Induced Voltage on Control Wire

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mull982

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A particular motor starter circuit has about (5) or so wires going back and forth to a control station and a starter. The voltage for these control wires are 120V.

The motor starter uses a solid state overload relay which uses some of the contorl wires from the field as inputs. On of these control wires is causing its solid state input to turn on when it shouldn't. When putting a meter on this wire it was found to have 28V. When measuring this voltage with a wiggy meter between this input and ground this 28V quickly disapears. When measuring with a fluke meter this 28V drops less rapidly.

The cable run is probably about 1000ft. I'm assuming that the voltage on this wire in question is being caused by an induced voltage or coupled capacitance with the long run.

If the 28V is a result of induction from the other cables, then I am trying to come up with an explanation as to what is going on. I'm assuming that the 28V drops to 0 when the wiggy is placed on it due to the fact that the wiggy has a low/high? impedance. If this is the case, then I am trying to figure out how this induced voltage should be represented. I am thinking it should be represented as a current source due to the fact that the voltage drops when a load it placed across it. Should this induced voltage be represented as a current source, or would it be a voltage source with a low VA and thus a large voltage drop when a load is placed across it?
 
mull982 said:
the wiggy has a low/high? impedance.

Should this induced voltage be represented as a current source, or would it be a voltage source with a low VA and thus a large voltage drop when a load is placed across it?
Click to expand...
mull -
The wiggy has a low input impedance compare with a DVM.

You can model the source either way:
a current source with a parallel impedance;
a voltage source with a series impedance

Either one works.

You did comment about a DVM slowly drawing down the voltage. You may have a DC offset as well.

Sounds like a fun science fair project.

Just for your interest, take a look at the overload spec sheet, for the input impedance - could be up in the megohms.

Could be that a 100k resistor across the input will kill the phantom voltage.

cf

cf
 
Cold Fusion said:
mull -


Could be that a 100k resistor across the input will kill the phantom voltage.



cf
Click to expand...

This is the part that I am trying to grasp. Why would this resistor kill this phantom voltage. At first I thought that this resistor would cause a voltage drop acorss the resistor but this would have to be in series. If the resistor is in parallel across the input as you mentioned, then wouldn't this 28V still drop across the resistor and thus be seen by the input?

What exactly happens when this resistor is placed?
 
Cold Fusion said:
mull -
The wiggy has a low input impedance compare with a DVM.
Click to expand...

Phantom Voltage is vaguely defined,
and solutions are difficult to prescribe exactly.

My Wiggy thing has a measured impedance of 3K Ohms,
and draws 40 milliAmps.
Mine draws down the phantom voltage predictably with 120 V.
But not always when the conduit is filled with 480 V.

I have had success placing 10 K Ohm resisters
across the leads, on a high impedance testor.
It was a a simple three prong receptical tester,
but that killed phantom voltage, again at 120V.

You could use an oscilloscope
and watch the phantom voltage vary as you change the load
(by tapping at the line with your Wiggy, like tap-tap-tap).
If there is a significant change (like 30%) when you tap-tap-tap,
then you are drawing down the phantom voltage.

Welcome to the forum.
 
090207-2202 EST

mull982:

Let us assume that the said one wire is your only problem.

Is the signal on this line a binary signal, meaning only two states true and false (0 and 1 or 0 and 120)? Assume it is. Then it is important to know what the threshold points are for true and false. It appears that these are below 28 V since you imply a problem with a 28 V noise level. If the control input that this wire connects to was well designed for a source of 0 or 120 V, then in this type of application it would be desirable for the threshold to be around 60 V, the mid point. But apparently it is not.

I discussed capacitive coupling in a different thread. See post #15 at http://forums.mikeholt.com/showthread.php?t=109919 A measurement on Anaconda's version of Romex I found 10 pfd per foot. At 60 Hz this calculates to a capacitive reactance of 265 megohms per foot. Assuming the same for your connection that would 0.265 megohms.

Your receiving input probably has some resistance to common. Try to measure it with no cable connected. Also see if you can determine the threshold voltages at this input to turn on and off the function. You could use a Variac to produce an adjustable voltage. At 120 V a 10 W resistor run at 5 W would have a resistance of 2880 ohms. So possibly put a 2500 ohm 10 W resistor from your problem input to common (probably neutral unless some other wire is run for logic common). Ohmite is a good source. Probably use an axial lead resistor or if a mounting problem then a tab type.

If the cable capacitance is as I assumed above, then the voltage across the 2500 resistor might be in the range of 1 to 2 V.

