Last Modified: 1/19/2000
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As you look through the examples you will notice that I've converted normalized equations , into more useful general equations.
I have tried to make sure the following formulas are correct, but I do not guarantee them.
Gas breakdown DC to approximately 1 microsecond is given by the following formula.
ElectircField: ( 24.5 pressure / atm + 6.7 sqrt( ( pressure / atm ) / Reff ) ) kV/cm as kV/m;
Where
Reff: 0.115 radius / gap " For Spheres ";
Reff: 0.23 radius / gap " For Cylinders ";
I need to have the definition of Reff checked as it was not clearly defined in my "Pulsed Power Formulary". If you can verify or correct the above please email me.
Following are examples of this gas breakdown formula
Case I & II: Have the same spacing and pressure for different shapes: spheres (I) and cylinders (II).
Case III: Same as Case II, but for higher pressure
Case IV: Same pressure as in case II, but different gap and radius
" Case I "
gap: 5 cm; pressure: 100 torr; radius: 2 cm;
Reff: 0.115 radius / gap " For Spheres ";
ElectircField as MV/m ~ ( 1.4 to 1.5 ) MV/m
" Case II "
gap: 5 cm; pressure: 100 torr; radius: 2 cm;
Reff: 0.23 radius / gap " For Cylinders ";
ElectircField ~ 1.1e3 kV/m
" Case III "
gap: 5 cm; pressure: 20 atm; radius: 2 cm;
Reff: 0.23 radius / gap " For Cylinders ";
ElectircField ~ 5.9e4 kV/m
" Case IV "
gap: 5 ft; pressure: 100 torr; radius: 6 in;
Reff: 0.23 radius / gap " For Cylinders ";
ElectircField as MV/m ~ ( 1.9 to 2.0 ) MV/m
Following is the formula for the duration of an air arc.
time: 88 sqrt( pressure / atm ) / ( ( resistance/ohm )^(1/3) * ( ElectircField / ( MV/m ) )^(4/3) ) * ns;
Following are examples
time: 88 sqrt( pressure / atm ) / ( ( resistance/ohm )^(1/3) * ( ElectircField / ( MV/m ) )^(4/3) ) * ns;
resistance: 3 ohms;
ElectricField: 500 kV/m;
pressure: 300 torr;
time as ns; ~ 12. ns
resistance: 3 ohms;
ElectricField: 500 kV/m;
pressure: 5 atm;
time as ns; ~ ( 3.0 to 3.1 ) ns
resistance: 273 ohms;
ElectricField: 50 kV/m;
pressure: atm;
time as ns; ~ ( 1.7 to 1.8 ) ns
Following is the formula for the resistive rise time of an oil switch.
riseTime: 5 sqrt(density/(g/cc))/ ( ( resistance/ohm )^(1/3) * ( ElectricField / ( MV/m) )^(4/3) ) * ns;
This formula was not clearly defined? What is the assumed units of the density? What are the assumed units of the result? Neither were given. Finally, what is the density of transformer oil? If you can help please email me.
Following are examples
" Example 1 "
density: 8/10 g/cc;
resistance: 3 ohms;
ElectricField: 500 kV/m;
riseTime as ns; ~ 7.81 ns
" Example 2 "
density: 9/10 g/cc;
resistance: 30 ohms;
ElectricField: 10 MV/m;
riseTime as ps; ~ 70.86 ps
The following drawing shows the general situation for liquid breakdown. Note in the following equations, time is the duration of the pulse above 63% of peak voltage.
Electrode
The electric field strenght for Liquid breakdown in transformer oil is give below. These equations only apply when the prestress is ( < 500V/cm ) DC across the gap.
BreakdownEFieldPositiveSide: 0.48 / ( ( time / microseconds )^(1/3) * ( StressedArea/ sq cm ) ^0.075 ) * MV/cm;
BreakdownEFieldNegativeSide: 1.41 * BreakdownEFieldPositiveSide * alpha;
Where
alpha: 1 + 0.12 sqrt( Emax/Emean - 1 );
Is the above drawing correct? Is there a good definition of stressed area? Also the text defines the gap distance (d) but doesn't use it in the equations. If you can give me better information please email me.
Following are examples for the transformer oil . breakdown formula.
" Example I "
BreakdownEFieldPositiveSide: 0.48 / ( ( time / microseconds )^(1/3) * ( StressedArea/ sq cm ) ^0.075 ) * MV/cm;
time: 100 ns;
StressedArea: sq in;
BreakdownEFieldPositiveSide as MV/cm ~ 0.9 MV/cm
BreakdownEFieldNegativeSide: 1.41 * BreakdownEFieldPositiveSide * alpha;
alpha: 1 + 0.12 sqrt( Emax/Emean - 1 );
Emax: MV/cm;
Emean: 1/2 MV/cm;
BreakdownEFieldNegativeSide as MV/cm ~ 1.4 MV/cm
" Example II "
time: 10 ns;
StressedArea: sq mm;
BreakdownEFieldPositiveSide as MV/cm ~ ( 3.11 to 3.19 ) MV/cm
Emax: MV/cm;
Emean: 1/2 MV/cm;
BreakdownEFieldNegativeSide as MV/cm ~ 5. MV/cm
The electric field strenght for Liquid breakdown in water is give below. These equations only apply when the StressedArea is greater than 1000 cm^2 ( 1.08 sq ft ).
