The corner that says nothing about an edge

An oscilloscope’s AC-coupled input is specified by one number, and the number is a frequency: a corner, usually somewhere between a tenth of a hertz and ten. It is an honest specification of a sinusoidal response and it is nearly useless for the thing the instrument is mostly used to look at, which is an edge.

Two quantities are wanted for a pulse, and neither is a corner frequency. How much the flat top decays over the pulse — the sag — and where the waveform sits relative to zero once a train of pulses has settled. Both come from the same two components and neither is on any data sheet.

Two components, one exponential #

A megohm and a microfarad. The time constant is a second and the corner is

$$f_c = \frac{1}{2\pi R C} = 0.159\ \mathrm{Hz}$$

which is the number the input is sold on.

Now send a pulse of duration $T$. The capacitor cannot hold the step, so the output jumps to the full amplitude and then decays as $e^{-t/RC}$ for as long as the pulse lasts. At the end of it the top has fallen to $e^{-T/RC}$ of where it started, and the sag is

$$\mathrm{sag} = 1 - e^{-T/RC}.$$

For a hundred-millisecond pulse that is 9.516%. The figure marches the netlist forward in time with the trapezoidal rule rather than evaluating the exponential, and gets 9.515% — the two routes agreeing to three parts in ten thousand, which is the integrator’s own step error and can be driven down by shortening the step.

The trailing edge is the other half of the same statement. When the drive returns to zero the output steps down by the full amplitude from wherever the top had got to, so it undershoots by exactly the sag and then recovers with the same time constant. A ten per cent sag is a ten per cent undershoot, always, and a scope trace that shows one without the other is showing something else.

The constant nobody quotes #

The useful question is the inverse one: given a corner frequency, how short must a pulse be for its top to be flat to one per cent?

$$T_{1\%} = -RC\ln(0.99) = 0.010\,05\,RC$$

so a one per cent top needs a pulse about a hundredth of the time constant. Expressed against the specified corner rather than the time constant, the pulse rate has to be

$$\frac{1}{T_{1\%}} = \frac{2\pi}{-\ln 0.99}\, f_c = 625.2\, f_c$$

and that factor of 625 contains no resistance, no capacitance and no amplitude. It is the whole of the translation between the specification and the measurement, and it is a fixed number the figure asserts across every setting of the slider.

Six hundred and twenty-five is a great deal more than anybody’s intuition. A 0.159 Hz corner sounds like a specification about very slow things; it says that a pulse must be shorter than ten milliseconds — a hundred hertz — before its top is flat to a per cent. A ten-hertz corner, which sounds like a lower-grade input, requires 6.25 kHz.

That is the essay’s first result and it is the reason the two numbers should never be substituted for one another. A corner frequency describes what happens to a sinusoid at that frequency. A pulse is not a sinusoid, its spectrum runs down to zero, and what a coupling capacitor does to it is decided by the ratio of its duration to a time constant.

Where the waveform sits #

Send one pulse and the sag is the whole story. Send a train, and something else happens that is more often the cause of an actual failure.

A capacitor in series passes no direct current, so once a repetitive waveform has settled its average must be zero. That is not a property of the filter’s passband, it is a conservation statement about charge, and it holds however slow the corner is: a settled AC-coupled square wave has as much area above the axis as below it.

For a pulse train of amplitude one and duty cycle $D$, that puts the top at $1-D$ and the bottom at $-D$. The height is untouched — nothing has been attenuated — and the whole waveform has slid down.

Those are measured by marching six hundred periods of a hundred-hertz train through the same megohm and microfarad — six time constants, which is what settling onto the new baseline takes — and reading the last two cycles. The top matches $1-D$ to three decimal places at every duty cycle and the height matches unity to two parts in a thousand.

Above fifty per cent duty the top never reaches half the amplitude. A comparator set at the middle of the waveform — which is the obvious place to set it, and where an automatic threshold puts itself — stops firing entirely. Not late, not jittery: it never fires.

That is a fault with an unusual signature. It appears when the pattern changes rather than when the signal does. Amplitude unchanged, frequency unchanged, edges unchanged, and a receiver that worked all day stops the moment the data happens to contain a long run. It is the reason serial protocols that must survive AC coupling are line-coded to bound their disparity, and the reason an oscilloscope in AC coupling is a poor instrument for looking at a duty cycle.

