Induced voltage (I think)

[drbond24]

I have what I believe is an induced voltage problem. I have this simple circuit:



The contact between e and f is a control system relay output that is rated for 220 VAC and has been verified to operate properly.

The coil between c and d is a 120 VAC operated contact in a valve positioner control cabinet. Anything greater than 22 VAC is sufficient to energize it, thereby opening the corresponding valve. Anything less than 8 VAC is de-energized which closes the valve. The area between 8 and 22 VAC is designated 'undefined' by the instruction manual, which I'm guessing means it could go either way.

The a side of the 120 VAC is the line and the b side is the neutral.

Now, when the contact is closed I have 120 VAC on the coil and the valve opens. No problem. However, when the contact is open I have 50 VAC on the coil which keeps the valve open. This 50 VAC manifests itself on the d side of the coil (referenced to ground). The voltage on the other side of the coil, using the same reference, is less than 1 VAC. We checked the contact and it is working. This voltage 'appears' on the conductor between e and d.

The problem, I think, is in the cable run between the 120 VAC and the contact. I can lift the wires completely on both ends and still measure 50 VAC on the cable. I did this with a Fluke 87V. I doubt this voltage has much 'umph' to it, but it is enough to keep the coil energized which, per the manual, requires 1.8 mA.

I cannot access the wire pull between these two points, but I know that it is in a cable vault with countless other cables. It is probably inside a large conduit with a bunch of other cables with various purposes. The cable is a shielded 3 conductor 16 AWG. The shield is grounded on one end.

My preferred solution at this point would be to put something across the coil terminals to counteract this voltage if that is possible, but I do not know what to use. I suspect an RC device but I have no idea how to go about sizing R or C. Other than pulling new wire via a different route, what can I do to remove or counteract this 50 VAC?

 

[gar]

121214-0829 EST

drbond24:

I believe you have a capacitively coupled current (voltage). But some of your description is confusing with respect to this conclusion.

Your sketch is hard to read because of size. I believe if you limited it to 640 pixels wide it would fit.

You are correct that there is an indeterminate region between on and off. This will be true of a conventional relay, solenoid, or electronic circuit.

One question --- is this valve device rated to operate at 120 V? If so I assume it contains electronics. A direct coil operated device that pulls in at 22 V would probably burn up at 120 V. But it is not impossible to design such a device. 1.8 mA at 22 V = 12,000 ohms, or 1.8 mA at 8 V is 4500 ohms. I suspect that you have a load (your valve device) with a 5000 ohm resistor in series with an optical coupler at the input.

What is confusing is that you have a two conductor shielded cable that your wires are within. I assume one wire is your b-c and the other is d-e, and the shield is connected to b-c. If this is the case, then the shield would prevent capacitive coupling from any wire outside the shielded pair and when d-e is opened at both ends there would be virtually zero voltage across (d-e) to (b-c). Can you confirm this?

.

 

[Jraef]

Since you have such a wide range of operating voltage to work with, I would start by putting a pot between d and e, then increase the value slowly until the problem goes away, remove the pot and measure the resistance value. Then put it back in and keep increasing it until the coil no longer operates. When you get there, read the resistance on the pot again and replace it with a suitable fixed resistor value that is half way between the two measurements.

 

[Phil Corso]

Dr. Bond...

1) Is there the possibility that the group of wires are in water?

2) Is the problem new?

3) Can you measure the current?

Regards, Phil Corso

 

[Rick Christopherson]

If you were going to have a problem with an induced current/voltage, it would be in the (de) run. But as was pointed out, if that is a shielded cable, it should not be where the problem lies.

We don't know anything about the relay contacts at (ef), but if it has some sort of debounce component, it could be sufficient to energize this small cd solenoid coil. You don't need debounce on a mechanical solenoid coil. Even though the contacts are rated for 220VAC, could they normally be intended for a DC control line?

This is just a wild shot in the dark, but try putting in a diode between the (e)-wire and the (e)-contact. Polarity shouldn't matter. If you want to test this idea out first, simply disconnect the (e)-wire from the (e)-contact. If the solenoid valve drops out, then it is a good sign that the contacts have a debounce circuit (and you also know it is not from induced current/voltage on the de line).

