Induced Voltage on Control Wire

[mull982]

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.
.

What if for example our meter impedence happend to be much lower than the capacitance coupled impedence. This would then lower the open circuit voltage reading we got with our meter becasue most of the voltage would drop across the coupled impedence and have less dropped across our low impedence meter. So from this I take it that we will read a different open circuit voltage or voltage on the lifted wire depending on the impedence of the meter. This would explain why when a "wiggy" which is a low impedence meter is placed across the input with the wire connected the voltage quickly drops from 28 to 0. This is because the low meter impedence in parallel with the input impedence causes a lower overall input impedence an therefore most if not all of the voltage is dropped across the coupled capactitance and does not show up on the input module.

090211-0736 EST
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 tested a couple of different scenarios when I was in the field. The first scenario was with the contractor pulled in in a local condition and this yieleded the maximum voltage. This give the 28V with the wire in the input and the 80V with the wire lifted from the input.

I then tested the circuit with the control station in the off position. Therfore there was no current flow, but just voltage out to the control station. From this I saw 7.7V with the wire landed and 25V with the wire in an open circuit state.

Lastly I tested the circuit with the contorl station in the local position but no current flow energizing the starter. From this I recorded 15.8V landed on the input and 48.8V in an open circuit state.

So it appears exactly as you said. The maximum coupling occurs when multiple wires are energized and there is current flowing.

Guess the magnitude of the source voltage also will determine load voltage. If we had 480V wire in the same conduit and were part of the coupling, then we could expect much larger input, and open circuit voltages than we saw.

090211-0736 EST
I definitely recommend that in addition to any shunt resistance you use that there should be shunt capacitance.
.

Is this just to deal with harmonics and noise as you mentioned.

 

[gar]

090211-1005 EST

mull982:

What if for example our meter impedence happend to be much lower than the capacitance coupled impedence. This would then lower the open circuit voltage reading we got with our meter becasue most of the voltage would drop across the coupled impedence and have less dropped across our low impedence meter. So from this I take it that we will read a different open circuit voltage or voltage on the lifted wire depending on the impedence of the meter. This would explain why when a "wiggy" which is a low impedence meter is placed across the input with the wire connected the voltage quickly drops from 28 to 0. This is because the low meter impedence in parallel with the input impedence causes a lower overall input impedence an therefore most if not all of the voltage is dropped across the coupled capactitance and does not show up on the input module.

Yes.

I think in your circuit that whether current is flowing or not is not important it is just the voltage present on the various wires and the capacitive coupling between the wires.

I do not know the value of the coupling capacitance in your cable, and it is variable. However, I previously guessed at 10 pfd per foot because that is the approximate value of a Romex cable.

A 1000 ft run would therefore be 10,000 pfd or 0.01 mfd. This has a capacitive reactance of about 265,000 ohms at 60 Hz. With 122.2 V across an 0.01 mfd and 10 meg series the voltage across the 10 megs is in the ball park of 121.5 based on an actual measurement.

But your wiring is not a single capacitor, a single voltage source, and a single 10 meg resistor. Rather it is several capacitors in parallel from a 120 V source, and another capacitor from the cable (the floating wire to the common wire, neutral). More realistically you have three 10 pfd/ft in parallel which equals 30 pfd/ft from a 120 V source and 10 pfd/ft to common. The equivalent circuit of this as a source is 120*0.75 = 90 V and a series capacitance of 40 pfd/ft. The 10 meg meter would read close to 90 V.

 

[gar]

090211-1021 EST

mull982:

I tested a couple of different scenarios when I was in the field. The first scenario was with the contractor pulled in in a local condition and this yieleded the maximum voltage. This give the 28V with the wire in the input and the 80V with the wire lifted from the input.

Your measurement of 80 V and my guesstimate of values producing 90 V are somewhat correlating.

Next change the distribution of capacitors. One 10 pfd/ft from 120 V and 30 pfd to common. Now 1/4 of 120 V would be expected, or 30 V.

I then tested the circuit with the control station in the off position. Therfore there was no current flow, but just voltage out to the control station. From this I saw 7.7V with the wire landed and 25V with the wire in an open circuit state.

This maybe the condition I just calculated and this again tends to correlate.

.

 

[gar]

090211-1027 EST

mull982:

Guess the magnitude of the source voltage also will determine load voltage. If we had 480V wire in the same conduit and were part of the coupling, then we could expect much larger input, and open circuit voltages than we saw.

