Fixing Capacitive Coupling (Phantom Voltage) issues with pull down resistor

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callenneff

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Seattle, WA
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Engineer
I am working on a DC controls system in which a reference voltage was used for speed control. There was a problem with the system of seeing a minimum 2V for the reference voltage instead of the approximately 0V which should have been seen for the lowest reference voltage. After much research me and an electrician discovered that there were phantom voltages on the system and somehow this was likely rectifying into a real voltage on our reference voltage.

In the end our fix was to toss a low impedance resistor (6.7kOhm) across the input to the isolated transmitter module we use in between the control circuitry and VFD used to drive this system. It completely fixed this problem but I'm looking for some help.

1. Does anyone have good reasoning behind how this is fixing my problem, as of now it seems like a pull-down resistor of sorts.
2. Is there any way to calculate the "proper" resistance value to bleed off this capacitive coupling or whatever the resistor is doing?

Any help is appreciated, thanks!
 
It absolutely is a pull down resistor.

There are two reasonable ways to calculate the value.
1) Figure out the impedance of the phantom 'source' and then select a resistance that is low enough to pull that source down enough to render the interfering signal negligible.
2) Figure out the drive capacity of your desired signal source, then select the lowest value resistance that doesn't overload the desired signal source.

Hopefully the value from 2) is lower than the value from 1).
 
I suspect that if he used a low impedance meter that voltage wouldn't exist. By the resistor fixing the problem, did it just get rid of the 2V phantom reading on his high impedance DVM or was there a real problem with the operation of the VFD?

-Hal
 
hbiss said:
I suspect that if he used a low impedance meter that voltage wouldn't exist. By the resistor fixing the problem, did it just get rid of the 2V phantom reading on his high impedance DVM or was there a real problem with the operation of the VFD?

-Hal
Click to expand...
There was a real problem of operation of the VFD, because of the minimum 2V we were unable to get the lowest speed control. After adding the resistor into the circuit the voltage was a perfect 0-8V instead of 2-8V.
 
The VFD control input is a very high impedance load between the control terminal and the signal common terminal. The resistor is placed in parallel with this load.
 
callenneff said:
I am working on a DC controls system in which a reference voltage was used for speed control. There was a problem with the system of seeing a minimum 2V for the reference voltage instead of the approximately 0V which should have been seen for the lowest reference voltage. After much research me and an electrician discovered that there were phantom voltages on the system and somehow this was likely rectifying into a real voltage on our reference voltage.

In the end our fix was to toss a low impedance resistor (6.7kOhm) across the input to the isolated transmitter module we use in between the control circuitry and VFD used to drive this system. It completely fixed this problem but I'm looking for some help.

1. Does anyone have good reasoning behind how this is fixing my problem, as of now it seems like a pull-down resistor of sorts.
2. Is there any way to calculate the "proper" resistance value to bleed off this capacitive coupling or whatever the resistor is doing?

Any help is appreciated, thanks!
Click to expand...
Just my thoughts, First try to eliminate the phantom voltage is possible. Maybe re-route the wiring or move components etc.
If that isn't possible and you add a resistor to drain the voltage. I would start with 10K and see what it does. When the signal is present make sure the voltage is still high enough to operate the end device, (input etc)
 
With previous responses I am going down the Pull down resistor option. Pulling new wires is the last resort.
Ordering 30W/550-ohms should dump the 40V but still stay cool
Thank you
 
Would you like to share how you calculated 550 ohms? That strikes me as an excessively low resistance but as I said I'm really only used to dealing with pull down resistors on signal level lines.
 
winnie said:
Would you like to share how you calculated 550 ohms? That strikes me as an excessively low resistance but as I said I'm really only used to dealing with pull down resistors on signal level lines.
Click to expand...
Trial & Error. Getting rid of the 43.8V phantom is not the only issue. Same line will have 120V normal so the heat dissipation is an issue.
Test #1: 43.8V(phantom) + 30K-ohms resistor = 28.3V(phantom residual
Test #2: 43.8V + 20K-ohms = 24V residual
Test #3: 43.8V + 15K-ohms = 21.4V residual
Test #4: 43.8V + 1K-ohms = 2.6V residual. This is acceptable. However it gets pretty hot with the 120V on it.
So a 550-ohms will take it down to 0V but again the heat issue at 120V.
Looks like the higher R values results in less phantom voltage drop so residual phantom voltage remains high
 
That is tremendously useful data. It lets you calculate the impedance of the leakage that is causing the problem.

