Hi everybody.
I´m trying to supply a circuit that requires only 500ma and I have a 24v switching power supply as input to the 7805. The thing is that I have a couple of detectors that are supplied with 24v and they have a single 7805 without heatsink and the 7805 does not get hot at all!!! but when I try to power mine with 24v it gets very hot and over time the output decreases. I would like to know if any of you have a suggestion or idea for preventing this and proper supply the 7805 with 24v.
I´ve noticed that some circuits have a small value resistance in series with the input of the IC, I´ve tried this also but the resistance gets hot too! if any of you have a circuit will be fine, o an idea to do this.
Hello, and welcome to the forum!
24VDC is way too high for input into a 7805 with no heat sink which is expected to supply 500mA. I don't have the formula in my head right now (It's sadly been a while since I've had to do this calculation so I hope I'm corrected if I mess up the actual numbers here) but I think that's something like 12 Watts [Edit: should be 9.5W] of wasted power which the 7805 must dissipate as heat. The 78xx series have a thermal resistance of 65°C/W above ambient from the junction to air, which IIRC means that with no heatsink, the junction's temperature will rise by 65°C above ambient for each Watt of power which must be dissipated. The maximum operating temperature is 125°C, so you can see that this is clearly being exceeded unless you're running the thing in a space which is -492.5°C. The 7805 won't like that and will go into thermal shutdown (it tries to protect itself by shutting down instead of just frying).
I think that the reason you don't see this with your other "detectors" (whatever they are) is that they draw almost no current through the 7805s. The more current you draw through it, the more power you need to dissipate as heat.
You need to bring down the input voltage to the 7805. The resistor you sometimes see on the input of a 7805 is for this purpose: to drop some voltage, and thereby shift some of the heat dissipation off the 7805 and onto the resistor. That resistor would need to be a power resistor and would probably also need to be heat sinked, at least with a 24VDC input. What I *do* know is that with a 24VDC input, you are not going to get 500mA out of a 7805. If you feed the 7805 with 9VDC instead of 24VDC you can do it.
I'd suggest using a more appropriate power source than your 24VDC. Or you could use one or more power resistors to drop the voltage, but then you're just moving the heat dissipation off the 7805 and you will need to dissipate it from the resistors (which may be a better idea). You could also put some other regulators (say, maybe a 7818 and 7812 in series) between the 24VDC power supply and the 7805, but that's a messy solution IMHO. [Edit: No matter what, you're going to need a good heat sink and heat sink compound.]
Also--you're including the input and output capacitors on the 7805, right? Without them, your 7805 might oscillate and heat up no matter what voltage you feed the thing with.
At any rate, good luck! I still suggest using a more appropriate power supply than the 24VDC one. Using that means that you have to do more work to keep things cool enough to work.
Regards,
Torben
[Edit: MikeMI got his numbers right and I got mine wrong, I think! You still have to dissipate way too much power, though.]