Re: Proper pull up resistors for optical limits

Okay, that gives me more to go on. The switches will be here
tomorrow so maybe there will be more specifics on current for the IR
and I can use the formula you gave me. Thanks so much for your
help, I'm glad there are knowledgable guys like you to keep us non-
electrical types out of trouble (ha,ha). Thanks again very much for
your help!

Frank

--- In CAD_CAM_EDM_DRO@yahoogroups.com, "Roy J. Tellason"
<rtellason@b...> wrote:

> On Thursday 13 May 2004 10:49 pm, fun2b432002 wrote:
> > The wiring diagram shows the resistor on the output side where

the

> > +5V comes off the connector pin (#4) on the servo amp,... it

shows

> > the resistor between the +5V output and the optical interrupter,
> > then from the optical interrupter it goes back to the isolated
> > ground on the connector on the servo amp (pin #6).
>
> Ok, then what you have here is *extremely* non-critical -- it's a

matter of

> the transistor on the output side of that opto connecting that

point to

> ground when it's turned on, and the resistor pulling the point to

+5 when

> the transistor is turned off. A minimum value is going to be

determined by

> how much current that transistor will shunt to ground when it's

on, so for

> example if you used a 1K resistor and assuming that +5 supply,

the maximum

> current is only going to be 5 mA. You could go somewhat lower,

perhaps, or

> higher, but going higher is going to give you less noise immunity.
>
> > The other side of the interrupter gets connected to limit in

pin (#5) and

> > then to the isolated ground (#6).
> >
> > On the optical switch: Pin 1= Anode, Pin 2= Cathode, Pin 3=
> > Collector, and Pin 4= Emitter. I'm unsure which gets connected

to

> > +5V, Ground, and Limit in, but I'm hoping there will be some
> > instructions with the switch. I'm very newbee to all of this.
> >
> > Hope that helps to make it a little clearer on what size

resistor I

> > need. Thanks
>
> The output side ("emitter" and "collector") are just two

connections to the

> transistor inside the device. Emitter would be grounded and the

collector

> would go to your pullup resistor and to the input pin (which

monitors the

> state of that junction).
>
> For the input side you need to light the LED in there. (It's

probably an

> infrared LED so don't expect to see it lit!) Figuring that

resistor is a

> matter of deciding first how much current you want to flow in

there. Values

> like 10 mA are pretty typical. The other thing you need to know

is the

> typical forward voltage drop of the LED. If that's not specified

for the

> part, just hook up a 1K resistor to +5, the other end to the

anode, and the

> minus side of your +5 supply (or ground) to the cathode. Measure

the voltage

> across cathode and anode, which will give you your voltage drop.

This stays

> fairly constant with varying current. Then subtract that from the

power

> supply voltage, and the difference is what needs to be dropped by

the

> resistor there, at the specified current. Say for example you're

driving

> the input with a +5 supply, the forward voltage drop measured is

1.6V, and

> you want to have 10mA of current flowing in there. The difference

in voltage

> is 5 - 1.6 = 3.4V, which gives you a 340 ohm resistor for 10mA.

A standard

> value of 330 ohms is close enough...
>
> If any of this isn't clear, let me know and I'll see if I can

explain it

> better.