Re: [CAD_CAM_EDM_DRO] Re: Proper pull up resistors for optical limits

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.