G0MDK
The Nature of ElectricityBy
Chuck Hobson g0mdk
G0MDK
INTRODUCTION
This “Nature of Electricity” presentation has been modified to include remarks normally used by a speaker as he proceeds through the slides. If you have any questions, you can email them to me and I will try to answer them. [email protected] (g zero mdk at tiscali.co.uk)
G0MDK
Elektrizität! Was ist es?
L'électricité! Qu'est-ce que c'est ?
Elettricità! Che cosa è esso?
Electricity! What is it?
Georg Simon Ohm 1787 - 1854
André Marie Ampère 1775 - 1836
Count Alessandro Volta
1745 - 1827
Michael Faraday 1791 - 1867
Joseph Henry 1797 – 1878
Nikola Tesla 1856 - 1943
Charles Augustin Coulomb 1736 - 1806
James Watt 1736-1819
CREDITS
G0MDK
Elektrizität! Was ist es?
L'électricité! Qu'est-ce que c'est ?
Elettricità! Che cosa è esso?
Electricity! What is it?
Georg Simon Ohm 1787 - 1854
André Marie Ampère 1775 - 1836
Count Alessandro Volta
1745 - 1827
Michael Faraday 1791 - 1867
Joseph Henry 1797 – 1878
Nikola Tesla 1856 - 1943
Charles Augustin Coulomb 1736 - 1806
James Watt 1736-1819
CREDITS
G0MDK
Elektrizität! Was ist es?
L'électricité! Qu'est-ce que c'est ?
Elettricità! Che cosa è esso?
Electricity! What is it?
Georg Simon Ohm 1787 - 1854
André Marie Ampère 1775 - 1836
Count Alessandro Volta
1745 - 1827
Michael Faraday 1791 - 1867
Joseph Henry 1797 – 1878
Nikola Tesla 1856 - 1943
Charles Augustin Coulomb 1736 - 1806
James Watt 1736-1819
CREDITS
G0MDK
Elektrizität! Was ist es?
L'électricité! Qu'est-ce que c'est ?
Elettricità! Che cosa è esso?
Electricity! What is it?
Georg Simon Ohm 1787 - 1854
André Marie Ampère 1775 - 1836
Count Alessandro Volta
1745 - 1827
Michael Faraday 1791 - 1867
Joseph Henry 1797 – 1878
Nikola Tesla 1856 - 1943
Charles Augustin Coulomb 1736 - 1806
James Watt 1736-1819
CREDITS
G0MDK
Elektrizität! Was ist es?
L'électricité! Qu'est-ce que c'est ?
Elettricità! Che cosa è esso?
Electricity! What is it?
Georg Simon Ohm 1787 - 1854
André Marie Ampère 1775 - 1836
Count Alessandro Volta
1745 - 1827
Michael Faraday 1791 - 1867
Joseph Henry 1797 – 1878
Nikola Tesla 1856 - 1943
Charles Augustin Coulomb 1736 - 1806
James Watt 1736-1819
CREDITS
G0MDK
Elektrizität! Was ist es?
L'électricité! Qu'est-ce que c'est ?
Elettricità! Che cosa è esso?
Electricity! What is it?
Georg Simon Ohm 1787 - 1854
André Marie Ampère 1775 - 1836
Count Alessandro Volta
1745 - 1827
Michael Faraday 1791 - 1867
Joseph Henry 1797 – 1878
Nikola Tesla 1856 - 1943
Charles Augustin Coulomb 1736 - 1806
James Watt 1736-1819
CREDITS
Note they all have units of
electricity named after them.
G0MDK
THE ELECTRON FOUND IN ALL MATTER
SOME PROPERTIES
• Radius < 10-15 metres
• Rest mass 9.1 × 10-28 grams
• Charge neg. 1.6 × 10-19 Coulombs
HOW SMALL? A thousand trillion electrons side by side measure 0.5m
HOW HEAVY? 1.2 thousand trillion trillion electrons weigh one gram
HOW POTENT? 6.25 million trillion electrons make a 1 Coulomb charge
One Coulomb flowing per second = one Ampere.
One gram of electrons contains 176,000,000 Coulombs of charge
HEART OF ELECTRICITY
G0MDKPUTTING THE ELECTRON TO WORK
Note: Ideal model used, Wires have zero resistance, light illuminates instantly and resistance is a fixed 100 ohms.
G0MDKPUTTING THE ELECTRON TO WORK
G0MDKPUTTING THE ELECTRON TO WORK
Note: If an oscilloscope and photo cell at the battery/SW end is triggered at SW closure, the photo cell & oscilloscope would see the light 6.67µs later.
