Confused about your circuit...  12V sig goes to opto LED, via a 2k=20
resistor.  Opto output collector has pullup to 5V and also goes to PIC=20
input.  Opto output emitter to ground.  But where is the extra LED?  At=20
the opto output?  If so, does it have its own resistor?

Cheers,
-Neil.


On 3/29/2013 11:29 AM, alan smith wrote:
> I have a 12V signal coming into a board, so I'm using an opto to isolate =
it (comes from a longish cable), so tied a 10K pup to 5V, and into the PIC,=
 but also tied a LED to the same signal.
>
> Thought of course is....opto comes on, drives the collector low, turns on=
 the LED and PIC sees it.   Well the LED does turn on, but the PIC wasnt se=
eing it low.  Looking just with a voltmeter, its sitting about 2.4V, so its=
 still a high.  OK, physics still work.
>
> I changed out the resistor driving the opto's LED from a 2K to a 1K, thin=
king that driving the LED harder will turn on the transistor harder...if it=
 wasn't fully saturated then its possible it can't drive all the way low.  =
No change.  Did a really quick LTspice, and it simulates the way I expect..=
..goes to ground.
>
>
> But now I'm thinking we might have a voltage divider between the pup and =
the LED circuit?  The other solution, that Ive done before and probably sho=
uld have on this was to put the LED in series with the opto input.   signal=
...resistor....LED.....opto's LED....ground.
>
> Just wondering what I can do with the existing circuit
>
>

--=20
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