3
\$\begingroup\$

As a simple project, I was trying to replicate an Arduino Nano, fabricating my own board. I'm using a UA78M05 linear regulator to feed 5V from a 9V battery (see image below) as external power supply.

schematic

simulate this circuit – Schematic created using CircuitLab

Now, this board uses an 6-pin IDC type connector for SPI programming. Since the 2nd and 6th pins of the IDC connector are, respectively, Vcc and GND, I'm applying a voltage to the output of the linear regulator. The problem is that the regulator immediately heat up (after maybe 10 seconds I can't touch it). Is it supposed to heat up like this and I simply forgot to put a heatsink? Or is there something wrong with the wiring? The rest of the board follows in the next image. All the parts were soldered by me and are SMDs on a board professionally frabicated (no home-made etching).

Arduino-uno-like board

Edit 1: Following the suggestion of Dwayne, I hooked up a 1N4007 diode across the output and input pins. The result was that now the heat distributes across the two components: for the same amount of time during which I keep the battery connected (or the IDC connector), if there's the diode, the regulator will heat less but touching the diode you can feel it heated. Is this an expected behaviour or there's something wrong with wiring/power dissipation?

\$\endgroup\$

4 Answers 4

5
\$\begingroup\$

You need to avoid backfeeding the regulator. This is done by diode ORing your two power sources (regulator and bus power). Note that [depending on your budget and requirements] this can be done with Schottky diodes or with ideal [IC, MOSFET-based] diodes, e.g. these.

The protection that Dwayne describes is useful when you cut the battery power to regulator (so its output may temporarily see higher voltage than its input), but is not a substitute for diode ORing when using multiple power sources that all have power at the same time.

Regarding heatsink: how much power are you drawing? At idle the Atmega328 shouldn't use more that 100mW so if you find yourself needing a heatsink at idle, something is probably wired wrong. That means 20mA at the most through the reg at idle. Times 4V (=9-5) it's about the same power dissipated in the regulator. Actually, which package of uA78M05 are you using? In the TO-220 you shouldn't be able to feel anything at this power, but with the smaller SOT223 package you might feet a small warm spot.

\$\endgroup\$
3
  • \$\begingroup\$ See electronics.stackexchange.com/a/49233/54580 for more on Atmega328 power consumption in various power saving modes. \$\endgroup\$ Nov 21, 2015 at 6:24
  • \$\begingroup\$ The problem arises even if just one of the power sources are connected. I sincerely didn't even try to plug both at the same time, for now. The Atmega is blank, so it does absolutely nothing. The only thing that draws power AFAIK should be the "power on" diode D1. So if the schematics are correct, I'm assuming the regulator maybe damaged? The regulator is a SOT223 package. \$\endgroup\$
    – user92300
    Nov 21, 2015 at 11:23
  • \$\begingroup\$ @mmassaro: If you can disconnect the atmega and leave just the LED and its series resistor as load for the regulator, you'd be able to know better what is going on. Something is damaged or shorted given what you say. \$\endgroup\$ Nov 21, 2015 at 11:48
1
\$\begingroup\$

The standard protection for your situation is to add a diode across the regulator. Anode to Regulator Output, Cathode to Regulator Input. 1N4005 or similar is good.

\$\endgroup\$
0
1
\$\begingroup\$

The TI LM7805 Family datasheet, which seems typical of LM78xx regulators, discusses back-feeding power in section 8.2 (p.13):

8.1.2 Raising the Output Voltage Above the Input Voltage

Because the output of the device does not sink current, forcing the output high can cause damage to internal low current paths in a manner similar to that just described in Shorting the Regulator Input.

And that description is:

8.1.1 Shorting the Regulator Input

When using large capacitors at the output of these regulators, a protection diode connected input to output (Figure 15) may be required if the input is shorted to ground. Without the protection diode, an input short causes the input to rapidly approach ground potential, while the output remains near the initial VOUT because of the stored charge in the large output capacitor. The capacitor will then discharge through a large internal input to output diode and parasitic transistors. If the energy released by the capacitor is large enough, this diode, low current metal, and the regulator are destroyed. The fast diode in Figure 15 shunts most of the capacitors discharge current around the regulator. Generally no protection diode is required for values of output capacitance ≤ 10 μF.

So it sounds to me as if the heat being generated when you back-feed the regulator is a very bad sign, and you are risking damaging the regulator.

From "Edit 1" in your question and your later answer it sounds as if the diode, by bringing the input voltage up to slightly below the voltage at the output, mitigated the problem in this case.

\$\endgroup\$
0
\$\begingroup\$

I found out the problem. It was a very silly problem of the diode that was reversed. Changing the orientation solved the issue.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.