Re: ParPort output low - hi-Z or ground?

--- In CAD_CAM_EDM_DRO@yahoogroups.com, "ghidera2000"
<ghidera2000@y...> wrote:

> --- In CAD_CAM_EDM_DRO@yahoogroups.com, "ghidera2000" > Oh and
> thanks for the 3.3 volt tip. I checked mine and the outputs
> > are 3.3v while the input high is a full 5v. Had to adjust the

led

> > resistors (7mA just ain't enough).
>
> Actually, I should never post before I have my morning coffee...
> 3.3V on outputs doesn't affect my on-state LED current...
>
> However, I'm supplying +5 from the floppy connector to the leds

and

> the port "off" state (off to my circuit operation anyway) is

3.3v...

> I still have current flowing to the parallel port... Hmmmmm

Consider the following condition: You supply +5 to the anode of the
opto led. The logic high from the parallel port is 3.3V With a
current limit resistor of some value a current will flow through the
led even with the cathode at 3.3V (parallel port pin "HIGH"). Now,
it takes a minumum amount of current to turn on the transistor but
there is a narrow zone of current flow that places the transistor in
analog mode (remember analog?). What that means is that the led
is "biased" up to a certain level even when it supposed to be off,
and that would make it more suspectable to being false triggered by
noise. It would work better if you were to either run the common
(anode) voltage at 3.3V and use the opto to do both isolation and
level translation or to buffer the pin output through a chip with
open collector outputs and the outputs at the +5 level. You want
the OFF condition to be one where the anode and cathode potentails
are the same voltage. If you do use the 3.3V as the source you will
have to recalculate the series current limit resistor to provide the
correct ON current.

IMO the opto should be on the motor controller and assume it is
going to be driven with a +5 signal. The breakout board should take
care of buffering and level translation. The buffering a level of
protection to your parallel port and can save having to replace an
entire motherboard.