If you think about the 2.6V across 1K ohms, this suggests that about 2.6 mA is flowing through the leakage. If this leakage comes from 120V, and just from capacitive coupling across the wire, my gut says that this would be several thousand feet of cable. If the cable is normal length, then I'd look for another source of leakage.

Is there any sort of light connected across the control switch? The sort that turns off when the switch is closed because it is powered by the controlled circuit? Have you checked for resistance or continuity across the switch when it is open?

Another reason to consider using an indicator rather than a resistor: if you use an old school _incandescent lamp_ as your indicator, you can take advantage of its dynamic resistance. A 130V 6 watt bulb has an operating resistance of 2.8K, but a cold resistance of about 300 ohms. So when the bulb is just pulling the 'phantom' voltage down it will have low resistance and not dissipate much power, and then when the circuit is on the bulb will light, heat up, and its resistance will increase and limit heat dissipation.

-Jonathan
 
winnie said:
That is tremendously useful data. It lets you calculate the impedance of the leakage that is causing the problem.

If you think about the 2.6V across 1K ohms, this suggests that about 2.6 mA is flowing through the leakage. If this leakage comes from 120V, and just from capacitive coupling across the wire, my gut says that this would be several thousand feet of cable. If the cable is normal length, then I'd look for another source of leakage.

Is there any sort of light connected across the control switch? The sort that turns off when the switch is closed because it is powered by the controlled circuit? Have you checked for resistance or continuity across the switch when it is open?

Another reason to consider using an indicator rather than a resistor: if you use an old school _incandescent lamp_ as your indicator, you can take advantage of its dynamic resistance. A 130V 6 watt bulb has an operating resistance of 2.8K, but a cold resistance of about 300 ohms. So when the bulb is just pulling the 'phantom' voltage down it will have low resistance and not dissipate much power, and then when the circuit is on the bulb will light, heat up, and its resistance will increase and limit heat dissipation.

-Jonathan
Click to expand...
You're correct. It gets better, lol. 4 Actuators(#38, #39, #40,#41) control wires #12, seven wires per actuator controls(3-open/off/close + power(L/N) + 2 indicators(position) and at least 3000ft each bundled into a single conduit. Control 120V is powered by internal 230/120V transformers per actuator so there in itself is at least a 40V floating neutral per actuator. Strangely enough, #40 & #41 do not have the 43.8V phantom issues. It only has 2.2V to Gnd/N.
There were incandescent position indicator lites but one bright the other very dim when valve position switched. These were replaced with LEDs the problem is due to the LED low threshold/forward bias voltage approx(2.5V) both are now of same brightness regardless of actuator position due to the 43.8V(phantom) present on the other position indicator line. Attempts have been made prior to address this issue(position Indicators) but I just inherited it. As mentioned earlier the 43.8V phantom voltage was not a critical issue with the old actuators as far as open/off/close position commands were concerned as the actuators were of the electromechanical type and wouldn't care less about the phantom voltages(except indicator lites). The electronics on this new actuator is sensitive to the phantom 43.8V. I have not been able to determine what the phantom voltage threshold is it will withstand but the fine print note said 19V(doesn't say actual or phantom voltage) can interfere with the operation of actuator.
Dang, I feel like I have just written a book and succeeded in confusing the issues, lol. Appreciate you all.
Note 1: The test results seems to confirm the Low Z meter use to verify it's a phantom voltage since the lower the resistor value the less of the phantom voltage measured unless I am missing something here.
Note 2: We are working on Actuator/valve #38. If we turn off the power(230V 3-phase) to the other three actuators then all the controls(Local/Remote/Remote for )this Electronic actuator works properly.
PS: A light bulb just went on in my head. I never checked if the phantom voltage still existed when the other 3 actuators were turned off. Got carried away by the fact that the actuator operates normally at the time) but that's my next test.
 
