Induced Voltage on Control Wire

[mull982]

I asked an electrician to measure the voltage on the contorl wire in question with the wire lifted from the input. He found that witht the control station in the off position in the field, he measured 18V between the wire and ground with the wire lifted from the input.

This tells me that the capacitance coupling is coming from the hot wire that is going out to the control station since this is the only energized wire during this test.

I am trying to figure out why we saw 18V on this open circuit? I would expect that we saw more because there is now no current flow. Do we have to take current flow through the meter and thus the impedence of the meter into condiseration?

Is it possible to have 18V backfeed though the output contact in the overload relay? If you look at the control schematic you can see that when in local we have voltage on the secondary side of the relay output contact. Is it possible that 18V will backfeed through this contact? (We actually wired this starter a little different than shown in the schematic to where a backfeed across this contact would provide a direct connection to the input in question)

 

[mull982]

090209-1359 EST

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.

Why are you putting 2 capacitors in series. Is one capacitor supposed to represent the source capacitance coupling and the other capacitor the input impedance? Why represnet the input impedance with capacitance?

090209-1359 EST
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.

.

OK so what you are saying is that if we had inductive coupling we would not see something in the magnitude of 28V. We would most likely see something less than a volt. unless there was somehow multiple loops of wire in a counduit or elsewhere in a circuit we are not seeing inductive coupline.

Is it safe to say that the majaroity of times these phantom voltages come about is due to capactive coupling as opposed to inductive coupling?

 

[gar]

090209-1540 EST

mull982:

If there is one 120 V energized wire in this cable or conduit, and your signal wire in question is unconnected at both ends, and you use a high impedance voltmeter (meaning 10 megohms input resistance), then most likely that is all capacitive coupling. But 18 V with a 10 meg meter is quite low for the cable length you have.

When other conductors in the cable are energized, then you can expect a higher voltage on the wire.

Since this is so confusing to you I have the following suggestion. Get a 2500 ohm 10 W power resistor and place it between your input and common. This should get your system to work. If not put a 25 W bulb there. With the system working, then monitor the voltage at this input with a Fluke DVM and see what the voltages are under operating conditions. When a signal is applied it should be close to your line voltage. When not (the logic zero state) the voltage should be fairly low, less than 10 V. This zero state voltage may vary depending upon what wires in the cable are energized.

If you start with a 25 W bulb do not assume this is a good final solution. When the bulb burns out then you are back to your current problem condition.

.

 

[gar]

090209-1601 EST

mul982:

Is it safe to say that the majaroity of times these phantom voltages come about is due to capactive coupling as opposed to inductive coupling?

YES.

Further if you use twisted pairs and a differential receiver, then magnetic coupling is greatly reduced.

.

 

[mull982]

090209-1540 EST

mull982:

If there is one 120 V energized wire in this cable or conduit, and your signal wire in question is unconnected at both ends, and you use a high impedance voltmeter (meaning 10 megohms input resistance), then most likely that is all capacitive coupling. But 18 V with a 10 meg meter is quite low for the cable length you have.

When other conductors in the cable are energized, then you can expect a higher voltage on the wire.

Since this is so confusing to you I have the following suggestion. Get a 2500 ohm 10 W power resistor and place it between your input and common. This should get your system to work. If not put a 25 W bulb there. With the system working, then monitor the voltage at this input with a Fluke DVM and see what the voltages are under operating conditions. When a signal is applied it should be close to your line voltage. When not (the logic zero state) the voltage should be fairly low, less than 10 V. This zero state voltage may vary depending upon what wires in the cable are energized.

If you start with a 25 W bulb do not assume this is a good final solution. When the bulb burns out then you are back to your current problem condition.

.

Gar

I understand the voltage divider application here and how adding resistance to the input terminals can lower the voltage seen across the terminals. I just dont see why we get 18V in an open circuit condition or what you meant in your first paragraph about it being all capacitive coupling.

 

[gar]

090209-1638 EST

mull982:

If you have a piece of wire hanging in free space connected to nothing and no magnetic field present. You add another conductor to the general vicinity, then you have formed a capacitor. A capacitor is two conductors with an insulator between them. This includes a pure vacuum.

A separate pair of wires on the floor form a capacitor. Two wires in a cable form a capacitor.

If you have an AC voltage source of some finite magnitude connected to a capacitor, then a current will flow thru the capacitor as a result of that voltage. If you connect a voltmeter in series with a capacitor and connect this series combination to an AC voltage source, then why would you not expect to get a reading on the voltmeter?

.

