➖ Differential Measurement & CMRR

A single-ended probe answers one question: how far is this node above ground? That is the right question for a GPIO line or a power rail, but it is the wrong question for a whole class of signals where the information is the difference between two wires and ground is irrelevant (or hostile). To read those, you need an instrument that subtracts two nodes and throws away whatever they have in common. The whole craft rests on one ordering idea: measure the difference you care about first, and reject the common-mode you do not, and a difference probe does both at once, in that order, and a “channel A minus channel B” hack does neither well.

By the end, you can

1. Explain what a differential measurement is and when a node-to-ground probe is the wrong tool
2. Separate a signal into its differential-mode and common-mode parts, and identify which one a difference probe keeps
3. Read a differential probe's spec sheet: bandwidth, differential range, common-mode range, attenuation, and CMRR
4. Calculate the residual common-mode leakage from a CMRR figure and a common-mode voltage
5. Choose between a true differential probe and a two-channel A minus B setup for a given measurement

Intuition first

Imagine two people standing in an elevator, and you want to know how much taller one is than the other. The dumb way is to measure each person’s height from the ground floor and subtract. But the elevator is moving: by the time you read the second person, the floor has risen, and your subtraction is polluted by the elevator’s motion. The smart way is to lay a tape measure between the tops of their heads and read the gap directly. Now it does not matter what floor the elevator is on. The gap is the same whether you are in the basement or the penthouse.

The two wires of a differential signal are the two people. Their height above the floor is the common-mode voltage, the level both wires ride on together. The gap between their heads is the differential voltage, the thing you actually want. The moving elevator is noise: electromagnetic interference, ground bounce, a motor rail sagging and surging. It lifts both wires together, so it lands entirely in the common-mode and not at all in the gap.

A single-ended probe measures from the floor, so the elevator ruins it. A differential probe measures the gap, so the elevator falls out of the answer. That falling-out is the entire point, and how completely it falls out has a name: the common-mode rejection ratio.

What a differential measurement actually is

A differential probe measures the voltage between two nodes, where neither node is ground. Call the two inputs $V_+$ and $V_-$. Any pair of node voltages can be rewritten as two new numbers that are far more useful:

$$V_\text{diff} = V_+ - V_- \qquad V_\text{cm} = \frac{V_+ + V_-}{2}$$

$V_\text{diff}$ is the differential-mode voltage, the difference you care about. $V_\text{cm}$ is the common-mode voltage, the average level both nodes share. A clean robot-hand CAN line sitting at 2.5 V recessive, with CANH at 3.5 V and CANL at 1.5 V during a dominant bit, has $V_\text{diff} = 3.5 - 1.5 = 2.0$ V and $V_\text{cm} = (3.5 + 1.5)/2 = 2.5$ V. The receiver only ever looks at the 2.0 V; the 2.5 V is along for the ride.

This decomposition is why two whole families of measurement require a differential probe rather than merely preferring one:

• Differential signaling. CAN, RS-485, Ethernet, USB, and LVDS all carry their information as $V_\text{diff}$ between a complementary pair of wires, with the $V_\text{cm}$ deliberately ignored by the receiver. Probing one wire to ground reads $V_+$ alone, which is neither the signal the receiver sees nor anything you can trust when the common-mode wanders.
• Floating and high-side measurements. Many nodes have no relationship to ground at all. The gate-source voltage of a high-side FET sits on top of the switched rail and slews violently with it. The voltage across a high-side current shunt rides on the full bus voltage. The voltage across an inductor in a buck converter swings between the input rail and ground every cycle. None of these has a grounded end, so a ground-referenced probe simply cannot read them, and clipping a ground lead onto one of those live nodes shorts it to earth.

Reading the spec sheet

Pick up any differential probe datasheet and the same five numbers decide whether it fits your measurement. Read them in this order:

1. Bandwidth. The highest frequency the probe passes without rolling off (the 3 dB point). A CAN bus at 500 kbit/s needs maybe 20 MHz to see clean edges; a USB 2.0 line needs hundreds of MHz. Too little bandwidth rounds your edges into mush.
2. Differential range. The largest $V_\text{diff}$ the probe can read before clipping. Pick one that comfortably spans your signal swing.
3. Common-mode range. The largest $V_\text{cm}$ the inputs can sit on while still working. This is the one people forget. A probe rated for a 30 V common-mode is useless for a 400 V high-side measurement no matter how good its other specs are.
4. Attenuation. How much the probe divides the signal before the scope sees it (for example 10:1, 50:1, 500:1). Higher attenuation buys a bigger range at the cost of resolution and added noise.
5. CMRR (common-mode rejection ratio). How completely the probe throws away $V_\text{cm}$. This is the headline figure, and it has a catch we will spend the rest of the lesson on.

