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.
 
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