Inductance or capacitance

[ptonsparky]

Gars Hypothetical problem.
120 volt.
I have a switch 1500‘ from my PLC input. The ON voltage level of the input is > 60 v, min 5 ma. OFF is < 20 v, max 2 ma.
Two wire control via a #12 UF cable with 2 conductors and an EG.
Closing the switch gives an input. Opening the switch does not remove the input. Why?
(Real life problem before PLC)


2nd scenario.

The above cable is now in the same raceway with 480 volt conductors that become energized at the above input. 100 amps.
How does this affect the previous problem?

 

[__dan]

1st scenario, it is possible the long run of control input wire has a capacitive effect of charge storage. The PLC input would have to have a super high input impedance to not drain off the cable Voltage quickly. You could add a shunt resistor across the PLC input to drain off the chage storage of the input cable. If the control input Voltage is 100 V and the on state has a 20 mA drain across the shunt resistor, a 5k Ohm shunt resistor would do that.

2nd secanrio, manufacturers always forbid running control wires in the same conduit as power (as you are well aware I'm sure). Current flow in power wires have a surrounding changing magnetic field. That changing magnetic field induces voltage in nearby wiring (continuously) and certainly enough to be read as an On state by the PLC. If you have to do it that way, power and control wire in the same conduit, probably shunt drain resistor across the PLC input would be necessary. But even with a shunt drain, the input cable could have a floating voltage above ground that the PLC may see. So, it may need a second resistor, either pull up or pull down depending on if the input is sourcing or sinking..

Once you get a balasting resistor method that works, you could add the resistors to both ends for redundancy.

 

[ptonsparky]

A typical run of cablecon will have control wires in it.

 

[gar]

210404-1145 EDT

ptonsparky:

Assume 15 pfd/ft of capacitance between two wires. Your 1500 ft is therefore a capacitor of about 0.023 mfd. At 60 Hz the capacitive reactance is about 100 k ohms. At 120 V to a 100 k ohm impedance the current is about 1 mA.

A simple experiment you may be able to perform is:
Connect a 0.022 mfd capacitor of 400 V rating in series with a 1 k ohm resistor. A 1 V reading across the 1 k resistor is 1 mA of current. I did this and got about 1 V. Why do I not just use the meter in mA range, because if there were some fault, capacitor failure for example, I could blow an expensive fuse in the meter.

What does 1 mA of current input to a PLC input do? Is it a logic 1 or a 0? You have to know the input characteristics to your PLC input.

In gage controls I built I used a 5 k resistor as the input to an optical coupler for 120 V AC inputs. When my input was supplied from a solid-state relay I would shunt the input to my input with a 1 k resistor so that SSR leakage current (actually snubber current in the SSR) was reduced to a smaller input voltage to my optical coupler input.

If capacitance between the control wires is greater than 15 pfd/ft, then you would have a greater input voltage (current). To make judgements on what to do requires more specific information on your specific equipment.

A way to solve this problem for virtually any length cable is to:
Put the signal wire in an electrostatic shield, and ground the shield. Then any capacitive currents from other wires flows to the shield and the common reference to which the shield is connected.

If you do not want to use shielded cable, then you need use sufficiently low shunt impedance at the PLC input to guarantee that the input does not turn on when it should not.

.

 

[gar]

210404-1404 EDT

ptonsparky:

I will work back from your figures. 20 V at 0.002 A is 10,000 ohms. The looks to be the input impedance at 20 V. 20 V is 1/6 of 120 V. The difference voltage between 120 and 20 V is 100 V. 100 V at 0.002 A is 50,000 ohms. But we really have a triangle to solve, not a straight line. The straight line approach gives a ballpark figure. Solving the right triangle gives us sq-root of 14400 - 400 = 118 V across the capacitor, or Xc = 118/0.002 = 59,160 ohms. At 60 Hz this is 0.046 mfd, or 30.7 pfd per foot. Might be possible, but if the EGC is positioned between the other conductors and grounded, then the capacitance between the other two conductors is less because the EGC acts as a partial shield.

Also you need to know more about the input circuit at the PLC input.

.

 

[gar]

210404-1547 EDT

ptonsparky:

I ran some additional experiments. Not real good correlation, but somewhere in the ballpark.

The experiments are on a crude roll of 50 ft of #14 Romex with ground. This is not the original tight roll, not a straight length, but my coiling up after it having been straight out. A previous experiment did not show significant difference between the original roll and straight.

Experiments with GR LRC bridge at 1 kH:
740 pfd or 14.8 pfd per foot. White to black with EGC floating.
1090 pfd or 21.8 pfd per foot. Black to EGC with white floating.

Fluke 27 on AC V range with test from 121 line voltage:
94.3 V on Fluke with Romex black connected to 121 V, and Fluke from white to neutral of 121 source.

Experiments with 121 line source 60 Hz applied across the 50 ft of Romex, and Fluke across a 47 ohm carbon comp resistor for current measurement.
2.2 mV, 47 micro-A, 121/47 = 2.6 megohms, 1000+ pfd, 20 pfd/ft, for white to black with EGC floating.
3.5 mv, 74 micro-A, 121/74 = 1.6 megohms, 1700 pfd, 34 pfd/ft, for black to EGC with white floating.
0.9 mV, 19 micro-A, 121/19 = 6.4 megohms, 400 pfd, 8 pfd/ft, for black to white with EGC to ground.

