Boxford TCL125. (Fresh start )

[John11668]

Not seeing any illumination on pin 12 as I rotate the disc past the single slot

[depronman]

Not seeing any illumination on pin 12 as I rotate the disc past the single slot

I suspect with no resistor you have blown the new optical sensors
Take the pin 12 and touch it to ground. See if this triggers the pin in mach3
If it does then you have likely popped the new sensor

Paul


Sent from my iPhone using Tapatalk

[John11668]

Are there optical switches with a better voltage tolerance ?

[John11668]

I suspect with no resistor you have blown the new optical sensors
Take the pin 12 and touch it to ground. See if this triggers the pin in mach3
If it does then you have likely popped the new sensor

Paul


Sent from my iPhone using Tapatalk

Have tried that and the yellow light goes out so sounds like I am stuck until I get some resistors .
Will look out for my mail early in the week.

Had a look on RSwww and literally pence each but with £4.95 postage

[inee]

https://www.ebay.co.uk/itm/RS-Infrar...4AAOSwFIVe2WWG

[John11668]

Will try to snag a couple of those inee

[John11668]

Ahh doddy !
So simple and straight forward. How could I possibly get it wrong?

So our friends at Boxford chose a 10v output without any apparent resistor in series? or where did they hide it
And I guess I will have to use 12v so maybe will have to add a resistor, or maybe 5v but I will need to go back to my 1968 physics A level notes if I really want to understand your calc .

I can manage most mechanical stuff , and have a reasonable grasp of electrical machines , single and 3 phase, Even managed to get a handle on electronics in the days of thermionic valves , but I have to say that when the world turned upside down, when the time came when current stopped flowing from positive to negative and it was decided instead that current was a stream of negatively charged particles careering towards a positively charged source then that was the time at which my "electronic "head started to hurt .

What I am saying is you have to bear with me a bit while this stuff sinks in . When I get my new switches, I may need to come back with the spec, and ask you to turn forward voltage and forward current into a suitable source voltage / resistor combination.
Hope you haven't got too sick of me by that time

[John11668]

OK so I have ordered new switches and in the meantime had a closer look at these





So this might be surprising but looks like Boxford have chosen to wire these lamps like xmas tree lamps ie in series .
So our green supply wire seems to ring around the switch commons then on to the two lamps in series. How do I add the forward voltages in this configuration??

Now as Inee points out from the Boxford wiring diagram the input voltage is 10v and there is no sign of a resistor . Is it possible that the lamp resistances stack up so that no external resistor is needed . and how would that be affected if i chose to supply 12v?

[Doddy]

Hmmm, you previously linked a reflective opto-coupler, (EDIT: Ah, I see what you did there - the linked datasheet has the slotted version later in the document) these are slotted opto-couplers (same premise, just a different physical configuration) - these more common with slotted wheels as shaft-encoders, etc.

For LEDs which typically outlive the service life of the machine, wiring in series is perfectly fine, and common practice. As you probably guess, you just need to add the forward voltage drop across the individual LEDs into the standard calculation for series resistor (the advantage being that you only need one resistor).

Where this conversation doesn't need to go, but I'll mention anyway, the forward voltage drop is not absolute - it'll vary across devices, temperature, forward current, day-of-the-week, and if you look at it funny. What that means is, you always design (calculate the series resistor) based on nominal values - and the slight variances fall by the wayside... what you can't do is drive an Led without a series resistor of some sort (or rather, a current limiting device or circuit) - as any slight variance in the forward voltage drop would have a significant swing on the forward current and could result in the destruction of the LED.

So, given a wet finger of 1.8V Vf for an IR LED, at 12V drive voltage and an If of 10mA, you'd end up with a series resistor value or 840 Ohm, or closest value of 820 Ohm in the E12 range of resistor values. But, you're overthinking things a little - you can wet-finger resistor values given a safe working forward current. So, whether it's one or two LEDs, and 10V or 12V, you'd probably just pick the ubiquitous 1k resistor. It's close enough. There's plenty of other factors in the system that will impact the performance more than 10-20% accuracy of resistor values.

I'm convinced to within a gnat's arse that there will be some method of current limiting/control in the Boxford board, regardless of the schematics. It's the done thing. And the cheapest (and most reliable) method is a series resistor. It *could* be that the original opto-isolators had integral series resistors for 5V operation (after all, once upon a time 5V was common logic supply voltage), but it's almost irrelevant here - you know the opto's you're buying, you can find the datasheet, you can determine the Vf, and then you can determine the series resistor. But 1k will do.

Your schematic is functionally okay, but remember both the LED and the associated photo-transistor are polarised devices - you need to get the wiring the right way around.

[JAZZCNC]

John if your thinking to go with Linux CNC then I would plan for it now because Mach3 can only use a Single index pulse or slot for the spindle encoder, Linux CNC can use multi-count encoder and will give you much better speed control.