# Does a Mobius resistor have zero inductance? How would you calculate the inductance?

Wikipedia describes a Möbius resistor as follows, and the Patent for this device gives a similar description.

A Möbius resistor is an electrical component made up of two conductive surfaces separated by a dielectric material, twisted 180° and connected to form a Möbius strip. As with the Möbius strip, once the Möbius resistor is connected up in this way it effectively has only one side and one continuous surface. Its connectors are attached at the same point on the circumference but on opposite surfaces. It provides a resistor that has no residual self-inductance, meaning that it can resist the flow of electricity without causing magnetic interference at the same time.

(a) Does such a Mobius resistor really have “no residual self-inductance” as claimed?

(b) If it does have a definite non-zero value of inductance, then how is this calculated?

(c) Are the claimed advantages of this resistor because it is constructed as a Mobius strip?

• Looks like a short circuit to me
– hft
Sep 1, 2022 at 2:49
• @hft The bulk of the device is non-conductive; only its surface is. Sep 1, 2022 at 14:00
• but there is only one surface so the + and - are directly connected by the conductor.
– hft
Sep 1, 2022 at 17:10
• This looks so gimmicky! Sep 1, 2022 at 17:30
• @hft yes, but the "conductor" has a resistivity that gives us the resistance we want between the two points. The 1966 patent properly explains that, Wikipedia leaves it implicit. Sep 1, 2022 at 17:50

The inductance can be calculated, but it is first necessary to look at the behavior at very fast timescales of a ns or so. Clearly the two faces of the strip form a transmission line and so, at short timescales, the resistor appears as two transmission lines in parallel. At short timescales, each line will look like a resistor of value equal to the characteristic impedance, Z, of the line. So at short time scales, this device looks like a resistor of value Z/2. But for typical construction of such a resistor with loop diameter of say 30 mm, this initial behavior will be gone in a ns or so, and then the resistance will be set by the resistivity, width and thickness and length of the conductive strips. I’m sure that when the inventor claims “no inductance”, then he means on time scales of more than a ns, after which the input to the device no longer looks like the input to a transmission line.

OK, so how might we calculate the inductance, valid for time scales greater than the transit time around the loop?

Referring to Fig1, which is a plan view, we see 2 identical loops in parallel, colored black and green, with the twist at the bottom of the sketch, near point P. The loops are identical, but the current travels in different directions around the loops. Thus the magnetic fields (almost) cancel, and the device looks (almost) non-inductive. One description is that the resistor is two anti-phased inductors wired in parallel, with a coupling coefficient almost equal to one, so that nearly all of the flux created by one coil passes through through the other. That’s fine as a qualitative description, but I asked how the inductance could be calculated. Perhaps there are a number of ways, but I offer the following.

Refer to point P on the Fig 1, where the inner and outer conductors “cross over” due to the twist. It is assumed that the length of twisted section is small compared to the overall length of the loop. Symmetry tells us that this point, on both conductors, is always at a potential equal to half of the voltage applied to the resistor. As P is always at the same voltage (half of applied voltage) on both conductors, then it follows that we could electrically connect the two loops together at this point, and this would have no effect on the operation of the circuit.

Fig 2 shows the equivalent circuit thus created by connecting the loops together at this point. Note that the small, horizontal link at P in Fig2 is not actually doing anything. From symmetry considerations there will be no current through this link – if there was then in which direction would the current be? And as there is no current through this link, then it can be removed without changing the operation of the circuit.

In Fig 3, this small link has thus been removed. Also, the two long, thin loops have been straightened, which will also not affect the operation of the circuit, noting that the Bfield from these thin loops is very localized in the small gap between conductors, so the field from one does not interfere with the other (or with anything else nearby), so we are free to straighten or bend these loops as we choose without affecting the operation of the circuit.

So we end up with the equivalent circuit of Fig3, where the resistor input terminals are still connected to two loops in parallel, but now the loops in question are long and very thin, magnetically separate, and their inductance is easily calculated with application of Ampere’s Law.

Fig 4 illustrates the method for finding the inductance of one of these long, thin loops. Provided the strip width and length are large compared to the separation, then to a good approximation, the Bfield is strong and uniform between the conductors, and zero everywhere else. Amperes Law states :-

$$\oint B.dl = \mu_0 I$$

Where Integral $$\oint B.dl$$ is a line integral is around a closed loop, $$I$$ is the current passing through the loop, and $$\mu_0$$ is magnetic permeability of free space

The line integral is easily calculated, because there is a constant field B along the bottom of the rectangular integration path, the vertical legs of the path are of negligible length, and the field is zero along the top of the path.

$$\oint B.dl = \mu_0 I$$

$$BW = \mu_0 I$$

$$B=\frac{\mu_0 I}{W}$$

Total flux through the loop = $$\Phi = BA = BTC = \frac{\mu_0 ITC}{W}$$

where $$T$$ is conductor separation and $$C$$ is loop length, in this case half of the total Mobius loop length.

Finally,

$$L \text{(Henry)} = \frac{\Phi}{I}$$ (can be found in any textbook)

$$L = \frac{\mu_0 TC}{W}$$ (for one loop)

$$\bf L_\text{mobius} = \frac{\mu_0 TC}{(2W)}$$ (because there are 2 identical loops in parallel)

How beautifully simple. So to minimize the inductance, you need small strip separation, small length of Mobius loop, and a wide strip. The Bfield from these narrow loops is very localized in the small gap between conductors, so the field from one does not interfere with the other (or with anything else nearby) and the Mobius loop can thus be bent into an ellipse or other shape without affecting the inductance seen between the terminals.

