Question

Question: 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!

Answer 1

Okay, this is a common and often frustrating problem in industrial control systems. Let's break down what's likely happening and why your solution worked.

Understanding the Problem: Phantom/Ghost Voltages and Capacitive Coupling Phantom Voltages

These are small voltage readings that appear on conductors that are not actively powered. They are typically caused by capacitive or inductive coupling between nearby energized conductors. They can be real voltage, meaning they can drive a very high impedance circuit.

Capacitive Coupling

Imagine two wires running parallel to each other for a long distance. They effectively form a capacitor. If one wire is carrying an AC voltage, the other wire will have a small AC voltage induced in it due to this capacitance. The higher the frequency and the closer the wires, the more coupling you'll see. Often, this is from the 60 Hz AC power wiring in your facility.

Rectification

Many high-impedance electronic circuits contain components (like diodes inherent in op-amps or semiconductors) that can inadvertently rectify these small AC phantom voltages into a DC voltage. This DC voltage then appears as your erroneous 2V minimum reference.

High Input Impedance of the Transmitter Module

Your isolated transmitter module likely has a very high input impedance (resistance to DC current flow). This is good for signal integrity in most cases, but it makes the input very susceptible to small currents produced by these phantom voltages. Think of it like this: a tiny current, when applied across a huge resistance, results in a significant voltage drop (Ohm's Law: V = IR).

Why Your Resistor Fixed It (Pull-Down Effect)

You're right, the resistor acts as a pull-down resistor, but the mechanism is subtle:1. Lowering the Input Impedance

The 6.7 kΩ resistor significantly lowers the effective input impedance of the transmitter module's input. Instead of a multi-megaohm impedance, you now have a combined impedance much closer to 6.7 kΩ.
2.

Bleeding Off the Phantom Current

The small AC current induced by capacitive coupling can now flow much more easily through the low-impedance path of the resistor. This prevents the rectification of the phantom voltage into a DC offset that your system was reading as a 2V minimum. The resistor effectively shorts out the voltage created by that current, by Ohm's law V = IR. So, for some small current (I), lowering the resistance (R) lowers the voltage (V).3.

Stabilizing the Input

The resistor provides a defined path to ground (or a common reference). This stabilizes the input voltage of the transmitter module, making it less susceptible to floating voltages and noise.

Calculating a "Proper" Resistance Value

This is tricky, and there's no single perfect calculation. It's often empirical (trial and error) with some educated guesses. Here's the thought process:1. Understand the Trade-Offs

Lower Resistance

More effectively suppresses phantom voltages, but...

Lower Resistance

...increases the current draw from your control circuitry. You don't want to load down the signal source so much that it affects the intended reference voltage signal.2.

Consider the Signal Source Impedance

Determine the output impedance of the source providing the reference voltage. This is often specified in the datasheet of the controller or DAC providing the reference. Let's call it `Z_source`.3.

Desired Loading Effect

Decide how much you're willing to load down the signal source. A general rule of thumb is that the input impedance of the load (the transmitter module with the resistor) should be at least 10 times the source impedance. This ensures that the load doesn't significantly affect the voltage signal. So, the target input impedance, `Z_in`, is:

Z_in >= 10  Z_source

4.

Calculate the Resistor Value

Since the resistor is in parallel with the (very high impedance) input of the transmitter module, we can approximate `Z_in` as just the resistor value. Therefore:R_pull_down >= 10 Z_source
5.

Empirical Adjustment

This is where you fine-tune. Start with a resistor value calculated as above, and observe the following:

Minimum Reference Voltage

Is the 2V offset still present? If so, decrease the resistor value (use a smaller resistance).

Signal Accuracy

Does the pull-down resistor noticeably affect the accuracy of the reference voltage at higher levels? If so, increase the resistor value (use a larger resistance).

Current Draw

Measure the current being drawn from the reference voltage source. Is it within the source's specified current limit? If not, increase the resistor value.

Example

Let's say your reference voltage source has an output impedance of 100 ohms.1. `Z_source = 100 ohms`2. `R_pull_down >= 10 100 ohms = 1000 ohms = 1 kΩ`So, you might start with a 1 kΩ resistor. If the 2V offset is still present, try 680 ohms, 470 ohms, or even lower, but keep an eye on the current draw from the signal source.

