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avoiding overheating for 7812 for 3A Answered

I've bought a 3A 12x2v tramsformer. . from half m getting about 18v output. applying bridge rectifier i filtererd into DC. But my problem is using LM7812 or LM7809 it gets extremly hot although i've used heatsink... so my question is
 
1. What can I do to avoid overheating ?
 
2. How can I use this 3A transformer as 1 or 1.5 A if i need?

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iceng

3 years ago

If it gets too hot you can add transistors that can pass more current by following the LM7812... Notice the fuse for the regulator is 1A but when adding the PNP a 3 or 5A fuse can be used...

I got the impression you did use much capacitance ?

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icengiceng

Answer 3 years ago

I'm asking you, to please comment on what capacitors you are using in your circuit ?

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Yonatan24

3 years ago

I don't think they're supposed to deliver so many amps

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-max-

3 years ago

To calculate the power lost (dissipated) Take the difference of the (average) input voltage and output voltage, and then multiply that difference to the current drawn. (18V - 9V) * 1.5A = 13.5W of heat that must be dissipated. That is about how much heat a small soldering iron might output.

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In order to dissipate that much heat, you need a heatsink that has a good thermal resistance between it and the air. Thermal resistance is calculated just like electrical resistance, if you treat temperature (in kelvin or celsius) like voltage and thermal power dissipated, (in watts) as current. And don't forget to factor in the temperature of the air, or ambient temp. (If you use a massive heatsink, it will also have some thermal "mass." That is, it takes a lot of heat to warm it up or cool it down quickly, you can treat this as a capacitor in electronics, but don't concern yourself with this unless you are pulling power from the 7809 for short periods of time and are trying to use the smallest possible heatsink for the job.)

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The way I do it quickly and easily is to make a "thermal circuit.", where you have ground, (or absolute zero) then connected to it is a temperature (voltage) source, and the temperature after this "part" is, say, 30*C. which is the ambient air temperature. Then above it are all your thermal resistances. From the air to the heatsink, from the heatsink to the 7809, and from the case of the 7809 to the magical silicon chip inside it. Each thermal resistance will increase the temperature of the silicon junction base on how much heat is being dissipated.

heat dissipated (treated as Current) will flow from the junction, through all those thermal resistances, and eventually to the ambient air. You can find ratings for the heatsink in *C/W as well as the datasheet for the 7805. You do not need to calculate the temperature rise of every specific thermal resistance if you don't want to, you can just add up all the thermal resistances because they are effectively in series. Say your heatsink has a rating of 10*C/W. This means that for every Watt of power dissipated, then the heatsink temperature will rise by 10 degrees, and I will assume that the sum of the rest of the thermal resistance (from junction to heatsink) about 1 *C/W, so about 11*C/W total. If you are dissipating 13.5W, then the temperature of the junction can reach 148.5*C. plus ambient air. Clearly that is too small a heat sink. The junction should NEVER exceed 120*C, let alone 100*C.

I will let you figure out what the smallest heatsink you can use is. Just do the calculations I did but in reverse.