The source end must be able to supply the resistor current which is 50 MA.

You might also have ground path (neutral) noise problems unless a separate common is used for the logic signal lines from the power lines.

You need to provide more information on the details of the circuit. How is common handled, what are the input thresholds, input impedances, capacitance per foot of the wiring, drive current available, and probably some more.

It is unlikely that you have an inductively coupled signal. The word induced usually refers to inductive coupling. Very likely you have a capacitively coupled signal. If this is the case, then a shielded cable would have been useful.

.
 
mull982 said:
Why would this resistor kill this phantom voltage.

If the resistor is in parallel across the input as you mentioned, then wouldn't this 28V still drop across the resistor and thus be seen by the input?

What exactly happens when this resistor is placed?
Click to expand...
The resistor's impedance is so low (relative to the high (capacitive) impedance of the "source") that, just as with two dissimilar resistors in series, most of the voltage is dropped across the higher impedance.
 
Mull -

Two things you need to help your understanding:
1. Lift the input lead and measure the open circuit voltage.

2. Check the relay spec, and see what the input impedance is.

I looked up an AB E3+, with a 120V input module. It was 2ma at 79V. For this discussion, I figured 40K ohms.

Also in the specs was the max off-voltage and the min on-voltage. For this one it was 20V off voltage and 70V on voltage (I think). So for any input voltage between 20 and 70, the input state is indeterminent. So with your input at 28V, that is a classic "I don't know".

My sketches look pretty bad, hopefully you can decipher.

For this example, I modeled the OFF circuit as a 100V source with a series resistance and a 40Kohm module input resistance. With 40k and 28V, the current = 28/40k. The voltage drop across the series resistance is 100-28 = 72V. So R = (72 x 40)/28 kohms = 100k (about).

Now start dropping different resistors across the input. See what that does to the voltage.

I did a 100k and a 50k.

When I suggested a 100k, I didn't know the input impedance would be so low (40K). glene could well be right and 10k is what you need. However, keep in mind the required dissapation when the input is on.

For 10K, P = (120^2)/(10 x 10^3) = 1.4W Little 1/10W or 1/4W won't work.

Read it, see what you think. If it's gobbledegook, I'll try again tomorrow.

cf
 
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gar said:
090207-2202 EST
It is unlikely that you have an inductively coupled signal. The word induced usually refers to inductive coupling. Very likely you have a capacitively coupled signal. If this is the case, then a shielded cable would have been useful.
.
Click to expand...

Is it unlikely to have inductive coupling in circuits. We have seen this phantom voltage thing appear in other cases in the plant, and it was always said that it was a result of inductance between wires. It sounds like you are saying, that most of the time it is a result of capacitance coupling. What would have to occur for inductive coupling.

In this particular circuit we only see this phantom voltage when current is flowing and the contactor is pulled in. That is why I was thinking that this phantom voltage was a result of current flow and was caused by inductance. Do you need this current flow for capacitance coupling or just the voltage? Is there a way to definitively distinguish between capacitance vs inductive coupling?

LarryFine said:
The resistor's impedance is so low (relative to the high (capacitive) impedance of the "source") that, just as with two dissimilar resistors in series, most of the voltage is dropped across the higher impedance.
Click to expand...

Would'nt this resistor be combined in parallel with the input resistance and then this overall resistance be in sereis with the source impedance (capacitave impedance). Are you saying that adding this resistor would cause more of the voltage drop to drap across the wire, and therefore not show up at the input? I'm having a hard time visualizing this one.
 
Cold Fusion said:
Mull -

Two things you need to help your understanding:
1. Lift the input lead and measure the open circuit voltage.

2. Check the relay spec, and see what the input impedance is.

I looked up an AB E3+, with a 120V input module. It was 2ma at 79V. For this discussion, I figured 40K ohms.

Also in the specs was the max off-voltage and the min on-voltage. For this one it was 20V off voltage and 70V on voltage (I think). So for any input voltage between 20 and 70, the input state is indeterminent. So with your input at 28V, that is a classic "I don't know".

My sketches look pretty bad, hopefully you can decipher.

For this example, I modeled the OFF circuit as a 100V source with a series resistance and a 40Kohm module input resistance. With 40k and 28V, the current = 28/40k. The voltage drop across the series resistance is 100-28 = 72V. So R = (72 x 40)/28 kohms = 100k (about).

Now start dropping different resistors across the input. See what that does to the voltage.

I did a 100k and a 50k.

