MK1 Steel router build log 1500x800x200

**Doddy** · 03-06-2018

Originally Posted by driftspin

Ok ,thoroughly does not even closely describe your commitment in helping me out.
Thank you for that.

Ok so...i decided to do the following based on this thread and other threads.

Since there seems to be 9.4k ish in line in the proximity switch emitter, i put a 4.7k resisitor in series connected to ground.

When connected to a 24vdc psu, this will give me about 8 volts at the point of connection at the bob. in the "open "position.

When in the " closed " position the emitter goes to 0v.. hope this will work.

I will report back when all is operational.

Doddy thanks again for doing research on this subject.

Grtz. Bert

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In honesty, I'm doing this for me, as much as anything - I will need to start thinking of wiring pretty soon, and you're bringing up the questions that I need to answer for myself.

I think the NPN sensors the black wire is actually the collector, not emitter (I guess the emitter is connected to blue/0v)

Re your proposal. Think carefully of what you're trying to achieve. With the NPN providing a switch to ground (current sink), by placing a resistor from the output to ground will give you a low(ish) resistance to ground even when the NPN transistor is OFF - the LED will still illuminate (you'll get around 1.5mA through it - enough to illuminate it, probably enough for the current through the opto's transistor to actuate the input to the UCx00 - the input impedance of that is something like 50k to ground (and 4k7 to +5).

Try it, by all means, but don't be surprised if the input is permanently on.

The easiest solution is to simply tack a small signal diode (1n4148 is probably the most common) across the pin 1/2 of each opto - in the opposite sense to the opto's LED (i.e. cathode to pin 1, anode to pin 2). That way you cannot reverse bias either diode beyond the forward conduction voltage of the opposing diode - max of 0.6V by the 1n4148 across the opto's LED, or around 1.5V by the opto's LED across the 1n4148. Essentially, this will limit the voltage at the input to the BoB to the V-Ref provided by the BoB's onboard regulator (~10V) plus 0.6V.

Of course, this (and the idea that you proposed) both serve to reduce the switching voltage to something around 10V, which is why I still think that just using a 12V supply to the BoB and to the sensor is an easy alternative.

**driftspin** · 03-06-2018

Originally Posted by Doddy

Try it, by all means, but don't be surprised if the input is permanently on.

.

Ok i get it.

So best would.be to raise voltage just over 9.3 or what ever comes out the bob with nothing connected.

I really like your idea of the reversed diode.

This will act as an indicator for status at the same time.

Grtz Bert.

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**driftspin** · 20-08-2018

Originally Posted by Doddy

In honesty, I'm doing this for me, as much as anything - I will need to start thinking of wiring pretty soon, and you're bringing up the questions that I need to answer for myself.

I think the NPN sensors the black wire is actually the collector, not emitter (I guess the emitter is connected to blue/0v)

Re your proposal. Think carefully of what you're trying to achieve. With the NPN providing a switch to ground (current sink), by placing a resistor from the output to ground will give you a low(ish) resistance to ground even when the NPN transistor is OFF - the LED will still illuminate (you'll get around 1.5mA through it - enough to illuminate it, probably enough for the current through the opto's transistor to actuate the input to the UCx00 - the input impedance of that is something like 50k to ground (and 4k7 to +5).

Try it, by all means, but don't be surprised if the input is permanently on.

The easiest solution is to simply tack a small signal diode (1n4148 is probably the most common) across the pin 1/2 of each opto - in the opposite sense to the opto's LED (i.e. cathode to pin 1, anode to pin 2). That way you cannot reverse bias either diode beyond the forward conduction voltage of the opposing diode - max of 0.6V by the 1n4148 across the opto's LED, or around 1.5V by the opto's LED across the 1n4148. Essentially, this will limit the voltage at the input to the BoB to the V-Ref provided by the BoB's onboard regulator (~10V) plus 0.6V.

Of course, this (and the idea that you proposed) both serve to reduce the switching voltage to something around 10V, which is why I still think that just using a 12V supply to the BoB and to the sensor is an easy alternative.

Hi doddy,

All in all i had to add 500 ohm to 4.7k to have the inputs switch reliably. I ended up at 8.65 volt. after switching the inputs would not drop..

Adding 500 raised it to 9.15 ish. which is ok
So you where right it needed to go near 9 and or slightly over.

0.5volts still illuminated the input opto's just enough to screw up somtimes

I still have a glitch in 1 of the inputs though..

Need to find out why.
maybe wiring issue or a slightly off 4.7k resistor.

ill try 1k instead off 500 first... maybe there is a switching effect or so.

Grtz Brt

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