PIC current drive

[Mosaic]

Hi, i am driving LEds from a pair of PIC pins. One pin sources and one sinks. With a 220 Ohm Resistor I get a current flow of about 9.7mA.
I see the max spec on a pin is 25mA. Now does this mean I can cut the resistance in half and safely drive the LED with about 20mA? This would be sinking 20mA from 1 pin and sourcing it from another, netting about 40mA thru the pic, 16F886.

 

[ericgibbs]

Hi, i am driving LEds from a pair of PIC pins. One pin sources and one sinks. With a 220 Ohm Resistor I get a current flow of about 9.7mA.
I see the max spec on a pin is 25mA. Now does this mean I can cut the resistance in half and safely drive the LED with about 20mA? This would be sinking 20mA from 1 pin and sourcing it from another, netting about 40mA thru the pic, 16F886.

hi Mosiac,
Look at this image, shows the individual pin ratings and the TOTAL rating for a PIC.

 

[Mosaic]

Thx Eric:

Pls have a look at this link:
My PIC Projects

It shows a PIC driving LEDs with no dropping resistor.
Is this a workable solution?

I do have a sequencer to design and not having dropping resistors would be good.

 

[colin55]

[FONT=&quot]Under normal conditions the output pin of a PIC chip will deliver 25mA and the FET pulls HIGH so that the voltage across it is less than 100mV. This produces a wattage-loss of 2.5mW
If you only allow the FET to pull HIGH to about 2v, it will have 3v across it and the wattage dissipated by the FET will be 75mW.
It may overheat and be destroyed.
You have to remember, the FET inside the chip is microscopically small.

[/FONT]

 

[MikeMl]

Thx Eric:

Pls have a look at this link:
My PIC Projects

It shows a PIC driving LEDs with no dropping resistor.
Is this a workable solution?

NO! It's very poor design practice. You can get away with it on a couple of pins, but read and understand the PIC data-sheet excerpt that Eric posted, and it clearly tells you why you should not be doing this.