Three places the translation is needed #

Looking at supply ripple under a switching load. The ripple is what AC coupling is for, and the load’s own current steps are what make the trace unreadable: each step shifts the baseline and the recovery takes seconds. The instrument is behaving exactly as specified and the trace is worthless, and the fix is a smaller coupling capacitor in front of the instrument rather than a different instrument.

Measuring a pulse width. A time interval read at fifty per cent of amplitude is read at fifty per cent of the displayed amplitude, and the display has slid down by the duty cycle. At a ten per cent duty the crossing is at 0.45 of the true amplitude rather than 0.5, which for a trapezoidal edge is a systematic width error of a few per cent of the rise time. Nothing is jittering; the answer is simply biased.

Anything with a burst structure. A packet radio’s envelope, a laser’s pulse train, a motor controller’s commutation — all have a duty cycle that changes with what they are doing, and all therefore have a baseline that moves with it. The signature is a measurement that depends on the traffic rather than on the signal.

Recovery, which is the other end of the same time constant #

A coupling capacitor’s virtue is that it removes a large direct component; the cost of removing a large one is that recovering from it takes the same time constant that made the removal good.

Drive the input above range for a moment — a probe touched to a rail, a relay closing, a supply coming up — and the capacitor charges to whatever it took. Returning to the signal leaves that charge to bleed off through the megohm at one time constant per second, so the trace walks back to the axis over several seconds while the instrument reads nothing useful.

The two requirements are exactly opposed and neither is negotiable: a lower corner means less sag and a longer recovery, in strict proportion, because both are $RC$. That is a design trade with no clever side, which makes it worth stating plainly rather than looking for one.

The same arithmetic gives the useful rule for choosing a corner in the first place. Decide the longest event whose top must be flat, multiply by a hundred to get the time constant, and the recovery from an overload will be a few times that — so a one per cent flat top on a hundred-millisecond event costs about a thirty-second wait after an overload.

What a probe does to the numbers #

The instrument’s coupling capacitor is not the whole network once a probe is on the front of it.

A ten-to-one probe is nine megohms in series with the instrument’s one, so the resistance the coupling capacitor discharges into is ten megohms rather than one — the time constant is ten seconds, the corner is 0.0159 Hz, and the sag at a hundred milliseconds falls from 9.5% to 1.0%. AC coupling is ten times better through a divider probe than through a direct lead, and for the same reason it attenuates ten times.

The compensation capacitor across the probe’s nine megohms complicates that at high frequency and not at low: the divider essay in this field measures the condition under which the two halves have the same time constant, and while it holds the probe is a flat ten-to-one divider and the arithmetic above is unchanged with $R$ multiplied by ten. When it does not hold, the probe has two ratios and the coupling capacitor is the least of the problems.

The third parameter is the one a designer usually has least of, and it is worth seeing moved on its own: the load resistance sets the corner exactly as the capacitance does, and the two enter the sag as a product.

Why the corner is the wrong number, said once more precisely #

It is worth putting the mismatch in spectral terms, because that is where the intuition that a corner should be enough comes from.

A rectangular pulse of duration $T$ has a spectrum that is flat up to about $1/T$ and falls after that, and — this is the part the intuition drops — it has a substantial component all the way down to zero frequency. A high-pass filter with a corner at $f_c$ removes everything below $f_c$, so it takes away a slice of the spectrum whose width is $f_c$ out of a total that runs to zero. The fraction removed is not $f_c$ over anything the pulse has; it is $f_c T$, because the pulse’s low-frequency content per unit bandwidth is proportional to its duration.

So the damage is proportional to the product of the corner and the duration, which is what $1 - e^{-T/RC}$ says to first order — $T/RC = 2\pi f_c T$. The corner alone is half a specification. Quoting it without the duration it is going to be used at is like quoting a gain without saying at what frequency, and the subject would not tolerate the second.

What this essay does not claim #

That AC coupling is a mistake. It is the only way to see a millivolt of ripple on a twelve-volt rail, and no amount of resolution substitutes for removing the part that is not wanted. The point is that its cost is a shape change with a computable size, and the size is not the specified corner.

That the sag is a filter’s fault. It is what a first-order high pass does, exactly, and every other first-order high pass does the same. Nothing here would be improved by a better capacitor. Making the corner lower reduces the sag proportionally and lengthens the recovery from an overload by the same factor, which is the actual trade.

