The probe that takes a tenth

The probe is part of the circuit opens this field with a trade and states it in one line: a ten-to-one probe costs a factor of ten in signal and buys exactly ten in bandwidth. The bandwidth half of that sentence is measured there, on two solved networks, and it is exactly right. The signal half is asserted, and behind it sits a second sentence that is asserted too — that a tenth of the signal is a tenth of the signal-to-noise ratio, so a small signal is where the trade stops being favourable.

Neither of those had a number. And under both of them is a smaller omission which is the reason: only two ratios have ever been built. A probe divider is a series resistance of $R_\mathrm{in}(n-1)$ with a trimmer of $C_\mathrm{load}/(n-1)$ across it, which is an expression in $n$ and not a pair of special cases, and every figure in this collection has evaluated it at one and at ten.

What a ratio actually is #

The divider is four elements. Nine megohms in series with the tip, a trimmer across it, then the instrument’s own megohm and the cable’s hundred picofarads to ground. At a general ratio $n$ the series arm is $R_\mathrm{in}(n-1)$ and the trimmer is $C_\mathrm{load}/(n-1)$, and the tip sees the trimmer in series with the cable, which is $C_\mathrm{load}/n$ — a series pair being dominated by the smaller of the two.

So the tip capacitance falls exactly as $1/n$. At two-to-one it is 57.5 pF, at ten-to-one 11.5 pF, at a hundred-to-one 1.15 pF. That is the whole mechanism and it is why the trade looks clean: the signal at the instrument falls as $1/n$ and the capacitance across the node falls as $1/n$, so a factor traded is a factor bought.

The condition under which that holds is stated in the rung below and it is worth restating as an equation rather than as a warning. The probe’s resistance is also across the node, and it is $n R_\mathrm{in}$, so the reading is low by $R_\mathrm{s}/(R_\mathrm{s} + n R_\mathrm{in})$ before any frequency is applied. On a two-kilohm source with a one-to-one probe that is 0.20 per cent — a fifth of the one-per-cent budget the bandwidth is defined against, and small enough to ignore.

Everything between one and ten is buildable and nothing sells it. A probe is one, ten, a hundred or occasionally a thousand, and the reason is packaging rather than physics: the ratio decides the printing on the barrel, the instrument reads that printing off a coding ring, and a two-to-one probe would need a scale factor no oscilloscope offers. The expression does not know that. It returns a series arm of a megohm and a trimmer of 115 pF at two-to-one, which is an ordinary divider, and the sweep below evaluates it at twelve values because the trade is a curve and had only ever been sampled at its two endpoints. Reading a trade off two points is how a linear relation gets assumed, and one of the two relations on this page is linear and the other is not.

The reproduction, which comes before the departure #

Two routes to the same edge, sharing the tip capacitance and nothing else. One is a bisection on two solved networks, taking the difference between what the node does and what the instrument displays. The other is $|1 - 1/(1+jx)| = 0.01$, solved for $x$ and divided by $2\pi R_\mathrm{s} C_\mathrm{tip}$ — a single pole, with no probe resistance in it anywhere.

On a fifty-ohm source they agree at every one of the twelve ratios and the first step from one-to-one to two-to-one buys 2.00. On three hundred ohms, 2.00. On a kilohm, 2.01. On two kilohms, 2.03, with the two routes never more than 1.8 per cent apart across the whole sweep. That is the calibration: where the probe’s own resistance is outside the error budget, the trade is exactly one factor for one factor, and both computations say so.

The agreement is what entitles the disagreement to be quoted, in the same way that the winding field solver’s full-window case entitles its quarter-fill one. Two routes that agree where they must are an instrument; two routes that agree everywhere have shared an assumption.

The departure in the bandwidth, which is the small one #

At ten kilohms the trade is not one-for-one and it is favourable in the unexpected direction. Buying a factor of two in signal returns 12.30 in bandwidth, because the probe being replaced had almost no bandwidth: its resistance had already spent the error the frequency was going to be measured against.

The reading to take from that is not that attenuation is a bargain on heavy sources. It is that the one-to-one probe stops being a measurement somewhere near ten kilohms, and the enormous apparent gain from the first step is the arithmetic reporting the collapse of the thing it is dividing by. Above about thirty kilohms there is no frequency at which a one-to-one probe is within one per cent, and the quantity being plotted stops existing rather than becoming small — the same kind of boundary that every model has an edge collects, with the model’s failure appearing as an undefined number rather than a large one.

The bandwidth half of the field’s opening claim therefore has a range, and the range is a source impedance: below about five kilohms it is exact, and above it the claim is conservative for a reason that is not a virtue.

