handle wild leg

A couple of things to consider on those solid state relays, You probably want ones designed to switch a DC load and you are going to have to meet the minimum load current.

All of the solid state relays I have used that are designed to switch an AC load will latch on once the control voltage goes up enough. They will switch off when the load current goes through zero on an AC load. If you are using them to control a signal into a computer, it will probably be a DC signal, so it will not switch off, even if the AC control signal goes away.

Another standard feature of solid state relays is that they have a minimum load current to switch on at all. I think it is related to how the load side of the optical isolation is powered. Driving a control signal into a computer input port probably will not provide that load. A pilot light probably would be enough, but look at the data sheet on the relay for sure.

BobH

Reply to
BobH
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I've been drill and tapping all day, mounting components in my refit mill drive control cabinet.

I'm wiring the machine so I'll have two hot legs when the phase converter is off. That way the control computer can run.

I want to drop Estop if power to the "wild" leg is missing. if I just use a small 110 volt coil relay , I think it will burn out with too much voltage. Correct? How would you handle?

Also, I'd like the computer to monitor all three phases through Opto

22 inputs:
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Again, the wild leg could go over 140. How do I handle it here?

Karl

Reply to
Karl Townsend

Use AC rated solid state relay?

Reply to
Ignoramus11321

You may have something there Iggy.

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If I'm reading right it will take up to 280 volts AC and go on at 85. Correct?

Reply to
Karl Townsend

I did not check that particular model, but yes, exactly that sort of thing. They are very forgiving of input voltage.

i
Reply to
Ignoramus11321

Karl Townsend wrote in rec.crafts.metalworking on Sat, 16 Oct 2010 16:53:20 -0500:

Could you add a resistor in series to the relay coil?

Reply to
dan

O.K.

Perhaps add a 120 VAC incandescent lamp in series -- one with about the same current drain as the relay coil (which you would have to either measure or calculate).

Hmm ... only voltage spec'd -- not current.

I've got some OPTO-22 modules -- but not AC input, so I could not check out the input impedance. However, I see that they also have G4IAC5A -- rated from 180-VAC to 280 VAC input.

Now -- I used to have a downloaded catalog which included schematics for the modules. IIRC, the AC Input one was:

High input terminal, current-limiting resistor, bridge rectifier, Opto-coupler (to the digital side of things), back to bridge rectifier, Low input terminal.

So -- with the resistor limiting current, the bridge rectifier making sure that it does not matter which direction the voltages is in at the moment, and the forward voltage drop of the LED in the opto-coupler, the only thing really voltage sensitive would be the resistor. More voltage equals more power dissipated in it -- and more current through the LED. I suspect that it would not care about short surges over 140 VAC. If you expect long exposure to a higher voltage, you need to add an external resistor in series, so the AC across the input terminals does not exceed 140 VAC.

Primary suggestion above is the series resistor between the AC and the input terminals. If you are willing to dissipate more waste power, make a voltage divider (two resistors in series -- one to high, one to low), and the terminals are connected across the low side resistor.

The Opto-22 inputs are designed to tolerate a very high voltage above ground, as long as both terminals are not that far apart in voltage.

Good Luck, DoN.

Reply to
DoN. Nichols

How about a current transformer in that leg? Not enough current flow and shut it down? You need a burden resistor and a small full wave bridge across the output winding, and a window detector to set your limits. A quad comparator and some simple logic will let you calibrate it for normal loads.

Reply to
Michael A. Terrell

most folks would use a comparator with a RC network on each leg, then use diode logic and OR them together to get a signal if any leg is out of tolerance - AND that with the off signal or you will never get started - one chip and about a dozen discretes - how hard can it be?

Reply to
Bill Noble

Like killing flies with napalm, but it'll work. "Off" leakage current is 14 ma. Have this Opto22 device switch 120VAC from line to a small relay, switch Estop with that relay.

Reply to
Don Foreman

I would use a phase loss detector on all three legs which will also detect phase rotation error too.

John

Reply to
John

Thanks for pointing out the load requirement.

OK as a separate issue, I'd like the control computer to know the phase is present. the spec on the G4 module say 90 -140 volt and 28K resistance. Is this a simple ohm's law thing where i could add 9.3K resistance to still have 120 volt trip this module and get up to 190 without damage?

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Karl

Reply to
Karl Townsend

Thanks for pointing this out. I didn't know this, makes sense.

karl

Reply to
Karl Townsend

I also used to apply a resistor across the input side to bleed off any charge buildup. Otherwise, they'd fire every once in a while. OK for heaters, not for motors. I also never used them in a safety application for that reason.

Pete Keillor

Reply to
Pete Keillor

Thanks for all the good advice. I'm thinking try a 9.3 K ohm resistor in front of the opto input module.

Karl

Reply to
Karl Townsend

say, sittin' here musing and drinking coffee.

What if i had an SSR fire a 220 coil relay? Then loss of any phase would drop Estop. correct? Makes it seem like less of a kludge to me anyway.

Karl

Reply to
Karl Townsend

I really don't understand why you are struggling with this - it is a little complex to make a true "phase loss" detector, but it is trivial to make a voltage monitor that will provide a signal when any of the three legs drops below a preset value with respect to any of the others for more than a half cycle. If you don't have the electrical knowledge, many of us do, you are talking $5 worth of parts and some time soldering stuff. Then you can do whatever you want with the signal - what you are doing here is using a pile of expensive already assembled parts to do something they were not intended to do - you might get it to work, but it is kind of like using a massey fergeson tractor to warm your swimming pool. all you need are three voltage comparators, which you diode OR together - one chip, some resistors, 6 diodes, 6 capacitors, a power supply chip and so on - this is a box that would be about the size of a cigarette pack, worst case - and your control circuitry probably already produces suitable power forms - you only need a fraction of a miliamp of power

Reply to
Bill Noble

Yes.

Reply to
Don Foreman

Karl doesn't like to assemble electronic stuff. You probably don't make much .308 ammunition. We all have our preferences!

Reply to
Don Foreman

Don even made a special trip out here to teach me circuit board soldering and had me going pretty good after an hour. But it didn't take, my next effort was useless. Now, I just found out a couple months ago that my neighbor of 15 years (he's still new here) is good at it. But, he's working 6-7 days a week, I won't ask for this.

Nothing wrong with using a sledge hammer to swat a fly if its not too expensive and its reliable.

Now, I am truly addicted to the .308 cartridge and especially the semi auto rifles that use them. Here's what Santa can get me for Christmas:

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Karl

Reply to
Karl Townsend

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