What am I missing here? The OP was Calleneff in January with a 2 V control signal, but suddenly it's a 43 volt something from Shuntme. Are there missing posts?
 
Shuntme said:
You're correct. It gets better, lol. 4 Actuators(#38, #39, #40,#41) control wires #12, seven wires per actuator controls(3-open/off/close + power(L/N) + 2 indicators(position) and at least 3000ft each bundled into a single conduit.
Click to expand...
3000 ft is in line with my estimate for how long the wires would need to be for the capacitance to cause that much leakage. (There were very wide error bars on my estimate, but at 3000 ft it is totally plausible that 120V could cause a couple of mA of capacitive coupling.)

Shuntme said:
Control 120V is powered by internal 230/120V transformers per actuator so there in itself is at least a 40V floating neutral per actuator.
Click to expand...
Are the 120V secondary coils grounded? The neutral should not be floating at 40V to ground unless you _intend_ to have an ungrounded system.
Shuntme said:
Strangely enough, #40 & #41 do not have the 43.8V phantom issues. It only has 2.2V to Gnd/N.
There were incandescent position indicator lites but one bright the other very dim when valve position switched. These were replaced with LEDs the problem is due to the LED low threshold/forward bias voltage approx(2.5V) both are now of same brightness regardless of actuator position due to the 43.8V(phantom) present on the other position indicator line. Attempts have been made prior to address this issue(position Indicators) but I just inherited it.
Click to expand...
With 3000 ft of wire, that is totally to be expected. The capacitive coupling would be more than enough to light LEDs.

Shuntme said:
As mentioned earlier the 43.8V phantom voltage was not a critical issue with the old actuators as far as open/off/close position commands were concerned as the actuators were of the electromechanical type and wouldn't care less about the phantom voltages(except indicator lites). The electronics on this new actuator is sensitive to the phantom 43.8V. I have not been able to determine what the phantom voltage threshold is it will withstand but the fine print note said 19V(doesn't say actual or phantom voltage) can interfere with the operation of actuator.
Click to expand...

Remember that 'Phantom voltage' is a _real_ voltage. 'Phantom' in this case doesn't mean that it is some ghost version of voltage, simply that this voltage is derived through a high impedance pathway, and goes away if you have a low impedance load or meter. Don't think of 43V of phantom voltage, think instead of 120V in series with a 46K capacitive leakage path. That leakage path forms a divider with your load impedance. This is why the old electromechanical actuators didn't have a problem; they were very low impedance and that swamped the leakage impedance. The new actuator has a high impedance input, which is why you are considering adding a parallel resistor to lower the total load impedance.

Shuntme said:
Dang, I feel like I have just written a book and succeeded in confusing the issues, lol. Appreciate you all.
Note 1: The test results seems to confirm the Low Z meter use to verify it's a phantom voltage since the lower the resistor value the less of the phantom voltage measured unless I am missing something here.
Click to expand...
Agreed.
Shuntme said:
Note 2: We are working on Actuator/valve #38. If we turn off the power(230V 3-phase) to the other three actuators then all the controls(Local/Remote/Remote for )this Electronic actuator works properly.
PS: A light bulb just went on in my head. I never checked if the phantom voltage still existed when the other 3 actuators were turned off. Got carried away by the fact that the actuator operates normally at the time) but that's my next test.
Click to expand...

Spitballing here:

1) If some of your 120V control circuits are ungrounded, then you need to think of the conduit as a possible source of phantom voltage. Every bit of metal in the conduit, including the conduit itself, forms a capacitor with every other bit of metal. If your neutral (which is also your signal reference voltage) is floating at 40V, then the conduit is 'pulling' your signal lines away from the neutral potential.

2) Will your new actuators function with _DC_ control signals? How about your LED indicators? DC would eliminate the capacitive coupling form of phantom voltage.
 
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