 

[Cold Fusion]

mull -
This answer first:
OC = open circuit
OCV = open circuit voltage
SSC = short circuit current

witht the control station in the off position in the field, he measured 18V between the wire and ground with the wire lifted from the input.

hummm .. OCV = 18V
Did the tech happen to measure the voltage before he lifted the lead?

So, as gar mentioned, something changed? Voltages aren't the same any more?

You have my sympathies. Troubleshooting by sending notes to the tech and then waiting for a return is hard to do.

cf

 

[mull982]

If you have an AC voltage source of some finite magnitude connected to a capacitor, then a current will flow thru the capacitor as a result of that voltage. If you connect a voltmeter in series with a capacitor and connect this series combination to an AC voltage source, then why would you not expect to get a reading on the voltmeter?

.

I would expect no reading on the voltmeter because you are not reading any difference in potential with the meter in series. Learning from this thread I would also suspect that if the resistane of the capacitor was much larger than that of the input resistance of the meter, then most of the voltage would drop across the capacitor. Is this where you were going with this?

mull -
This answer first:
OC = open circuit
OCV = open circuit voltage
SSC = short circuit current


hummm .. OCV = 18V
Did the tech happen to measure the voltage before he lifted the lead?

So, as gar mentioned, something changed? Voltages aren't the same any more?

You have my sympathies. Troubleshooting by sending notes to the tech and then waiting for a return is hard to do.

cf

The only thing that changed was that the control station was switched to the off position so instead of having a source wire and return swich wire energized at 120V (local operation), we only had the source wire at 120V and thus had one energized wire as opposed to two before.

I will personally make my way to the starter tomorrow and measure the voltage at the input with the contactor operating in local and see that I get the 28V, and then disconnect the input and verify this voltage.

 

[gar]

090210-0758 EST

mull982:

If I have a meter and a capacitor in series and connected across an AC voltage source of fixed voltage and frequency, then the smaller the capacitor the larger is the voltage drop across the capacitor and the lesser the voltage across the meter.

Get some components and experiment. A 1000 or 2000 pfd (0.001 or 0.002 mfd) capacitor 400 V DC rating, an 0.1 and 1.0 mfd Mylar capacitor at 400 V DC rating, two sockets and 100 W 120 V bulbs connected in series, a 2500 ohm 10 to 25 W Ohmite fixed resistor, a 10,000 ohm 5 W power resistor, a 27 K 2 W carbon composition resistor, a 47 K 1 W carbon comp, a Simpson 260 or 270 meter, a Fluke 27 or 87 or similar meter, a 100 or 200 ft roll of Romex, and a Variac.

With this you can experiment with various combinations.

It is virtually impossible to do an efficient troubleshooting job at a remote location relative to where the problem is located. 50 ft away can fit my definition of remote in some cases. You need the correct tools, technical knowledge, and accurate input data to solve a problem.

This is now 29 posts and 3 days after this thread was started. From the start the problem was basically identified, and one or more possible solutions were obvious.

You could have had the system operating the first day.

A 100 W 120 V bulb will have a resistance of about 107 ohms at 60 V as compared to 144 ohms at 120 V and 9.8 ohms at 0 volts. Two 100 W bulbs in series as a load for your troublesome wire would be near 20 ohms for the logic 0 state and about 200 ohms at a logic 1 state. The life of these bulbs at 50 % of voltage rating would be extremely long even in a rapid on-off application.

.

 

[steve66]

I'm not familiar with this control card, but I'm still wondering why it looks like there are two outputs wired together and connected to the transformer output?

This isn't an open collector output card is it?

Steve

 

[mull982]

hummm .. OCV = 18V
Did the tech happen to measure the voltage before he lifted the lead?

So, as gar mentioned, something changed? Voltages aren't the same any more?

You have my sympathies. Troubleshooting by sending notes to the tech and then waiting for a return is hard to do.

cf

OK. I personally went back and re-created the condition we first had giving the numbers we had in my origonal post. This condition was that the control station was in the local position, and the contactor was pulled in, in local control. I observed the following.

1) Measured 28V on the input in question, with the wire landed on this input. This was the voltage that I saw origonally in this condition.

2) When the wire was lifted off the input we read 80V. We read this 80V between the wire and ground.

This 80V seems to make sense. I attribute this value to being that because the input impedance of the meter is much larger than that of the capacitively coupled impedance then most of the voltage will drop across the much higher impedance of the meter and thus read 80V.

So assuming 100V was our souce in CF's example, then 20V will drop across the capacitively coupled impedance, and the other 80V is dropping across the meter. Does this make sense to you guys?

