Inductorsand Magnetic fields

BITX20 bidirectional SSB transceiver

BITX20 bidirectional SSB transceiver

LO BFO

Mic

Mixer

MixerIF FilterRF Filter

Antenna

Transmit direction shown

The Colpitts oscillator

See the BITX20 circuit

LO: Local Oscillator

BFO: Beat frequency Oscillator

R3

R2

R1

+12 V

0 V

C1

C2

L1

C3

Discharge of an Inductor

R

+10 A

0 V

Switch I

L

Graph of inductor discharge from 10AR=1 Ohm, L=1 Henry

0.5 1 1.5 2Seconds

2

4

6

8

10Amps

0.5 1 1.5 2

Seconds

5

10

15

20

25

Amps

The same discharge from 27.18A

10*2.718

10/2.718

10

e=2.718

Exponential decay

The decay time constant = L / R

If R is in Ohms and L in Henries the time is in seconds

Every time constant the voltage decays by the ratio of 2.718

This keeps on happening (till its lost in the noise)

This ratio 2.718 is called “e”.

Exponential decay

It’s a smooth curve. We can work out the current at any moment.

The current at any time t is: I = I0 / e

(t*R/L)

I0 is the current at time zero.

t*R/L is the fractional number of decay time constants

For e( ) you can use the ex key on your calculator

Fields

• Electric fields– Capacitors

• Magnetic Fields– Inductors

• Electromagnetic (EM) fields– Radio waves– Antennas– Cables

Construction of inductors

www.germes-online.com

Key to diagrams

Red rectangle = Outline of a Coil

Blue Rectangle = Outline of a Core

Red shading = Positive value

Blue shading = Negative value

Stronger shading is more positive / negative

An air cored coil

Magnetic potential in air

Magnetic potential is measured in Amps!

One often talks about Ampere turns but what counts is the total amps round a closed circuit.

The magnetic potential between 2 points on an iron bar is equal to the current in a loop round the bar between those points

A coil on an iron bar

Magnetic potential with an iron bar

A coil on a closed iron core

Magnetic potential for the closed core

Field strength H X component => Y component =

>

Flux density B X component => Y component =

>

Magnetic field strength H is measured in Amps per metre

Since magnetic potential is in amps the field strength H must be in amps per metre.

Magnetic flux density B ismeasured in Webers per square

metre

(Or Tesla)

Permeability

• Magnetic field strength H (Amps/Metre)

• Magnetic flux density B (Webers/m2)

• B= μ * H (like Ohms law but for magnetics )

• Permeability μ = μ0 * μr

• μ0 is 4 Pi*10-7 Henries per Metre (by definition of the Amp)

Induced VoltagesA moving magnet near a coil of wire will induce a voltage in the coil. This is due to the varying magnetic flux through the coil not the motion itself.

The voltage will be:

• Voltage = Magnetic flux change per second times number of turns in the coil.

• We can calculate the magnetic flux (in Webers) from the flux density B and the area.

Inductance

When a current flows round a coil it produces a magnetic field.

The magnetic field H produces a magnetic flux density B.

Some or all of the flux (in Webbers) passes through the coil.

If the current is varying then the magnetic flux varies.

The varying magnetic flux causes a back EMF in the coil.

We can calculate the inductance from the geometry and the permeability before making the coil.

Inductance of a toroid

Toroids are the easiest to calculate since one can assume that their magnetic flux is uniform and only passes round the core.

Magnetic field strength H = Amps * turns / circumference

Magnetic flux density B = H * permeability

Magnetic flux = B*cross section of toroid.

Induced voltage = turns * Magnetic flux /second

So induced voltage = (Amps /second)*turns*cross section* permeability* turns/circumference

Inductance of a toroid

So for a Toroid (from previous slide):

Induced voltage = (Amps /second)*turns*cross section* permeability* turns/circumference

But for any Inductor:

Induced voltage = (Amps/second)*Inductance

So for any Toroid:

Inductance = turns*turns*cross section* permeability /circumference

A real toroid exampleFor a T37-2 toroid (all dimensions must be in metres)

Mean circumference = 22.87*10-3 metres

Cross section = 6.4*10-6 square metres

Relative permeability = 10

So the permeability is = 12.57 * 10-6

So inductance = Turns squared * 3.51*10-9 Henries

Or Turns squared * 3.51 nano Henries

The manufacturers quote a value of: Turns squared * 4.3nH

What approximations did we make?

A T37 toroid has an inner diameter of 5.21 mm and an outer of 9.35 mm Almost a 2:1 ratio.

We assumed the flux was uniform across the cross section. In fact it will be almost double on the inner surface due to the higher magnetic field strength on the shorter path.

We assumed the flux in the air was negligible. However this core has a relative permeability of only 10 so the flux in the air could be significant. (However by symmetry it should be small if the coil is wound evenly)

Questions