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I'm using an LM7805 in voltage regulator mode.

  • Vin is 15V
  • Vout is 5V (regulated)
  • The attached load draws 200mA

LM7805

How to calculate how much power the LM7805 would dissipate in this setup (and whether or not I need a heat sink)? Couldn't understand it from looking at the datasheet.

P.S. I'm asking because it gets quite hot, which I didn't expect. Checked the attached load, it is around 0.2A, so well within the limit of 1.5A.

rustyx
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    For a linear regulator, Iout is roughly equal to Iin. All voltage dropped across the pass device is dissipated as heat. – M D Mar 30 '19 at 10:55
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    Wait, this has been asked multiple times, see [here](https://electronics.stackexchange.com/questions/18478/my-linear-voltage-regulator-is-overheating-very-fast). – Unknown123 Mar 30 '19 at 12:38
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    Also there is more detailed datasheet [here](http://www.ti.com/lit/ds/symlink/lm340.pdf) – Unknown123 Mar 30 '19 at 23:37

1 Answers1

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Take a look at package thermal data section in your datasheet page two.

Depending on the package of the regulator you are using, lets say the thermal junction-ambient coefficient \$\theta_{JA}\$ is roughly \$20°C/W\$.

You've got \$ P = V \cdot I = (15\ \text{V} - 5\ \text{V}) \cdot 0.2\ \text{A} = 2\ \text{W} \$ dissipated as heat.

If your ambient temperature is \$25°C\$, then the regulator would heat up more or less into \$65°C\$.

It is quite hot for sure.

Community
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Unknown123
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  • 2W is about the limit for use without a heatsink. – Jasen Слава Україні Mar 30 '19 at 11:48
  • Yes, I'm very linearizing here for simplification. The ambient will also heat up which eventually also increase the regulator temperature. Also, in the `NOTE 1` below the package thermal table in the datasheet OP linked there are maximum power dissipation formula and careful warning statement which would be useful for OP while learning this topic. – Unknown123 Mar 30 '19 at 11:56
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    Different manufacturers have [*very* different](https://www.petervis.com/electronics/Voltage_Regulator_Heatsink/Heatsink_for_TO-220_Voltage_Regulator.html) thermal resistances. For TI it's `19 °C/W` (--> 63 °C) for STM `50 °C/W` (--> 125 °C), for Fairchild even `65 °C/W` (--> 155 °C). The latter two will probably die without a heat-sink (see absolute maximum ratings). – Suuuehgi May 03 '20 at 15:40
  • As far I remember, I assumed 20 as a best case for quite hot. +1 for clarifying. – Unknown123 May 04 '20 at 12:40