Concept

Compensated divider — where it appears

A resistive divider with a capacitor across each arm, whose division is the same at every frequency only while the two time constants match. A mismatch of a per cent shows as an overshoot or a droop on an edge, and as nothing at all on a slow signal.

Named by 5 essays across 2 fields — each of them below, with the objects they name alongside it.

A 10:1 divider with 12.8 pF across its top resistor. computed 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.

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.

instruments · Compensation
A 1.0% doublet: 0.078 dB in the magnitude, 36× the settling time. computed by solving, not by drawing. Above, the magnitude of a fast circuit followed by a pole and a zero that were meant to cancel and miss by 1.00%, against the same circuit with the cancellation exact: the worst disagreement anywhere up to the fast corner is 0.0777 dB. Below, the error left in the step response, in units of the tail's own amplitude of 0.909%. Settling to 0.10% takes 245.2 fast time constants against 6.9 with the cancellation exact, and the closed form τ·ln(A/B) gives 245.2 — a time that contains nothing of the fast circuit at all.

The cancellation that leaves a tail

A pole and a zero placed on top of each other disappear from the response. Miss by one per cent and the magnitude changes by 0.078 decibels, which no measurement would report as a fault, while the time to settle to a thousandth goes from 6.9 time constants to 245 — thirty-six times longer. The settling time has a closed form containing neither the fast circuit nor the doublet's separation as such, and its consequence is blunt: settling to a part in ten thousand needs a cancellation good to a part in ten thousand, however fast the amplifier in front of it is.

transients · Doublet
A 10:1 divider into 100 pF and 15 pF more behind 1.0 kΩ, trimmed for the calibrator, on a 1 ns edge. computed by solving, not by drawing from the residues of its four poles, on a time axis from 0.1 ns to 1 ms. The trimmer is set to 12.8 pF, where a 1 kHz square wave reads flat: the upper arm then matches the lower arm's total of 115 pF. On a 1 ns edge the output overshoots to 1.128 times where it settles, because for the first few nanoseconds the late 15 pF is not yet part of the divider, and falls back over that branch's own time constant, 15 ns, within 1% only by 35.5 ns. With a second branch in the upper arm, 9.0 kΩ and 1.7 pF — a scaled copy of the late branch — the same edge is flat to 7.0e-11 (dashed).

The time constant one trimmer cannot reach

A ten-to-one divider whose lower arm is 100 pF of instrument input and 15 pF more reached through a kilohm reads flat on a 1 kHz calibrator with its trimmer at 12.8 pF — the same setting as a lower arm of 115 pF lumped — and overshoots 12.8% on a 1 ns edge, settling within 1% only after 35.5 ns. No setting of the one trimmer is flat: the best judged on a fast edge is 6.1%, at 11.9 pF, and a trimmer set by eye so the fast edge lands flat leaves the calibrator's square wave 9.9% low with a 114 µs slope on its top. Flat at every time needs the upper arm to be a scaled copy of the lower one — a second branch of 9 kΩ and 1.7 pF — and then a 1 ns edge is flat to 10⁻¹⁰, with a window of −18 to +24% on that resistor and ±6% on its capacitor.

instruments · Compensation
A ten-to-one divider with 15 pF of its arm behind a kilohm, stepped through 1 kΩ: one trimmer overshoots 0.00936% where an ideal source shows 13.2%, and the second branch that was exact no longer settles first. computed by solving, not by drawing: the step of the divider of the earlier essay, 9 MΩ and a trimmer over 1 MΩ with 100 pF direct and 15 pF behind a kilohm, through a source resistance of 1 kΩ, by residues. One trimmer at its calibrator setting (solid) overshoots 0.00936%, and the second branch that made the arm a scaled copy (dashed) rises without overshoot on the source's own lag. An ideal source (dotted) shows a 13.2% hook with one trimmer and nothing with two. A source resistance charges the input capacitance, 11.5 pF, so its lag and the hook act on one edge, and when they are comparable the hook's excess goes into shortening the lag.

The source the hook hides behind

A compensated divider with a late branch in its lower arm overshoots 13.2 per cent on an ideal step, and every measurement of that hook has been made from a source with no resistance. Through a real source the input capacitance charges first, and the hook and the source's lag act on the same edge. The overshoot is 10.5 per cent from 50 Ω, 4.13 from 300 Ω and gone by a kilohm, where the source's lag equals the hook's. The excess has not gone away: it is in the edge, which the screen shows 15 per cent faster than the source and the probe's capacitance allow. And the second branch that made the divider exact only makes its ratio exact, so from a kilohm the exact divider settles in 65.6 ns and the one with the hook in 44.

instruments · Compensation
The same 15 pF spread along 2 kΩ of line, on a 1 ns edge: with 1 of its branches copied the output is flat to 1.07%, against 11.2% with one trimmer alone. computed by solving, not by drawing: the response to a 1 ns edge of a ten-to-one divider whose lower arm is 100 pF direct and 15 pF along a 2 kΩ resistive line, with one trimmer (dashed) and with the first 1 branch of the line's exact partial-fraction expansion copied into the upper arm (solid). The expansion's time constants are 12.2 ns, 1.35 ns, 486 ps and so on, as 1/(2k − 1)². Inverted numerically by Talbot's method, and the line itself checked against a netlist of sections.

The branches a square root needs

A resistive cable's late charge is a continuum and not one branch, and the partial fractions of its tanh are an exact description of it: 81.1 per cent of the capacitance behind a time constant of 12.2 ns, 9.01 per cent behind a ninth of that, 3.24 behind a twenty-fifth, all through the same kilohm. So a divider's upper arm can copy it one piece at a time. On a 1 ns edge one trimmer alone is 11.2 per cent out, one copied branch 1.07, two 0.252 and four 0.0353. The first branch does most of the work and the uncopied capacitance falls only as one over the count; what saves the rest is that a branch faster than the edge is averaged away. Faster edges ask for more branches, and the tolerance on the first ends the series before the arithmetic does.

instruments · Compensation

Named alongside it

The objects these essays reach for when they reach for this one.

ResiduesStep responseTime constant matchingAttenuator probeTrimmerModel rangeOvercompensationComponent toleranceDesign tradeoffDistributed elementInput capacitancePoles

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