Linear regulators overheating

Question

I've designed and built a small board for supplying about 2A at 12V (regulated) from a 18V unregulated wall wart for a project of mine. It uses two LM7812 linear regulators to supply 1A each to separate loads. My problem is that it is overheating.

Below is the schematics.

Below is the board design.

And here is one picture of the prototype.

My question is: What's wrong with my board?

Is it overheating because of the undersized heatsink, or is there a problem with the design? Or should I just scrap it and throw away all linear regulators in my bins and design a new one using switching regulators?

Answer 1

This is a great example of why you need to do the math and nost just blindly build something. It should have been obvious from the start your regulators would get hot just from the basic physics.

You say you have about 18 V in and that each regulator is producing 12 V at 1 A. That means (18 V - 12 V) * 1 A = 6 W is being dissipated by each regulator. That is way more than a TO-220 case can handle in free air.

There are two obvious options:

1. Put a heatsink on each regulator. Make sure it can handle the 6 W of power and still keep the part cool enough at whatever your ambient temperature is. This time do the math up front. Make sure to take into account not just the thermal resistance of the heat sink but also the thermal resistance of the die to the case. The total must be low enough so that 6 W into the die won't exceed its operating temperature.

   For example, let's say you need to keep the die at 125°C or below (I just made that up, it's your job to look at the datasheet and find the real number) and that your ambient won't exceed 25°C. That means you can afford up to 100°C rise in the die above ambient. Since 6 W of power is being dissipated, that means your combined die to case and case to ambient thermal resistance needs to be 100°C / 6 W = 16.6°C/W.

2. Use a swithing regulator. At these voltages, you should be able to find something that can do 2 A out. The single switcher will replace the two linear regulators. It will also be smaller and cheaper since it won't have to deal with getting rid of 18 W of heat.

Answer 2

Firstly, you should be looking at what the voltage is doing with an oscilloscope, on both sides of the regulators, under load. The DC measurement of 19.5V isn't accurate because it's not pure DC. Your multimeter probably does RMS measurements assuming sinusoidal waveforms, which is moot because that's a feature of its AC measurement mode.

Your unregulated power supply's voltage causes too much drop out. Find one that puts out around 15V. You can go as low as 14.5. See this question: 7812 minimum input voltage? Minimizing the drop-out will knock Watts off the heat production.

The wall-wart may not have enough of a reservoir capacitor inside. Or maybe none at all: just a transformer and bridge.

The 0.33 uF capacitor on the input side of your regulators is almost certainly not enough to sufficiently reduce ripple. Ripple has to be reduced at least to the point that the voltage on the input side never dips below that minimum 14.5V needed for the regulator to maintain regulation.

The 0.33 uF capacitor value from the datasheet is not intended to be a reservoir capacitor. It is not required if you have a reservoir capacitor close to the regulator; it's there for stability and should be designed if the reservoir capacitor is far from the regulator (for instance, if it is inside the unregulated supply).

Putting two and two together, it should be obvious that the reservoir capacitor also helps you minimize the input voltage: the smaller the peak-to-peak ripple you can achieve, the closer the voltage can be to the minimum.