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sconisbee
07-17-2010, 05:48 PM
So I have a need to momentarily activate a relay when a signal wire goes from +24v to 0v. And i have no idea how to wire a standard relay to do that.

Basically I'm wiring in spindle control to my cnc lathe that currently has manual spindle control but the control is capable of spindle control its just not wired in.

The control outputs are all 24v and always high in a resting state and go low when a command is triggered.

For example when the control reads a M03 command it pulls output card pin 9 low. When this happens I need a relay to close to trigger the spindle contractor but i cant figure out how to wire a relay this way.

Any Ideas? if any of this makes sense?

I dont have a wiring diagram for the control as of yet, but in theory I could examine the wiring for the M00 relay and the Coolant Relay but they are both quite buried in the cabinet.

The Artful Bodger
07-17-2010, 05:56 PM
Does your signal have enough current capacity to handle the coil relay? If so, connect one end of the coil to the signal line and the other to 24+.

You should probably put a diode across the coil to prevent inductive spikes damaging the signal source.

If you signal line does not have the current capacity you could use a simple tranistor configuration to control the relay.

Be aware though that in matters such as this it is a good idea to have someone who knows what they are doing involved in the modification, unless you want to put the electronics card at risk.

EVguru
07-17-2010, 06:17 PM
If you want a momentary action from a relay, wire an electrolytic capacitor in series with the coil. The relay will pull in whilst the capacitor charges. You may also need a bleed resistor for re-triggering.

winchman
07-17-2010, 06:44 PM
If I understand correctly, you can use a normally-closed relay or the normally-closed contacts of a double-throw relay. The contacts will close when the voltage falls below what's needed to energize the relay, and open again when the voltage goes high.

You can delay the action of the relay with a capacitor in parallel with the coil. The contacts won't close until the energy stored in the capacitor is bled off through the relay coil.

The only problem with this approach is that the contacts will be closed when the system is off. That may cause the motor to try to start turning before the voltage comes up enough to open the contacts. This could be solved by having the system set up so the control system has to be up and running before power is available at the motor.

sconisbee
07-17-2010, 06:51 PM
Does your signal have enough current capacity to handle the coil relay? If so, connect one end of the coil to the signal line and the other to 24+.

Be aware though that in matters such as this it is a good idea to have someone who knows what they are doing involved in the modification, unless you want to put the electronics card at risk.

I believe this may well be how the M00 and coolant relays are operating as there is no sign of any other means of triggering the relay such as transistor circuits.

I had wondered about connecting it up so the signal wire was connected to the coil and the other to +24 but I cant get my head around how this works? I noticed the same setup with some monostable circuits i've seen.

I have a contact at the company that made these old controls and he is looking for a diagram for me and I'll be able to confirm any modifications with him before carrying them out, but its nice to have a plan of action before calling as it makes things go a little smoother.

I had also thought of the method that winchman suggested but I know for a fact the existing relays dont work this way and I would rather match things compared to whats there already at this stage.

MaxHeadRoom
07-17-2010, 07:10 PM
I am assuming you want to simulate the start button that probably has a retaining contact on the contactor, this is why you are asking for a momentary?
Why not make the M03 relay maintained if the output is on all the time during M03 and do away with the retaining contact on the contactor?
This way you could use another relay with a N.C. contact in series to operate the M05, which I assume you want also?
This is also assuming the control cancels the M03 output automatically when issuing a M05?
You should also have a means of E-stopping the spindle or opening the circuit with a push maintained E-stop button in series.
If you want to use the outputs as they are, you should be able to feed the live side of the relay with the same 24vdc supply, the other side of the relay to the output and when it goes low, it will trigger the relay.
Max.

sconisbee
07-17-2010, 07:45 PM
I am assuming you want to simulate the start button that probably has a retaining contact on the contactor, this is why you are asking for a momentary?
Why not make the M03 relay maintained if the output is on all the time during M03 and do away with the retaining contact on the contactor?
This way you could use another relay with a N.C. contact in series to operate the M05, which I assume you want also?
This is also assuming the control cancels the M03 output automatically when issuing a M05?
You should also have a means of E-stopping the spindle or opening the circuit with a push maintained E-stop button in series.
If you want to use the outputs as they are, you should be able to feed the live side of the relay with the same 24vdc supply, the other side of the relay to the output and when it goes low, it will trigger the relay.
Max.



