The resistor turned round

The resistor that is a helix found that where a film resistor’s capacitance goes decides whether a simple model of it survives. Capacitance between adjacent turns builds the three-element part exactly, to a part in a trillion. Capacitance from the turns to the part’s mounting does not: its current returns through only part of the film, and it counts for a third of itself at the terminal. That essay used the third as a fact to arrive at and moved on, and it ended by naming a bench measurement it could have made: mount a part with one end nearer a ground pour, and it has two flat resistances, one for each end grounded. Measuring the difference separates the two capacitances.

Following that up shows something less specific and more useful than a bench trick. A picofarad of stray capacitance on a resistor is not a quantity of capacitance in the sense that the schematic symbol implies. It is a distribution, and every reading of the part takes a different average of it. The average that a terminal measurement takes is not the one a transfer takes, and neither of them is the total. This essay drops the lead inductance and the helix, so that nothing is left but a film and its stray, and it measures how many different numbers one stray can be.

A resistor with a far end #

The arrangement is what an LCR meter makes of a resistor on a board. One lead goes to the meter’s driven terminal and the other to its guard or ground, and the part sits at some distance over a ground plane or a chassis. Its film is a resistance $R$, distributed along its length, and at every point $x$ along it, measured from the driven terminal, there is a small capacitance $c(x)\,dx$ to the mounting. The far end is at the same potential as the mounting, which is the important condition. Everything here is solved on 200 sections and checked against the exact line where one exists. The total stray $C = \int c\,dx$ is 1 pF. That is a generous value for a small resistor over a ground plane and a modest one for a large part, and the arithmetic scales in proportion.

The voltage along the film falls linearly from the terminal to the far end: at low frequency $v(x) = V(1 - x)$, since the stray’s current is much smaller than the film’s. So a section of stray at $x$ has a voltage $V(1 - x)$ across it and draws $j\omega c(x)\,V(1 - x)\,dx$. That current enters at the terminal, flows along the film to the section, leaves through the stray, and a charge-accounting argument gives the admittance at the terminal. By the energy stored in the capacitance, which is what a capacitance reading measures,

$$C_{in} = \int_0^1 c(x)\,(1 - x)^2\,dx.$$

The square appears because the stored energy goes as the voltage squared. It means a stray next to the terminal is seen whole, a stray at the middle is seen as a quarter of itself, and a stray next to the grounded end is not seen at all. For a uniform stray the integral is $C/3$, the helix essay’s third. For a straight taper $c(x) = C[1 + a(1 - 2x)]$, with $a = 1$ meaning all of it towards the terminal, the integral is $C(1/3 + a/6)$.

One part, two capacitances #

Turning the part round replaces $a$ with $-a$. Nothing about the part or its total stray has changed, and the capacitance read is now $C(1/3 - a/6)$. With $a = 0.6$ that is 0.433 pF one way and 0.233 pF the other. With $a = 1$, the extreme of a part whose stray tapers to nothing at one end, it is 0.500 and 0.167, a ratio of three. On a megohm at 10 kHz those appear as 1.56° of angle and 0.84°, and as the slider shows the angles scale with the taper while the resistance reads the same. A careful user who turns a suspect part round and gets a different phase has discovered something real about its mounting, and it is not a fault.

The solved value matches the first-order weight within two parts in a thousand at every taper, and the small residue is the section count. This is a good example of a formula doing more than the calculation it was derived to perform, because the weight is the entire result: the sections add nothing to it. It is also the same thing right in size, wrong in angle measured from the other side. That essay found that a resistor’s angle goes as the first power of frequency and its size as the second, so an angle is the first thing to leave. The same statement here is about which capacitance an angle is reading.

The weight, at every position #

To see the weight itself, reduce the stray to a single lump and walk it along the part.

The terminal’s curve is the square law: 1 pF at the terminal is 1 pF, 0.25 pF at the middle and nothing at the far end. But a second curve appears as soon as a different question is asked. A resistor is often not measured at its terminal. As the upper arm of a divider, or as a shunt feeding a virtual earth, the quantity of interest is the current that arrives at the far end. To first order, that current is reduced by an amount set by a different weight,

$$I_{out} = \frac{V}{R}\left(1 - j\omega R \int_0^1 c(x)\,x(1 - x)\,dx\right),$$

in which the charging current is taken from the film. The weight $x(1 - x)$ has its largest value, a quarter, at the middle of the part. It is zero at both ends, and that is intuitively clear. A stray at the terminal draws its current directly from the source, without any of it passing through the film, so the far end does not see it. A stray at the far end has no voltage across it. Only a stray in the middle takes current away from the film that would otherwise reach the far end, and its average over a uniform stray is $C/6$.

