How hot does a voltage regulator get?

Question

Since I've always did electronics through my Arduino and I feel I'm pretty much a beginner, I've never used a voltage regulator before. I want to create this circuit, but I want to hide it in my printer so I want it to work without a battery.

http://hacknmod.com/hack/beginner-spy-tutorial-your-first-diy-mini-fm-bug-transmitter/

I am planning to put a 3 V voltage regulator on it, but I wonder how hot a voltage regulator can get if the maximum voltage is used. And should I be worried about the heat, as in should a try to cool it?

Answer 1

The heat produced by a voltage regulator isn't a function of the voltage, it's a function of the power.

You can calculate the power the voltage reg will dissipate by taking the voltage-drop across the regulator, times the current flowing through the regulator.

E.G: $$Power = Volts * Amps$$

In other words, if you have a 30V source, and your device is running off 3V, you have \$30V - 3V = 27V\$ across the regulator. However, if your device is only drawing ~3 mA: $$27V * 0.003A = 0.081W$$

You would only have a dissipation of 81 milli-watts, which wouldn't even get too warm to the touch.

However, if you have a 5V input, with a 3V output (giving 2V across the regulator), yet your deice is drawing 1A, you have: $$2V * 1A = 2W$$

You have a power dissipation of 2 watts!

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Basically, there is more involved in evaluating heat production then just the voltages

Answer 2

Given the head from Fake Name, you need to figure the temperature. The parameter Theta JA, \$\Theta_{JA}\$, tells you for every Watt the part dissipates, the chip will go up that many degrees above ambient. The part's data sheet will tell you this parameter. JA means from Junction (chip die) to Ambient. However you will probably see it get hotter since it is in an enclosed are. Another one you will see is Theta JC, Junction to Case. That's mainly used with heat sinked part.

The part's data sheet will also tell you the maximum junction temperature. For regulators, that is often 125 C.

Answer 3

I think it should be stressed that a linear voltage regulator, which acts as a variable resistance in series with the supply, is dissipating most heat when it is supplying its maximum current at its minimum voltage. In this case the voltage drop across the regulator is a maximum. If there is any choice, use an input voltage, that at its lowest level, is just above its "dropout" voltage. i.e. the minimum the regulator can work at.