Yes the voltage would be proportional.

A solution to all of this capacitive coupling is to run the floating wire as the center wire of a shielded cable with the shield connected to common.

Quote:
Originally Posted by gar
090211-0736 EST
I definitely recommend that in addition to any shunt resistance you use that there should be shunt capacitance.
.

Is this just to deal with harmonics and noise as you mentioned.

Basically yes. But if you do not need a DC path, then a capacitor alone is maybe a better choice because there is negligible power dissipation. You already have a DC path thru your input and maybe there is no need for a DC path anyway. What you choose to use as a load may well depend upon what parts you normally stock in combination with what is needed to solve the problem.

.

 

[mull982]

Cold Fusion and Gar thanks for the help with this one. I have a much better understanding of what is going on with this circuit and will be able to troubleshoot it better and explain it to others.

Thanks again.

 

[Cold Fusion]

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

Good you checked - I'm neither above nor below screwing up the math

cf

 

[Cold Fusion]

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.

Whoops - see the above post

cf

 

[mull982]

Basically yes. But if you do not need a DC path, then a capacitor alone is maybe a better choice because there is negligible power dissipation. You already have a DC path thru your input and maybe there is no need for a DC path anyway. What you choose to use as a load may well depend upon what parts you normally stock in combination with what is needed to solve the problem.

.

I am going back and revisiting this capacitive coupling issue for it was put on the backburner for a bit.

I am getting ready to order the shunt resistor to put in parallel with the load when I went backed and looked through this thread again. I noticed above that Gar mentioned using a shunt capacitor instead of a resistor. Ignoring any harmonic or frequency issues, is the only reason for using a shunt capacitor over a resistor the fact that the resistor will dissipate heat and the capaciotor will not? So if choosing between a resisotor and capacitor it is best to use a capacitor?

If I am going to add a 10k shunt impednece as calculated in this tread then I need to calcualate a capacitor value that would give a 10k impedence. I do this by:

Z = 1 / jwC

so

C = 1 / jwZ

using w=2*pi*f = (2)(3.14)(6)=377 and Z = 10,000 we get

C = 1 / (j * 377 * 10,000) = 2.65 E-7 or .265 E-6

So this would tell me that I need a .26 microfarad capaciotor to give me the required 10k impedence. Is this correct?

 

[gar]

090414-2030 EST

mull982:

Your calculation is correct.

To experiment and verify that capacitive coupling is the problem a resistor is the least expensive means.

But a good capacitor to use is the following:

See http://www.vishay.com/docs/26507/f1773200.pdf

Either a 0.22 or 0.27 MFD will work. These capacitors are designed for across the line connection.

See Mouser http://www.mouser.com/Search/Refine.aspx?Keyword=75-F17734222000

.

 

[mull982]

090414-2030 EST

mull982:

Your calculation is correct.

To experiment and verify that capacitive coupling is the problem a resistor is the least expensive means.

But a good capacitor to use is the following:

See http://www.vishay.com/docs/26507/f1773200.pdf

Either a 0.22 or 0.27 MFD will work. These capacitors are designed for across the line connection.

See Mouser http://www.mouser.com/Search/Refine.aspx?Keyword=75-F17734222000

.

Thanks Gar

Do these capacitors have a wattage rating such as must be met with a resistor. In this example we said that if using a resistor it must be rated for 1.4W. I would think this does not matter in the case of a capacitor due to to fact that it stores and releases energy and therefore does not dissipate any heat or power.

 

[gar]

090415-0620 EST

mull982:

Not in the same sense as a resistor. The resistive component of internal impedance of a capacitor is a power loss and becomes important in some applications such as switching regulators.

The capacitor I pointed to in my previous post is designed for across the line connection at 50 to 60 Hz, and up to whatever voltage was specified in the datasheet, 253 V RMS (357 peak), 630 VDC. This is a very good capacitor for these frequencies.

.

 

[mull982]

090415-0620 EST

mull982:

Not in the same sense as a resistor. The resistive component of internal impedance of a capacitor is a power loss and becomes important in some applications such as switching regulators.

The capacitor I pointed to in my previous post is designed for across the line connection at 50 to 60 Hz, and up to whatever voltage was specified in the datasheet, 253 V RMS (357 peak), 630 VDC. This is a very good capacitor for these frequencies.

.