By the way, if disconnecting the contact at (e) doesn't cause the solenoid valve to drop out, then you do have some sort of induced voltage/current on the de line. If that's the case, we'll need to know a few more things.

 

[Rick Christopherson]

By the way, if disconnecting the contact at e doesn't cause the solenoid valve to drop out, then you do have some sort of induced voltage/current on the de line. If that's the case, we'll need to know a few more things.

Since you have such a wide range of operating voltage to work with, I would start by putting a pot between d and e, then increase the value slowly until the problem goes away......

When I said we would need more information, I hadn't considered using a pot to experiment with first. The pot is a good idea, but it wouldn't go between (d) and (e). That is already a low-resistance path.

The pot could be used to test out the size of a pull-down resistor. It should be connected between either (d) or (e) on one end, and to either (b), (c), or (gnd) on the other end.

However, as I said above, if disconnecting the contact at (e) causes the solenoid to drop out, then this would not be a necessary avenue to pursue.

 

[jim dungar]

If the controlling contact can be a 2-pole device, then simply break both side of the solenoid.

 

[Besoeker]

This is just a wild shot in the dark, but try putting in a diode between the (e)-wire and the (e)-contact.

If the coil is rated for AC that would be a bad idea.

 

[Rick Christopherson]

If the coil is rated for AC that would be a bad idea.

That would be an issue for a transformer, but I don't believe it is an issue for a solenoid. A DC solenoid may not like AC power, but typically, an AC solenoid is not going to have a problem with half-wave rectified power.

As a matter of fact, I have seen cases where residual magnetism in the plunger can cause an AC solenoid to chatter or loose pull-in power. Adding a diode can correct this without needing to saturate the coil. I had that very problem on a pneumatic solenoid on a compressed air system. The coil was 24VAC, and it was giving me fits. So I switched out the power supply to 24VDC and the fits went away.

If I am wrong about this, I am open to hearing your reasons. :thumbsup:

 

[SG-1]

If the issue is capacitive coupling a 10k resistor between c & d ( in parallel with the coil ) may be enough to drop the voltage below the pickup. If the control relay has a normally closed contact & it is break before make ( at some point both contacts are open together ) it could be placed between c & d.

 

[Besoeker]

A DC solenoid may not like AC power, but typically, an AC solenoid is not going to have a problem with half-wave rectified power.

Typically, you are incorrect.

 

[drbond24]

121214-0829 EST

drbond24:

I believe you have a capacitively coupled current (voltage). But some of your description is confusing with respect to this conclusion.

Your sketch is hard to read because of size. I believe if you limited it to 640 pixels wide it would fit.

You are correct that there is an indeterminate region between on and off. This will be true of a conventional relay, solenoid, or electronic circuit.

One question --- is this valve device rated to operate at 120 V? If so I assume it contains electronics. A direct coil operated device that pulls in at 22 V would probably burn up at 120 V. But it is not impossible to design such a device. 1.8 mA at 22 V = 12,000 ohms, or 1.8 mA at 8 V is 4500 ohms. I suspect that you have a load (your valve device) with a 5000 ohm resistor in series with an optical coupler at the input.

What is confusing is that you have a two conductor shielded cable that your wires are within. I assume one wire is your b-c and the other is d-e, and the shield is connected to b-c. If this is the case, then the shield would prevent capacitive coupling from any wire outside the shielded pair and when d-e is opened at both ends there would be virtually zero voltage across (d-e) to (b-c). Can you confirm this?

.

Here is a smaller version of the diagram:



Yes, it is rated for 120 VAC. The valve positioner control enclosure, which is where my problem is, simply takes this 120 VAC as a logical '1' and sends a different signal to open the valve. This 120 VAC is not opening the valve, it is simply generating the command to open the valve. The 50 VAC mystery voltage is also being interpreted as a logical '1' though.

Actually, the 120 VAC and the contact input 'cd' are physically next to each other. The contact 'ef' is on the other side of the room, so the cable run is for 'af' and 'de'. The shield is connected at the 'c' end of 'bc'.

