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Author Topic: Mach3, Delta ASDA-B2, ESS. 12 straight hours working on this. could use assist..  (Read 1195 times)

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I have drawn a somewhat simplified circuit diagram for one pulse input of your drive.

With reference to the attached diagram:

If you apply +24V at pin35 and 0V at pin 37 current will flow through the 2k parallel resistor R4, and through R3, R1, the photodiode, R2.
Note that the forward bias voltage drop of the photodiode is 2V.
The current is:
I= 24/2000 (through R4) +(24-2)/(2000+75+75) (through the forward biased photodiode)

If you powered the photodiode by applying +24V to pin39 and 0V to pin 37 the current would be:
=147mA and the photodiode would be destroyed within micro seconds.

If you power the photodiode by applying +5V to pin 39 and 0V to pin 37 the current is:
 =20mA which is entirely acceptable.

The termination of pin 39 allows the use of differential signal which can turn the photodiode on and off rather faster
than using 24V at terminal 35.

What must be avoided is the application of greater than 5V between pins 39 and 37 or the photodiode will blow.
My contention is that the first arrangement, namely using the in-built 24V supply at pins 17 and 14, and pins 35 and 37
of the photodiode input circuit is easier and more forgiving than differential signalling and even 200k pulses per second is
still very fast indeed.

Maybe at a later date you will want to try differential signalling, if you are demanding even faster axis speeds and/or higher resolution,
but for now stick with single ended signalling.

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