Why does analysis indicate that ground lifts do not fix ground loops?

Question

Note: "Ground loop" in this question refers to a ground loop caused by linked magnetic fields. (Sometimes ground loop refers to slightly different scenarios.)

Recently I've been looking into ground loops a lot since I've encountered them in circuits I've set up. I've been trying to analyse ground loops using circuit theory to understand why they happen and how to fix them, however my analysis produces results opposite to what is suggested by Wikipedia (as of 14/09/2021) and many pages on the internet. Where am I wrong?

Below is a schematic of a typical ground loop situation, with mesh currents \$i_s\$ and \$i_g\$ circulating around their respective loops. \$R_L\$ models the input impedance to some amplifier/oscilloscope/multimeter/etc. We can assume that the contents of the two boxes are unaffected by any EMI in the environment (maybe they connect off via twisted pairs far, far away). The signal that will end up getting amplified/read will be \$i_s R_L\$. In the above diagram: the top wire is the signal wire, the middle wire is the signal cable's ground shield, and the bottom wire is the ground connection via the mains wiring.

Each bit of conductor has some resistance, modelled as \$R_1\$ and \$R_2\$. We assume that \$R_L\gg R_1\$ and \$R_L\gg R_2\$ (the amplifier has a high input impedance, say a few megaohms - way higher than any resistance the wires might have). The loops of wire will also have some self-inductance - but we will model that directly using Faraday's law later on. Note that I didn't include any resistance in the top signal wire - this is because it is negligible compared to \$R_L\$ which it is in series with.

Finally, we assume that the area of the loop for \$i_s\$ is negligible compared to the area for \$i_g\$ (you could imagine that the signal and ground wires overlap exactly, or are a twisted pair).

Now we can solve for the loop currents using mesh analysis. For the ground loop, we will just apply Faraday's law directly rather than trying to include the parasitic mutual and self inductances into the model so that we can use KVL. For the signal loop, there is no linked flux per our previous assumption, which means Faraday's law just simplifies to KVL around the loop: $$ v_s-i_sR_L-(i_s-i_g)R_1=0\tag{1} $$ For the ground loop, we assume that it links a magnetic flux caused by EMI in the environment and by the loop's own magnetic field, which is proportional to the current flowing around it (this is the loop's self inductance). By Faraday's law, we get an EMF around the loop equal to: $$ \oint_\text{clockwise ground loop}\mathbf{E}\cdot d\mathbf{l}=-\frac{d\Phi_\text{total}}{dt}=-\frac{d}{dt}(\Phi_\text{noise}+Li_g)=-\frac{d\Phi_\text{noise}}{dt}-L\frac{di_g}{dt} $$ We will call the noise term \$v_n\$ - it's sort of like a voltage source is distributed around the loop. Thus: $$ \oint_\text{clockwise ground loop}\mathbf{E}\cdot d\mathbf{l}=v_n-L\frac{di_g}{dt} $$