When I suggested a 100k, I didn't know the input impedance would be so low (40K). glene could well be right and 10k is what you need. However, keep in mind the required dissapation when the input is on.

For 10K, P = (120^2)/(10 x 10^3) = 1.4W Little 1/10W or 1/4W won't work.

Read it, see what you think. If it's gobbledegook, I'll try again tomorrow.

cf
Click to expand...

CF

The solid state overload relay that I am referring to is indeed an 120V E3+ relay so your input impedance value of 40kohm may be correct.

Your drawings definitely help visualize the situation however there are still a few questions in my mind. The 100k impedance that you show after the voltage source, is this the impedance of the cables that you back calculated using the the 28V. In an ideal situation we would want all of the phantom voltage to drop across this 100k so that no voltage was at the imput terminal.

Why are you assuming 100V as the source? Would this capacitively coupled voltage be of this magnitude?

What would you suspect that I see as an open circuit voltage giving these conditions. I will go check, but what should happen?

Basically you are saying that by adding a resistor in parallel with the 40k the oveall input impedacne is lowered. This input impedance can then be represented as a voltage divider with the 100k source impedance to see how much voltage will be dropped across the input. The goal is to get it below the 20V specified for "turn-off".

What would happen with this voltage drop when the input was actually supposed to be on? Would we represent the source with 120V rather than 100V?
 
mull982 said:
Are you saying that adding this resistor would cause more of the voltage drop to drap across the wire, and therefore not show up at the input?
Click to expand...
Not across the wire; across the weak coupling that's between the current-carrying wire(s) and the problematic control wire.
 
LarryFine said:
Not across the wire; across the weak coupling that's between the current-carrying wire(s) and the problematic control wire.
Click to expand...

Ah ha! I see the 100k source impedance in CF's diagram represents the impedance across the coupling as you described.

So basically we are creating a voltage divider with the source impedance and the parallel load impedance?

Its starting to become less fuzzy:D
 
We had a similar problem when 220vac control wires (multicore) cable which was laid for 1 km.

We shifted from 220 vac to 24 vdc and we never had any problem again.

cheers.
 
mull982 said:
What would you suspect that I see as an open circuit voltage giving these conditions. I will go check, but what should happen?

100V?
Click to expand...

I think I thought of the answer to my question.

With an open circuit no current would be flowing and thus there would be no voltage drop across the source impedance. Therfore lifting the wire off the load an measuring voltage, you should see the full 100V on this wire for this case. Sound right?
 
I agree with what's been said about phantom voltages and a resistor across the input to dissipate this voltage. But it would be unusual for a phantom voltage to cause a false input into a PLC unless you have 100's of feet of wire.

I would leave the wiggy across the input (it will act as the lower resistance across the input), and see if you still have an issue. If you do, you must have another problem.

For example, why does it look like there are 2 outputs tied together going to the switches? And is it OK to leave one "AC Common" open?

Or it could be a programming issue.

Steve
 
Just coincidence, our signal was also meant for a PLC input. We had to source the 24 vdc from the PLC panel and get the return to energise 24vdc relay and the dry contacts of the dc relay were used for the 220 vac control.
 
mull982 said:
The solid state overload relay that I am referring to is indeed an 120V E3+ relay so your input impedance value of 40kohm may be correct.
Click to expand...
Pure accident, but nice

mull:
Why are you assuming 100V as the source? Would this capacitively coupled voltage be of this magnitude?

cf:
I just picked out some numbers that would give the values we had. We knew the input impedance (40K) and we knew the voltage at the input terminals with no signal applied (28V). It's an ohm's law series circuit problem. I picked a source voltage of 100V and then calculated a source impedance - came out around 100k. As for magnitude or coupling method - I don't know. The model I am using does care what the coupling method is. I know it's there, I can measure it, and see the effects - the input is turned on.

What would you suspect that I see as an open circuit voltage giving these conditions. I will go check, but what should happen?

Your DVM will have an input impedance of 10+megohms. So when you measure the OC voltage, the source impedance is invisible - lost in the 10megs input impedance. Every voltage source can be defined by the OC voltge and the short circut current. The source voltage is as measured OC. The source series impedance is OCV/SSC.

So, if you lift the leads and measure, you will see the source OC voltage. You could also short out the input lead and measure the short circuit current. Then you would kow exactly what the source impedance is.