G0MDK
A QUESTION
1. 3.3µs was arrived at using a Radar type calculation
2. Velocity x time results in distance travelled (v x t = d)
3. In Radar v = c, the speed of light- 300 million metres per seconds.
4. Radar range (c x t = d) 300 x 106ms-1 x 3.3 x 10-6s = 1000m (one way)
5. In our case we know d (1000m) and c, so we calculate t
or 3.3µs
Now consider current flow again
WHERE DID THE 3.3µs COME FROM?
s3.3x10s0.33x10ms3x10
m10
c
dt 65
18
3
G0MDKCURRENT FLOW
100
100
100
100
G0MDKCURRENT FLOW
100
100
100
100
G0MDKCURRENT FLOW
100
100
100
100
Current measures 1A in fig 3. What about the current in figures 1 & 2?
G0MDKCURRENT FLOW
100
100
100
100
The battery doesn’t see the 100 ohm load in figures 1 & 2 All it sees is the characteristic impedance of the pair of wires In our example, this impedance is assumed to be 400 ohms.
G0MDK
ANOTHER QUESTION
1. No way Jose!
2. Why not?
Reason:
1. Electrons are particles with mass as previously stated
2. As particles approach c their masses increase enormously
3. This is in accordance with Einstein’s “Special Relativity”
4. This has been demonstrated at Cern and SLAC
5. Cern and SLAC use GeV’s to reach near c velocities
6. No particles including the electron have ever been accelerated to c
Can electrons travel through wire at the speed of light ( c ) where c = 300,000,000 metres per second?
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ELECTRON VELOCITY-1
ELECTRON VELOCITY IN A VACUUM TUBE
G0MDK
ELECTRON VELOCITY-1
ELECTRON VELOCITY IN A VACUUM TUBE
Formula from A-Level Physics
Calculation
x2x20001.759x10v 11
= 26.5 million metres per second (final velocity at the anode)
Let’s increase voltage on the Anode of the tube and calc. velocities
2Vm
ev
o
G0MDK
ELECTRON VELOCITY-1
Anode voltage (kilovolts)
Electron Velocity million metres/s
Velocity to speed of light ratio
8.00 53. 0.177 (12.5%)
16.00 75 0.250 (25%)
64.00 150 0.500 (50%)
128.0 212 0.701 (70%)
256.0 300 1.000 (100%)
512.0 424* 1.410 (141%)
SOMETHING WENT WRONG! Electrons CANNOT exceed c!
Increase in mass with velocity (relativistic mass) was not taken into account
G0MDK
ELECTRON VELOCITY-2
cv
1
1
When a mass velocity approaches the speed of light its mass increases
This is in accordance with Einstein’s theory “Special Relativity”
That is to say: relativistic mass (mr) = gamma ( ) times mo
Relativistic mass calculations are done using the following formulas:
or m x m #1
#3
#2
#4
where
2r v0.5meV
eV = electron volt is a unit of energy used in particle physics
2Vrm
ev velocity
2erm 2v
V
G0MDK
ELECTRON VELOCITY-2
cv
1
1
When a mass velocity approaches the speed of light its mass increases
This is in accordance with Einstein’s theory “Special Relativity”
That is to say: relativistic mass (mr) = gamma ( ) times mo
Relativistic mass calculations are done using the following formulas:
or m x m #1
#3
#2
#4
where
2r v0.5meV
eV = electron volt is a unit of energy used in particle physics
2Vrm
ev velocity
2erm 2v
V
G0MDK
ELECTRON VELOCITY-2
cv
1
1
When a mass velocity approaches the speed of light its mass increases
This is in accordance with Einstein’s theory “Special Relativity”
That is to say: relativistic mass (mr) = gamma ( ) times mo
Relativistic mass calculations are done using the following formulas:
or m x m #1
#3
#2
#4
where
2r v0.5meV
eV = electron volt is a unit of energy used in particle physics
2Vrm
ev velocity
2erm 2v
V
Known as the Lorentz Transform
G0MDK
ELECTRON VELOCITY-2
cv
1
1
When a mass velocity approaches the speed of light its mass increases
This is in accordance with Einstein’s theory “Special Relativity”
That is to say: relativistic mass (mr) = gamma ( ) times mo
Relativistic mass calculations are done using the following formulas:
or m x m #1
#3
#2
#4
where
2r v0.5meV
eV = electron volt is a unit of energy used in particle physics
2Vrm
ev velocity
2erm 2v
V
Known as the Lorentz Transform
G0MDK
ELECTRON VELOCITY-2
On the last entry, notice the significant mass increase (x 17,219)
Velocity m/s Gamma
100,000,000 1.225
200,000,000 1.7320
250,000,000 2.4490
290,000,000 5.4770
299,000,000 17.320
299,999,000 547.72
299,999,999 17,219
G0MDK
ELECTRON VELOCITY-2
On the last entry, notice the significant mass increase (x 547723)
Velocity m/s Gamma Voltage
100,000,000.000 1.2250 34.8kV
200,000,000.000. 1.7320 197kV
250,000,000.000 2.4940 435kV
290,000,000.000 5.4770 1.31MV
299,000,000.000 17.320 4.4MV
299,999,000.000 547.72 140MV
299,999,999.000 17,219 4.4GV
299,999,999.999 547723
Voltages shown in the 3rd column required to obtain velocities in the first column.
