Graig,
If the output of the encoder has a totem pole with 4.7K in series it won't pull the C10 input down to a reliable level.
If the output has an open collector pulling down 4.7K it will work with the zener. To test that you could add an additional pullup and see if the output is still near zero.

John

Hi,
not a lot to it, just place a Zener across each input, it will limit the max voltage input to 4.7V, ie within the BoBs sweet spot.
Don't need a circuit board just the the Zener tail into the input screw terminal with the other lead going to COM. Note the band (polarity) of the Zener.

Lots show resistors used to?

Don't really need another resistor because of the output impedance of the AB outputs....don't you just love Thevenin's Theorem!

Craig

Hi MN300,

If the output of the encoder has a totem pole with 4.7K in series it won't pull the C10 input down to a reliable level.

You might be right......but if OP tries it my guess is it will work as is. If it does not work reliably, especially when spinning the MPG at speed
then a buffer transistor would be indicated. My pendant needs to signal 24V, active low, consequently I fed the A/B inputs direct into the base of a small signal NPN transistor
and the collector pulled the 24V from the BoB low....and its perfectly reliable.

My guess is that OP would rather not use a transistor if it can be avoided......it represents a complication that non-electronic people shy away from.

Craig

Thevenin's Theorem also applies to current source and a resistor to ground. A 1V source and with a 1 ohm resistor in series looks the same as a 1 amp current source through 1 ohm.
A magazine puzzler asked how to tell them apart if the power source was hidden and all you could see was the resistor. - Answer, the one with the current source would get warm with no load.

I agree, the physically simpler solution is better if possible.

Hi,
not a lot to it, just place a Zener across each input, it will limit the max voltage input to 4.7V, ie within the BoBs sweet spot.
Don't need a circuit board just the the Zener tail into the input screw terminal with the other lead going to COM. Note the band (polarity) of the Zener.

Don't really need another resistor because of the output impedance of the AB outputs....don't you just love Thevenin's Theorem!

Craig

Sweet. No clue who that man is but clearly he was smarter then me 😂😂.

As per diagram:

Why are the Zeners on the 0V to the common?

I thought they would be on the A and B?

Or, is what I am looking at that the Zener draws some of the power from A and B dumping it into the ground?

So if A puts 5.2V from encoder to Bob the Zener will take 0.5V away from that to the ground allowing 4.7V to continue to the board..?

And for the drawing, do I just sodler the junctions (little green circles)?

You might be right......but if OP tries it my guess is it will work as is. If it does not work reliably, especially when spinning the MPG at speed
then a buffer transistor would be indicated. My pendant needs to signal 24V, active low, consequently I fed the A/B inputs direct into the base of a small signal NPN transistor
and the collector pulled the 24V from the BoB low....and its perfectly reliable.

My guess is that OP would rather not use a transistor if it can be avoided......it represents a complication that non-electronic people shy away from.

Craig

I like simple. Simple to make, simple.to repair lol.

Now, I'm all for what will work reliable and best. I learn fairly quick and am not afraid to do what's needed as long as I can get some direction haha!

Kyle

After you make the circuit with the zener measure the high and low logic levels again. The highs will be the zener voltage (4.7V). The lows should be under 1.6V to meet the minimum spec. The actual threshold may well be higher but you can't count on that - design for worst case.

Hi,

Or, is what I am looking at that the Zener draws some of the power from A and B dumping it into the ground?

Yes that is correct, it sort of 'diverts' the extra voltage from getting into the BoB....we don't want to put any stress on the ESD diodes in the ICs.

And for the drawing, do I just sodler the junctions (little green circles)?

You can if you wish, I would just poke the wire in with the wire from the A output in the same screw terminal. The negative (COM) side hopefully you have a COM screw
terminal within reach and you can poke it into there.

Craig

Hi,

Quote

Or, is what I am looking at that the Zener draws some of the power from A and B dumping it into the ground?

Yes that is correct, it sort of 'diverts' the extra voltage from getting into the BoB....we don't want to put any stress on the ESD diodes in the ICs.

Quote

And for the drawing, do I just sodler the junctions (little green circles)?

You can if you wish, I would just poke the wire in with the wire from the A output in the same screw terminal. The negative (COM) side hopefully you have a COM screw
terminal within reach and you can poke it into there.

Craig

Well that sounds easy.

So I could take the Zener and basically insert it like a jumper from. Port 2 to common and then port 3 to common and just make sure its correctly orientated?

Amazing drawing attached.