Ohm's law can be stated more generally as: $$ \int_\text{resistor}\mathbf{E}\cdot d\mathbf{l}=iR $$ Thus, we can split the closed loop integral into integrals over each resistance. This gives: $$ (i_g-i_s)R_1+i_gR_2=v_n-L\frac{di_g}{dt}\tag{2} $$ Now we have the two loop equations, equation 1 and equation 2. There's a derivative in there, which is a bit annoying (makes the whole thing a first order ODE in \$i_g\$). However, we can just take this to the Laplace domain to turn it into an algebra problem. (If you're not used to the Laplace transform, \$sL\$ is essentially the same as \$j\omega L\$ used for phasors.) $$ V_s-I_sR_L-(I_s-I_g)R_1=0\tag{1a} $$ $$ (I_g-I_s)R_1+I_gR_2=V_n-sLI_g\tag{2a} $$ Solving equation 2a for \$I_g\$ gives us: $$ I_g = \frac{V_n+I_sR_1}{R_1+R_2+sL} $$ Substituting this into equation 1a gives us: $$ \begin{align*} V_s-I_sR_L-I_sR_1+I_gR_1&=0\\ V_s-I_sR_L-I_sR_1+\left(\frac{V_n+I_sR_1}{R_1+R_2+sL}\right)R_1&=0\\ V_s-I_sR_L-I_sR_1+\left(\frac{R_1}{R_1+R_2+sL}\right)V_n+\left(\frac{R_1}{R_1+R_2+sL}\right)I_sR_1&=0 \end{align*} $$ $$ \begin{align*} V_s+\left(\frac{R_1}{R_1+R_2+sL}\right)V_n&=I_sR_L+I_sR_1-\left(\frac{R_1}{R_1+R_2+sL}\right)I_sR_1\\ V_s+\left(\frac{R_1}{R_1+R_2+sL}\right)V_n&=\left(R_L+R_1-\left(\frac{R_1}{R_1+R_2+sL}\right)R_1\right)I_s \end{align*} $$ Therefore: $$ I_s=\frac{V_s+\left(\frac{R_1}{R_1+R_2+sL}\right)V_n}{R_L+R_1-\left(\frac{R_1}{R_1+R_2+sL}\right)R_1} $$ The signal that we read will be equal to \$I_sR_L\$. Thus, we multiply by \$R_L\$ to get: $$ V_L=I_sR_L=\frac{V_s+\left(\frac{R_1}{R_1+R_2+sL}\right)V_n}{R_L+R_1-\left(\frac{R_1}{R_1+R_2+sL}\right)R_1}\cdot R_L $$ We can apply some algebra to finesse the denominator into the following: $$ V_L=\frac{V_s+\left(\frac{R_1}{R_1+R_2+sL}\right)V_n}{R_L+\frac{(R_1)(R_2+sL)}{R_1+R_2+sL}}\cdot R_L $$ The fraction in the denominator is just the impedance of \$R_1\$ in parallel with \$R_2+sL\$, which can be notated as: $$ V_L=\frac{V_s+\left(\frac{R_1}{R_1+R_2+sL}\right)V_n}{R_L+(R_1)||(R_2+sL)}\cdot R_L\tag{3} $$ Now, since we assumed that \$R_L\gg R_1\$ and \$R_L\gg R_2\$, the denominator is approximately equal to \$R_L\$, which gives: $$ V_L\approx\frac{V_s+\left(\frac{R_1}{R_1+R_2+sL}\right)V_n}{R_L}\cdot R_L $$ This gives our final expression for \$V_L\$: $$ V_L=V_s+\left(\frac{R_1}{R_1+R_2+sL}\right)V_n $$ But this doesn't match what Wikipedia says about ground loops. The above equation implies that increasing \$R_1\$ will make the noise worse, not better. However, the Wikipedia page suggests putting a 10 ohm resistor on the ground shield of the signal cable to mitigate ground loops, but this equation suggests that doing that should increase the noise seen at the load!

Furthermore, we can take the limit as \$R_1\rightarrow\infty\$, which models the case of an open circuit. To do this, we should use equation 3 (which is before we approximated that \$R_L\gg R_1\$). Taking equation 3 to the limit of \$R_1\rightarrow\infty\$ gives us: $$ V_L=(V_s+V_n)\frac{R_L}{R_L+R_2+sL} $$ This implies that you would see the full noise of the loop, except scaled down by a voltage divider (which would have approximately unity gain if \$R_L\$ was very large). And this even makes sense if you look at the circuit without the middle ground path in it!!! However, this contradicts common practice (according to the internet), where a "ground lift" is often used to mitigate ground loops by open circuiting the ground shield of the signal wire.

Furthermore, if you open circuit \$R_2\$, you get: $$ V_L=V_s\frac{R_L}{R_L+R_1} $$ which is similar to before, except with a slightly different voltage divider and no noise. If \$R_L\$ is very large, we get \$V_L=V_s\$.

What's going on here? Why does this analysis seem to indicate that much of the internet is wrong about how to mitigate ground loops? In fact, I could even argue that the above result makes sense. People say that putting resistance in there prevents the ground loop currents from flowing. But who cares about current! We care about voltage (if the input impedance of the load is very high, it is basically just a voltmeter). While increasing \$R_1\$ makes it harder for current to flow overall, it increases the coupling between the loops for \$I_s\$ and \$I_g\$.

Help! Either there is an error in my calculations (which I doubt would affect the final answer, since the final answer makes intuitive sense to some degree). More likely is that I am missing something important in the model. A core assumption is wrong.

Answer 1

Note: When I use the term "ground lift" I am referring to when the ground of one end of a signal cable is disconnected. The term applies to other scenarios too, such as a ground lift switch that isolates the primary and secondary windings of an isolation transformer or other types of isolators, or a (probably dangerous) cheater plug.

I have come to the conclusion that my analysis was correct given the assumptions. I was able to reproduce the core predictions of my analysis in an experiment (see my description of it here), which would suggest that the assumptions are reasonably accurate (at least in my case).

What some get wrong

Ground loop noise from electromagnetic induction is not necessarily related to the current. Electromagnetic induction means that there are effectively parasitic voltage sources across the signal line, ground shield, and ground wiring. Even if no current is flowing, these will still create voltage differences.

In most circuits, increasing a resistance will increase the voltage drop across it, even though the current goes down (the simplest example is a voltage divider - this can be extended to other circuits using Thevenin's theorem). Changing the resistance of parts of the ground loop (e.g., by inserting \$10\Omega\$ resistor, or making its resistance \$\infty\$ by open circuiting the shield) will simply change how the noise voltages are distributed, but will not remove them.