 

[mull982]

090210-0758 EST

mull982:

If I have a meter and a capacitor in series and connected across an AC voltage source of fixed voltage and frequency, then the smaller the capacitor the larger is the voltage drop across the capacitor and the lesser the voltage across the meter.
.

I understand this concept and see it as a basic voltage divider like I mentioned before. I just had a hard time seeing it when you started throwing various capacitors in there. I understand that the meter has an impedance that has to be taken into consideration when in series. I appreciate the example.

090210-0758 EST

This is now 29 posts and 3 days after this thread was started. From the start the problem was basically identified, and one or more possible solutions were obvious.

You could have had the system operating the first day.

.

This is not a system that is in operation rather it is a new system that is being commisioned. There is no rush to get this system online and therfore this issue is not presenting a major hold up.

I wanted to understand completely in theory what was going so that I could explain it correctly to others, as well as be able to troubleshoot something similar in the future. I didn't want to simply throw a quick fix on the problem and walk away from it without gaining a full understanding of what was happening.

I apprecaite the help you have given.

 

[gar]

090210-1907 EST

mul982:

This 80V seems to make sense. I attribute this value to being that because the input impedance of the meter is much larger than that of the capacitively coupled impedance then most of the voltage will drop across the much higher impedance of the meter and thus read 80V.

So assuming 100V was our souce in CF's example, then 20V will drop across the capacitively coupled impedance, and the other 80V is dropping across the meter. Does this make sense to you guys?

Yes.

But it is slightly more complex. If you get a second meter also of high impedance and place this across the cable capacitance from the hot wire to the floating wire the reading will not be 20 V, but something higher. This experiment should be done with a larger capacitor (0.1 mfd 100 VDC rating) and a 27 k ohm 1/2 W resistor (selected so the reactance = the resistance connected across sine wave generator with low harmonic distortion of possibly 20 V output.

I just now ran an experiment with 22 K and 0.1 mfd with about 10 V excitation.

At 60 Hz --- Source voltage 10.54 V, capacitor voltage 8.17 V, and resistor voltage 6.81. Calculated capacitive reactance Xc = 1/(2*Pi*60*0.1*10^-6) = 1/(376.8*10^-7) = 10^7/376.8 = 26.54 k ohms. 22/26.54 = 0.828 or 39.7 degrees. Sin 39.65 = 0.638, and cos 39.65 = 0.770. 10.54*0.77 = 8.12 and 10.54*0.638 = 6.72. Thus the calculated values are close to the measured values.

Next I determined the frequency where the capacitive reactance equals 22 k. This is f = 1/(2*Pi*Xc*C) = 1/(6.2832*22,000*0.1*10^-6) = 10^6/(6.2832*2200) = 1,000,000/13823 = 72.34 Hz. Thus I set the oscillator to 72 Hz and the results were:

At 72 Hz --- Source voltage 10.36 V, capacitor voltage 7.31 V, and resistor voltage 7.32. From 10.36 the calculated values would be 10.36*0.707 = 7.32 V.

Based on my previous experiment of last night my line voltage has too much harmonic content to get a good measurement and thus the reason for the test with an oscillator. If you do the experiment with line voltage, then the capacitor has to have a sufficiently large voltage rating, probably 400 VDC is a good rating. The resistor needs an adequate power rating.

Yes you have a voltage divider when using the capacitor and resistor, but the capacitor's current is 90 degrees shifted from the voltage across it. The resistor's voltage is in phase with the capacitor's current. So the two voltages are 90 degrees apart.

.

 

[gar]

090210-2105 EST

Missing paren.

But it is slightly more complex. If you get a second meter also of high impedance and place this across the cable capacitance from the hot wire to the floating wire the reading will not be 20 V, but something higher. This experiment should be done with a larger capacitor (0.1 mfd 100 VDC rating) and a 27 k ohm 1/2 W resistor (selected so the reactance = the resistance connected across sine wave generator with low harmonic distortion of possibly 20 V output.

Should read:

But it is slightly more complex. If you get a second meter also of high impedance and place this across the cable capacitance from the hot wire to the floating wire the reading will not be 20 V, but something higher. This experiment should be done with a larger capacitor (0.1 mfd 100 VDC rating) and a 27 k ohm 1/2 W resistor (selected so the reactance = the resistance) connected across a sine wave generator with low harmonic distortion of possibly 20 V output.

.

 

[Cold Fusion]

mull -

This 80V seems to make sense. I attribute this value to being that because the input impedance of the meter is much larger than that of the capacitively coupled impedance then most of the voltage will drop across the much higher impedance of the meter and thus read 80V.

you got it

So assuming 100V was our souce in CF's example, then 20V will drop across the capacitively coupled impedance, and the other 80V is dropping across the meter.

