What happens if I draw too much current from ATMEGA IO pin?

Question

I revisited an older device that I made and discovered, that I had an 8 Ohm speaker connected directly to ground and an IO pin of an Atmega328P. According to my calculation that means that the pin would have to source 625 mA, which is way above the 40 mA that any single port should be able to source and even above the 200 mA that the Vcc and GND support. Yet it somehow worked and the chip seems to be working just fine. The application uses a duty cycle of 50%, so in the best case it would use maybe 312.5 mA, which is a little better but still way above the ratings.

So my question is, what happens if the output pin is asked to source more than it's rating?

I've seen chips that just release their magic smoke while others just can't keep up with supplying that kind of voltage, so only the voltage drops without any other problems. How does the Atmega328P handle this?

The Atmega clearly didn't burn through, but what other kind of damage should I expect?

Answer 1

There's a few crude calculations you can do using some numbers from the datasheet.

This graph is in the datasheet:

It shows the voltage dropped by the high-side MOSFET driver in the IO pin for different amounts of current draw. From that we can get its resistance R(DSON).

For instance at 10mA, at 25C, the output voltage would be 4.75V. That's a 0.25V drop from the 5V supply voltage. So we can calculate the resistance as:

• R=V/I = 0.25V / 0.01A = 25Ω

Now we have that resistance we can use it to calculate the actual current from the IO pin with an 8Ω load. Simply add the two together:

• 25Ω + 8Ω = 33Ω

Now we can calculate the current through that circuit:

• I=V/R = 5V / 33Ω = 0.152A

That is, of course, way above what the IO pin can actually supply (or so the datasheet says).

So what happens? Well, two things, basically:

1. The voltage on the IO pin will be below the 4.2V specification for the minimum of an output pin when HIGH, and
2. It will get hot.

The actual voltage will be, of course, V=I×R = 0.152A × 25Ω = 3.8V droped, so 5 - 3.8 = 1.2V

What is more critical, though, is the heat. 152mA through a 25Ω resistance generates P=I²R = 0.152 × 0.152 × 25 = 0.5776W. That's quite a lot. Enough to melt the internal circuits in the chip and release the Magic Smoke™.

Running at 50% duty cycle, though, makes a difference. That's 50% of the power, which makes a big difference to the heat.

What would the actual allowable heat be? Well, at 40mA, with 25Ω, that is 0.04 × 0.04 × 25 = 0.04W.

So you're generating more than 14 times the amount of heat you are allowed to. Even at 50% duty cycle, that's over 7x the limit.

Will it kill the chip? Well, possibly, possibly not. It depends how rugged the chip is and how well it can withstand the heat. Short bursts may be OK since the heat doesn't build up enough to melt things. Extended use, though, will make it get too hot, and things will go pop.