Self inductance vs loop area of inductance. How to combine them/which to pick in which case?

Question

When it comes to inductance in circuits, it is a rule of thumb to minimize the loop area. This implies a return path (i.e. two wires). There is also self inductance of a single wire.

I am not sure how they play together: Do I need to add them? Or do I need to choose one, depending on the geometry? If so, which one to pick?

Example: Voltage source to sink, connected by two 1mm diameter wires in parallel (3mm apart) at a distance of 1m (one wire is supply, the other is return/ground).

A calculator for Parallel Wire Inductance tells me the inductance of this configuration is 705nH.

Another calculator for Wire Self Inductance tells me that the inductance (self inductance) of each wire is 1.51uH.

Note that if the two wires are completely adjacent to each other (distance 1mm instead of 3mm), 705nH becomes 0H which makes sense because the loop area is zero. Do we still need to consider self inductance in this case?

What is the correct equivalent circuit? Do I need to consider the first inductance ("loop area") or the second one ("self inductances") or both?

^{simulate this circuit – Schematic created using CircuitLab}

Answer 1

There is no such thing as 'the inductance of a single straight wire'.

Inductance is exclusively a property of a current loop.

This a frequent misunderstanding propagated by the numerous online calculators that do suggest that one can associate a well defined inductance with a single straight current carrying wire. The magnetic field from a single straight current would extend infinitely far away and would require infinite energy, so inductance would be infinite as well.

In practice, the current must return somewhere, an only then, when current goes in a closed loop, loop inductance is well defined.

There is also the concept of net partial inductance. Here, the total loop inductance is distributed among all the segments of the loop in a sensible way. It is useful for reasoning what parts of a circuit contribute most to the loop inductance. But still, the net partial inductance of any component depends on the entire loop. Taking the wire segment out of the circuit would immediately invalidate its partial inductance.

Answer 2

What is the correct equivalent circuit? Do I need to consider the first inductance ("loop area") or the second one ("self inductances") or both?

Neither.

The correct equivalent circuit for two 1mm dia wires spaced 3mm (between centers) and 1m long is a transmission line, characteristic impedance 211ohms, delay 3.33ns.

To calculate the impedance, see online calculators like:

www.rfwireless-world.com/calculators/Twin-wire-line-calculator.html

or en.wikipedia.org/wiki/Twin-lead#Characteristic_impedance

If your frequencies are low so that 3.3ns is insignificant, then you can replace the transmission line with a lumped element equivalent, which is a series inductance of 703nH and a shunt capacitance of 15.8pF.

What are you trying to model?

Edit:

But why does your answer not include 2x 1.51uH from the self inductance

There is no 'self inductance', there are only two parallel lines for which there is an exact solution, for example see

www.ece.rutgers.edu/~orfanidi/ewa/ch11.pdf section 11.5

This exact solution includes all the magnetic fields and gives you the inductance exactly.

The 'self inductance' you calculated from that website quoted a formula from Grover which comes from a paper by Rosa g3ynh.info/zdocs/refs/NBS/Rosa1908.pdf which is an attempt to calculate some form of inductance for an isolated wire. Why would you think your should add it to an exact solution for the two wire problem?

but I am explicitly interested in the inductance aspect only

with respect, if you have an easily accessible exact solution you should use it, you may decide to ignore the capacitance, but you it never hurts to know it's value.

To summarise, at low frequencies and for impedances where you can ignore the capacitance, the equivalent circuit is a single 703nH inductor.

If you are modelling a single wire above a conducting plane, then this is a little different - if the centre of the conductor is 'h' above the plane, then solve the 2 wire problem for 2 conductors spaced at 2h and halve the characteristic impedance.

Answer 3

Note that if the two wires are completely adjacent to each other (distance 1mm instead of 3mm), 705nH becomes 0H which makes sense because the loop area is zero.

Impossible, because the distance is "between" the centers, so it must be > ~ 1.01 mm. They can't "touch" each other.

What is the correct equivalent circuit?

EDIT: and insert from paper Rosa1908 p311, for a square loop ...
Thanks to @Tesla23.

These wires have self-inductance, they are also "coupled" as a "transformer". No capacitive effect in simulation. If yes, then this "system" is a "transmission line".

Here is what happens when coupling varies. Good coupling, no self?

Corrected inductance (external+internal) calculated as this.
It is why we twist each wire ...