Gar

Thanks for the help, and for pointing out those capacitors

 

[gar]

090415-0730 EST

mull982:

You might want to put a 470 1/2 W carbon composition resistor in series with the capacitor to reduce peak inrush current when the controlling switch closes at a voltage peak. This would limit peak initial current to about 170/470 = 0.36 A. The time constant at 0.27 mfd would be 170 microseconds. Average power dissipation in the resistor under steady state conditions would be about 0.07 W.

.

 

[mull982]

090414-2030 EST

mull982:

Your calculation is correct.

To experiment and verify that capacitive coupling is the problem a resistor is the least expensive means.

But a good capacitor to use is the following:

See http://www.vishay.com/docs/26507/f1773200.pdf

Either a 0.22 or 0.27 MFD will work. These capacitors are designed for across the line connection.

See Mouser http://www.mouser.com/Search/Refine.aspx?Keyword=75-F17734222000

.

GAR

When looking into the capacitor that you recommended I came across a number of ceramic type capacitors in my shop. Will one of these ceramic capacitors work for this application. Are they rated high enough?

 

[gar]

090422-1025 EST

mull982:

Ceramic is OK if the voltage rating is adequate. Precise capacitance is not a factor.

I would probably use a 400 to 600 V DC rated unit. Note: for a 120 V sine wave the peak voltage is about 170 V.

An across the line use increases the power dissipation in a capacitor in comparison to an applied DC voltage with a small supeimpose audio component even when the DC component is equal to the across the line AC peak.

.

 

[mull982]

090422-1025 EST

mull982:

Ceramic is OK if the voltage rating is adequate. Precise capacitance is not a factor.

I would probably use a 400 to 600 V DC rated unit. Note: for a 120 V sine wave the peak voltage is about 170 V.

An across the line use increases the power dissipation in a capacitor in comparison to an applied DC voltage with a small supeimpose audio component even when the DC component is equal to the across the line AC peak.

.

I'm having trouble determining the capacitance ratings of the ceramic capacitors that I have. (No meter to measure them)

For example, one of them has printed on it .01M followed by Z5U, followed by 1kV. How do you convert the .01M into a capacitance value?

 

[gar]

090422-1316 EST

mull982:

See:
http://www.aaron.co.kr/pdf/capactitor/application.pdf

This is probably a 0.01 mfd at 1000 V. It would take 22 of these to make an 0.22 MFD, or in other words the impedance is 22 times larger.

For a quick test put a small pencil type soldering iron on the circuit as a load. 50 W at 120 = 14400/50 = 300 ohms. A 25 W incandescent would be an even lower resistance when not powered, maybe 40 to 60 ohms.

.

.

 

[nawao]

"On of these control wires is causing its solid state input to turn on when it shouldn't.

"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."

You only want an explanation as to what is going on or want to solve the issue?, because to avoid the trouble you just need to wire the control cable with a shieded pair and ground it at the reception side.

 

[gar]

090423-0723 EST

nawao:

Running 1000 ft of shielded cable is an unrealistic approach from a cost point of view, and probably physically as well. Other less costly solutions will work.

.

 

[mull982]

While waiting for my capacitors to come I experimented with a 20k resistor that I had. When running in the local position the 27V that I referenced before was seen on input terminal #1. When connecting the resistor as shown in the attached schematic, it was only able to drop the voltage at this terminal to about 22V which was still enough to make the input turn on once in a while. I decided to stop there and wait until I got the proper size caps to continue experimenting.

At this point I left the resistor in the circuit as shown. However when the pump was put back in the remote position I had a problem with the coil staying pulled in even when the command from the OUT A contact was opened.

To set up the scenario, when the switch was put into the remote position we had 120V on input #1 (normal) and 0V on terminal #2 and 65V on terminal #5. I want to note that with the wires lifted from terminals 2 and 5 the wires were both at about 85V open circuit voltage. I guess the low impedence of the coil brought the voltage on terminal 2 down to zero when landed.

The contactor pulled in fine and all voltages looked o.k. The problem was when in the remote position, once the OUT A contact in the E3 module opens the contactor should drop out. This was not happening when this contact was opened. Something was holding the contactor in, and even with the OUT A open we measured 120V across coil and at terminals 2 and 5. After a while troubleshooting we decided to remove the resistor, and Wallah!! the contactor dropped out. I am sitting here looking at this schematic baffaled as to how this resistor shown in the position that its in could have caused the coil to stay pulled in. Any ideas????

I put the resistor in this location to bleed of the capacitance coupled voltage on this input talked about earlier in this thread.