CMRR, and why it has a frequency

An ideal differential amplifier outputs $A_d(V_+ - V_-)$ and nothing else. A real one leaks a little of the common-mode through:

$$V_o = A_d\,(V_+ - V_-) + A_\text{cm}\,\frac{V_+ + V_-}{2}$$

The common-mode gain $A_\text{cm}$ should be tiny next to the differential gain $A_d$. CMRR is the ratio of the two, quoted in decibels with the 20-log rule:

$$\text{CMRR}_\text{dB} = 20\log_{10}\!\left(\frac{A_d}{\lvert A_\text{cm}\rvert}\right)$$

A precision instrumentation amplifier with laser-trimmed, better-than-0.1%-matched resistors can reach 100 dB, sometimes 130 dB. But here is the part that bites: CMRR is highest at DC and degrades as frequency rises. The tiny parasitic capacitances inside the probe and its leads do not match as perfectly as the resistors do, and their impedance falls with frequency, so more common-mode sneaks through as the signal speeds up. A probe that boasts 80 dB on its cover may be down to 50 dB or worse at the megahertz where your switching noise actually lives. Always read CMRR as a curve versus frequency, never as the single best-case number.

The residual-leakage formula

CMRR is abstract until you turn it into a voltage. The common-mode that survives the probe and shows up as a phantom signal at the output is:

$$V_\text{leak} \approx \frac{V_\text{cm}}{10^{\,\text{CMRR}_\text{dB}/20}}$$

Read it as: every 20 dB of CMRR divides the leak by ten. At 80 dB the common-mode is attenuated by a factor of $10^{80/20} = 10^4$. So a 5 V common-mode leaks through as

$$V_\text{leak} \approx \frac{5\ \text{V}}{10^4} = 0.5\ \text{mV}$$

Half a millivolt of error riding on your measurement. If your differential signal is hundreds of millivolts, that is invisible. If you are trying to read a 2 mV drop across a high-side shunt while a noisy 24 V common-mode swings on top, that 0.5 mV (and it grows as the noise climbs in frequency and CMRR sags) is a real chunk of your reading. This single formula is how you decide whether a given probe is good enough before you trust the trace.

See it / Try it

The simulator below puts the elevator on your bench. The left model is a true differential probe; the right is a single-ended probe reading one node against ground. Start with a small differential signal and slide the common-mode V_cm up: watch the single-ended reading drown as the common-mode floods it, while the differential reading holds steady. Then drag the CMRR slider down from 120 dB toward 60 dB and watch the common-mode leak grow on the differential side. The leak readout is the formula above, live.

Two things to take away from playing with it. First, the differential probe’s reading barely moves as V_cm climbs, because the common-mode rides out, exactly the elevator falling away from the gap. Second, CMRR is not free: lower it and the leak creeps back in, and at any fixed CMRR a bigger common-mode means a bigger leak. The note under the sim repeats the warning that matters on real hardware: clipping a single-ended ground clip onto a live, non-ground node shorts it through earth.

?You need to measure the gate-source voltage of a high-side N-channel FET in a motor driver. The source slews between 0 V and 24 V every switching cycle. What tool is correct?

• A A standard passive probe with its ground clip on the FET source
• B A true differential probe across gate and source, with adequate common-mode range
• C A single channel referenced to board ground, reading the gate node
• D A multimeter on DC volts across gate and source

?A differential probe is spec'd at 80 dB CMRR at the frequency of interest. With a 10 V common-mode present, roughly how much common-mode leaks into the reading?

• A About 1 V
• B About 100 mV
• C About 1 mV
• D Zero (CMRR means the common-mode is fully rejected)

Lab: prove rejection on the bench

Take the robot hand’s CAN harness and a differential probe. First, with the bus idle, inject a known common-mode: tie a small AC source (or just let a nearby motor run) so CANH and CANL both ride on a wandering level. Read the differential trace; it should sit flat at the recessive level while the common-mode swings. Now do the naive thing on purpose, on a low-voltage node where it is safe: single-ended probe CANH to ground, and watch the common-mode pollute the trace. Then sweep the common-mode source up in frequency and watch your differential probe’s rejection degrade as you climb toward its CMRR roll-off. Log the differential reading, the leak you can see, and the frequency where rejection visibly falls apart. That last number is the real bandwidth of your trust in the probe, and it is almost never the headline CMRR on the box.

↘ Why 'two channels, A minus B' is a trap, and the math of mismatch

The tempting substitute for a real differential probe is to put channel A on $V_+$, channel B on $V_-$, and let the scope compute $A - B$. It looks equivalent and it is not, for three reasons that compound.

Channel mismatch. Each channel has its own gain, offset, and probe attenuation, and its own frequency response. The scope subtracts the two post-acquisition, so any mismatch between the channels appears directly as common-mode error. If channel A has gain $g_A$ and channel B has gain $g_B$, the computed difference is

$$g_A V_+ - g_B V_- = \frac{g_A + g_B}{2}\,(V_+ - V_-) + (g_A - g_B)\,\frac{V_+ + V_-}{2}$$

The second term is pure common-mode leakage, and its size is set by the gain mismatch $g_A - g_B$. A 1% mismatch caps your effective CMRR at only about 40 dB no matter how perfect each channel is on its own. Matched passive probes drift apart with temperature and cable bending, so even a carefully nulled pair degrades.

