The resistor that is a helix

The resistor that is right in size and wrong in angle took the three-element model of a resistor — its resistance, eight nanohenries of lead inductance in series, 0.4 pF of end-to-end capacitance across the film — and found that its angle leaves before its size does, at every value. One number sets the angle’s edge, the time constant $L/R - RC$, and at one resistance that number is zero: $R = \sqrt{L/C}$, 141.42 Ω, where the impedance’s angle has no first-order term at all and rises only as the cube of frequency. That is the resistance to choose where phase matters, a current shunt’s sense resistor or an attenuator’s arms.

That essay ended with the doubt that a three-element model invites. The capacitance of a real film resistor is not across its ends. A film resistor is trimmed by cutting a spiral groove into a cylinder of film, so the resistance runs along a helix, and the capacitance is spread along the cut: between each turn and the next, and between the turns and whatever the part is mounted over. Near $\sqrt{L/C}$, the one place where the angle is small enough for the distinction to show, a distributed part and a lumped one should disagree. The question is whether $\sqrt{L/C}$ is still the answer, and it turns out to depend entirely on where the capacitance goes.

Where the capacitance goes #

The helix here has $N$ turns between a terminal and ground. Each turn carries its share of the eight nanohenries and its share of the film, and the 0.4 pF is split in a stated proportion between two places: across each turn’s own film — the capacitance between one turn and the next — and from the middle of each turn to ground, the capacitance to the board or chassis under the part. The totals are always the three-element part’s. Only the split and the distribution change.

At 141.4 Ω the three-element part is flat: its angle at 100 MHz is 0.0027°, the cube law’s residue. A helix with the same eight nanohenries and the same 0.4 pF, all of it to its mounting, is 1.37° out at the same frequency — five hundred times more — and its angle rises in proportion to frequency, as if $\sqrt{L/C}$ had never been special. The slider on the figure at the head of the page moves the capacitance from the turns to the mounting in quarters: a quarter to the mounting gives 0.34°, a half 0.67°. And with none to the mounting the two curves are one curve.

Between the turns, nothing changes #

The last case is the surprise, and the reason is short enough to write down. A turn of the helix is a piece of inductance $L/N$ in series with a piece of film $R/N$, and the capacitance between that turn and the next bridges the piece of film. Its value, if the whole part has $C$ between its ends, is $NC$ per turn, since $N$ equal capacitances in series make $C$. So one turn’s impedance is

$$\frac{j\omega L}{N} + \frac{R/N}{1 + j\omega (R/N)(NC)} = \frac{1}{N}\left(j\omega L + \frac{R}{1 + j\omega R C}\right),$$

and $N$ turns in series are the three-element part, exactly, at every frequency.

The solves confirm it to rounding, from 1.8 × 10⁻¹⁵ at one turn to 7.8 × 10⁻¹³ at sixty-four, over three resistances and five decades of frequency. So distributing a capacitance along the film is not in itself a departure from the lumped model. Capacitance between adjacent turns is capacitance across the film however finely it is cut up, and a helix whose only capacitance is between its turns has its angle flattest at $\sqrt{L/C}$ exactly. The three-element model is not an approximation to such a part; it is the part.

To the mounting, a third #

Capacitance to the mounting is different in kind. It does not bridge the film; it hangs from a point partway along it to ground, and the current it draws flows in through the part of the film between the terminal and that point and not through the rest.

The flattest resistance moves with the split. Bisected on the measured slope of the angle, it runs from 141.4 Ω with none of the capacitance to the mounting to 245 Ω with all of it — $\sqrt3$ times $\sqrt{L/C}$ — and between, it follows $\sqrt{L/(C_{across} + C_{ground}/3)}$ to a part in a thousand. The capacitance to the mounting counts a third.

The third is a weighted average, and the next figure measures the weights. A shunt admittance $Y$ hung from a point with resistance $R_u$ above it and $R_d$ below changes the two-terminal impedance by $-R_d^2 Y$ to first order: the only part of it that shows at the terminal is the part the downstream resistance puts a voltage across, and that voltage is proportional to $R_d$. Along a uniform film, $R_d/R$ runs from one at the driven end to zero at the grounded end, and the average of its square is a third.