The point of this exercise is to show that it is virtually impossible to get the electron to move at the speed of light.
G0MDK
A QUESTION REVISITED
ELECTRONS DIDN’T TRAVEL AT c BUT THE SOMETHING DID
1. How about the “nudge” theory (cue ball effect etc.)?
Electrons out of the negative terminal nudge the next one on etc. The end result could be electrons at the light bulb 3.3µs later (?)
There have been many arguments on this issue since the 1920’s A paper on this notion was submitted to the 1997 IEE.
G0MDK
RECTANGULAR PULSE
WHAT IS HAPPENING DURING INTERVALS: (A – B), (B – C), (C – D)?
(A – B) and (C – D)? Nothing! There is NO voltage or current B – C? 100V and 0.25A
Note: Z of the transmission line pair = 400 Ohms. What would the situation be in 5.66µs?
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RECTANGULAR PULSE
Negative pulse moving back to the Pulse Generator
G0MDK
SINGLE RECTANGULAR PULSE EXAMINATION
RECTANGULAR PULSE Time Domain:
Viewed on an Oscilloscope
a2
a2tj)(
1d(t)Af(t) e
RECTANGULAR PULSE Frequency Domain: Viewed on a Spectrum Analyser
Pulse reconstruction formula Fast Fourier Transform formula
)1)(
)1
f
fff ( ( sincsin)A(
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PULSE RECONSTRUCTION-1
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PULSE RECONSTRUCTION-2
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PULSE RECONSTRUCTION-3
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PULSE RECONSTRUCTION-4
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RECAP
1. The nudge (cue ball) explanation of conduction unresolved
2. Electrical energy travels ~ speed of light over wires to a load.
3. Likewise, pulses travel ~ the speed of light over wires.
4. Single pulses are made up of wide spectrums of frequencies
5. Pulse (spectrum of frequencies) travel as TEM signals at ~ c
6. In the circuit comprising a 100V battery, switch & light bulb: the leading edge of a pulse occurs at 100V switch on and the trailing edge of a pulse occurs at 100V switch off
7. Very long pulses have same properties as very short pulses
8. AC signals to µ-wave frequencies) travel as TEM modes
9. Note that the wave length of 50Hz = 6 million metres
G0MDK
CONCLUSION
1.Electrical energy travels from source to load over wires as TRANSVERSE ELECTROMAGNETIC WAVES (TEM mode)
2.Current drift* (Amperes) is a consequence of EM Waves NOT THE OTHER WAY AROUND This may be difficult to visualize in a pair of wires, but if you consider EM microwaves travelling down a wave guide, there will be surface currents in the wave guide walls. These are also drift currents. They are also the consequence of EM energy
* A sample calculation of current drift is shown in appendix 1 of this presentation.
G0MDKAppendix 1 ELECTRON DRIFT
The current drift rate through a conductor is in the order of mm/s. The drift rate of 1A through a 1mm diametre copper wire is worked out as follows:
Current density J = amperes per unit area (J = I/A)
so J = 1Amp./(pi x r2) = 1/(3.14 x 0.00052) = 1.6 x 106
J can also be expressed as J = nevd
Transposing: vd = J/(ne)
Copper has an electron density n of 8.47 x 1028 m-3
With e = 1.6 x 10-19 coulombs of charge: ne = 1.4 x 109
Thus: vd = J/(ne) = (1.6 x 106) / (1.4 x 109) = 1.14mm/s
G0MDK
That is the nature of Electricity as I
perceive it.
Thank you for attending Chuck
Hobson BSc(hons) BA
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