If I disconnect both ends of 'de' I have 50 volts on each end. There are a couple of other spare cables pulled from the same starting and end points, presumably via the same unknown route, and they each have 50 VAC on both ends and they aren't connected to anything at all. One is shielded, the other is not. I will re-verify all of this tonight just to be sure I am not missing something, because I know that does not make any sense. I'm on night shift presently, by the way.

 

[drbond24]

Since you have such a wide range of operating voltage to work with, I would start by putting a pot between d and e, then increase the value slowly until the problem goes away, remove the pot and measure the resistance value. Then put it back in and keep increasing it until the coil no longer operates. When you get there, read the resistance on the pot again and replace it with a suitable fixed resistor value that is half way between the two measurements.

Interesting idea. I will likely give this a try tonight.

 

[drbond24]

Dr. Bond...

1) Is there the possibility that the group of wires are in water?

2) Is the problem new?

3) Can you measure the current?

Regards, Phil Corso

Water would be very improbable, but I have no way to verify one way or the other. The cable run is in a cable vault that never sees the light of day and has no exterior walls. I see no way water could be introduced, but I cannot say it is impossible.

Yes, the problem is new. This is a new installation we are attempting to commission.

I should be able to measure the current, yes. I will do that tonight as well.

 

[drbond24]

If the controlling contact can be a 2-pole device, then simply break both side of the solenoid.

!!!!

Now that is an idea. I could produce such an arrangement, too. Not the most effective use of our limited number of contacts so I do not think I'll go here first, but I will definitely keep it in mind.

 

[Rick Christopherson]

Typically, you are incorrect.

If I am wrong about this, I am open to hearing your reasons. :thumbsup:

Gee. Instead of being a jackwaggon, why not explain it, like I suggested you openly could.

I don't think it is wrong for a solenoid, but I am open to hearing reasons why it could be, if you could stop being snarky for a minute.

 

[Phil Corso]

Rick...

Firstly, puttng a diode in seies will, of course, change the voltage to a half-wave DC cirsuit. Now you have an additional problem; the RMS value of the current is considerabley reduced!

Secondly, while in a measuring-mode, determine the conductor-to-ground capacitance!

Regards, Phil Corso

 

[Rick Christopherson]

Firstly, puttng a diode in seies will, of course, change the voltage to a half-wave DC cirsuit. Now you have an additional problem; the RMS value of the current is considerabley reduced!

The OP already established that the solenoid is functional at the lower end of the voltage range.

 

[hurk27]

Yes, it is rated for 120 VAC. The valve positioner control enclosure, which is where my problem is, simply takes this 120 VAC as a logical '1' and sends a different signal to open the valve. This 120 VAC is not opening the valve, it is simply generating the command to open the valve. The 50 VAC mystery voltage is also being interpreted as a logical '1' though.

Actually, the 120 VAC and the contact input 'cd' are physically next to each other. The contact 'ef' is on the other side of the room, so the cable run is for 'af' and 'de'. The shield is connected at the 'c' end of 'bc'.

If I disconnect both ends of 'de' I have 50 volts on each end. There are a couple of other spare cables pulled from the same starting and end points, presumably via the same unknown route, and they each have 50 VAC on both ends and they aren't connected to anything at all. One is shielded, the other is not. I will re-verify all of this tonight just to be sure I am not missing something, because I know that does not make any sense. I'm on night shift presently, by the way.

While the diagram really doesn't make sense as to your new description, I now am thinking of a PLC input that controls an output that controls the valve?

If the above is correct, it is possible you have a floating input that has not been address as pull up or pull down, if you apply 120 volts to the input you have to have a pull down resistor to force a low when the contacts open, most PLC's will incorporate these resistors if pre addressed, they may be a set of dip switch's or other method to address the input, this will also force a "0" when ghost voltage is encountered.

Also if the relay is a electronic type there might be a small leakage current if the control is a switching transistor instead of dry contacts.

 

[Phil Corso]

Rick...

An AC-Coil may become polarity sensitive, i.e., reverse force direction, with DC applied, if the armature material is not that normally found in a DC Solenoid!

Phil