Basically you are saying that by adding a resistor in parallel with the 40k the oveall input impedacne is lowered. This input impedance can then be represented as a voltage divider with the 100k source impedance to see how much voltage will be dropped across the input. The goal is to get it below the 20V specified for "turn-off".

you got it:smile:

What would happen with this voltage drop when the input was actually supposed to be on? Would we represent the source with 120V rather than 100V?

You will need a different model for the ON state. That one will likely be a 120V source and a 6 ohm (2000ft #14 wire) impedance.

Todays sketch shows an ON state and OFF state.

edit to add: I got interupted a few times this morning while I was working on this - you have this all figured out already :)

cf
 
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Cold Fusion said:
Your DVM will have an input impedance of 10+megohms. So when you measure the OC voltage, the source impedance is invisible - lost in the 10megs input impedance. Every voltage source can be defined by the OC voltge and the short circut current. The source voltage is as measured OC. The source series impedance is OCV/SSC.

So, if you lift the leads and measure, you will see the source OC voltage. You could also short out the input lead and measure the short circuit current. Then you would kow exactly what the source impedance is.

cf
Click to expand...

What does OC stand for in your above explanation, I cant seem to figure this out. And basically what you are saying is that your meter does send out a small current, but because the impedence of the meter is so high you will not read much of a drop across the source impedence and will therefore see the 100V.

This is all starting to make sense now, and it appears that we are dealing with not much more than a basic voltage divider. That makes it easy to visualize.

Now trying to understand and model how the capacitive coupling works. Basically as you said between the energized wire, and unenergized wire there is a series of capacitors. The longer the cable run the larger the capacitor plates become and thus the larger the capacitor. Can this be represented by adding small capacitors in series as you mentioned to come up with an overall larger capacitor.

Since the current through the capacitor is i = C dv/dt then, because the voltage is a sin wave this is allowing this current to flow throught the capacitor (100k source impedance) and thus through to our overload input impedeance. It is this current that flows through the capacitor that flows through our circuit impedances and causes the voltage drops we have been discussing.

As others have mentioned if our source voltage were DC then once the capacitor charged we would no longer have any current flow through the capacitor and thus not have any current flowing through our impeduences causing the voltage drops. If that was the case, then the secondary side of the capacitor would have a 0 voltage when fully charged and thus have 0V on the input line? Is this why the DC voltage would prevent this problem? What happens when the capacitor is fully charged?
 
090209-1359 EST

mull982:

Earlier had a substantial response, and while still working on it ZAP and it disappeared. This happens moderately often. Do others have this problem?

I will start over but with a different attack. Get back to basics.

Two conductors insulated from each other form a capacitor. Therefore your isolated wire, for test purposes we have it connected to nothing else, has some composite capacitance to all the other wires in this cable or conduit. That capacitance can be measured. It might change from time to time depending upon what happens with the other wires. But assume it does not change.

If you know the capacitance, then you can calculate the capacitive reactance.
Xc = 1/(2*Pi*F*C)
1200 pfd has a reactance of 1/(377*1200*10^-12) or 10^12/452160 = 2.21*10^6 = 2.21 megohms. Apply 120 V 60 Hz to this capacitor and the current is 54.3 microamps. When I did this the experiment did not work well. The harmonic content on the AC voltage was too high. So erase this method as an approach.

New approach. Consider a capacitive voltage divider. Now I put a 0.1 mfd capacitor in series with the 1200 pfd unit. I measured 1.48 V across the 0.1 mfd capacitor. This is a ratio of 120/1.48 = 81. The ratio of 100,000 to 1200 is 83. This is a good correlation.

In your case get two capacitors. 1.0 mfd and 0.1 mfd. Your not going to expect much voltage, but get at least 200 V DC rating. You use your 10 megohm input impedance meter to measure the voltage across the load capacitor and from this and an assumed voltage on the wires in the cable calculate the cable capacitance.

Knowing the cable capacitance, then use an appropriate load at the input from the input terminal to the signal common. I believe I previous suggested about 2500 ohms for a resistor. Obviously a 10 to 20 W unit.

You had earlier ask about inductive coupling. This comes from a changing magnetic field coupled to a coil. This can be a one turn loop. The induced voltage will be proportional to current in the inducing circuit. If I create a 1 turn loop in a rectangle 5 feet by 40 feet and place it on the ground below my overhead power lines I can see millivolts of signal from the current in those lines. I do not see 10s of volts. Put many turns in the loop and I would see 10s of volts. Make the loop 6" by 40 feet and I will reduce the signal by a factor of 10.

You are seeing capacitively coupled noise and not inductive.

.
 
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