Important Considerations

Shielded Cables

Using shielded cables for your reference voltage signal can significantly reduce capacitive coupling. Make sure the shield is properly grounded at one end (usually the source end) to prevent ground loops.

Twisted Pair Cables

Using twisted pair cables can also reduce the effects of capacitively coupled noise by keeping the impedance balanced.

Wiring Practices

Keep signal wires separated from high-voltage power wires. Run them in separate conduits or cable trays. Avoid running them parallel for long distances.

Grounding

Ensure proper grounding throughout your control system. Ground loops can exacerbate phantom voltage problems. Use a star grounding configuration where possible.*

Isolation

You are using an isolated transmitter, that's good. Be sure that the ground of your 24VDC power supplies are all tied together.

In summary, your 6.7 kΩ resistor is acting as a pull-down resistor, lowering the input impedance and bleeding off the phantom current that was being rectified into a DC offset. A more "proper" value can be estimated by considering the source impedance and the desired loading effect, and then fine-tuned empirically. Flag for review

Answer 2

Okay, this is a common and often frustrating problem in industrial control systems. Let's break down what's likely happening and why your solution worked.

Understanding the Problem: Phantom/Ghost Voltages and Capacitive Coupling Phantom Voltages

These are small voltage readings that appear on conductors that are not actively powered. They are typically caused by capacitive or inductive coupling between nearby energized conductors. They can be real voltage, meaning they can drive a very high impedance circuit.

Capacitive Coupling

Imagine two wires running parallel to each other for a long distance. They effectively form a capacitor. If one wire is carrying an AC voltage, the other wire will have a small AC voltage induced in it due to this capacitance. The higher the frequency and the closer the wires, the more coupling you'll see. Often, this is from the 60 Hz AC power wiring in your facility.

Rectification

Many high-impedance electronic circuits contain components (like diodes inherent in op-amps or semiconductors) that can inadvertently rectify these small AC phantom voltages into a DC voltage. This DC voltage then appears as your erroneous 2V minimum reference.

High Input Impedance of the Transmitter Module

Your isolated transmitter module likely has a very high input impedance (resistance to DC current flow). This is good for signal integrity in most cases, but it makes the input very susceptible to small currents produced by these phantom voltages. Think of it like this: a tiny current, when applied across a huge resistance, results in a significant voltage drop (Ohm's Law: V = IR).

Why Your Resistor Fixed It (Pull-Down Effect)

You're right, the resistor acts as a pull-down resistor, but the mechanism is subtle:1. Lowering the Input Impedance

The 6.7 kΩ resistor significantly lowers the effective input impedance of the transmitter module's input. Instead of a multi-megaohm impedance, you now have a combined impedance much closer to 6.7 kΩ.
2.

Bleeding Off the Phantom Current

The small AC current induced by capacitive coupling can now flow much more easily through the low-impedance path of the resistor. This prevents the rectification of the phantom voltage into a DC offset that your system was reading as a 2V minimum. The resistor effectively shorts out the voltage created by that current, by Ohm's law V = IR. So, for some small current (I), lowering the resistance (R) lowers the voltage (V).3.

Stabilizing the Input

The resistor provides a defined path to ground (or a common reference). This stabilizes the input voltage of the transmitter module, making it less susceptible to floating voltages and noise.

Calculating a "Proper" Resistance Value

This is tricky, and there's no single perfect calculation. It's often empirical (trial and error) with some educated guesses. Here's the thought process:1. Understand the Trade-Offs

Lower Resistance

More effectively suppresses phantom voltages, but...

Lower Resistance

...increases the current draw from your control circuitry. You don't want to load down the signal source so much that it affects the intended reference voltage signal.2.

Consider the Signal Source Impedance

Determine the output impedance of the source providing the reference voltage. This is often specified in the datasheet of the controller or DAC providing the reference. Let's call it `Z_source`.3.

Desired Loading Effect

Decide how much you're willing to load down the signal source. A general rule of thumb is that the input impedance of the load (the transmitter module with the resistor) should be at least 10 times the source impedance. This ensures that the load doesn't significantly affect the voltage signal. So, the target input impedance, `Z_in`, is:

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