Measurement, which is a circuit on a circuit

A divider with two ratios

Put capacitance in a resistive divider and it divides by resistance at direct current and by capacitance at high frequency, and those are two different numbers unless one equation holds. The adjustable trimmer on every oscilloscope probe exists for that single equation, and the square wave on the instrument's front panel is a display of which of the two ratios is currently winning.

A resistive divider has one ratio. Put a capacitor across each resistor and it has two, and they are different numbers.

At direct current the capacitors do nothing and the ratio is R₂/(R₁ + R₂). At high frequency the resistors do nothing and the ratio is C₁/(C₁ + C₂) — the top capacitance over the sum, which is the reciprocal-looking arrangement that follows from a capacitor’s impedance being inverse in its value. Those two expressions are equal only when RC₁ = RC₂.

A 10:1 divider with 12.8 pF across its top resistorcomputed by solving, not by drawing. The divider's resistors set a ratio of 0.10000 and its capacitors 0.10000; the step starts at the second and relaxes to the first over 115 µs. The balance R₁C₁/R₂C₂ is 1.0000, and the edge lands 0.00% away from where it settles. At 12.8 pF the two ratios are the same number and the response is flat.00.0500.10002.5057.5010time, in network time constantsoutput ÷ inputboth ratios: 0.1000the divider9M12.8 pF1Msolved, then checked — two ratios, one network0.0% out at 12.8 pF
Fig. 1 A ten-to-one divider’s step response at a trimmer value that satisfies that equation, with three other settings drawn behind it. The response is flat: the edge lands exactly where the response settles, because both ratios are 0.10000. Move the trimmer and the edge lands somewhere else and relaxes to the settled value over the network’s own time constant. The slider is the trimmer.

What the step response is showing

The shape is the whole diagnostic and it reads directly.

The edge lands at the capacitive ratio, because at the instant of a step the capacitors are the only components that matter — a resistor’s current is proportional to voltage and a capacitor’s to rate of change, and at an edge the rate of change is unbounded.

It then relaxes to the resistive ratio over the network’s time constant, which is (R₁ ∥ R₂)(C₁ + C₂) — 115 µs for the values drawn.

So an edge that lands above where it settles means the capacitive ratio exceeds the resistive one, which means C₁ is too large: over-compensated. An edge that climbs from below means C₁ is too small: under-compensated. The direction of the error is legible at a glance, which is why the adjustment is made against a square wave rather than against a meter.

The measured numbers make it concrete. At the balancing value of 12.78 pF, the edge lands at 0.10000 and settles at 0.10000, and the overshoot is 0.000%. At 19 pF the balance ratio is 1.487, the edge lands at 0.14179 and the overshoot is 41.3%. At 9 pF the balance is 0.704, the edge lands at 0.07258 and the response climbs, 27.1% short.

The settled value is the same in every case, and that is the first thing to check on any of these traces. Whatever the trimmer is set to, the divider settles at 0.10000 — because at direct current the capacitors are not there and the resistors decide, and no capacitor value changes a resistive ratio. A probe that is badly compensated reads correctly on a steady voltage and incorrectly on everything else, which is exactly the kind of error that survives a calibration check.

The time constant does move a little with the setting: 108.9 µs at 6 pF, 115.0 µs at balance, 126.0 µs at 25 pF. That is (R₁ ∥ R₂)(C₁ + C₂) with C₁ varying, and it is a reminder that the trimmer is a real component in the network rather than a correction applied to the answer.

A 10:1 divider with 19.0 pF across its top resistorcomputed by solving, not by drawing. The divider's resistors set a ratio of 0.10000 and its capacitors 0.14179; the step starts at the second and relaxes to the first over 121 µs. The balance R₁C₁/R₂C₂ is 1.4870, and the edge lands 41.27% away from where it settles. At 12.8 pF the two ratios are the same number and the response is flat.00.0500.1000.15002.5057.5010time, in network time constantsoutput ÷ inputresistive: 0.1000the divider9M19.0 pF1Mcapacitive: 0.1418solved, then checked — two ratios, one network41.3% out at 19.0 pF
Fig. 2 The same divider over-compensated at 19 pF. The edge overshoots to 0.148 and falls to 0.100 over about 120 µs. On a square wave at a few kilohertz this is the familiar spike-and-droop that means the trimmer needs turning down, and the amount of overshoot is a direct reading of the ratio RC₁/RC₂.

