Right, I am having first of all a few problems concerning with efficiency of a 5V voltage regulator... It is heating up like crazy (connected to a 22V supply, 10xAA 1.2V) and that is before it is supplying any power. Before long it switches itself off from overheating.

I have it connected up as

|    |    |
+V GND OUT

Like as you would... But is there a way to stop it heating up insanely? Its max voltage is 35V i am so to believe - so i am well within the boundaries there...

---

Oh and also, say i have 6x LED in series - would the current dissipated be 30mA or 6x that? I know ofcourse a voltage drop of each one is 3V - so the required voltage would be 15V. Just wondering because I want to use a resistor to make them as bright as physically possible without burning them out

Thanks

 

6 LEDs in series: current is 30mA or whatever goes through one LED.

However, you calculate the power dissipation of your regulator as such:
\(I_{out}\times(V_{in} - V_{out})\)

So the power dissipation is: 0.51 watts plus various inefficiencies, call it 0.6 watts.

For anything over about 0.5 watts you'll probably need a simple clip on heatsink.

Though for 0.6 watts it still shouldn't be THAT hot...

You also need to be aware that there are no shorts on the output and the regulator is properly grounded. Also check battery polarity, there is a diode which shorts out reversed batteries to protect the regulator I think.

 

Yeah but how can I calculate Ic if it isn't even supplying anything yet?

Edit:
Just realised even as small as 100mA, it is 2 watts i think i ought to have a heatsink with this

Edit again:
To save all the hassle i might as well use a PP3 to power this part of the circuit with the regulator... It is only a PIC, so that isnt going to gobble up a load of power anyway

Thanks again

 

Ten 1.2 volt batteries make 12 volts...not 22.
A 5 volt regulator should not heat up when it's not supplying any power to a load. I expect you have made a mistake.
Your pinout description means nothing without the exact number of the regulator. The manufacturers have not made them all the same.
Milliamps is not a dissipation, it's a current flow. Milliwatts is a dissipation. 3 volts times 30 milliamps is 90 milliwatts.
6 of the 3 volt LEDs in series is 18 volts, not 15.

Now that we have the math corrected, we can get to the circuit. Please post a drawing which includes the part number of your regulator.

 

If you have not used caps on the input and output of your 3-pin regulator, you most likely have oscillations occuring, which will heat the sucker up good and right.

As for using 12v and dropping it to 5v, You will have to dissipate much more heat to drop the 12v to 5v then if you were dropping 9v to 5v. If you use an LDO, you can even go from 6v (5 1.2v AA) to 5v and it would be MUCH more efficient.

If you are using 10 AAs for the current, then make 2 banks of 5, connect the 2 banks in parallel to double the available current, while keeping the voltage low enough to not lose the efficiency in the regulator.

 

OP does not make any sense to me.
Either his regulator is faulty or he is doing it wrong.

12V applied to a 7805 will not heat up at 30mA load no matter what

 

If he has no caps, it could.

 

If it were a rectified DC, I would not debate on caps issue, but u see, OP is using AA cells.
Without caps would the reg heat up?

 

Ten 1.2 volt batteries make 12 volts...not 22.
A 5 volt regulator should not heat up when it's not supplying any power to a load. I expect you have made a mistake.
Your pinout description means nothing without the exact number of the regulator. The manufacturers have not made them all the same.
Milliamps is not a dissipation, it's a current flow. Milliwatts is a dissipation. 3 volts times 30 milliamps is 90 milliwatts.
6 of the 3 volt LEDs in series is 18 volts, not 15.

Now that we have the math corrected, we can get to the circuit. Please post a drawing which includes the part number of your regulator.

Well sorry about the mistakes but it was too late into the night to be thinking straight...

No i have no caps on the regulator, and yes the regulator is a 7805.

I also don't have the circuit diagram as I was only testing out on breadboard

 

If you read the datasheet for the 7805, it will tell you that you NEED output and or input caps on the regulator. Add them, and see if you still have the problem.

Remember, these are NOT shown in 90% of schematics or wiring diagrams. They are a GIVEN.

 

I've also seen regulators oscillate if the recommended capacitors are not used.

These caps have nothing to do with filtering ripple from the DC source; these caps are for preventing internal oscillation of the regulator.

It would be best to follow the manufacturer's recommendation, but it wouldn't hurt to try sticking a couple of 0.1 ceramic caps or maybe 10 uf tantalum caps into the breadboard as near the regulator legs as possible to bypass both the input and the output and see if that drops the unloaded current drain down to near zero amps.

 

I've also seen regulators oscillate if the recommended capacitors are not used.

These caps have nothing to do with filtering ripple from the DC source; these caps are for preventing internal oscillation of the regulator.

It would be best to follow the manufacturer's recommendation, but it wouldn't hurt to try sticking a couple of 0.1 ceramic caps or maybe 10 uf tantalum caps into the breadboard as near the regulator legs as possible to bypass both the input and the output and see if that drops the unloaded current drain down to near zero amps.

Right on. As close as possible. I would avoid tantalum if you are new to electronics, as an accidental reversal causes fireworks.

But the internal oscilliations WILL cause problems. Including overheating.

 

Doesn't nearly every electronic circuit oscillate and get hot when made on a lousy old breadboard?
Make a proper pcb or use stripboard with the tracks cut short.