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[on:]

I made a 7805 based 5V power supply for my ATmega chip. Connected LED via resistor and hooked it up to 12 VDC power supply. LED lit fine, voltage was around 5V on the out pin and everything seemed ok. I disconnected 12V supply.
Then I connected FTDI USB header to program the chip. LED lit up again (which surprised me at first, but then I realized that it provides power from USB so I don't even need to connect 12V supply to the jack), and I burned Bootloader to the chip just fine, but few minutes later I noticed that LED went off. I touched LM7805 and it was super hot, and apparently burned out.  Nothing else was damaged.
This took me by surprise, I didn't even used my 12VDC external power supply in the setup.  What did I do wrong?  Do I need put a diode between LM7805 and +5V line?
Here's a basic schematic (doesn't include ATMega, and other components)
 

[Reply #1 on:]

Feeding a voltage into the wrong end of the regulator is not recommended, however I have never come across a case of this burning out the regulator.

Was the 12V supply left connected up but not powerd up. That could cause a short on the input side and produce what you saw. If this is the case then a diode would prevent this from happening.

[Reply #2 on:]

What did I do wrong?  Do I need put a diode between LM7805 and +5V line?

Typically a reverse-connected diode between 7805's input and output.

[Reply #3 on:]

Feeding a voltage into the wrong end of the regulator is not recommended, however I have never come across a case of this burning out the regulator.

Was the 12V supply left connected up but not powerd up. That could cause a short on the input side and produce what you saw. If this is the case then a diode would prevent this from happening.

Nope 12V wasn't connected at all
I must've wired somewhere wrong somewhere... Can't tell now...

What did I do wrong?  Do I need put a diode between LM7805 and +5V line?

Typically a reverse-connected diode between 7805's input and output.

Thanks! You mean like this?

Would it drop my output voltage noticeably?

Also I was wondering, if I'm using my own power supply, I don't really have to connect FTDI's VCC to the circuit at all right? Just the ground and RX/TX?

[Reply #4 on:]

Thanks! Would it drop my output voltage noticeably?

No, it just forces any 'reverse current' around the regulator thus protecting it.

Also I was wondering, if I'm using my own power supply, I don't really have to connect FTDI's VCC to the circuit at all right? Just the ground and RX/TX?

That is correct and probably your best choice. However I would wire two 1k ohm series resistors between the Rx and TX pins from the FTDI pins to pins Rx and Tx pins on the AVR chip so that if your supply is powered up but the USB cable is not plugged into the PC that some reverse current won't try and back into the unpowered FTDI chip.

Lefty

[Reply #5 on:]

If you follow dhenrie's advice you short everything out through the diode and end up frying that first. He has a certain reputation on this forum.

As well as RX and TX you also have to connect the DTR line to get the auto reset.

[Reply #6 on:]

I'd put the diode on the 12v input side of the 7805 regulator.

[Reply #7 on:]

I'd put the diode on the 12v input side of the 7805 regulator.

Yes that is what I meant, in seriese with the supply and the regulators input. I know dhenery recommends the reverse bypass diode but it just shifts any problem into the diode.

[Reply #8 on:]

If you look at the schematics for the UNO and Duemilanove boards, they have circuits
so that power jack input and USB can co-exist without fatalities.

You're lucky you didn't fry the USB port on your PC. Last week I actually hooked up my
FTDI cable [same onboard I/F as shown here] to a board powered by the power jack, and
it shutdown my notebook. Luckily the USB port survived.

I discovered the idiots who designed the pcb had a TINY TINY 6-8 mil trace covered by
solder mask that was shorting across 2 pads that you wire together to power the board
from the FTDI header.

Who does something like that? Either you the user solders in a heavy jumper or not. Makes
zero sense to have a pre-existing invisible 6 mil trace bringing power in. How's about 50 mils,
guys!

[Reply #9 on:]

If you look at the schematics for the UNO and Duemilanove boards, they have circuits
so that power jack input and USB can co-exist without fatalities.

I don't trust the arduino onboard power setup. I powered my arduino with 12v thru the barrel jack and measured 8v+ on the board 5v pin. I noticed the board LEDs were brighter when powered via the external power jack and decided to check the board voltage with a multimeter.

[Reply #10 on:]

I don't trust the arduino onboard power setup.

Hmmm, I assumed those things had been fully tested.

[Reply #11 on:]

If you look at the schematics for the UNO and Duemilanove boards, they have circuits
so that power jack input and USB can co-exist without fatalities.

I don't trust the arduino onboard power setup. I powered my arduino with 12v thru the barrel jack and measured 8v+ on the board 5v pin. I noticed the board LEDs were brighter when powered via the external power jack and decided to check the board voltage with a multimeter.

Well if that was a true and accurate measurement then it should have destroyed both the AVR chip and the USB serial converter chip as 8vdc is well above their absolute maximum safety value. Now if one was to measure between the Vin pin and the 5v pin with a voltmeter then 8vdc could be a perfectly valid reading. But if the on board power led was indeed much brighter then normal then of course you have a true over voltage problem and would almost have to be a defective on board +5vdc regulator. That would be an unusual failure but certainly possible.

Lefty

[Reply #12 on:]

USB has resettable fuse But the computer will remember what tripped it on some of them and shut the USB power off if you hook it back up.

Had that happen on one pc had to uninstall the device.

I would use the same setup as the uno like the pic