stovetop. Your stove is at say 100C. Your pan temp will not immediately rise to 100C. There will a gradual rise in temperature until they match.

I'm planning on using a much more powerful chip anyways so I can keep junction temps close to ambient, but I'm still extremely curious about this.

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    \$\begingroup\$ More manufacturer data is needed for this but I think I saw someone on here do rough calculation using a tiny cube of silicon and it's thermal properties such as specific heat capacity and conductivity as a model. But even if you could, hot spots due to defects in the material put limits on how much you can shorten the pulse length while raising the pulse magnitude because it concentrates the heat when the pulses are intense enough because heat can't leave fast enough to distribute evenly in the material. hundreds of ms is a long time though. \$\endgroup\$ Commented Nov 4, 2022 at 16:25
  • \$\begingroup\$ You need a much better, finer thermal model for your device than what you have. You need to thermally model the high current/power parts of the chip (the big transistors, I assume), where those devices are located on the die, the thermal characteristics of the die out to the package, etc. It's not very easy of straightforward. \$\endgroup\$ Commented Nov 4, 2022 at 17:01
  • \$\begingroup\$ We've done such modeling for pulsed RF amplifiers (like are used in a radar application), but these were amplifiers we designed in house and so knew all the fine design details. \$\endgroup\$ Commented Nov 4, 2022 at 17:03
  • \$\begingroup\$ Only half-kidding, but that device has built-in thermal shutdown with a dedicated pin to flag when it occurs. In the absence of thermal response data from TI, you could test it yourself and see if you trip the shutdown. \$\endgroup\$ Commented Nov 4, 2022 at 17:20
  • \$\begingroup\$ Can you provide a schematic? \$\endgroup\$ Commented Nov 4, 2022 at 18:15

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