The departure in the noise, which is not small #

The other half of the claim is that a tenth of the signal is a tenth of the signal-to-noise ratio. That is a statement about noise and nothing in this collection had computed it, because computing it means asking what the probe contributes rather than what it attenuates.

The probe is four elements and two of them are resistors of nine megohms and one megohm. A resistor is a noise source — the floor a resistor sets is the whole of that argument — and these two sit at the instrument’s own input, on the far side of the attenuation. So their noise is not divided by ten. It is multiplied by ten, along with everything else at that node, when the instrument scales its reading back up.

A factor of 31.3, not a factor of ten. The signal-to-noise ratio at the tip falls by three times more than the signal does, and it does so at every ratio on the sweep: 8.50 µV at two-to-one, 26.3 at five, 114.5 at twenty, 584.4 at a hundred.

Where 31.3 comes from, which is a closed form #

The solved answer has a shape, and the shape has no source resistance in it and no measurement bandwidth either.

The two divider resistors sit at a node whose capacitance is the cable’s hundred and fifteen picofarads plus the trimmer’s $C_\mathrm{load}/(n-1)$, and whose resistance is the parallel combination of $R_\mathrm{in}$ and $R_\mathrm{in}(n-1)$. Integrate white noise through a single pole made of those two and the resistance cancels: the density rises as its square root and the noise bandwidth falls as its inverse, which is exactly the reciprocity the total that has no resistor in it measures over five decades. What is left is $\sqrt{kT/C}$ on the node’s own capacitance, multiplied by $n$ because the instrument multiplies:

$$v_\mathrm{tip} = \sqrt{n(n-1)\,kT/C_\mathrm{load}}$$

The solved network agrees with that to 0.4 per cent at every ratio from two to a hundred. And the expression’s zero at $n = 1$ is not a curiosity — it is the statement that a one-to-one probe has no divider to be noisy, which is what the solve independently reports when the probe’s contribution comes out at exactly zero volts and the entire 1.782 µV belongs to the source.

That is the second route this page rests on. One computation is four resistors, each replaced in turn by itself in series with a source, integrated over a grid; the other is three constants and two capacitances. They share the netlist’s topology and not one arithmetic step.

The grid is part of the first of those and is worth naming, because it was wrong twice before it was right. A noise total is an integral, the integral is taken on the trapezoidal rule over whatever frequency points it is handed, and the two obvious choices each fail at one end. Points spaced evenly in frequency step straight over a corner near a kilohertz and overstate the answer by thirteen per cent; points spaced evenly in the logarithm resolve the corner and under-resolve the flat stretch above it. The grid used here is the union of the two, and the check that it is fine enough is that the answer stops moving — four parts in ten thousand between three hundred and sixty points and fourteen hundred, against a departure of thirteen per cent between the two grids that are wrong. A closed form that agrees with a converged integral is a result; one that agrees with an unconverged one is a coincidence waiting to be quoted.

The narrower the measurement, the worse the trade #

The factor of 31.3 is a property of the band it was measured in, and the direction it moves is the opposite of the one intuition offers.

The source’s own noise grows as the square root of the bandwidth. The probe’s own noise does not, because it is $\sqrt{kT/C}$ and that number is already the whole of the frequency axis — the pole made of nine hundred kilohms and a hundred and twenty-eight picofarads is near a kilohertz, so a measurement wider than a few kilohertz collects all of it and nothing more. The penalty is therefore a ratio of a constant to something that grows.

In a one-kilohertz band the source is worth 178.8 nV and the ten-to-one probe 35.33 µV: a penalty of 197.6. In ten kilohertz, 94.6. In a hundred kilohertz, 31.3. In a megahertz, 12.2, and it would keep falling. So the attenuator is least affordable exactly where a slow, careful, narrow-band measurement is being made — which is where a small signal is usually looked at, and where the quadrature reasoning about bandwidth in the instrument’s own rise time has nothing to say, because that page is about how fast the front end is and this is about how quiet it is.

The habit worth taking from it is that a noise penalty quoted without a bandwidth is not a number. The same pairing that the bandwidth noise sees insists on for a filter’s corner applies here to an instrument’s: the band is not a detail of the measurement, it is half of the answer.

The second axis, which is amplitude #

Two things decide which probe to use and the field has only ever drawn one of them. The source impedance decides what each ratio buys. The amplitude decides which ratios are available, because the attenuation divides the signal and $\sqrt{n(n-1)kT/C}$ does not divide with it.

The curve is the answer to a question that is usually settled by habit: how much bandwidth a measurement can have. It is not set by the instrument’s specification and it is not set only by the node’s impedance. On this node, at this resolution, the bandwidth is bought with amplitude, and the exchange rate is 2.52 decades of signal for 2.01 decades of frequency rather than one for one, because the noise per unit of ratio rises from 1.782 µV to 5.844 µV across the sweep.

