jswayze
- 28th May 2005, 02:51
I'm trying to power an "ultrabright" LED with a PIC12F629. Each PIC output is rated at 25ma, and the LED lights up just fine. However, I have an LED that requires about 45ma, and it's fairly dim on just one output. To solve this, I figured I could take another PIC output and tie it together with the first one. You get 5v out and twice the current.
Here's what happens:
With the first output connected to the LED and powering it, I attached a lead from the second output to the LED. The brightness increased and all was well.
Next, my program shut off the LED then turned it back on again. This time, nothing. I measured 0 volts at the combined output. If I took one output away I'd get my 5v again and the LED would light.
It seems that while the LED is on, if you add the additional output everything is OK, but if you turn it off and back on again, something happens to wipe out the voltage.
With my feeble brain and minimal experience I thought that maybe the current from one output was being diverted into the other output and somehow canceling them. I thought that a diode might do the trick, but I don't have one (will go to Radio Shack tomorrow and try)
Is it obvious what's happening here? I'm sure I'm violating some basic rule of electrical design, I just don't know what it is.
Thanks!
Jeff
Here's what happens:
With the first output connected to the LED and powering it, I attached a lead from the second output to the LED. The brightness increased and all was well.
Next, my program shut off the LED then turned it back on again. This time, nothing. I measured 0 volts at the combined output. If I took one output away I'd get my 5v again and the LED would light.
It seems that while the LED is on, if you add the additional output everything is OK, but if you turn it off and back on again, something happens to wipe out the voltage.
With my feeble brain and minimal experience I thought that maybe the current from one output was being diverted into the other output and somehow canceling them. I thought that a diode might do the trick, but I don't have one (will go to Radio Shack tomorrow and try)
Is it obvious what's happening here? I'm sure I'm violating some basic rule of electrical design, I just don't know what it is.
Thanks!
Jeff