>
> Sure, I understand the voltage divider but still don't see the limits
> of the OOPIC line. Am I right to assume that the divider gets it in
> the 0-5volt range and then through mathematical calculation the
> correct value is achieved with the OOPIC?
Right! The absolute limit of voltage on the input pin comes from the
PIC16F877 Data Sheet - find it at Microchip. For convenience, let's
assume it is 5V (you MIGHT get by with 5.2V but no more).
The two resistors in series total 24k ohms. The bottom end (visually
think of the string as vertical) of the 2k resistor connects to logic
ground. The center point connects to the IOLine and probably that
Schottky diode while the top end connects to the 60V line. The OOPic
will see 0-5V when the signal varies from 0-60V.
The math for the voltage divider
I = 60V / 24k ohms = 0.0025A or 2.5 mA.
V (across 2k resistor) = 2k * 0.0025A = 5V
V (across 22k resistor) = 22k * 0.0025A = 55V
So, the 22k resistor drops 55 of the 60V and leaves 5V at the pin.
Remember what I said earlier about the 60V not going above that level.
If there is ANY chance the voltage can be higher, change the voltage
divider.
Or, use much high value resistors and feed an Op Amp connected as a
voltage follower. The Op Amp can't exceed the supply voltage of 5V
regardless of the input voltage. You can try 220k and 20k. You might
want to play around with these numbers because an Op Amp won't go all
the way to 5V. Rail-to-rail op amps come close - TI TLV2474 (quad)
come close.
As to plugging and un-plugging the serial cable: that's what the rest
of us have to do! If there was another solution, we'd know about it.
You know, you are only switching 2 wires. Maybe you want to add a
DPDT toggle switch and hack together a couple of DB-9 connectors.
Richard
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