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Contents - AP Technology

Contents - AP Technology

Contents - AP Technology

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Basic electrical-engineering formulas<br />

Cross-section and diameter calculation of flexible leads<br />

A = lead cross-section in mm2 A = d<br />

Z = lead diameter in mm<br />

n = number of individual wires<br />

Z = 1,34·n·d d = individual wire-Ø in mm<br />

2 ·0,785·n<br />

Conductor resistance<br />

R =<br />

G =<br />

·L<br />

1<br />

R<br />

S<br />

L<br />

R =<br />

·S<br />

=<br />

Characteristic wave impedance<br />

R = electrical resistance in <br />

G = electrical conductivity in S<br />

S = conductor cross-section in mm 2<br />

L = length of conductor in m<br />

= specific resistance (Rho)<br />

= conductivity (Kappa)<br />

Example given L = 800 m, R = 100 , S = 0,15 mm2 required = Conductivity<br />

Calculation route L<br />

=<br />

R·S<br />

Ohm's Law<br />

I =<br />

U<br />

R<br />

1<br />

<br />

I = electrical current in A<br />

U = electrical voltage in V<br />

R = electrical resistance in <br />

Example<br />

U = 220 V ; R = 980 <br />

U 220 V<br />

I = =<br />

R 980 <br />

I = 0,22 A<br />

L<br />

Z =√<br />

C<br />

Z<br />

L<br />

=<br />

=<br />

characteristic wave impedance in <br />

inductance in H<br />

C = capacity in F<br />

For coaxial cables r = dielectric constant<br />

60<br />

Z = In<br />

r<br />

Effective capacitance conductor/mass<br />

C=<br />

r ·10 3<br />

D<br />

d<br />

18·ln D<br />

d<br />

800 m<br />

=<br />

100 ·0,15 mm2 = 53,3<br />

m<br />

·mm 2<br />

ln = natural logarithm<br />

D = Ø above dielectric<br />

d = Ø of inner conductor<br />

C = capacity in pF/m<br />

r = dielectric constant<br />

D = Ø above dielectric<br />

d = Ø of inner conductor<br />

ln = natural logarithm<br />

Resistance/Temperature<br />

Technical Guidelines<br />

R W = R K (1+·) R K = cold resistance at +20°C in <br />

R W = hot resistance in <br />

R W = R K +R R = change in resistance in <br />

= temperature changes in °C<br />

R = ·R K · = temperature coefficient<br />

=<br />

R W -R K<br />

R K · <br />

Installation in series of ...<br />

Cu = 0,0039 1/°C<br />

Alu = 0,00467 1/°C<br />

Example<br />

= 70 °C<br />

R K = 100 <br />

= 0,0039 1/°C<br />

R W = R K ·(1+·)<br />

R W = 100 W(1+0,0039·70)<br />

R W = 127,3 <br />

Resistors R g = R 1 + R 2 + R 3 + ...<br />

Capacitors<br />

1 1 1 1<br />

+ + + ...<br />

L g L 1 L 2 L 3<br />

Inductances L g = L 1 + L 2 + L 3 + ...<br />

Installation in parallel of ...<br />

Resistors<br />

Two Resistors<br />

Capacitors C g = C 1 + C 2 + C 3 + ...<br />

Inductances<br />

Powers of ten<br />

1 1 1 1<br />

= + + + ...<br />

R g R 1 R 2 R 3<br />

R1 · R2 Rg =<br />

R1 + R2 1 1 1 1<br />

= + + + ...<br />

L g L 1 L 2 L 3<br />

10 12 Tera T 1 000 000 000 000<br />

10 9 Giga G 1 000 000 000<br />

10 6 Mega M 1 000 000<br />

10 3 kilo k 1 000<br />

10 2 hekto h 100<br />

10 1 deka da 10<br />

10 0 1<br />

10 -1 dezi d 0,1<br />

10 -2 centi c 0,01<br />

10 -3 milli m 0,001<br />

10 -6 mikro µ 0,000 001<br />

10 -9 nano n 0,000 000 001<br />

10 -12 piko p 0,000 000 000 001<br />

16.085

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