Ground lifts with single-ended connections

Applying a ground lift to a cable shield is like taking the shield resistance to infinity. It causes all of the noise drop to occur across that gap - allowing the grounds on either end to float at completely different voltages, which will result in lots of noise added across the line. It will also prevent ground loop currents from flowing.

Whether this causes the noise seen at the receiver to improve overall depends on where most of the noise was coming from in the first place. If the noise was predominantly coming from the voltage across the shield, then it will make it worse. On the other hand, if the noise was primarily due to ground loop currents creating \$IR\$ drops within sensitive signal circuitry in the transmitter or receiver, then it will make it better. I suspect the former is more likely, since you can avoid the latter with good circuit design by ensuring that the signal ground is not placed in the path of ground loop currents (and it would tend to require larger trace resistances, high gain circuitry, or low signal levels). These parasitic \$IR\$ drops would also be problems for balanced lines.

However, even if internal \$IR\$ drops were the dominant factor, a ground lift could never completely reject the noise, since opening the ground shield puts the noise voltages directly in series with the signal (referring to the symbols in my question: \$V_s\$ would be directly in series with \$V_n\$ - so the ground lift will result in an SNR of \$V_s/V_n\$).

Ground lifts with other connections

If a signal cable is able to reject the ground noise across the line as a common-mode signal (e.g., balanced line fed into differential receiver), or if the noise can be tolerated for some other reason, then lifting the ground of a cable shield will improve things overall, since ground currents are stopped (preventing undesirable voltage drops elsewhere), while issues from noise voltages across the disconnected shield are avoided.

Current ratings

Another issue with induced ground currents is that it prevents loop currents from circulating through ground conductors that may not be rated for those currents (I don't know in practice how often ground currents can be that large and have only heard about it being a theoretical issue from other people).

Common-impedance coupling

People sometimes use the term "ground loop" to refer to issues where common-impedance coupling through mains ground conductors causes there to be undesirable voltage drops between the mains ground of two devices due to multiple devices pulling current through a shared ground resistance. This causes the devices to couple: current sunk by one device will by seen as a voltage shift in the ground of another device. This is a different phenomenon from what was discussed in the question.

In this case, the ground shield is in parallel with the rest of the ground wiring and helps to reduce the total resistance between the two grounds, resulting in a lower voltage drop (despite a slightly higher current). In this case, opening the ground shield will make this voltage drop larger and most likely worsen overall performance (unless, as before, parasitic drops within signal circuitry were more dominant). Ground currents from another device exceeding the rating of the ground shield is a possibility, but is a separate issue to noise.

Conclusion

Ground lifts have limited effectiveness when applied to the ground of single-ended lines, and will most likely make the noise worse. Even when they can improve the noise, they are limited in how effective they are. They make much more sense for balanced lines (in which case a ground lift on the receiver side is preferable to minimize filtering due to the line capacitance - as suggested by Handbook for Sound Engineers, 4th ed).

If the noise across the line is the dominant factor (and not internal \$IR\$ drops), then a ground lift will worsen the noise, and lowering the resistance of the ground shield will improve the noise. However, this is also of limited effectiveness in practice since it would be hard to get a cable with a shield resistance much below the mains wiring resistance (limiting noise attenuation to ~6 dB according to the formulas I derived in my question). It also has the disadvantage of increasing the loop current, increasing potentially problematic voltage drops elsewhere. To conclude:

Lifting the ground on one end of a single-ended cable will make the noise worse across that cable - it may prevent noise in other lines, and may actually improve the noise overall if parasitic \$IR\$ drops elsewhere were more dominant, but the full loop noise will still appear across that cable, which a single-ended cable cannot reject.

Answer 2

CORRECT

"Professional audio equipment intended for use with balanced lines may have a ground lift switch for the cable shield." Ref Ground Lift

Example:

(Image source: Audio Science Review)

and

(Image source: Shure)

and

(Image source: ProSoundWeb)

The diagram in the other wikipedia page you reference, Common_ground_loops, is incorrect.

This is not a balanced line:

WRONG

(Image source: Wikipedia - Ground loop (electricity))

And as you analyzed and confirmed in experiments, it does not reduce induced noise. Perhaps this is your chance to make a useful and well-founded wikipedia edit.

Oddly, as you point out, the wrong diagram showing a break of the ground in a single-ended connection, is prevalent on the internet.

Lincoln said it best: "These days in America it is more difficult than in any other period in history to verify the truth of what a man reads on the internet" --- Abraham Lincoln

Answer 3

The option of a ground lift exists mainly in connection with an audio transformer, typically in the context of a DI. Without a ground lift, the ground of the unbalanced signal source is connected with the ground of the (typically balanced) signal recipient. The ground loop current of any changing magnetic field, assuming near-zero resistance of the ground loop, will compensate any net magnetic field flow change through the loop while running through the devices themselves and inducing secondary voltages there.