The ramp, and why a true step would have been wrong

There is a detail in how the figure is computed that started as a bug and ended as a correctness condition, and it is worth setting out because the symptom was so unhelpful.

The drive is a ramp with a rise time of a three-hundredth of the network’s time constant, not a discontinuous step. The reason is that C₁ couples the input directly to the output, so a true step makes the output jump at t = 0 — while the integrator, like every initial-value solver, starts every state at zero. Those two statements contradict each other, and the trapezoidal rule resolves the contradiction by ringing.

The first version of this figure reported a settled value of −1.5×10⁻⁸ for a divider that settles at 0.1, and an “overshoot” of −1.3×10⁷ per cent. Nothing about that was subtle once seen; what made it worth recording is that the shape of the trace looked like a plausible oscillation, and a figure that had merely been drawn rather than asserted on would have shipped.

A ramp three hundred times faster than the network is a step as far as the network is concerned, and a real generator’s edge is finite anyway. The residual cost is that the measured landing value is about one per cent past the capacitive ratio — the network relaxes measurably during the ramp — which is why the figure asserts that agreement to three per cent rather than to the arithmetic.

What a probe costs an edgecomputed by solving, not by drawing across 61 probe speeds. Rise times combine in quadrature, so a probe as fast as the signal inflates the measurement by 41.4% and one three times faster by 5.4%. Ten per cent — usually the most anybody will accept — needs the probe to be 2.2 times faster than the edge it is watching.00.250.500.751100m110the probe's own rise time ÷ the signal'sinflation of the measurementequal speeds: 41% too slowten per centprobe must be 2.2× fastersolved, then checked — quadrature, not addition10% costs a 2.2× faster probe
Fig. 3 What a probe costs an edge, from the previous essay. The ramp used to drive the divider here is a version of the same idea from the other side: a generator’s edge is finite, the network’s response to it is what a measurement actually shows, and treating either as instantaneous is the assumption that has to be checked rather than made.

How precisely the trimmer has to be set

The balancing value needs no search: it is RC₂/R₁, one division, and for the values here it is 12.7778 pF exactly.

What is worth measuring is the window — how far from that value the setting can be before the edge is visibly wrong — because that is what decides whether an adjustment is needed at all.

The setting window for a 2% flat edgecomputed by solving, not by drawing at 51 trimmer settings. The balance point is 12.8 pF exactly — it is R₂C₂/R₁ and needs no search — and the edge lands within 2% of the settled value from 12.5 pF to 13.1 pF, a window 4.5% wide. Standard capacitor values step by about ten per cent, so no fixed part reliably lands in it.00.2000.400101520trimmer capacitance (picofarads)how far the edge lands from settled2%exact at 12.8 pF4.5% of a windowsolved, then checked — the window measured, not quoted4.5% wide at 2%
Fig. 4 The departure of the edge from the settled value, against trimmer capacitance, with the window inside which it stays under two per cent shaded. The window is 12.49 pF to 13.07 pF: 4.52% wide. The slider is the flatness accepted, and at half a per cent the window narrows to 1.13%.

Four and a half per cent, for two per cent of flatness. One and a bit per cent, for half a per cent. At ten per cent of flatness — which is a badly adjusted probe, visibly wrong on a square wave — the window is 22.6% wide.

The scaling in those three numbers is worth reading. The window widens roughly in proportion to the flatness accepted: 1.13%, 4.52% and 22.6% for 0.5%, 2% and 10%. That is the linear relation from the paragraph below stated as a window rather than as a slope, and it says something practical about the adjustment — halving the acceptable overshoot halves the range of settings that achieve it, so each further factor of two in flatness costs the same effort as the last. There is no point at which the adjustment becomes suddenly harder or suddenly easier.

Standard capacitor values step by about ten per cent, so no fixed component reliably lands in either window — and that, rather than any argument about manufacturing tolerance, is why the adjustment exists. It is not there to compensate for drift or for temperature; it is there because the target is narrower than the grid of available values.

The second thing the shape says is that the error is linear through zero rather than quadratic. A five per cent error in the trimmer gives about five per cent of overshoot. There is no flat region around the optimum, which is why the adjustment is fiddly and why it can be made accurately: a quantity with a sharp minimum is easy to find and a quantity with a broad one is not.