Clearly just the use of the EGC as a shield is advantageous, but not as effective by any means as a full shield around the shielded wire would be.

.

 

[ptonsparky]

210404-1404 EDT

ptonsparky:

I will work back from your figures. 20 V at 0.002 A is 10,000 ohms. The looks to be the input impedance at 20 V. 20 V is 1/6 of 120 V. The difference voltage between 120 and 20 V is 100 V. 100 V at 0.002 A is 50,000 ohms. But we really have a triangle to solve, not a straight line. The straight line approach gives a ballpark figure. Solving the right triangle gives us sq-root of 14400 - 400 = 118 V across the capacitor, or Xc = 118/0.002 = 59,160 ohms. At 60 Hz this is 0.046 mfd, or 30.7 pfd per foot. Might be possible, but if the EGC is positioned between the other conductors and grounded, then the capacitance between the other two conductors is less because the EGC acts as a partial shield.

Also you need to know more about the input circuit at the PLC input.

.

The input does use an Optical Isolator. Nothing else I can tell you from the information I have.

 

[gar]

210405-0845 EDT

ptonsparky:

I am curious as to how you know that an optical coupler is not used. However, it probably makes no difference how electrical isolation is provided at the inputs. If no isolation is used, then I could expect potential major problems in the real world.

Whatever the input is we need to know its characteristics.

We can assume it is a moderately high input impedance. Possibly higher than 5 k to 10 k.

We need to do tests on one input as a starter.

I know not what you have with which to do such tests, but first how is this input wired?
"
120 volt.
I have a switch 1500‘ from my PLC input. The ON voltage level of the input is > 60 v, min 5 ma. OFF is < 20 v, max 2 ma.
Two wire control via a #12 UF cable with 2 conductors and an EG.
Closing the switch gives an input. Opening the switch does not remove the input. Why?
(Real life problem before PLC)
"
We can estimate the input impedance at around 60/0.005 = 12,000 ohms, or 20/0.002 = 10,000 ohms. Seems logical.

Is this an isolated input from from any internal logic, and/or from other inputs, or outputs? I would assume yes. If yes it means some sort of isolator.

Since you only need to supply a contact closure as an input, implied from your description, this means the input is supplying current or voltage. However, your statement of input voltage and current levels might imply that an external voltage source its required. This needs to be clarified.

Suppose that your PLC input is the voltage source, then you have a much more difficult problem to solve because the cable capacitance is like a partial switch closure. In this case you would need to use a guard voltage technique.

If you externally supply voltage to your input switch, then you can shield the input wire, or shunt the leakage current at the input.

.

 

[ptonsparky]

Gar, optical isolation is used.
Control power for inputs is not supplied by PLC. PLC is 24v DC.

This is a hypothetical problem.
Given the wires are pulled, the system starts at contact closure, but remains on when they are opened.

Shunting at the input is the quick, easy solution. Yes?

 

[winnie]

ptonsparky, if control power for the inputs isn't coming from the PLC, I assume you have a power supply near the PLC providing 120V, and the switch 1500 feet away is closing to connect the local 120V supply back to the PLC input, correct?

In general loading the input would be the way to go, to deal with any capacitive coupling along the long control cable.

-Jon

 

[ptonsparky]

No different than what we did before PLC.

 

[gar]

210405-1129 EDT

ptonsparky:

I am still on your first circuit.

I don't care what powers the PLC. I need to known about the input circuit. There has to be power to the input. Where does that come from? Are the PLC inputs DC or AC? IF AC, then DC is also likely to control them, but there could be capacitor isolation, and then AC would be required.

If the inputs responds to DC, then you can put a low pass filter at the input with a rollover at as low a frequency as you want that does not interfere with the response time of the system that is required.

If the input has no internal power source, then you have to supply switched power externally. That means that the input control wire can be shielded.

.

 

[don_resqcapt19]

This type of thing can even be an issue with standard control circuits that operate starter coils when the control wires are very long. Take a look at this SquareD document.

undefined | Schneider Electric

www.se.com

 

[ptonsparky]

Post 1:. 120 volt.
I failed to say AC.

 

[ptonsparky]

This type of thing can even be an issue with standard control circuits that operate starter coils when the control wires are very long. Take a look at this SquareD document.

undefined | Schneider Electric

www.se.com

Yes, We dealt with it by loading up the circuit. One starter coil or interposing relay may lock in, but adding a second as a dummy load solved the problem.

The low requirements for an ON input current of a PLC would make it considerably worse.

 

[ELA]

For you Gar,
https://cdn.automationdirect.com/static/specs/c008na.pdf

Additional loading often required depending upon usage.

 

[gar]

210405-2046 EDT

ptonsparky and ELA:

A common solid-state input is made by OPTO-22. I have used these as well as many of my own design.

The specifications for the IAC5 and IDC5 can be seen at:

http://www.farnell.com/datasheets/1913633.pdf

IDC5 --- 1.5 k, for OFF less than 3 V or 1 mA, for DC ON 10 to 32 V, for AC ON 12 to 32 V.
IAC5 --- 28 k, for OFF less than 45 V or 3 mA, for DC or AC ON 90 to 140 V.

These values may not be uncommon for typical PLC inputs.

Also there are other variations on these basic parts.

.