OK. So the formula shows that the inductance is not inherently zero as claimed, but just how small will it be for typical construction? Assume the following dimensions.

$$T = 0.05 mm = 0.05\times 10^{-3} m$$

$$C = 47 mm = 0.047 m$$ (corresponds to D=30mm)

$$W = 10 mm = 0.01 m$$

$$\mu_0 = 1.26\times 10^{-6}$$ (for free space, as no magnetic materials are present)

.

$$L = uTC/(2W)$$

$$L = 1.48\times 10^{-10}$$ Henry = 0.148 nH

That is a very low inductance to be sure, ideal for fast current sensing. To exploit such low inductance will require Kelvin sense terminals (4-wire resistor) which is easily accomplished, otherwise the inductance of the lead-in wires will far exceed that of the resistor itself. To put the value of 0.15 nH in perspective, the package inductance of the mosfet source connection on a TO220 power semiconductor package is about 5 nH, measured 6 mm from the die, so 0.15 nH is an absurdly small value of inductance for an electronic component.

I will later add more text to show a better method of constructing low-inductance resistors, that I have been using for decades with good results.

• Nice explanation! I suppose the next thing to look at would be the parasitic capacitance. I suppose you could get at that from the transmission line approach, but another way might be to look at the surface charge distribution. Sep 1, 2022 at 15:02
• This reminds me of the technique of making wire wound resistors by winding wire doubled back on itself (i.e. "fold" the wire in two, then wind that) Sep 3, 2022 at 7:21

To answer the third question :-

(c) Are the claimed advantages of this resistor because it is constructed as a Möbius strip?

The fact that the resistor is made from a Möbius strip certainly puts an interesting twist on the problem, but is ultimately just a gimmick. The 180 degree twist plays no part in producing the low inductance. The low inductance is as a result of having a pair of wide, thin conducting strips, separated by a small distance, with the current flowing in opposite direction on each strip. This form of construction provides near perfect cancellation of magnetic fields, which in turn is responsible for the low inductance, while the 180 degrees Möbius twist is an irrelevant distraction.

There is a much better way of producing low-inductance resistors that I have been making and using for decades with good results, and that is shown in Fig 3 and Fig4, where there is no pesky 180 degree Möbius twist, but the low inductance performance is identical.

Personally I build these non-Möbius resistors with only a single folded loop, that is, just one of the two parallel connected loops shown in Fig 3, as in the single loop of Fig 4. Yes, you can half the inductance with the 2 loops, but you can achieve the same result more simply by halving the length of a single loop. I use very thin (as thin as 0.025 mm) brass shim for the conductive strips, separated by a single layer of Mylar or Kapton insulating film. If anything, the resistance tends to be lower than you would wish without making the loop length excessively long, in which case using 2 parallel loops only makes that problem worse. So unless you need a very low resistance indeed, a single folded loop is the better choice.

Thus freed from the 180 degree Möbius twist, the strip resistor as described above is far superior because it can be packaged much more compactly and elegantly. One elegant packaging solution is to wrap the strip around the circumference of a short, solid metal cylinder (eg D=25mm, L=10mm) made from copper or aluminium. The strip can be mechanically and thermally bonded to the outside surface of the cylinder, with thin insulating film between of course, providing excellent heat conduction from the heat-dissipating resistive strip to the solid metal cylinder. By choosing appropriate strip thickness, strip width, and cylinder diameter, it is easy to arrange for the strip length to be just less than the cylinder circumference. To dissipate more heat, the end face of the solid metal cylinder is screwed to the face of an off-the-shelf finned heat-sink.

Or a planar package can be arranged by clamping the folded strip between 2 rectangular copper plates, thus providing an efficient thermal path from both sides of the heat-dissipating strip. The plates can be screwed together, glue-bonded together, or even soldered together. With both forms of packaging, the very close proximity of the highly conductive metal cylinder or plates does not significantly degrade the fast-performance of the resistor, because the magnetic field is almost entirely between the conductive strips. I have used both packaging methods with good results, for making ultra-low-inductance, high-current, current sense resistors.

Summary. The Möbius strip construction does make for a low inductance, but not because of the 180 degree Mobius twist, which is a useless gimmick that makes efficient packaging difficult.

• You mean the resistance tends to be lower than you would wish? If the resistance is too high, two parallel loops would halve it. Sep 2, 2022 at 15:49
• Yes, what I meant to say is that the resistance tends to be lower than you would like, even if I carelessly said the opposite :) Thank you. I have edited accordingly. Sep 2, 2022 at 21:28
• The Moebius stripe has only one face! How can you say " Clearly the two faces of the strip form a transmission line " ? You (and Wiki) both draw currents but how do you know these currents are anything real? Where did the two opposite currents go in fig 4? Sep 2, 2022 at 23:59
• @Shaktyai At the connection point, or actually at any point on the loop, there are two conductive strips, separated by a thin insulating film, and these two strips form a transmission line. Why would the currents not be "real"? As per Fig 1, it is a very simple circuit comprising 2 loops in parallel. In Fig4, only one of the 2 long, thin loops are drawn, for the purpose of calculating the inductance of a single loop. The inductance of both loops in parallel is of course half of that. Sep 3, 2022 at 11:21
• @Shaktyai Because of the symmetry, if the twist is exactly opposite the injection point, no current flows there, so we can cut the strip there and it has no effect. (What if the twist isn't exactly opposite the injection point? Surely we can just move it until it is, but idk how to prove this has no effect on the current flow) Sep 3, 2022 at 14:28