core24 min studyLesson 959 in Physics

Ground loops, shielding, and common-mode rejection

Follow unwanted currents and common-mode voltages through a connected measurement system instead of treating ground as an ideal universal node.

Before you begin

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The question for today

What you’ll understand

Separate differential and common-mode input components.
Convert a stated CMRR into a residual input-referred error.
Identify shared-return pickup and distinguish shielding from isolation.

Ground is a circuit connection with resistance, inductance, and geometry. Two instruments attached to different points can therefore disagree about the voltage called zero, especially when power or digital currents share the measurement return. Differential inputs and shields can reduce particular couplings, but they do not turn the complete system into an ideal isolated voltmeter. A useful investigation follows actual current paths and identifies whether an unwanted voltage reaches the amplifier as common mode, differential mode, or an overload outside its permitted range.

Define the input decomposition and rejection ratio

Let the two input voltages relative to the amplifier reference be v+ and v−. Their difference is the desired differential coordinate, and their average is the common-mode coordinate. In a small-signal linear model, both can contribute to the output through different gains. The common-mode rejection ratio compares differential gain with common-mode gain at a stated frequency and operating condition. A decibel value is an amplitude ratio using twenty times the base-ten logarithm.

Input-referred common-mode leakage equals the common-mode voltage times the ratio of common-mode to differential gain. Even a large rejection number can leave an error bigger than a microvolt-scale signal when the common-mode voltage is large. The inputs must still remain inside their common-mode range and the output inside its linear range; CMRR is not a promise of correct operation beyond those limits.

v_d=v_+-v_-,\qquad v_{\mathrm{cm}}=\frac{v_++v_-}{2},\qquad v_o=A_dv_d+A_{\mathrm{cm}}v_{\mathrm{cm}},\qquad \mathrm{CMRR}_{\mathrm{dB}}=20\log_{10}\left|\frac{A_d}{A_{\mathrm{cm}}}\right|

vd and vcm are voltages, and the two gains are dimensionless linear voltage gains at the frequency of interest. Frequency dependence and phase can matter. The formula describes common-mode leakage while the amplifier remains within its specified operating domain.↔ Scroll wide equations horizontally.

A return current can create a differential error directly

If two grounded instruments create a closed conductive path, a difference of their reference potentials can drive current around the loop. Shared impedance in a sensitive return then creates a voltage drop that is added to the measurement. Once this contribution is differential at the amplifier pins, common-mode rejection does not remove it. Increasing CMRR alone therefore cannot solve every mains-frequency pickup problem.

Balanced input impedances and routing reduce conversion of common-mode interference into differential voltage. A mismatch in cable capacitance or source resistance can spoil high-frequency rejection even when the amplifier itself has excellent low-frequency CMRR. Twisting a signal pair reduces loop area and magnetic pickup; an electrostatic shield addresses electric-field coupling. A conductive shield is not a universal remedy for low-frequency magnetic fields or for current intentionally flowing through a shared signal return.

I_{\mathrm{loop}}=\frac{\Delta V_g}{Z_{\mathrm{loop}}},\qquad V_{\mathrm{pickup}}=I_{\mathrm{loop}}Z_{\mathrm{shared}},\qquad v_{\mathrm{err,input}}=v_{\mathrm{cm}}\frac{A_{\mathrm{cm}}}{A_d}

The loop equations are a lumped linear model with impedances at the interference frequency. The practice example uses real resistances. ΔVg is a reference-potential difference, and Zshared is the portion whose voltage drop enters the differential signal.↔ Scroll wide equations horizontally.

Choose connections from the coupling mechanism and frequency

Shield termination depends on frequency, cable geometry, isolation, and the intended return path. A low-frequency single-point electrostatic connection and a high-frequency low-inductance chassis connection solve different problems. Blanket instructions to connect every shield at one end, or every shield at both ends, omit those conditions. Guarding can reduce leakage or effective input capacitance when appropriately driven, but it also requires a stable circuit and a defined signal reference.

Practical remedies can include rerouting return currents, reducing loop area, improving balance, or using a suitable galvanically isolated measurement interface. Protective-earth connections remain part of the equipment's intended protection and are not removed as a signal-processing technique. Verify a change with the relevant frequency and common-mode conditions, and preserve the experiment's physical signal path. A quieter trace is useful evidence only when the connection change has not also attenuated or disconnected the wanted measurand.

Work it through

A large CMRR can still leave a large residual

At a specified low frequency, a hypothetical linear differential amplifier has Ad=100 and CMRR=80 dB. A 10 µV differential signal is accompanied by a 1.0 V common-mode voltage. Assume all input and output ranges are respected. Find Acm, the desired output, and the leakage expressed both at the output and referred to the input.

The gain ratio is $10^{80/20}=10000$ , so $|A_{\mathrm{cm}}|=100/10000=0.010$ . Take the gains positive for this numerical comparison.
The desired differential output is 100×10 µV=1.0 mV. The common-mode leakage is 0.010×1.0 V=10 mV, ten times that desired output.
Dividing the leakage by differential gain gives 100 µV input-referred error. A rejection specification must therefore be evaluated against both the common-mode amplitude and the required signal accuracy.

The result

The common-mode gain is 0.010; leakage is 10 mV at the output, equivalent to 100 µV at the input, despite the 80 dB rejection ratio.

The living notebook

Explore the model

Interactive lab

Interactive lab · waves

Cancellation is sensitive to matching

For each fresh comparison, Reset and set BOTH Phase difference=180° and Amplitude ratio. Compare ratios 1.00 and 0.80. The two fixed-frequency copropagating waves have resultant amplitudes 0 and 0.20. Interpret this only as a cancellation-matching analogy. The wave instrument has no wires, ground impedance, differential amplifier, CMRR, shielding, or magnetic coupling.

SumFirst componentSecond component

Sum at x = 0: 0.866
Resultant amplitude: 1.732
Phase difference: 60°

Phase difference60°Amplitude ratio1

t = 0 s

Two waves share the same frequency and wave number. Playback advances t in sin(2x − t) + a sin(2x − t + phase). Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forA small mismatch can leave a substantial residual when large contributions were expected to cancel.

Pause & reason

Check your understanding

Interactive checks need JavaScript. You can still read the explanation.

Read the checkpoint explanation

CMRR rejects common-mode input within its operating domain. A voltage drop added differentially through shared impedance is amplified along with the wanted differential signal.

Your turn

Try it yourself

Use a resistive low-frequency loop model with reference difference ΔVg=20 mV and total loop resistance 2.0 Ω. A 0.10 Ω shared return lies in the measurement path. Find loop current, differential pickup, and the unwanted output of an otherwise ideal gain-100 amplifier. Does increasing its CMRR address this particular differential term?

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Compute the loop current first and then its drop across the shared return. The problem explicitly places that drop in the differential measurement path.

Compare with the solution

The current is 20 mV/2.0 Ω=10 mA. The shared-return drop is (10 mA)(0.10 Ω)=1.0 mV, producing an unwanted 0.100 V output at gain 100. Increasing CMRR does not reject this already differential contribution. Changing the return-current geometry or using a suitable isolated interface can address the mechanism, subject to the full system requirements; removing protective earth is not the proposed remedy. The values are a declared lumped example, not a universal model of every interference path.

Keep the idea

Take it with you

CMRR must be evaluated with common-mode amplitude, frequency, balance, and operating range.
A ground-loop voltage can become differential before the amplifier sees it.
Shielding, twisted pairs, return routing, and isolation address different coupling mechanisms.

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