At a tenth of the resolution the floor drops to 17.82 µV and the hundred-to-one probe becomes usable at 5.847 mV; at three hundred to one the floor is 534.5 µV and a hundred-to-one probe needs 175.4 mV. The trade is the same shape at each and the whole staircase slides along its axis, which is what a resolution requirement does to a dynamic range — a floor and a ceiling is the general form, with the floor here belonging to the instrument rather than to the circuit.

The node that is a capacitance #

There is a third case and it is the one where the field’s whole vocabulary stops applying. Everything above treats the node as a resistance, so the probe’s capacitance makes a pole and the error grows without limit above it. A node that is itself a capacitance behaves completely differently, and the netlist has been able to build one from the beginning.

The consequence is a change of kind rather than of size. On a resistive node the error is a function of frequency and cannot be corrected by scaling; on a capacitive one it is a fixed fraction at every frequency, which is a gain error and can be calibrated away by a single multiplication. A reading that is 25.84 per cent low everywhere is a worse-looking number and a better measurement than one that is one per cent low at 69.2 kHz and forty per cent low a decade above it.

That is the node where the trouble is at the input describes, where a photodiode’s junction capacitance is not a parasitic but the price of its area — and it is the case in which a probe is at its least harmful, for a reason nothing about a probe’s specification would suggest.

A quantity that ceases to exist is the more useful failure of the two, because it cannot be quoted by mistake. The reason the ceiling falls with the node’s capacitance is arithmetic — a tip of 11.5 pF is a smaller share of 3.3 nF than of 33 pF — and the reason it matters is that the ratio of the two capacitances, rather than any frequency, is the model’s range here.

What it does not say #

It does not say the ten-to-one probe is a poor choice. On a signal of a few hundred millivolts the noise penalty is invisible against the instrument’s own quantisation, the bandwidth is genuinely ten times better, and the divider is the reason a probe can be used on a node at all — a divider with two ratios shows what it costs in adjustment rather than in signal, and that cost is a trimmer setting held to a few per cent.

It also does not say the noise figure here is the whole instrument’s. The model contains the source, the probe and nothing else: the oscilloscope’s own input-referred noise is not in it, and on a real front end that is a comparable contribution which would be multiplied by the same $n$. Adding it would move every number on the noise axis up and would not change the shape, because it enters exactly where the divider’s noise does.

And the ground lead is still absent, as it is in the rung below. A fifteen-centimetre return wire is of the order of a hundred and fifty nanohenries in series with the measurement, and none of these figures contains an inductance. Every bandwidth quoted above is therefore optimistic by whatever that resonance contributes, and the honest statement is that the boundary computed here is the loading boundary and not the only one.

Where the same trade appears without a probe #

The shape of this trade — divide before the noisy thing and pay in signal, or do not divide and pay in loading — is not about probes, and two other pages in this collection meet it.

Two terminals measure the leads as well is the version where the repair is structural: the offending resistance is moved into a path where it does not enter the answer, and the price is two more wires rather than a factor in signal. The ammeter that is a resistor is the version where the two errors point in opposite directions and the best arrangement sits at their geometric mean. This page is the version where the two costs point the same way and the trade is genuinely a trade, which is why it has an axis rather than an optimum.

The distinction decides what a design can do about it. A geometric-mean optimum can be reached: there is a shunt resistance at which the two errors are equal and no other value is better, so the answer is a number. A structural repair can be bought: two more conductors remove the error and nothing about the measurement is given up. Neither is available here. Bandwidth and resolution improve in opposite directions of the same parameter, so all a designer can do is choose a point on the curve, and choosing it needs both axes drawn — which is why the amplitude figure above is the one of the eight nearest to being advice. The other move is to change the node rather than the probe, since the whole curve slides in proportion to the source impedance: a follower ahead of the measurement point is worth more than any choice of ratio, and it is the same repair the millivolts in the wire reaches for about a shared return and rejects, because there the interfering current is in the conductor whether the instrument is there or not.

The number worth carrying #

A ten-to-one probe buys exactly ten in bandwidth and costs 31.3 in signal-to-noise ratio, in a hundred-kilohertz band on a two-kilohm source. In a one-kilohertz band it costs 197.6.

The habit that goes with it is one substitution. The sentence “a tenth of the signal is a tenth of the signal-to-noise ratio” treats an attenuator as a thing that only divides, and an attenuator built from resistors is a thing that divides and adds. Whenever the division happens before the noise source rather than after it, the two factors are not the same factor, and the way to find out which is which is to put every resistor in the netlist and drive them one at a time.