Why the response is flat when the two agree, not merely equal at the ends

The equation RC₁ = RC₂ makes the ratio at zero frequency equal the ratio at infinite frequency. That is what has been shown so far, and it does not by itself say the ratio is the same at every frequency in between. It could equally describe a response that departs and returns.

It does not, and the reason is worth a paragraph because it is the algebraic heart of the arrangement.

Each arm of the divider is a resistor in parallel with a capacitor, so each has an impedance of R/(1 + sRC). The divider’s ratio is the lower impedance over the sum, and when the two time constants are equal the factor (1 + sRC) is the same in both arms and cancels completely. What is left is R₂/(R₁ + R₂), a constant with no s in it anywhere.

So the response is not approximately flat or flat over a band. The pole and the zero of the divider land on top of each other and annihilate, and the network has no frequency dependence at all. That is a much stronger statement than “the two ends agree”, and it is why the step response goes immediately to its final value rather than reaching it.

It is also why the failure is a single time constant rather than something more complicated. Away from balance the pole and the zero separate but there is still only one of each, so the departure is always a single exponential relaxation from the capacitive ratio to the resistive one — which is exactly what every trace in the figures does, at every setting.

What the divider is for

It is worth stepping back to say why anybody would put capacitance into a divider deliberately, since the whole essay is about the trouble it causes.

They would not. The capacitance in a probe’s lower arm is not a design choice; it is the instrument’s input capacitance and the cable’s, and it is there whether or not anyone wants it. The trimmer in the upper arm is the fix: given that C₂ exists and cannot be removed, adding a C₁ that satisfies RC₁ = RC₂ makes the divider’s two ratios agree and gives it a flat response.

That framing changes what the equation means. It is not a constraint the design must respect; it is a repair, and what it repairs is a frequency response that would otherwise fall away above 1/2πRC₂ — about 1.4 kHz for the values here. A ten-to-one probe with no trimmer at all would be a low-pass filter with a corner in the audio band.

A 10 kΩ + 10 kΩ divider, solved with its loadThe unloaded answer is 6.00 V. It is 1% low at a load of 495 kΩ and 4.00 V at a load equal to the divider's own resistance. The ratio does not predict any of this; the magnitude does.02461001k10k100k1M10Mload resistance across the output (ohms)output voltage, solved with the load in place6.0 V with nothing connected1% low at 495 kΩthe circuit12 VR₁R₂R_Lsolved, then checked — the load swept over six decadesthe ratio is 1% wrong below 495 kΩ
Fig. 5 The uncompensated argument, from the networks field: a divider’s ratio depends on what is across it. This page is the same statement with a frequency axis. What is across the lower arm is the instrument, its capacitance is unavoidable, and the compensation is what stops that capacitance from turning the ratio into a function of frequency.

What the adjustment does not fix

Three things, and they are worth separating from the one it does.

It does not change the tip capacitance. The trimmer’s value is what makes the tip present C_load/ratio, so compensating the divider and minimising the loading are the same adjustment — which is a genuinely fortunate coincidence and is why the previous essay and this one share a number. But minimising is not eliminating: 11.5 pF is what a correctly compensated ten-to-one probe presents, and the probe is part of the circuit is about what that costs.

It does not survive a different instrument. The balancing value depends on C₂, which is the instrument’s input capacitance plus the cable’s. Move a compensated probe to a different oscilloscope and C₂ changes, the equation no longer holds, and the probe needs re-adjusting. That is why the adjustment is on the probe rather than inside the instrument, and why the ritual is performed on connection rather than once at the factory.

It does not extend to high frequency. Everything here is a two-element model of each arm, and a real probe has inductance in its leads, distributed capacitance along the resistor, and a cable that is a transmission line rather than a lumped capacitance. Above a few tens of megahertz the flat response this adjustment produces stops being flat for reasons no trimmer addresses, which is why a fast probe has more than one adjustment in it.