That the baseline shift settles instantly. It takes several time constants, so a waveform whose duty cycle changes shows a slow drift of the whole trace towards its new baseline — seconds, at a 0.159 Hz corner. That transient is neither the sag nor the shift; it is the approach to the shift, and it is what makes the fault look intermittent.

That the numbers transfer to a differential input. They do for each side. What they do not cover is the common-mode path, where two coupling capacitors that are not matched turn a common-mode step into a differential one, and the mismatch that matters there is a tolerance rather than a value.

The other three things a coupled input does #

An alternating-current coupling is sold on a corner frequency and delivers four behaviours, of which the corner is one. A divider with two ratios is the same network with the capacitor deliberately trimmed, where the two ratios are made equal rather than left to disagree. The instrument’s own rise time is the other end of the same bandwidth, and it is quoted just as misleadingly. The current the instrument draws is the error the coupling capacitor is often fitted to avoid, and The current that does not reach the input is the leakage that discharges it. The millivolts in the wire is the offset a coupling removes and the one it does not.

The gate #

Two routes to the sag — a trapezoidal march of the netlist and $1 - e^{-T/RC}$ for the same two components — asserted to agree to three parts in a thousand at every pulse width the slider offers. The residue is the integrator’s step error and shrinks with the step, which is what makes it a tolerance rather than a disagreement.

The 625 is asserted as a constant, not computed at the drawn setting: the ratio between the pulse rate a one per cent top needs and the specified corner is required to equal $2\pi/(-\ln 0.99)$ to a part in $10^9$, at every value of R and C.

The settled top is asserted against one minus the duty cycle at every duty cycle on the slider, and the height against unity — because “the waveform moves and does not shrink” is the claim, and only the pair of assertions says it.

And the threshold claim is stated as a distance rather than as a side. At fifty per cent duty the top is the threshold, 0.5023 against 0.5, so asking whether it is crossed has no answer there. The assertion tests the clearance within two per cent of half duty and the side elsewhere, which is what a claim about a boundary has to do at the boundary.

A note on the march itself. The pulse starts a fraction of its width into the run rather than at zero. A trapezoidal integrator begins from a state, and a source that is already high at $t=0$ is a state the initial one contradicts — the first edge is then integrated rather than stepped and the whole trace is displaced by a unit. Nothing on this site had met that before, because every earlier caller drives a shunt element whose output starts at zero either way. One step, computed twice is where the rule’s own error is established — falling by a factor of four every time the step is halved, which is a claim about a method and can be watched — and the displacement described here is not that error but an initial condition, which no refinement of the step size removes.

The other three instruments with the same complaint #

An AC-coupled input is one of four places in this field where a specification written for one kind of input is quoted about another, and the four make the same argument in four quantities.

The instrument’s own rise time is the closest relative and the one that answers the same question about the fast end of the band. Rise times add in quadrature, so ten per cent of inflation needs an instrument 2.18 times faster than the edge — a constant with no instrument in it — and measured on the solved network it is 2.79 for a one-pole front end, 3.97 for two, 4.87 for three and 5.62 for four. The rule is optimistic at every pole count, which is the wrong direction for a rule of thumb to err in, and it is optimistic for the same reason the corner on this page is: a bandwidth is a statement about sinusoids and an edge is not one.

The probe is part of the circuit makes it about impedance rather than about bandwidth — a hundred and fifteen picofarads across a two-kilohm source being one per cent wrong at 6.8 kHz, with nothing inaccurate anywhere — and the corner the instrument has no part in makes it about rejection, with a 95 dB instrument delivering 84 dB at a kilohertz because of an imbalance in the source.

Four specifications, four quantities, one shape: each is a true statement about the instrument alone and none of them is a statement about a measurement. The sag computed here — 9.5 per cent on a hundred-millisecond pulse through a 0.159 hertz corner, with a one per cent flat top needing a pulse rate 625 times the corner — is the version of that with the fewest excuses, because a corner frequency and a pulse duration are both printed on the front of the instrument.

The 625 is worth keeping for the same reason the 2.79 is: it contains no component. A ratio of a pulse rate to a corner frequency is dimensionless and is the same number for a millisecond pulse through a 1.6 hertz corner as for a hundred-millisecond one through 0.159, so it transfers between instruments and between applications without recomputation. What does not transfer is the duty cycle result above it, which is a statement about the waveform rather than about the coupling — and above fifty per cent duty no corner frequency at all makes an AC-coupled pulse reach half its own height, which is a refusal rather than a tolerance.