Electronics advice needed for G540 spindle control.

[TXFred]

I'm working on a way to run the X2's spindle from the Gecko G540's VFD drive.

It looks like the motor control will work just by hooking it up to the G540. The part that is tricky is the switch that's built into the potentiometer. This is used to reset a spindle fault on the mill.

The easy solution is to just have a switch to activate the spindle. But I'm after more. I've designed this simple circuit which will trigger an emergency stop on the G540 whenever there is a spindle fault. That way, if the spindle stalls, the stepper motors won't keep moving and cause more damage.



When the spindle stalls, the Spindle Fault line goes to +110VDC. This activates the relay or opto isolator, which opens and triggers an emergency stop on the G540.

The momentary switch is to reset the spindle fault. The resistor is to limit the current through the relay, so that the relay side of the circuit won't reset the fault. This is similar to what is on the mill now. I've just swapped out the orange light for the relay/opto.

It should be noted that this circuit maintains the isolation of the G540 from the X2 spindle control so long as a mechanical relay or an opto-isolator is used. A solid state relay would eliminate this isolation.

The problem I see is this. I haven't found a good relay or opto that can handle the 110v that the X2's board is putting out. At least, I haven't found a cheap and small one.

Does anyone know of a relay or opto that will work here?

Alternately, what's a good way to take that 110v and reduce it to 12 volts so I can use a regular automotive relay?

Cheers,
Fred

[SCzEngrgGroup]

I'm working on a way to run the X2's spindle from the Gecko G540's VFD drive.

It looks like the motor control will work just by hooking it up to the G540. The part that is tricky is the switch that's built into the potentiometer. This is used to reset a spindle fault on the mill.

The easy solution is to just have a switch to activate the spindle. But I'm after more. I've designed this simple circuit which will trigger an emergency stop on the G540 whenever there is a spindle fault. That way, if the spindle stalls, the stepper motors won't keep moving and cause more damage.



When the spindle stalls, the Spindle Fault line goes to +110VDC. This activates the relay or opto isolator, which opens and triggers an emergency stop on the G540.

The momentary switch is to reset the spindle fault. The resistor is to limit the current through the relay, so that the relay side of the circuit won't reset the fault. This is similar to what is on the mill now. I've just swapped out the orange light for the relay/opto.

It should be noted that this circuit maintains the isolation of the G540 from the X2 spindle control so long as a mechanical relay or an opto-isolator is used. A solid state relay would eliminate this isolation.

The problem I see is this. I haven't found a good relay or opto that can handle the 110v that the X2's board is putting out. At least, I haven't found a cheap and small one.

Does anyone know of a relay or opto that will work here?

Alternately, what's a good way to take that 110v and reduce it to 12 volts so I can use a regular automotive relay?

Cheers,
Fred

The LED on any opto, and the "coil" on any SSR, can operate from pretty much any voltage, as long as it has a suitable series resistor to limit the voltage the LED sees. Figure out what current you need in the LED, and divide the input voltage by that current to get the required series resistor value. The problem will be the required power rating of that resistor, due to the large voltage across it. The phototransistor side is more voltage sensitive, since it will see the full supply voltage when the opto is turned "off". That shouldn't be a problem in this case.

Regards,
Ray L.

[jalessi]

These are cheap.

http://skycraftsurplus.com/110vdcrelaydpdt10amp1.aspx

http://tinyurl.com/yf5pggq

[Al_The_Man]

If I understand correctly the logic is opposite of the run of the mill opto, the output requires on when the input is off?
A 4N35 would require about a 2.7k res. at about 5w, (a power waster!)
Al.

[TXFred]

If I understand correctly the logic is opposite of the run of the mill opto, the output requires on when the input is off?
A 4N35 would require about a 2.7k res. at about 5w, (a power waster!)
Al.

That's correct, if I used an opto I'd need to add additional circuitry to invert the signal. This is tricky, since both sides of the circuit have issues with being grounded. The mill side cannot have a clear path to ground since that will prevent the spindle fault from ever being set in the first place. That's bad if the mill motor does stall.

The G540 side of things has its own problems. The G540 sees an open circuit as an emergency stop. If there is an additional circuit that's doing the inversion, it's got to be grounded. That ground may prevent the G540 from seeing an emergency stop signal.

So if I went with an opto, I'd have to have an intermediate circuit with its own independent power supply and ground to do the inversion. I was hoping to find an inverting opto, but so far I've found nothing.

All in all, the relay looks simpler, although it will not operate quite as fast as an opto.

Fred

[Crevice Reamer]

Hi Fred. I don't know what you consider cheap, but here's a relay that will work:

https://www.surpluscenter.com/item.a...tname=electric

CR.

[Al_The_Man]

Hi Fred. I don't know what you consider cheap, but here's a relay that will work:

CR.

The only thing is that has an AC coil. He needs DC.
Al.

[Crevice Reamer]

The only thing is that has an AC coil. He needs DC.
Al.

It's just a coil. Volts is Volts. DC can energize an AC coil. It should work fine.

CR.

[jalessi]

Crevice Reamer,

Often, a DC relay will have a 'flywheel rectifier' [FR] placed reverse-biased across the input terminals, so as to prevent a high back-EMF as the magnetic field collapses upon de-energizing.

Plugging a DC relay with a FR into an AC circuit may cause an overload as half the AC cycle could be passing through the FR as a short-circuit. Incorrectly connecting the DC supply will cause a similar effect. Sometimes a second rectifier is installed in the casing of the relay so as to overcome the problem of reverse polarity.

An AC relay will be wound with different wire as the inductance of the wire will offer some limitation to the current that the relay draws when energized. Instead of an FR it may be possible to place a small capacitor or a neon indicator across the terminals of the relay to minimize any back-EMF effects as the magnetic field collapses, should the need arise, if indeed the relay does not already have these components fitted inside its casing.

So, in summary, they are two different animals. While one may get away with interchanging DC and AC relays on occasions, it is a practice that involves risk to the relays themselves and to the equipment being controlled by their contacts, so it is a practice best avoided.

Jeff...

[SCzEngrgGroup]

It's just a coil. Volts is Volts. DC can energize an AC coil. It should work fine.

CR.

Driving an AC relay with DC of the same voltage will burn up the coil due to rapid over-heating. The current when driver with DC will be considerably higher, and constant (the coil will be always saturated), rather than cycling as with AC, so the average power (current sqaured times resistance) will be MUCH higher, with a resultant increase in self-heating.

Regards,
Ray L.

[Al_The_Man]

It's just a coil. Volts is Volts. DC can energize an AC coil. It should work fine.

CR.

No, the DC resistance of an AC coil is too low to use it on DC, the AC coil relies on inductive reactance to limit the current.
This is why a AC solenoid or relay will burn up quickly if the armature does not move over, due to the decrease in inductance, just leaving a very low resistance.
Al.

[Crevice Reamer]

Okay, then this would seem to be your best bet:

http://parts.digikey.com/1/parts/752...-11dg-110.html

CR.