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