Two probes on a 2.0 kΩ sourcecomputed by solving, not by drawing twice per frequency: the node alone, and the node with the probe's elements across it. The one-to-one probe's 115.0 pF makes the reading one per cent wrong at 6.79 kHz. The ten-to-one probe puts 12.8 pF in series with the cable, so its tip sees 11.5 pF and the same error arrives at 69.2 kHz — 10 times further up, bought with a factor of ten in signal — the two edges stand in the ratio of the tip capacitances, 10.00. At direct current neither probe is capacitive at all and the ten-to-one still reads 0.02% low, because 10 MΩ across 2.0 kΩ is a divider.-6-4-201101001k10k100k1M10M100Mfrequency (hertz)log₁₀ of the error in the reading1× probe, 115.0 pF at the tip10× probe, 11.5 pFone per cent1× is 1% out at 6.79 kHz10× at 69.2 kHzsolved, then checked — the node with and without the probe1% wrong at 6.79 kHz with a 1× probe
Fig. 6 The loading a compensated probe still imposes, from the previous essay. The trimmer’s job is done and the response is flat; what remains is 11.5 pF across the node, and the edge at which that makes the reading one per cent wrong is the number the whole field is about.

The same arrangement, in three other places

The pole-zero cancellation above is not a probe technique. It is a general one, and recognising it elsewhere is most of what makes it worth understanding rather than memorising.

Every attenuator in a wideband instrument. A switched attenuator is a chain of dividers, and each of them has stray capacitance across each arm. Each therefore carries a trimmer, and the alignment of an instrument’s attenuator is the same adjustment repeated once per range. A step response that changes shape when the range is switched is a divider somewhere in the chain whose time constants no longer match.

High-voltage dividers. A hundred-to-one divider for a kilovolt supply has a large resistance in its upper arm, which makes even a picofarad of stray capacitance across it consequential. The same equation applies, and the physical construction of the upper arm — the spacing, the shielding, the grading rings — is largely an exercise in controlling C₁ so that it can be matched.

Lead compensation in a feedback loop. A network that places a zero on top of a pole to cancel it is doing algebraically what the trimmer does: making a factor appear in both numerator and denominator so that it disappears. The feedback field meets it as a way of buying phase margin, and it has the same fragility — a cancellation that is exact only when two independently drifting quantities stay equal.

That last one is the useful warning. A cancellation is a coincidence between two components, and it holds only as well as the coincidence does. A trimmer set to a twentieth of a per cent by a careful adjustment is set to that precision on the day; a resistor and a capacitor with different temperature coefficients will not keep it, and nothing about the flat response gives any warning as it drifts.

Two ratios, one equation

The result is unusually clean and worth restating in its shortest form.

A divider with capacitance has a ratio at zero frequency and a ratio at infinite frequency. They are equal if and only if the two arms have the same time constant. The step response shows which is larger. The setting window for two per cent of flatness is 4.5% wide, so the adjustment is a trimmer rather than a part.

Every one of those sentences is a measurement on a solved network, and the last one is the reason the component exists.

What makes this one of the more satisfying corners of the subject is how completely the picture and the arithmetic agree. The equation is one line. The failure is one exponential. The diagnostic is a shape on a screen that says both how far off the setting is and in which direction. And the window — the quantity that decides whether the adjustment is needed at all — is a number that falls out of the same solve, and it turns out to be narrower than the spacing of the parts anybody could buy.

Very little in this collection lines up that neatly. Most models here are excellent inside a range and wrong outside it, with the boundary somewhere inconvenient; this one is exact inside a window and the window is measurable. The difference is that the failure has one degree of freedom, and a one-dimensional problem can be adjusted rather than merely characterised.

One step response, computed twice: from the poles, and by walking the network forwardA damping ratio of 0.22, so the overshoot is 49.2%. The two curves are drawn on top of each other; the panel below is the difference between them, which is the trapezoidal rule's error at 500 steps and reaches 1.70e-3 V.00.50011.50024output (volts), for a 1 V step inthe final valuesolid: from the poles · dashed: stepped forwardtime (milliseconds) above · the same span as a fraction, belowgap between the two routes (volts)1e-71e-61e-51e-41.0m10m1.0e+2m1solved, then checked — residues against 500 trapezoidal stepsthe numerical route is out by 1.7e-3 V
Fig. 7 A step response computed from the poles by residues and by walking the network forward in time, from the transients field. The second of those routes is what draws every trace on this page, and its behaviour on a network with a direct feedthrough — where it rings unless the drive is ramped — is the gotcha this essay records.