Read from the angle’s slope, the weight is 0.250 for one turn — the single ground capacitance sits at the middle of the film, where $R_d = R/2$ — then 0.3125, 0.3241 and 0.3281 for two, three and four, and $(4N^2 - 1)/(12N^2)$ in general, which is a third less $1/(12N^2)$. The measured weights match the sum to two parts in a thousand. Five turns are enough to reach the continuous limit to a per cent.

So the characteristic impedance of the lumped model is the answer for a distributed part only when the capacitance is between turns. When it is to the mounting, the answer is the characteristic impedance with a third of that capacitance, and for a real part, which has both, the flattest resistance is $\sqrt{L/(C_{turn} + C_{mount}/3)}$ — larger than the lumped $\sqrt{L/C}$ by up to $\sqrt3$.

Which end is grounded #

The third has a condition hidden in it that is worth stating, because it is a property of the circuit rather than the part. It assumes one end of the resistor is at the return and the other is driven, so the film’s voltage runs from the whole to nothing. A resistor used as a shunt to ground, a termination or the lower arm of a divider is mounted that way, and the rule applies. A resistor floating between two nodes that both move — the upper arm of a divider, a feedback resistor — has a voltage along its film that runs between two nonzero values, and its capacitance to the mounting then loads both of those nodes as well as bridging the part. It is no longer a two-terminal element, and no single flat resistance describes it.

The practical reading is that $\sqrt{L/C}$ is a property of the part only for the capacitance that belongs to the part, the capacitance between its own turns. The capacitance to the mounting belongs to the part and the board together, and its effective value depends on which end the board grounds. The same resistor has a different flat resistance mounted as a termination and as a series element, and a data sheet’s single end-to-end capacitance cannot say which.

A divider’s two arms #

The condition about which end is grounded has a sharp consequence in the circuit where a resistor’s phase matters most, the compensated divider. A divider with two ratios found that a resistive divider with capacitance divides by resistance at direct current and by capacitance at high frequency, and that the two ratios agree only when each arm’s time constant is the same. An oscilloscope probe’s trimmer exists to enforce that single equation.

The lower arm of such a divider has one end at the return, so its capacitance to the mounting counts a third of its value across the arm. The upper arm floats between the input and the tap, and its capacitance to the mounting does something else entirely: it loads the tap, where it adds directly to the lower arm’s capacitance, and it loads the input, where it is simply a load. So the same distributed capacitance enters the compensation equation with a weight of a third across one arm, and in the other as a capacitance from the tap to ground whose weight depends on where along the arm each piece of it sits — the square-of-position rule above, measured from the other end. A scope probe’s tip resistor, a floating upper arm of several megohms, is the case where the distinction is largest, and the probe is part of the circuit measured how much capacitance that end of a probe presents to the node it touches. A divider built from two identical resistors on the same board, compensated on the assumption that each is a three-element part, is compensated for a circuit it is not. The trimmer finds the right setting anyway, because it is adjusted against the square wave rather than calculated; the calculation is what this changes.

What a degree costs #

The angles here are small, and it is fair to ask what 1.37° at 100 MHz is worth. Used as a current shunt, a resistor’s angle is the phase error between the current and the voltage read across it — the angle between two of the arrows three voltages that close on one draws — and a power measurement multiplies the product of the two by the cosine of the angle between them. An error of $\delta$ in that angle changes a reading taken at a power factor $\cos\varphi$ by about $\delta\tan\varphi$: at a power factor of a half, 1.37° is 4.1 per cent of the reading, where the three-element part’s 0.0027° is eighty parts per million. The ammeter that is a resistor found the shunt’s other errors, its burden and its amplifier’s offset, both independent of frequency; this one grows in proportion to frequency, and a shunt chosen at $\sqrt{L/C}$ for its flatness has had its flatness removed by the board it sits on.

For an attenuator or a termination, where the reflection is what matters, the angle turns into a reactive part of the load. A 141 Ω termination with all of its capacitance to the mounting is not the matched, purely resistive load its value suggests, in the same way that the capacitor that is an inductor found a bypass capacitor turning into what it was not above a frequency its own geometry chose.

Where the third stops #

A weight of a third is a statement about the first order in frequency, which is the order the angle’s flatness is decided at. It is not a claim that the helix is a lumped part with a third of the capacitance at every frequency.

