I have an external DC power supply with a 7805 voltage regulator.
Input voltage for the 7805 is 19V, output is exactly 5V.
I want to power my Arduino using this. I cannot feed this into the Arduino's power jack, since this goes to
the onboard voltage regulator that requires at least 7V. You can power through USB though and that is 5V.
So I hacked a usb cable: the connection end goes into the Arduino USB port, the stripped end is attached to
my 5V power supply. The Arduino powers up nicely and works fine (just running the blink sketch). Power
consumption is 5mA max.
However as soon as I power the Arduino this way, the 7805 in my power supply goes burning hot.
It goes from room temperature to over 50 °C in a matter of seconds.
I'm very uncomfortable with this, as this feels like a short circuit somewhere.
Any guesses about what's going on here? Is there something in Arduino's internal power management that prevents this setup?
(the external power supply stays exactly at room temperature when loaded with a pure resistive load that draws the same current as the Arduino does.
The heating up is definitely not default behavior for the 7805).
No guesses required, it makes perfect sense. The 7805 dissipates power according to (approximately) (Vin-5)*Iout, which in your case has Vin=19V and Iout ~= 30mA. So your power dissipation is (19-5)0.03 = 0.42W, and assuming 65 C/W thermal resistance for your 7805 suggests it should heat up by 650.42=27.3C above room temperature, which would take it to about 50C as you observe.
I don't have an explanation for why this doesn't happen with a resistive load other than a) you have it hooked up wrong, b) the resistance isn't what you think it is.
--
The Ruggeduino: compatible with Arduino UNO, 24V operation, all I/O's fused and protected
If the arduino under normal circumstances draws 5ma then your 7805 regulator has to dissipate 5x(19-5)/1000 watts = 0.085watts which under normal circumstances would cause no over=heating. It is clear therefore that you are drawing far too much current through your regulator and I suspect you have a fault in your 'hacked" USB cable.
For testing purposes I've powered my arduino with a 7805 by connecting the 7805 +5v to the arduino +5v pin and connecting the arduino and 7805 grounds together.
zoomkat:
For testing purposes I've powered my arduino with a 7805 by connecting the 7805 +5v to the arduino +5v pin and connecting the arduino and 7805 grounds together.
Nothing wrong with that, actually for more than testing purposes. I have one here that's been running similarly off an external 5V regulated supply for months and months. It consists of an Uno, an Ethernet shield, a 4x20 LCD and a couple other little things that don't amount to much, current-wise. The regulator on the Uno gets pretty warm if I try to run everything from it.
Thanks to your responses I figured it out, I appreciate the help!
When measuring the current the Arduino uses, I simply misread the digits.
It uses 50 mA, not 5 mA.
I tested the 7805 with a resistive load that used 5 mA, and according to the heat dissipation formula,
in that case the 7805 temperature will only rise with a few degrees, as observed.
However, the Arduino really uses 10 times as much current as that:
with 50 mA, it is indeed perfectly normal that the 7805 temperature should be hot, its temperature is 65 °C when measured, which is
what ruggedcircuits calculated. The hacked usb cable works fine.
I think I will search for an alternative because the 7805 is consuming 1W, of which 1/4 is used for the Arduino, and 3/4 is now converted to heat...
If the wasted 3/4ths of a Watt are a problem, I can think of two solutions -
Lower voltage into the regulator. If you built the power supply yourself, changing to a 6VAC transformer will give you about 9VDC after rectification and filtering. That will bring your efficiency from ~25% to ~50%.
A switching regulator is more complicated to build, but it can be ~90% efficient.
If you were running off a battery, the wasted energy will reduce battery life. But in most applications, the wasted energy is not a problem, especially if you are running off of "wall power". With appliances, heating, and lighting consuming thousands of watts, an additional watt is not going to add anything measurable to your electric bill. In the winter it essentially adds nothing since any heat generated subtracts from the heat needed from you furnace. (Except electric heating is usually more costly than gas or oil.) The same is true in an automotive application. You are not going to notice any loss of power or any loss of gas mileage with an additional one-Watt load on the engine. Your car lights are over 100W total, and you don't notice any difference in performance when driving at night...
A valuable lesson there, rubbish data (5ma) gets rubbish answers (wiring fault).
A cheap and cheerful way of preventing the 7805 overheating is to dump some of the supply voltage before it gets to the regulator.
If you are confident that the supply will remain at 19 volts and the consumed current will remain at 50 ma then you can drop around 9 volts across a series resistor of 180 ohms. This will then present around 10 volts to the regulator and so reduce its dissipation down to around 0.25 watts, rather than your original 0.7watts. The 180 ohm resistor dissipation will be around 0.5 watts so a unit rated at least 1 watt will be required.