I am assuming you want to simulate the start button that probably has a retaining contact on the contactor, this is why you are asking for a momentary?

Yes and no, the main reason for momentary action is because the M03 signal is momentary.


This way you could use another relay with a N.C. contact in series to operate the M05, which I assume you want also?

M05 is already in place and in effect I will be copying the way this is setup. But without dissasembling half of the electronics cabinet it would be hard to trace all the wiring off the M05 Relay base

macona
07-17-2010, 11:36 PM
use a delay-off timer

J Tiers
07-18-2010, 12:00 AM
is the output you refer to a "driven" output, or is it just connected to +24 and activated by disconnecting and allowing it to "drop'? (same as if connected to a switch)

A driven output is easy, an output that is disconnected is harder.

One way for a 'disconnecting contact" uses a more sensitive relay.....

Wire from +24, through relay coil, through a diode and then a capacitor in parallel with a high value resistor (1) to the contact. Resistor (2) from contact to common. Diode "points" away from coil.

Now, when contact is closed, relay has no voltage across it, and capacitor is discharged by the high value resistor (1).

When the contact opens, resistor (2) "pulls" the relay coil down through the capacitor closing it, until the capacitor charges up, at which time the relay opens again, because resistor (1) is too large to hold it in.

When contact closes, there is no voltage across teh relay coil, and no change of state, so the relay is un-affected. With the diode there to block any possible reverse current through the capacitor, it definitely will be unaffected..

A "sensitive" relay, because resistor (2) has to pull it down and close it.

Time of closing set by relay coil and capacitor.

Time of "resetting" set by capacitor and resistor (1).

No need for catch diodes etc, since the relay coil never has a fast change of current, the charging of the capacitor is relatively slow.

Mark McGrath
07-18-2010, 04:16 AM
Wiring a relay so that the coil always has a live feed is unsafe.A short anywhere on the other side of the coil energises the relay.
That is particularly clever on something like a spindle drive (NOT).

Mark

sconisbee
07-18-2010, 06:42 AM
Wiring a relay so that the coil always has a live feed is unsafe.A short anywhere on the other side of the coil energises the relay.
That is particularly clever on something like a spindle drive (NOT).

Mark

I agree Mark thats whats got me confused about how they have done it, as that is the way the controller is wired for m00 and the coolant and according to ANC thats how the spindle drive is done also. Before doing anything Ill be checking with ANC first.

J Tiers Its a driven output coming straight from the card that drives all of the M functions on the control. I'll try to pull the coolant and M00 relays tommorow and see how they are wired.

J Tiers
07-18-2010, 10:10 AM
Wiring a relay so that the coil always has a live feed is unsafe.A short anywhere on the other side of the coil energises the relay.
That is particularly clever on something like a spindle drive (NOT).

Mark

Can be..... so you use caution. A lot of relay drives are like that though, low side drive.

It really is about the same..... a stuck contact or shorted driver on a high side drive is just the same problem.

I trust wire insulation more than I trust a solid-state drive component, or even a relay contact. And UL, etc agree with me.

I was going to suggest a circuit for a driven output, but it has begun thundering here, and time to unplug the computer.

MaxHeadRoom
07-18-2010, 10:12 AM
Yes and no, the main reason for momentary action is because the M03 signal is momentary.