So there are at least three answers to how much capacitance a uniform resistor has: the $C$ a geometry calculation gives, the $C/3$ a meter gives at its terminal, and the $C/6$ a divider’s transfer sees. The resistor that is only a resistor drew a single capacitance across the part, and that drawing is honest about the between-turn capacitance and for no other part of the total. Whichever number a design uses, it should match how the part is used.

What two readings recover #

The bench measurement the helix essay named is then a straightforward calculation. Two readings of the part, one each way round, are $C(1/3 + a/6)$ and $C(1/3 - a/6)$.

Half their sum is $C/3$ for any taper, so a sum of two readings gives the total stray, which no single terminal reading can. Their difference is $Ca/3$, so it gives the taper. Both are exact to the second order for a straight taper, and that recovers two numbers about a part’s mounting from one component and one meter, without seeing the board. Two ports from two one-ports makes the corresponding point for a network: enough one-port readings can identify things a single reading cannot, and turning a part round is one more of them.

There is a limit, and it follows from the weights. A non-straight distribution cannot be recovered from two numbers. Two readings know two moments: $\int c(x)[(1 - x)^2 + x^2]\,dx$ and $\int c(x)(1 - 2x)\,dx$. A hump in the middle has the same sum as a slightly larger uniform stray, and no reading at the two ends tells them apart. That is an intrinsic limit of a terminal measurement and not something a better meter can overcome. The sum’s weight, $(1-x)^2 + x^2$, is a half at the middle and one at the ends, so a stray in the middle reads half as much in the sum as the same stray at an end, and two readings understate it. The stray that the transfer weight cares most about is the one the terminal readings report least accurately. To learn that, the part has to be measured as a transfer as well, which is the measurement a divider actually needs.

Two lumps, and a line that is neither #

All of this has been first order, and for a uniform stray there is a closed form to test it against. A uniform RC line’s terminal admittance is $Y_{in} = (\gamma/R)\coth\gamma$ with $\gamma = \sqrt{j\omega RC}$, and the transfer is $(\gamma/R)\,\mathrm{csch}\,\gamma$. These are exact at every frequency.

The lump of $C/3$ across the part agrees with the line to within a degree up to $\omega RC = 1.38$, where both read about 24°. That is a wide range. It is much the widest band any lumped model of this part has, and it is why “a third” is the right number for practical work at the terminal. The whole $C$ is out by 27.3° at $\omega RC = 1$, and it never agrees at any frequency above the one where it does not matter.

Far above that, the line and any lump part company in kind. A capacitor’s phase goes to 90°, and the line’s stops at 45° and stays there, because its admittance goes as $\sqrt{j\omega}$. There is no frequency at which a distributed stray behaves as a capacitor. The same half-order shape appeared in the time constant one trimmer cannot reach in connection with a resistive cable, and it is the signature of any line whose loss is not small. A real part’s lead inductance takes over long before that on low-value resistors, as the first essay on this part found. On a 10 MΩ part with a picofarad along it the half-order region begins near 100 kHz, and there the half-order behaviour is the part.

The corner a divider actually has #

The transfer weight’s practical consequence is large enough to quote on its own. A high-value resistor’s bandwidth, in a divider or in a current-to-voltage stage, is commonly estimated by putting the stray across it and calculating $1/2\pi RC$. That is wrong by nearly an order of magnitude.

The current reaching the far end of a uniform line is 3 dB down at $\omega RC = 9.18$, whereas a lump across the part would put it at 1, and the first-order weight of a sixth would put it at 6. So a 100 MΩ arm with 1 pF along it, a common enough upper arm for a high-voltage probe, has its real corner at 14.6 kHz where the lump says 1.59 kHz. Against the first-order weight the line is flatter than one pole and falls faster afterwards: it is 1 per cent down at $\omega RC = 1.35$, and beyond the corner it falls exponentially in $\sqrt{\omega}$. That is what a diffusion does.

