# UnitMath Example: ( Pulsed Power Diagnostics ) Last Modified: 12/16/1999

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The following examples show how to use UnitMath in common Pulsed Power diagnostic calculations. Much of this information is from North Star Reasearch Corporation's "Pulse Power Formulary". I know all this information is available in other sources, but if there are any issues with me posting this information please send me email.

I have tried to make sure the following formulas are correct, but I can not and do not guarantee them.

### Current Loop ( Not Integrated ) TOP

The following figure shows an unintegrated square (rectangular) current loop. The voltage generated by that loop is given below is give below.

Voltage: µo length turns / ( 2 pi ) ln( outer / inner ) dI/dt;

Following are examples for the unintegrated rectangular current loop formula.

" Example 1 "
Voltage: µo length turns / ( 2 pi ) ln( outer / inner ) dI/dt;

dI: kA;
dt: 100 µs;
length: 5 inch;
outer: ft;
inner: 4 in;
turns: 3;
Voltage ~ 0.84 V

" Example 2 "
dI: 1 kA;
dt: 100.0 µs;
length: 5.0 inch;
outer: 1.0 ft;
inner: 4.0 in;
turns: 3;
Voltage ~ ( 0.78 to 0.89 ) V

Example 2 shows that given the above uncertainties a perfect 1 kA current change could give an output voltage out of anywhere from ( 0.78 to 0.89 ) V. The following calculation answers the question: What conclusion can I draw form a given output voltage? In this case an exact voltage of 84/100 V could mean a current change of anywhere between ( 940. to 1,076. ) amps again based on the uncertainties in example 2.

dt: 100.0 µs;
length: 5.0 inch;
outer: 1.0 ft;
inner: 4.0 in;
turns: 3;
Voltage: 84/100 V;
dI: amp to 5 kamp; ~ ( 2,500.5 ± 2,499.5 ) amp
solve( Voltage = µo length turns / ( 2 pi ) ln( outer / inner ) dI/dt; dI: amp ) ~ dI: ( 940. to 1,076. ) amp;

### Current Loop ( Integrated ) TOP

The following figure shows an integrated square (rectangular) current loop. The voltage generated by that loop is given below is give below.

Voltage: µo length turns ln( outer / inner ) / ( 2 pi R C ) * deltaCurrent;

Following are examples for the integrated rectangular current loop formula. Example 1 assumes that all the variables are exact ( not realistic ), Example 2 adds some uncertainty, Example 3 is the most realistic.

" Example 1 "
Voltage: µo length turns ln( outer / inner ) / ( 2 pi R C ) * deltaCurrent;

deltaCurrent: kA;
length: ft;
outer: ft;
inner: 4 in;
turns: 3;
R: 50 ohm;
C: 25 microF;
Voltage ~ 0.16 V

" Example 2 "
deltaCurrent: 1 kA;
length: 1.0 ft;
outer: 1.0 ft;
inner: 4.0 in;
turns: 3.0;
R: 50 ohm;
C: 25 microF;
Voltage ~ ( 0.14 to 0.18 ) V

" Example 3 "
deltaCurrent: 1 kA;
length: 1.0 ft;
outer: 1.0 ft;
inner: 4.0 in;
turns: 3.0;
R: 50 ohm * (1 pm 10%);
C: 25 microF * (1 pm 10%);
Voltage ~ ( 0.12 to 0.22 ) V

Example 3 shows that given the above uncertainties a perfect 1 kA deltaCurrent could give an output voltage out of anywhere from ( 0.12 to 0.22 ) V. The following calculation answers the question: What conclusion can I draw form a given output voltage? In this case an exact voltage of 16/100 V could mean a deltaCurrent of anywhere between ( 715. to 1,369. ) amps again based on the uncertainties in example 3.

length: 1.0 ft;
outer: 1.0 ft;
inner: 4.0 in;
turns: 3.0;
R: 50 ohm * (1 pm 10%);
C: 25 microF * (1 pm 10%);
Voltage: 16/100 V;
deltaCurrent: amp to 5 kamp; ~ ( 2,500.5 ± 2,499.5 ) amp
solve( Voltage = µo length turns ln( outer / inner ) / ( 2 pi R C ) * deltaCurrent; deltaCurrent: amp ) ~ deltaCurrent: ( 715. to 1,369. ) amp;

### Rogowski Coil ( Not Integrated ) TOP

The following figure shows an unintegrated Rogowski Coil. The voltage generated by a Rogowski Coil is given below. Note: The Voltage is independent of the coil position so long as the current source is more than 2 turn spacings away from the coil. This formula is normally writen in terms of radius not diameter, but diameter is more useful as it is what you actually measure. For more information see Use Raw Data in Your Calculations.