M05 is already in place and in effect I will be copying the way this is setup. But without dissasembling half of the electronics cabinet it would be hard to trace all the wiring off the M05 Relay base

If your trigger signal is already momentary, why not then wire it in parallel with a N.O. contactor auxiliary contact as is done in a normal push button start circuit?
I read your first post to mean you Wanted a momentary?
How does the M05 operate now? You should be able to integrate a N.C. contact on it in series with the M03 contact & coil circuit to drop the contactor out, it is still advisable to still include the e-stop in the series string.
Max.

sconisbee
07-18-2010, 10:26 AM
How does the M05 operate now?


Ill try and describe best I can how M05 operates not sure how to describe it but ill try.

The M05 output is a driven output on a output driver card in the machine, and it drives Pin 4 on the board.

Pin 4 is constant at 24v DC (not the best regulated 24 volt either, its almost only half rectified)

When the controller see's M05 then Pin 4 drops to 0vdc at this point the M05 relay energizes and breaks the latch circuit for the spindle contactor thus allowing the contactor to open.

Now in this situation what Mark is worried about doesn't really matter as if there's a short then the spindle will just stop however I share his concerns over a spindle start in this way.


One more thing I can say is there is a wire coming down from the output card that is common between the coolant and the m05 relays, I originally thought that this was a common ground but after checking it turns out that this is the same constant 24vdc signal, so all I can surmise is that the relay is connected between the constant 24vdc line and the signal line, and when the signal line drops to 0 then it allows the relay to energize, would this be correct?

sconisbee
07-18-2010, 10:31 AM
Only just saw J Tiers' 2nd reply, It seems to becoming clear how the machine works and I think all that needs to be done is to add 2 more relays wired up to be low side drive like the M05 and wire them between the 24vdc line that goes to the m05 and coolant relays and then the other side of the coil to the respective signal line.

Then have the relays double up on the manual start push buttons as this is how the tech at ANC said most of their controls would be done as it keeps the E-stop in the loop.

And as i cant see any other driver circuits that are separate to the driver board I am assuming that the outputs can drive a relay direct just like the M05 one. Now i just have to get a couple of suitable relays and sockets and run through it with ANC and see what they say.

J Tiers
07-18-2010, 01:10 PM
What UL generally requires for a "control circuit" in equipment I do controls for is that clearances to all other circuits be equal to that for the highest voltage in the box. Insulation ditto.

that isn't universal, so refer to the specific standard if you need to know.

The effect of that requirement is to make inadvertant shorts very unlikely. it is recognized that nothing is perfect. The standards don't require perfection, they require reasonable construction to prevent problems.

It is nearly always possible to envision a series of failures which collectively create a hazard, all you do is assume that every precaution is nullified.... That brings you back to where you started.

The assumption is, right or wrong (and in many shops, its wrong) that once a failure of one type becomes known, it will be fixed before it can combine with future failures to cause a hazard.

Most standards are "one failure", some assume two. Knowing what goes on, two may be better, but the fact is that it only takes one more than you plan for, no matter what. And the better-protected, ironically the worse the safety may be.... if it allows one failure to be ignored, or worse yet, unknown.

Mark McGrath
07-18-2010, 04:20 PM
What UL generally requires for a "control circuit" in equipment I do controls for is that clearances to all other circuits be equal to that for the highest voltage in the box. Insulation ditto.

that isn't universal, so refer to the specific standard if you need to know.

The effect of that requirement is to make inadvertant shorts very unlikely. it is recognized that nothing is perfect. The standards don't require perfection, they require reasonable construction to prevent problems.

It is nearly always possible to envision a series of failures which collectively create a hazard, all you do is assume that every precaution is nullified.... That brings you back to where you started.

The assumption is, right or wrong (and in many shops, its wrong) that once a failure of one type becomes known, it will be fixed before it can combine with future failures to cause a hazard.

Most standards are "one failure", some assume two. Knowing what goes on, two may be better, but the fact is that it only takes one more than you plan for, no matter what. And the better-protected, ironically the worse the safety may be.... if it allows one failure to be ignored, or worse yet, unknown.


You will get an award from the plain English society for that post Jerry. :D