A designer who adds a compensating capacitor to the arm has to know which of these they are compensating. A divider with two ratios showed the equation that makes a divider’s two ratios agree, and it is an equation for lumped parts. A trimmer across a 100 MΩ arm whose stray is distributed along it cannot make the arm’s response flat at all, because no single capacitor reproduces a coth. It can only match it over a band, which is the situation the time constant one trimmer cannot reach measured for a different distributed element, and high-voltage dividers handle it in the usual way, by surrounding the resistors with a metal screen at an intermediate voltage so that the stray to ground never develops in the first place. That guard is not a shield in the usual sense, since it does not reduce any capacitance. It sets the voltage on the stray.

Why a guard changes the weight #

The formula also explains what a guard ring is doing from a different direction. The weight is $(1 - x)^2$ because the voltage on the stray at $x$ is $(1 - x)$. A guard that holds the mounting not at ground but at a potential following the film along its length removes the voltage from every section, so the stray draws no current at all, whatever its value. The guard that is part of the return analyses the same idea at an amplifier’s input. A conductor driven to the film’s own voltage, or graded along its length with its own divider, removes both weights without removing a picofarad. It is the only cure on this page that does not need any capacitance to be reduced, and it also ends the difference between the two readings.

A partial version is common and cheap. Mounting a high-value part with its driven end away from any ground plane, so that the taper’s sign makes the stray heaviest where the voltage is least, minimises the terminal reading. The figure of a third is then a pessimistic estimate, and the same part mounted the other way round is pessimistic in the other direction. In a precision high-value divider, part orientation over the board is a design variable, which is an odd thing to say about a two-terminal part with no polarity.

What this model leaves out #

The film here is uniform, and a helix’s capacitance has a between-turn component that would add to this one exactly as the helix essay measured. The part’s resistance is a constant, with no voltage coefficient, which high-value resistors have enough of to change the voltage distribution along them. Lead inductance is dropped, and that limits this to parts well above $\sqrt{L/C}$: at the megohms and tens of megohms where a stray to a mounting is what matters, the inductance is many decades out of the way. And the mounting is treated as a single perfect ground, so that a stray’s current returns with no impedance of its own. On a real board that return shares a path with something else, which is a concern of the source that is not a source, and a part mounted over a split plane is no longer described by one taper.

Dielectric loss in the stray is also absent, and it is large for the materials a resistor is usually mounted over. A lossy stray adds a part in phase with the current, which the angle cannot distinguish from a smaller value of resistance. That is one reason a very high-value resistor’s apparent resistance falls with frequency in ways the film does not explain.

The numbers worth keeping #

A stray from a film to its mounting counts as $(1 - x)^2$ of itself at the terminal and as $x(1 - x)$ in the current through the part. A uniform stray is therefore $C/3$ and $C/6$. A tapered 1 pF on a megohm reads 0.433 pF one way round and 0.233 pF the other, half the sum of two readings is a third of the total stray whatever the taper, and their difference is the taper. $C/3$ across the part is a one-degree model up to $\omega RC = 1.38$; the line’s phase then settles at 45° and never reaches a capacitor’s. The current at the far end is 3 dB down at $\omega RC = 9.18$, which puts a 100 MΩ arm with a picofarad along it at 14.6 kHz, not at 1.59.

Still open: coupled turns, the mixed stray, and a shunt’s own heat #

Turns that couple. With mutual inductance between adjacent turns, the helix’s inductance is distributed as a coupled structure rather than as series pieces. The same kind of weighting question applies to it, and the answer may not be a square law, since a mutual couples each section to all of its neighbours. Whether the coupled helix still reproduces the three-element part at any stray split is the helix essay’s first question, and it is still unanswered.

A shunt whose sense loop shares its current path. A four-terminal shunt’s phase error is set by the mutual inductance between its current path and its sense loop. By analogy with this essay’s weights, a sense connection partway along the part should weigh the inductance by where it is attached, and the same kind of formula should exist. It would put a phase error directly on the placement of two sense vias.

A resistance that moves with its own current. A shunt carrying a large current warms, and its resistance and time constant follow, so its phase error is a function of the current being measured. That second nonlinearity would be measured against the thermal fixed points already solved for other parts, and the question is how large a current has to be before the part’s angle is no longer a property of the part.