Modest CMRR even at best. A trimmed instrumentation amplifier reaches 100 dB because the matching happens on one silicon die with laser-trimmed resistors. Two independent scope channels through two independent cables cannot come close. The math above shows why: their rejection is bounded by component matching you do not control.

It cannot measure a floating node safely. Both scope channels still share the scope’s grounded inputs. If neither $V_+$ nor $V_-$ is at ground, A-minus-B does not rescue you from the fundamental problem that you have grounded both ends through the chassis. A true differential probe floats both inputs at high impedance, which is the only thing that makes a high-side or mains-referenced measurement safe.

So A-minus-B is acceptable only when the common-mode is small, slow, and benign, and when a momentary ground clip on either node is harmless. The instant the common-mode is large, fast, or hot, you need a probe built to reject it, not a subtraction done after the damage is already in the data. (Note one conflict with the grounding extract worth flagging: differential signaling by itself only buys headroom, while common-mode noise rejection is strictly a property of the balanced, matched interface plus a differential receiver. The AUTHORITATIVE framing here treats the differential probe as the receiver whose CMRR and matching deliver the rejection, which is the right mental model for measurement even though signaling and balance are formally independent ideas.)

Grounded in Wikipedia: “Common-mode rejection ratio”, “Differential signalling” (CC BY-SA).

Key takeaways

• ▸A differential probe measures the voltage between two nodes, neither of which is ground, and rejects whatever the two share.
• ▸Any pair of nodes splits into differential-mode ($V_+ - V_-$, the signal) and common-mode (the average level); a difference probe keeps the first and throws away the second.
• ▸Differential signaling (CAN, RS-485, Ethernet, USB, LVDS) and floating / high-side measurements require a differential probe, not a ground-clipped one.
• ▸Read five specs: bandwidth, differential range, common-mode range, attenuation, and CMRR, and read CMRR as a curve, because it is high at DC and degrades with frequency.
• ▸Residual leak $\approx V_\text{cm} / 10^{\text{CMRR}_\text{dB}/20}$: every 20 dB divides the common-mode error by ten.
• ▸A-minus-B on two channels is a poor substitute: channel mismatch caps CMRR low and it still cannot safely read a floating node.

Practice 1 warm-up

A differential bus has $V_+ = 3.5$ V and $V_- = 1.5$ V. Compute the differential-mode voltage and the common-mode voltage. Which one does a differential probe report?

↘ Show worked solution

$V_\text{diff} = V_+ - V_- = 3.5 - 1.5 = 2.0$ V. $V_\text{cm} = (V_+ + V_-)/2 = (3.5 + 1.5)/2 = 2.5$ V. A differential probe reports the differential-mode voltage, $2.0$ V, and rejects the $2.5$ V common-mode. A single-ended probe on $V_+$ alone would read $3.5$ V, which is neither the signal of interest nor a number that stays put when the common-mode moves.

Practice 2 core

You are reading a 2 mV drop across a high-side current shunt that sits on a 24 V rail. Your probe has 100 dB CMRR at DC. How large is the residual common-mode leak, and is it a problem? Now suppose at the switching frequency of the converter the CMRR has degraded to 60 dB. Recompute the leak and comment.

↘ Show worked solution

At DC: $V_\text{leak} = V_\text{cm} / 10^{\text{CMRR}/20} = 24\ \text{V} / 10^{100/20} = 24 / 10^{5} = 24\ \mu\text{V}$ ($0.024$ mV). Against a 2 mV signal that is about 1% error, comfortably small.

At the switching frequency with CMRR fallen to 60 dB: $V_\text{leak} = 24\ \text{V} / 10^{60/20} = 24 / 10^{3} = 24\ \text{mV}$. That is now twelve times larger than the 2 mV signal you are trying to read. The DC spec looked fine and the real measurement is buried in common-mode leak. This is exactly why CMRR must be read as a curve versus frequency, not as the headline DC number.

Practice 3 stretch

A colleague proposes measuring a differential signal by putting two matched passive probes on $V_+$ and $V_-$ and using the scope’s A-minus-B math. Both probes are spec’d to 1% gain tolerance. Estimate the best-case effective CMRR of this setup, and give one reason it could still be the wrong choice even if that CMRR were acceptable.

↘ Show worked solution

The common-mode leakage from gain mismatch behaves like a common-mode gain of about $\lvert g_A - g_B\rvert$ against a differential gain of about $1$. With 1% tolerance the worst-case mismatch is roughly $0.01$ in the gains, so the effective CMRR is about $20\log_{10}(1/0.01) = 20\log_{10}(100) = 40$ dB. That is far below the 80-130 dB of a real differential probe, and it ignores offset and frequency-response mismatch that make it worse.

Even if 40 dB were enough for the signal, the setup is wrong whenever the node is floating or hot: both scope channels share the chassis ground, so neither $V_+$ nor $V_-$ can sit safely off ground. A high-side or mains-referenced measurement would short a live node to earth through a probe clip. For those, only a truly floating differential probe is safe, regardless of the math.