Theoretical analysis of the proposed EPMC

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Preliminary
March 13, 2013
Theoretical analysis of the proposed EPMC
I. Plain EPMC
The source of the electro-magnetic field is a PCB conductor
Current conductor
I
F  nqVB
F  ILB
F
I
 2k '
IL
R
Tm
k '  107
A
I
B  2  107
R
B
B
R  The distance from the current conductor to a
Magnetic field
secondary coil.
Let us assume R  2mm
I  Electrical current, for our calculation we
assume the most popular 3A current
I  3A
H  5 mm, L  5 mm
L
I
B  2  107
H
B
PCB conductor
Boris Fradkin
I
R
2  107  3
 3  104 Tesla  3 Gauss
0.002
The receiver of the magnetic field is the secondary coil put on the PCB conductor.
Secondary coil
L
L  H  5mm
0.2 inch
n  1000
R
R  1 H  2.5 mm
2
0.1inch
The secondary coil is an air coil, therefore
0  4 k '
B
n
I
L2  4 R 2
B  4 k '
0
I 
B
From this equation:
L2  4 R 2
4 k '  n
0
I 
3  104
25  106  4  6.25  106

4  3.14  107  1000
3  104  34  106
3  5.8  103

 1.38mA
12.56
4  3.14  104
I  Isec ondary  1.38mA
Let us calculate the electrical parameters of the secondary coil: Resistance and Inductance:
R
Inductance:
L
H
 NR 
2
9 R  10 H

1000  0.1
2
9  0.1  10  0.2
 2.63 mH
X L  wL  2    f  L  2  3.14  50  2.63  103  0.825
Resistance:
We will use a 0.1mm wire. This wire resistance is 2.176  m .
The length of one turn is 6  6  4  4  20 mm
Total secondary coil length is:
1000 turns  20 mm  20 m
Resistance R  20  2.176 44 
Say total secondary coil impedance is 50 
Boris Fradkin
Let us use a standard for current sensors 100  load resistor.
The secondary voltage to be amplified shall be:
U out  I 
Rload
100
 1.38  103 
 0.092V
Rlod  R
100  50
Conclusion: The output voltage after amplification is big enough to make necessary
measurements and obtain needed control voltage.
2. RCD (leakage current device)
The residual current to be detected is 6mA (not 3A as in previous case). By using the
same calculation technique we will obtain that the secondary voltage to be amplified
will be approximately 0.18 mV.
Conclusion: There is no problem to amplify this low frequency signal to obtain the
necessary control voltage.
References:
1. Transformer and Inductor Design Handbook
Col.Wm.T.McLyman
Jet Propulsion Institute of Technology, Pasadena, California
2. Electricity Part B
Dr. D.Singer, David Rehgulad Publ. House Ltd
3. Magnetism
Open University Library #19240, 28914
Boris Fradkin
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