CNC 3020 Enclosure/Upgrade

[Pointy]

I think this machine will be the death of me! I am on my 5th or 6th version of the charge pump circuit now, and 4th or 5th version of the IO board. The charge pump part is working beautifully. A five pin connector takes the 5v, gnd, switched gnd, estop switch position to the IO board and also supplies the charge pump signal. (Previous version I had forgotten the IO board needed power to send the charge pump signal, and was switching the 5v. DOH!) Without the switched ground from the mosfet, the spindle relay transistor wont do anything, meaning the the spindle wont work until a valid charge pump signal is applied. The estop position is just fed back to parallel port pin 15 so that Mach3 also knows when an Estop has occurred. All of this this works great, I just have one niggle with the IO board.

In my previous version of the IO board I had forgot the Limit,Home inputs are NC, which meant the LED's were on all the time. I added a transistor, and now the LED works but the parallel port pin is held low. The strange thing is I tested this with my breadboard before I made the new board, but somehow had managed to test and make something completely different. Here's that part of the schematic...

Attachment 11722

This was my thinking...

When X6-1 & X6-2 are connected to the Home switches, pin 2 of OK1A is grounded, meaning PP 13 is held low. By grounding the base of Q3 LED 1 should be off.
When X6-1 & X6-2 are open, R14 will supply base current to Q3, lighting up LED 1, and also hold pin 2 of OK1A high, making PP13 go high.

It's the last part that doesn't work, the LED lights up but pin 2 stays low, thus Mach3 doesn't see the switch activate.

I am guessing it has to do with the base of Q3, but don't know to be honest, I am better with digital electronics.

Any help or suggestions much appreciated.

Regards,

Les

[irving2008]

I think this machine will be the death of me! I am on my 5th or 6th version of the charge pump circuit now, and 4th or 5th version of the IO board. The charge pump part is working beautifully. A five pin connector takes the 5v, gnd, switched gnd, estop switch position to the IO board and also supplies the charge pump signal. (Previous version I had forgotten the IO board needed power to send the charge pump signal, and was switching the 5v. DOH!) Without the switched ground from the mosfet, the spindle relay transistor wont do anything, meaning the the spindle wont work until a valid charge pump signal is applied. The estop position is just fed back to parallel port pin 15 so that Mach3 also knows when an Estop has occurred. All of this this works great, I just have one niggle with the IO board.

In my previous version of the IO board I had forgot the Limit,Home inputs are NC, which meant the LED's were on all the time. I added a transistor, and now the LED works but the parallel port pin is held low. The strange thing is I tested this with my breadboard before I made the new board, but somehow had managed to test and make something completely different. Here's that part of the schematic...

Attachment 11722

This was my thinking...

When X6-1 & X6-2 are connected to the Home switches, pin 2 of OK1A is grounded, meaning PP 13 is held low. By grounding the base of Q3 LED 1 should be off.
When X6-1 & X6-2 are open, R14 will supply base current to Q3, lighting up LED 1, and also hold pin 2 of OK1A high, making PP13 go high.

It's the last part that doesn't work, the LED lights up but pin 2 stays low, thus Mach3 doesn't see the switch activate.

I am guessing it has to do with the base of Q3, but don't know to be honest, I am better with digital electronics.

Any help or suggestions much appreciated.

Regards,

Les

Your diagram doesn't show, but I'm guessing from your description you've connected the input of the bob ok1a pin2 directly to the base of Q3. This can't work as the base will always be 0.7v above the emitter, which is grounded.

Assuming the ok1a pin2 input is a logic input and not an optoisolator, to make it work, either:

- connect ok1a pin 2 to the collector then reverse your logic, switch closed= high

or

- put another resistor between the pullup-ok1a junction and the base of Q3. I don't know what values you've used so I've shown my calcs and you can run them for yourself. Assuming your led needs 10mA and q3 has a gain of 50 then the current into the base needs to be10/50mA= 0.2mA so the total resistance on the base to 5v rail is (5-0.7)/0.2k=21.5k. If your pullup is say 4.7k then the new resistor needs to be 21.5-4.7=16.8k, the value isn't that critical so use next lowest standard value of 15k. As a check, the ok1a input will see a voltage of (5-0.7) * 15/(15+4.7)+0.7v =4v when the switch is open.

If ok1a pin2 is optoisolated then option 1 may not work and a different calc is needed for option 2.

Hope this helps.

[Pointy]

Thank you so much for replying Irving, this all makes sense now. I also think that I inadvertently did your second option when I tested it., I was using a 4.7k pullup and a 4.7k current limiting resistor on the transistor base. I will test this out as soon as I get a chance, but it probably won't be for a few days.