At 245 Ω, the flat resistance for a part whose capacitance all goes to its mounting, the helix and the three-element part with a third of the capacitance agree to a per cent below 300 MHz. They are ten per cent apart by 1.19 GHz, and above that the helix follows its own curve: it is a lossy line shorted at the far end, a series impedance and a shunt capacitance distributed together, and no lumped capacitance of any value follows it. That is the same statement the comparison of a ladder network with a line made in the time domain, arriving here from the other side: a line can be replaced by lumped elements up to a frequency and not beyond it.

The helix is the internal version of the ceiling the resistor that is only a resistor found for the package as a whole, where the body stops being small against a wavelength. There it was the part’s outside dimensions; here it is the film’s own inductance and capacitance, distributed along the cut, that stop being small against the frequency. They are the same physics seen from two ends of the part, and each sets a frequency above which three lumped elements describe nothing.

What a designer should take #

For a phase-critical resistor, the resistance to choose is $\sqrt{L/C}$ only if the part’s capacitance is between its own turns. If it is mounted over a ground plane with one end grounded, weight the capacitance to the plane by a third and choose $\sqrt{L/(C_{turn} + C_{mount}/3)}$; with all of it to the plane the answer is $\sqrt3$ larger. Measuring a part’s impedance with one end grounded and again with the other end grounded is the way to separate the two capacitances, since between-turn capacitance does not care which end is grounded and capacitance to the mounting does.

And treat any three-element model as valid to first order in frequency. It is exact for the capacitance between turns, correct with a weight of a third for the capacitance to the mounting, and wrong above the frequency at which the film becomes a line, which for a small film part is around a gigahertz.

How the numbers were obtained #

Each helix is a netlist of $N$ turns: an inductance of $L/N$, the turn’s film as two resistances of $R/2N$ with a node between them, a capacitance of $N$ times the between-turn share across the turn’s film, and the turn’s share of the mounting capacitance from its middle node to ground; the last turn ends at ground and the impedance is read at the first. The three-element part is its closed form, $j\omega L + R/(1 + j\omega RC)$, which the earlier essays checked against its netlist. Flat resistances are bisected in log resistance on the sign of the angle’s slope at 100 kHz, sixty iterations; effective capacitances are that slope’s departure from $L/R$. The impedances are solved at thirty to forty points a decade.

What it leaves out #

The mutual inductance between turns. Each turn’s inductance here is its own. Real turns of a helix couple magnetically, which is where most of a spiral resistor’s inductance comes from; coupled turns make the inductance distributed in a way that is not a sum of pieces, and whether the between-turn result still holds exactly with coupling is not tested.

Lead inductance at the ends. Every turn here carries its share of the inductance. In a leaded part much of it is in the leads, lumped at the two ends, and only the film’s own share is distributed. That does not change the first-order weights: a capacitance’s effect at the terminal depends on the resistance downstream of it, and an inductance in series contributes no resistance. It would change where the line behaviour starts, since a lumped lead inductance and a distributed film make a different structure from a uniform helix above a gigahertz.

Non-uniform cuts. A laser-trimmed film is cut to its value, so its pitch and its capacitances need not be uniform along it. The weight of a ground capacitance is the square of its position along the film, so capacitance near the driven end matters nine times as much as the same capacitance two-thirds of the way down.

The dielectric. The film sits on ceramic and is coated, and both have loss. A lossy capacitance adds an in-phase part to the angle that depends on the loss tangent rather than on $L/R - RC$, and near the flat resistance it would set the floor.

Still open: coupled turns, the end that is grounded, and the shunt’s sense loop #

Turns that couple. With mutual inductance between turns, the helix’s inductance is distributed as a coupled structure rather than as series pieces. Solving a helix with a coupling coefficient between adjacent turns would say whether the between-turn capacitance still reproduces the three-element part, or whether coupling makes the inductance itself carry a weight the lumped model misses.

The same part, turned round. A resistor whose capacitance to the mounting is not uniform — heavier near one end, as a part on its side or with one end nearer a ground pour — has two flat resistances, one for each end grounded. Measuring the difference is the bench test that separates the two capacitances, and computing it for a stated non-uniformity would say how large it is in practice.

The shunt whose inductance is a loop. A four-terminal current shunt’s phase error is set by the mutual inductance between its current path and its sense loop as much as by its own inductance, which makes its time constant a question of layout. That is the second of the earlier essay’s open questions, and the helix’s lesson — that a distributed parasitic weighs by where along the part it sits — should carry over to where along the shunt the sense loop is attached.