I would likely ignore the meter impedance and just call it an open circuit. So the 80V is the OCV.

If you do that, the coupling impedance comes out about 74K.

Interesting note, using 10Meghoms for the meter input impedance gives the coupling impedance 75K.

I attached a couple of sketches showing what I think you will see.

gar's rigorous description is absolutely correct. However I believe you can see the concepts and get answers that will work with a little less rigor

cf

 

[gar]

090210-2144 EST

Cold Fusion:

What you have illustrated is an approximation and it will give rough results, but it is not the actual circuit.

My only problem with this analysis is that persons not familiar with electrical circuit theory will propagate the idea that the applied voltage to a series RC circuit equals the sum of the voltage magnitudes of the components of the RC circuit.

.

 

[Cold Fusion]

What you have illustrated is an approximation and it will give rough results, but it is not the actual circuit..

Absolutely true

My only problem with this analysis is that persons not familiar with electrical circuit theory will propagate the idea that the applied voltage to a series RC circuit equals the sum of the voltage magnitudes of the components of the RC circuit.

Yes, you are correct, That absolutely could be problem.

cf

 

[mull982]

mull -

you got it


I would likely ignore the meter impedance and just call it an open circuit. So the 80V is the OCV.

If you do that, the coupling impedance comes out about 74K.

Interesting note, using 10Meghoms for the meter input impedance gives the coupling impedance 75K.

I attached a couple of sketches showing what I think you will see.

gar's rigorous description is absolutely correct. However I believe you can see the concepts and get answers that will work with a little less rigor

cf

This makes sense. Instead of the 100V source that you used origonally and the 100V that I referenced the source is 80V and just about all of this 80V drops across the high impedence of the meter since it is large compared to the coupling impedence. You say to ignore the meter impedence and consider the OCV however I'm assuming you still mean that this voltage drops across the high impedence of this meter.


What is the typical input impedence of a meter?

090210-2144 EST

Cold Fusion:

What you have illustrated is an approximation and it will give rough results, but it is not the actual circuit.

My only problem with this analysis is that persons not familiar with electrical circuit theory will propagate the idea that the applied voltage to a series RC circuit equals the sum of the voltage magnitudes of the components of the RC circuit.

.

Gar

I understand what you are saying about the capacitive element causing a current phase shift, and therefore causing the votlage drops to be out of phase. Thanks for brining that to my attention.

 

[mull982]

mull -

you got it


I would likely ignore the meter impedance and just call it an open circuit. So the 80V is the OCV.

If you do that, the coupling impedance comes out about 74K.

Interesting note, using 10Meghoms for the meter input impedance gives the coupling impedance 75K.

I attached a couple of sketches showing what I think you will see.

gar's rigorous description is absolutely correct. However I believe you can see the concepts and get answers that will work with a little less rigor

cf

CF

Thanks for the schematics they make things much easier to visualize.

In looking through your "loading resistors" schematic I arrive with slightly different values using a 10k load resistor.

For Vi across the input I get.

Vi = 80V * (8k / 74k + 8k) = 7.8V

Also for power across the resistor in the "on state" which I belive is what you calculation for Pd was I get.

Pd = (120)^2 / 10k = 1.4W

I used the on state voltage of 120V for the On State condition where it looks like you used 12V for the what should be off state condition.

 

[gar]

090211-0736 EST

mull982:

My Fluke 87 input impedance is 10 megohms and less than 100 pfd in parallel. My typical HP (Hewlett-Packard) instruments are 10 megohms, and unknown capacitance. Typical 10X scope probes are 10 megohms and about 10 pfd shunt capacitance.

In a multi-conductor cable or bunch of wires where there is one floating wire of concern (meaning no connection at either end) and other hot wires that may be connected to fairly low impedance loads and from time to time are energized from a low source impedance voltage source constitutes a varying amount of capacitive coupling to the floating wire. As a result the equivalent source voltage and composite coupling capacitance to the floating wire will change.

My guess is that the greatest coupling condition to the single floating wire occurs when all other wires are energized with the maximum voltage and the same phase. This would be the test condition to use in determining the resistance and/or capacitance to load the floating wire. I definitely recommend that in addition to any shunt resistance you use that there should be shunt capacitance.

Assuming about 20 V is your threshold to define a zero state, then under the conditions that cause maximum coupling I would add enough load to the floating wire to reduce the coupled voltage to maybe 1 V if this can be done with a reasonable load. Capacitive loading can be very useful because it is more effective than resistance in reducing high frequency noise.

Note: any unwanted signal can be put under the general classification of noise, even if it is an unwanted pure sine wave.

.