When both instruments (like a keyboard and a mixer or amp) are already grounded, the potential of a balanced or unbalanced input cannot float to levels problematic for the input circuitry so one can safely separate the grounds.

An alternative for an amp with unbalanced input is a "poor man's ground lift" separating signal ground and earth via the AC inputs of a high voltage high current bridge rectifier that has the DC outputs short-circuited solidly so that it will withstand mains shortcircuit current long enough to trip a breaker.

Since the bridge rectifier will require at least 1.2V across to start conducting, it usually works as a ground lift while still preventing fatalities.

You can also introduce a "hum break" resistor (about 10Ohms, small enough to trigger a GFCI breaker that must be present) in the signal ground line to the first preamp stage. So that this actually helps in any way, the reference ground of the first input stage has to be on the side of the signal source rather than the amp. That way any input hum common to both signal and signal ground is not amplified by the first input stage. It is not eliminated either, but in comparison to the amplified signal after the first input stage it will be considerably less prevalent.

Most discrete or opAmp-implemented input stages are comparatively easy to modify in that manner.

Those approximations of a ground lift for unbalanced signals can help making an amp or similar more robust against ground loop hum. An actually balanced signal/connection in combination with an explicit ground lift separating redundant (non-signal) ground connections will be preferable.

A ground lift only makes sense when the arriving signal is balanced (trivially when generated with a transformer, but possible also by routing both signal and return signal voltages through equal resistors, even when the return signal voltage is not actually the inverted main signal but just signal ground) and the differential between signal and return is what is getting amplified.

In that case it breaks the ground loop while not affecting the differential between signal and signal return.

Answer 4

It is an undeniable fact known to many users of audio equipment that mains frequency hum or buzz may appear in a sound system at the moment a ground loop is created, and may disappear at the moment that ground loop is broken.

The effect may occur with either balanced or unbalanced connections.

The question above describes a model in which hum or buzz is NOT ameliorated by breaking a ground loop. Such a model is helpful in understanding why grounding a cable shield on one side only sometimes doesn't improve matters, but fails to explain why ground lifts sometimes work.

Ground lifts can ameliorate mains frequency hum or buzz when the equipment to which the shielded cable is connected is, either through oversight, or through faulty reasoning, or even intentionally, not designed to properly handle ground current.

The designer may have assumed that it is the end user's responsibility to avoid or break ground loops. Or the a designer may not have understood the consequences of his or her design choices with respect to ground loop induced noise. But, for whatever reason the equipment

1. Has a nontrivial impedance between the cable shield and the protective mains ground, and
2. The voltage developed across that impedance by ground loop current is coupled to the circuit and amplified along with the desired signal.

Equipment may be tested for such conditions by applying a mains frequency (but not mains voltage!) current between the cable shield connector to the safety ground connector. If this creates a hum or buzz in the output, there is nontrivial impedance between those points, and some of the voltage developed across that impedance is coupled with the signal and amplified.

A "quick fix" solution to hum and buzz in such equipment is to break the ground loop with a ground lift switch. Alternatively, the ground loop may be broken by using an transformer, or other galvanic isolation technique. But perhaps the best solution is to fix the equipment by soldering a wire with sufficiently low resistance from the connector's shield pin directly to the safety ground connection within the equipment, bypassing any amplification circuitry.

Note that modern PCB design practices, such as the use of ground planes, will tend to prevent such situations in the first place.

---

Edit: I am adding an example of model of a system with an unbalanced signal and with a ground loop, which benefits from a ground-lift.

^{simulate this circuit – Schematic created using CircuitLab}

The following shows the behavior of the circuit.

The first part of output is with the ground lift switch open, and the second part with the ground lift switch closed. As can be seen, the ground lift switch benefits the signal to noise ratio quite a bit. The model doesn't include the wire resistances in the signal line, or in the ground loop (except between the potentiometer "ground" and the amplifier "ground". As a result, the ground loop current is not realistic. This is done to make the issue apparent in a graph of the output signal. A signal to noise ratio of, say, 40 dB, will be noticeable to the ear. However, it would be hardly noticeable in a linear-scale graph of the combined signal and noise with time.

The model can be corrected to include resistance in the cable shield, and also resistance in the mains ground wire. This resistance can significantly improve the S/N ratio. Similarly, a ground lift resistor between the cable shield and the potentiometer "ground" will also improve S/N ratio.

Although making the model more realistic will (usually) improve the S/N ratio, the principle remains.. Ground current within this particular receiving equipment negatively affects the signal to noise ratio, and breaking the ground loop with a ground lift switch which disconnects the cable shield, or adding a resistor in series with the cable shield, improves the S/N ratio.