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So after the resistor is in place from the output wire to the +5v, if the switched wire goes to Gnd, does that mean the 5v will go to ground also ? (sorry for my lack of understanding)
Ok, I understand now. I wasnt sure if the resistor was enough to stop the voltage going to Gnd and damaging something.
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Resistor overheating isn't the limiting factor. With 200 ohms you'd have 5/200=25mA. The parallel port can (apparently) only source/sink about 14mA, hence to avoid damaging the port if connecting directly I'd use something higher. So I'd agree to sticking with at least 1k just to be on the safe side.
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Some specs from the BOB Manual
Attachment 9053 Attachment 9055
Attachment 9054 Attachment 9056
The Sensors are NPN.
mycncuksucks
assuming 1.2v forward voltage on the opto means that the existing series resistor is (5 - 1.2)/.007 =542, so probably 560R. the suggested series resistors of 1,5k for 12v and 3k for 24v give currents of 6.9mA and 6.4mA respectively.
dropping the 'on' current to 5mA will need a pull up resistor of 220R and the 'off' current through the 220R would be (5-.4)/220=21mA and 0.1W dissipation, needing a 0.25W resistor. I guess 5mA through the opto will be ok and not have noise immunity issues.
It would be better to use a higher supply rail, e.g. 12 or 24v and use the suggested 1k5 or 3k resistors 'off' current/power being 7.7mA/0.1W or 7.9mA/0.2W respectively.
Ok... I have a spare PC PSU. 12V (Model LC B200SFX)
If I use this to power the sensor, I connect the PSU -0V to the sensor and to the Gnd terminal on the BOB as shown "right ?" plus the +12V to the sensor input. Using the 1K5 to limit the input to the BOB.
Do I still need a pull up resistor ?..... Its getting a bit confusing for me.