Need ideas for project

[Machine Doctor 13]

I repair older machine tools. I am currently trying to fix a problem on an older Mazak turret lathe. It has an old stepper relay that is mechanically worn out and nobody makes new stepper relays. The lathe has 8 tool stations on a turret. When the CNC computer control program calls for a tool change a signal is sent to a solenoid to unclamp the turret and rotate to a different tool. This operation works fine.

The CNC computer control also sends a +24 VDC signal to the stepper relay to energize the relay coil and cause the wiper to advance to the appropriate output (there are 8 outputs, one for each tool station) as feedback to the computer control as to which tool station is currently active.

I want to replace the stepper relay with something more modern, and am thinking of possibly using a multiplexer circuit. I have looked at some multiplexers but it seems they use 3 inputs to determine which output is selected. Different combinations of high signals on the 3 inputs determine which of the 8 outputs is selected. I am dealing with 1 input and each time it goes high (+24 VDC) the next output needs to be selected.

Any ideas would be greatly appreciated.

 

[JimB]

My first thought is that you need a counter.

A three bit binary counter will count up from 0 to 7 (8 steps), you can then use the multiplexer to convert the binary count to a 1 to 8 count to actuate the turret.
Logic circuits cannot provide enough current to drive relays/solenoids, you will have to use a driver transistor of some kind on the outputs to the turret.

JimB

 

[rjenkinsgb]

Look at a 4017 or 4022 CMOS ICs, those are a combined counter/decoder.
Data: https://www.ti.com/lit/ds/symlink/cd4017b.pdf

The outputs are low current but you can use such as a darlington driver array chip to boost them so they can switch 24V relays.
A ULN2803 would be suitable with the counter operating on 5V:

https://www.ti.com/lit/ds/symlink/uln2803a.pdf

You may need a debounce circuit at the input as mechanical contacts often give several pulses as they close, when seen at logic speeds (when things can count at megahertz frequencies).

If it's operating from a solid state PLC output then just a resistive divider and a series resistor from that to the logic input should be OK; eg. 3k9 and 1K0 for the divider then a 10K or higher from the divider junction to the IC clock input.

 

[Machine Doctor 13]

I really appreciate the advice. I don't know how to use a 3 bit binary counter for this application because there is only one input. Every time a new tool is called for the CNC control sends a 2 second +24 VDC signal to the stepper relay. I don't have to send outputs to drive solenoids for the turret. That function is being handled just fine. I only need to send feedback to the CNC control to tell it what tool station is currently on.

 

[JimB]

I don't know how to use a 3 bit binary counter for this application because there is only one input.

That is OK, there is only one input to a 3 bit binary counter, but there are three outputs ie three bits.
But lets not get hung up on that...

The suggestion by rjenkinsgb to use a combined counter decoder is a good one, simplifies the circuit a bit.

JimB

 

[Les Jones]

Does the stepper relay replacement have to remember it's last position when power is removed or does it need to reset to a fixed starting position (0 ?) when power is applied ? If it does need to remember it's last position when power is removed then a small microcontroller may be a better solution as it could save the last position in it's internal EEPROM when powered down.

Les.

 

[rjenkinsgb]

Does the stepper relay replacement have to remember it's last position when power is removed

Good point - but a CMOS counter/decoder can be backed up with a small battery and last years, with no programming needed.

 

[Machine Doctor 13]

Yes, it has to remember it's last position. I am not familiar with a counter decoder but will definitely look into it.
Thanks to all.

 

[Pommie]

Do you know the current requirement of the turret? If not too high then a 4017 driving a UDN2981A might be a good combination.

Mike.
P.S. is there another PNP darlington that is readily available.

 

[Machine Doctor 13]

The turret rotates to another tool just fine as is. I need to replace the stepper relay which only provides feedback to the CNC control as to what tool is now operational. There is a wiper on the stepper relay that is connected to +24 VDC and this wiper advances to 1 of 8 output contacts (each one representing one of the 8 tools). An absolute pulse coder would work but there is no place to mount one.

 

[rjenkinsgb]

Note that you can still get hold of Ledex switches; some places on ebay have "new old stock" ones.
eg. https://www.ebay.co.uk/itm/LEDEX-EL...Shorting-28VDC-Part-No-8111-1-pc/152096416303

They take standard kit-type rotary switch wafers so you can make the contacts up as you wish.

For 8 way you would have to add a wafer and pulse circuit to step it past the four unused positions, any time it got there.


Or uniselectors - again, there are several on ebay:
**broken link removed**

As with the Ledex switch, use contacts beyond the needed ones to cycle it back to the first contact position.

 

[Machine Doctor 13]

I'm looking at a 4017 Decade Counter. This might work.

 

[Pommie]

The 4017 can only supply a tiny current - not enough to drive your turret. You need something like the chip I posted above to supply a higher current at 24V. However, that might still not supply enough - are you able to measure the current that the turret takes?

Mike.

 

[Machine Doctor 13]

The turret is driven by a separate circuit and that works just fine. The stepper relay I'm trying to replace just supplies feedback to the CNC as to what tool is currently being used. So maybe the hot set up is the 4017 driving the UDN 2981A. Thanks Mike.

 

[unclejed613]

i had a bit of an "off the wall" thought when i read the first post. a lot of old pinball machines used stepper relays for the digit counters. replace the 120V coils with a 24V coil, and you're in business...

 

[Machine Doctor 13]

I constructed the attached circuit on a breadboard and it works. This is for the tool changer on a lathe. The clock pulse to the 4022 is + 15 VDC and the outputs from the 4022 are also + 15 VDC. The corresponding outputs from the 2981A are +15VDC too, but I need to convert them to + 24 VDC for the CNC control board.
Any ideas would be greatly appreciated.

 

[Nigel Goodwin]

Simply feed the 2981A from 24V, job done.

 

[Machine Doctor 13]

Do you mean put + 24 V to pin 9 of the 2981A? However, if you mean use + 24V for inputs to the 2981A how do I convert the + 15V from the 4022?

 

[rjenkinsgb]

The 4022 supply and the signal inputs to the 2981A can be as low as 5V without any problem.

The "VS" supply to the 2981A is completely independent, voltage wise, and can be anything up to 50V.
Just feed that from the 24V supply and the outputs will be 24V, or very close to that.

Using a lower supply for the 4022 part could also make it easier to rig up a battery backup for that, so it retains its state when the main machine power is off.
Just use a resistor divider to reduce the input to the clock pin to slightly below its supply voltage.

 

[Machine Doctor 13]

Thanks so much for this very helpful advice. I'm 70 yrs old and repair older CNC machines. I looked at this project as a challenge, but knew I would need some help.
Thanks a lot.