Voltage: µo turns ( minorDiameter/2 )^2/ majorDiameter * dI/dt;

Following are examples for the unintegrated Rogowski Coil.

" Example 1 "
Voltage: µo turns ( minorDiameter/2 )^2/ majorDiameter * dI/dt;

dI: 1 kA;
dt: 100 µs;
majorDiameter: 1/3 ft;
minorDiameter: inch;
turns: 40;

turnSpacing: pi majorDiameter / turns as cm; ~ 0.80 cm
mustBeMoreThanOne: ( majorDiameter / 2 - minorDiameter/2 ) / ( 2 * turnSpacing ) ~ 2.39

Voltage ~ 0.80 V

" Example 2 "
dI: 1 kA;
dt: 100.0 µs;
majorDiameter: 0.33 ft;
minorDiameter: 1.0 inch;
turns: 40;

turnSpacing ~ 0.8 cm
mustBeMoreThanOne ~ 2.33 to 2.43

Voltage ~ ( 0.72 to 0.90 ) V

Example 2 shows that given the above uncertainties a perfect 1 kA deltaCurrent could give an output voltage out of anywhere from ( 0.72 to 0.90 ) V. The following calculation answers the question: What conclusion can I draw form a given output voltage? In this case an exact voltage of 80/100 V could mean a deltaCurrent of anywhere between ( 885. to 1,117. ) amps again based on the uncertainties in example 2.

dt: 100.0 µs;
majorDiameter: 0.33 ft;
minorDiameter: 1.0 inch;
turns: 40;
Voltage: 80/100 V;
dI: amp to 5 kamp;
solve( Voltage = µo turns ( minorDiameter/2 )^2/ majorDiameter * dI/dt; dI: amp ) ~ dI: ( 885. to 1,117. ) amp;

### Rogowski Coil ( Integrated ) TOP

The following figure shows an integrated Rogowski Coil. The voltage generated by that loop is given below is give below.

Voltage: µo turns ( minorDiameter/2 )^2/ (majorDiameter R * C) * dI;

Following are examples for the integrated Rogowski coil.

" Example 1 "
Voltage: µo turns ( minorDiameter/2 )^2/ (majorDiameter R * C) * dI;

dI: 1 kA;
majorDiameter: 1/3 ft;
minorDiameter: 2 inch;
turns: 40;
R: 50 ohm;
C: 10 microF;

turnSpacing: pi majorDiameter / turns as cm; ~ 0.80 cm
mustBeMoreThanOne: ( majorDiameter / 2 - minorDiameter/2 ) / ( 2 * turnSpacing ) ~ 1.59

Voltage ~ 0.64 V

" Example 2 "
dI: 1 kA;
majorDiameter: 0.33 ft;
minorDiameter: 2.0 inch;
turns: 40;
R: 50 ohm * (1 pm 1% );
C: 10 microF * (1 pm 5% );

turnSpacing: pi majorDiameter / turns as cm; ~ 0.8 cm
mustBeMoreThanOne: ( majorDiameter / 2 - minorDiameter/2 ) / ( 2 * turnSpacing ) ~ 1.51 to 1.64

Voltage ~ ( 0.57 to 0.73 ) V

Example 2 shows that given the above uncertainties a perfect 1 kA change in current could give an output voltage out of anywhere from ( 0.57 to 0.73 ) V. The following calculation answers the question: What conclusion can I draw form a given output voltage? In this case an exact voltage of 80/100 V could mean a deltaCurrent of anywhere between ( 885. to 1,117. ) amps again based on the uncertainties in example 2.

majorDiameter: 0.33 ft;
minorDiameter: 2.0 inch;
turns: 40;
R: 50 ohm * (1 pm 1% );
C: 10 microF * (1 pm 5% );
Voltage: 70/100 V;
dI: amp to 5 kamp;
solve( Voltage = µo turns ( minorDiameter/2 )^2/ (majorDiameter R * C) * dI; dI: amp ) ~ dI: ( 888. to 1,317. ) amp;

### Current Transformer TOP

Current Transformer

Under construction: 12/16/1999