Physical properties of transformers

Question

In class, we were given equations for the transformer in terms of number of turns primary/secondary and coupling coefficient K, however effect of distance between coils, presence/absence of core, and composition of core were not covered.

Of particular interest: What is the equation for 2 parallel straight wires? Clearly turn ratio is 1:1, but how to account for distance between the two?

Answer 1

About the core

Consider an inductor, and the magnitic field that it generates:

it creates loops all around the windings, with a direction given by the direction of the current (see Ampère's law).

If you put two inductors close to each other, the changing ^{(Credits to Curd)} field generated by one will induce (sorry for the pun) a current in the other one, with the proportion given by the number of windings (because of Faraday's law of induction). But this coupling will be limited to the portion of the field which falls in the area of the second inductor, which can or cannot be a limited portion of the total.

Using a core, you force whe magnetic flux over a closed path, and the great part of it will follow that path:

This translates in a higher efficiency of the coupling, as almost all the magnetic field is induced in the secondary, as opposed to the previous case.

What happens with two wires?

If you have a single ideal wire, it also generates a magnetic field, again described by Ampère's law:

Since this field is distributed through all the wire, it will be much weaker than using windings, because these have the effect to concentrate this field into the inner space (where is the core).

As for inductors, distance reduces the portion of the magnetic field that the wires share, and with it the power transferred. Note that with the core, in the ideal case in which all the flux is convoyed the distance doesn't matter, as all the flux is into the core.

Answer 2

Imagine 2 water hoses.

One with water pointed to another hose to catch the water.

Without efficient coupling, you have a leaky hose and none is transferred. In that case the primary impedance is just that of the primary hose.

Knowing you are an engineering student with a brilliant imagination, and you know how iron filings collect around a wire to show the magnetic flux, you need to use laminated iron steel to couple the high power low frequency power to the secondary wire, or a ceramic ferrite core for higher frequencies that have smaller size and less loss than iron. The higher the F, the lower the permeability that can be used to become low loss.

But 50/60Hz can use big fat hoses or high µ magnetic common cores to join the primary secondary wires.

There are many shapes and sizes depending on whether you need a fire hose or a garden hose or a sewer hose.

But if you own a soldering gun, you have a transformer with many primary turns and only 1 secondary turn that will heat the heavy copper tip enough to melt solder in a second.

Now if this analogy is not clear, you could be a programmer. ;)

Answer 3

With no core, distance decreases the coupling inductance and increases the leakage inductance. In the limiting case of two wires far apart, they each have only their inherent inductance, which is all leakage inductance if you view them together as a transformer.

In a perfect transformer, the leakage inductance is zero. If you were to put a resistor accross the secondary, the primary would appear to be just that resistance divided by the square of the turns ratio. For example, if you had a perfect 1:2 transformer and put 1 kΩ accross the secondary, the primary would look like a 250 Ω resistor and nothing else.

As coupling is decreased, such as by increasing distance, more of the primary's impedance comes from leakage inductance and the series component that is the resistor accross the secondary reflected to the primary gets less and less. Eventually at no coupling you have only the leakage inductance, which is the primary acting as a stand-alone inductor.

Answer 4

From Grover's Inductance Calculations Working Formulas and Tables, pg 31, the mutual (coupling) inductance betwen two equal parallel straight filaments is

\$M=0.002l \left[log_e\left(\frac{l}{d} + \sqrt{1 + \frac{l^2}{d^2}} \right)- \sqrt{1+\frac{d^2}{l^2}} + \frac{d}{l}\right]\$

\$l\$ is the length of the wire and \$d\$ is the distance between the wires.

In this expression the dimensions \$l\$ and \$d\$ are centimeters and the inductance is microhenrys.