BreakdownEFieldPositiveElectrode: 0.23 / ( ( time / microseconds )^(1/2) * ( StressedArea/ sq cm ) ^0.058 ) * MV/cm;
BreakdownEFieldNegativeElectrode: 0.56 / ( ( time / microseconds )^(1/3) * ( StressedArea/ sq cm ) ^0.070 ) * MV/cm;
Following are examples for the water breakdown formula.
" Example I "
BreakdownEFieldPositiveElectrode: 0.23 / ( ( time / microseconds )^(1/2) * ( StressedArea/ sq cm ) ^0.058 ) * MV/cm;
BreakdownEFieldNegativeElectrode: 0.56 / ( ( time / microseconds )^(1/3) * ( StressedArea/ sq cm ) ^0.070 ) * MV/cm;
time: 100 ns;
" Example II "
Given the circuit at the right with the capacitor charged to Vi and the switch closed at time = 0, find the following:
time: 0 s to 1 microSec;
time: 0 s to 1 ms;
StressedArea: 5 sq ft as sq cm; ~ 4,645.15 sq cm
BreakdownEFieldPositiveElectrode as MV/cm ~ ( 0.43 to 0.46 ) MV/cm
time: 10 ns;
StressedArea: sq m as sq cm; = 10,000 sq cm
BreakdownEFieldPositiveElectrode as MV/cm ~ ( 1.31 to 1.38 ) MV/cm
BreakdownEFieldNegativeElectrode as MV/cm ~ ( 1.35 to 1.38 ) MV/cm
RLC Circuit TOP
w: sqrt( 1/ ( L * C );
Q: w * L/R;
Z0: sqrt( L/C);
w0: w * sqrt( abs( 1 - 1/ ( 4 sq Q ) ) );
ReverseVoltage: 0 V;
PeakCurrent: Vcap exp( -R * PeakTime / ( 2 * L ) ) * sinh( w0 PeakTime )/ ( w0 * L );
ReverseVoltage: -Vcap exp( -( pi * R ) / ( 2 * w0 * L ) );
zeroTime: pi/w0;
PeakCurrent: Vcap exp( -R * PeakTime / ( 2 * L ) ) * sin( w0 PeakTime )/ ( w0 * L );
Example I ( Q > 1/2 )
R: 3/2 mOhm;
C: 12 mF;
Vcap: 19 kV;
Q: w * L / R; ~ 1.361
Z0: sqrt( L/C) as mOhm; ~ 2.041 mOhm
w0: w * sqrt( abs( 1 - 1/ ( 4 sq Q ) ) ) as kHz; ~ 37.969 kHz
ReverseVoltage: -Vcap exp( -( pi * R ) / ( 2 * w0 * L ) ) as kV; ~ -5.492 kV
zeroTime: pi/w0 as µs; ~ 82.740 µs
PeakCurrent: Vcap exp( -R * PeakTime / ( 2 * L ) ) * sin( w0 PeakTime )/ ( w0 * L ) as Mamp; ~ 5.806 Mamp
time: 0 s to 100 microSec;
Vcap exp( -R * time / ( 2 * L ) ) * sin( w0 time )/ ( w0 * L ) as Mamp; ~ ( -1.361 to 4.476 to 5.806 ) Mamp
Example II ( Q <= 1/2 )
L: 50 nH;
R: 3 Ohm;
Vcap: 19 kV;
Q: w * L / R; ~ 0.022
Z0: sqrt( L/C) as mOhm; ~ 64.550 mOhm
w0: w * sqrt( abs( 1 - 1/ ( 4 sq Q ) ) ) as MHz; ~ 29.972 MHz
ReverseVoltage: 0 V;
PeakCurrent: Vcap exp( -R * PeakTime / ( 2 * L ) ) * sinh( w0 PeakTime )/ ( w0 * L ) as kamp; ~ 6.314 kamp
Vcap exp( -R * time / ( 2 * L ) ) * sinh( w0 time )/ ( w0 * L ) as kamp ~ ( 0 to 6.252 to 6.314 ) kamp
Example III ( Q > 1/2 & uncertainty)
L: 50.0 nH;
R: 1.5 mOhm;
Vcap: 1.9 kV;
Q: w * L / R; ~ 1.314 to 1.411
Z0: sqrt( L/C) as mOhm; ~ ( 2.036 to 2.047 ) mOhm
w0: w * sqrt( abs( 1 - 1/ ( 4 sq Q ) ) ) as kHz; ~ 38. kHz
ReverseVoltage: -Vcap exp( -( pi * R ) / ( 2 * w0 * L ) ); ~ ( -593.162 to -507.621 ) V
zeroTime: pi/w0 as µs; ~ ( 82.089 to 83.418 ) µs
PeakTime: ( 31.207 to 31.719 ) µs
PeakCurrent: Vcap exp( -R * PeakTime / ( 2 * L ) ) * sin( w0 PeakTime )/ ( w0 * L ) as Mamp; ~ 0.6 Mamp
Vcap exp( -R * PeakTime / ( 2 * L ) ) * sin( w0 PeakTime )/ ( w0 * L ) as Mamp ~ ( 0.557 to 0.581 to 0.605 ) Mamp