Can High LED Current Actually Burn Up Output Pins?

erco

Senior Member
******************* Worst practice observation follows. *****************************

I slammed together an 5V 20M2 RGB LED project lately without current limiting resistors and it works just fine. Three output pins drive a parallel PAIR of 10mm RGB LEDs. TTYTT I did put one resistor onto green just to balance the color intensity and get a better white. No PWM, just full power high/low. Working great, even though each pin is outputting some 50-60 mA, occasionally all 3 pins on at once for white. I'm fully aware I'm pushing it and operating outside the recommended parameters. I was in "git her done" mode to film a video. I had good intentions to rectify (!) the LED problem later and add resistors before something burned up but it's still working, if only for a few minutes at a time. Someday I'll wire up a similar non-critical circuit and do a continuous destructive test.

I used to get my knickers in a twist worrying that a Picaxe pin can sink more current than it can source and build the circuit accordingly. I'm starting to wonder if it matters. The internal impedance self-limits current at some point. Of course a dead short would led more current flow, but any LED will drop 2-4 volts and the internal impedance can drop the rest. Deal?

Of course I'm pushing my luck and Hippy and Technical can't condone my actions, I'm just curious if anyone has actually burned up an output pin from too much current.
 

jscottb

Active member
Most modern light-duty LED's should be ok, maybe. I will still throw a buffer chip or Darlington array on the board even today if I am driving more than a few LED's though. I do *try* to follow the datasheet specs, but some times it's "who cares?" ;-)

Along these lines, I have noticed in my working of late with Chinese made uC's (like Attmega clones) have what seems a good bit higher sourcing ability than the real chips. No science behind that, just a casual observation.
 

AllyCat

Senior Member
Hi,

Is that sourcing or sinking the current? I don't believe the M2 (FET output driver) pins are (normally) capable of sourcing that much current, even at the maximum supply voltage. The N-channel (pull-down) FET might sink 50 mA, but its well above the claimed capability of the chip. The pin driver stages are characterised towards the end of "section 31" in the M2 base PIC Microchip Data sheets (for three different supply voltages).

The "Absolute Maximum" current rating of the pin output drivers is 25mA, but Microchip are notoriously cautious in their specifications (note that it's the output protection diodes that are rated to only 20mA). There are two likely failure modes: the first is local overheating of the chip, but silicon is quite a good thermal conductor (and you're probably not operating anywhere near the upper "ambient" temperature limit). Driving 60 mA into a short-circuit might dissipate 300 mW, but probably less than 150 mW into a (RGB) LED. If the FET does fail, you might be "lucky" and lose the functionality of just that pin, but if the whole output stage is damaged then there might be a high through-current that takes out the whole chip. Which leads to the second failure mode:

It seems that many ICs are still manufactured using "bond wires" (rather than "solder bumps" or COB, etc.) which can act as a tiny "fuse". I guess all the PIC bond wires have a similar dimensions and fusing current (some semiconductors use multiple bond-wires for particularly high currents), so they will have a similar rating to the Supply and GND pins, i.e. about 80 mA in the data sheets. Thus the likely failure mode is if you over-drive multiple outputs for a significant length of time, when you might "lose" the whole chip.

Cheers, Alan.
 

inglewoodpete

Senior Member
It's just a guess but could it be that most high-brightness LEDs have a forward voltage of around 3.5 volts. The PIC's output MOSFETs could have been saved by their smaller P = E x I power loss.
 

rq3

Senior Member
******************* Worst practice observation follows. *****************************

I slammed together an 5V 20M2 RGB LED project lately without current limiting resistors and it works just fine. Three output pins drive a parallel PAIR of 10mm RGB LEDs. TTYTT I did put one resistor onto green just to balance the color intensity and get a better white. No PWM, just full power high/low. Working great, even though each pin is outputting some 50-60 mA, occasionally all 3 pins on at once for white. I'm fully aware I'm pushing it and operating outside the recommended parameters. I was in "git her done" mode to film a video. I had good intentions to rectify (!) the LED problem later and add resistors before something burned up but it's still working, if only for a few minutes at a time. Someday I'll wire up a similar non-critical circuit and do a continuous destructive test.

I used to get my knickers in a twist worrying that a Picaxe pin can sink more current than it can source and build the circuit accordingly. I'm starting to wonder if it matters. The internal impedance self-limits current at some point. Of course a dead short would led more current flow, but any LED will drop 2-4 volts and the internal impedance can drop the rest. Deal?

Of course I'm pushing my luck and Hippy and Technical can't condone my actions, I'm just curious if anyone has actually burned up an output pin from too much current.
Erco, probably not. It's trivial to measure the source/sink current capability of a particular pin. Set the pin HIGH, and pull it to ground with a potentiometer or decade resistance box. When the pin voltage gets pulled to 1/2 of the supply, the pot or decade box value will be equal to the internal pin resistance (source current). With the pin set LOW, and the variable resistance between the pin and the power supply, the same applies, but you'll be measuring the sink current pin resistance.

Because the internal impedance is pretty much self limiting when driving things like LEDs, the real issue is when you ask multiple pins to SOURCE current. In that case, all of the current must be supplied by the single Picaxe supply pin, and may indeed fuse its wire bond(s).

Other reasons for limiting total device current include thermal gradients across the device's silicon die, which may lead to things like out of bound variations of voltage references (for the A/D functions, for example). Factors like these are included in the manufacturer's spec sheet. They try to cover all bases, perhaps to the detriment of other, simpler, requirements.

An LED may actually appear brighter to the human eye if pulsed at high current for a brief period, rather than driven with pure DC (High/Low, think of an iPhone flash, a very fast high current pulse). It still appears to be completely on, but thermally the driver (the Picaxe) is well within its average current ratings. In other words, an LED may appear to be brighter when driven directly by a 10% duty cycle pin at 250KHz, than the same LED driven by the same pin set LOW with the "correct" resistor in series. Both methods may be electrically and thermally safe for the Picaxe, but the pulsed method lets you drive 10 LEDs (or more) without violating the chip current and thermal ratings.
 

geoff07

Senior Member
For several years I helped out at a school, where a different brand of very similar chips were programmed by GCSE D&T pupils. Many of them forgot/didn't bother with the led dropper. I don't recall any failures. They were never on for long so time may be a factor. I even had chips connected in reverse polarity that survived (long enough at least).
 

erco

Senior Member
Yes I've also had students mis-wire circuits in numerous and creative ways and never had a failure. These Picaxes are tough little soldiers and can certainly take their fair share of electrical abuse.

"Tune for minimum smoke."
 
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