0
\$\begingroup\$

How would I calculate the mutual inductance (Lm) between parallel wires if I know the inductance per length of the wires and the separation between the wires?

enter image description here

\$\endgroup\$
  • \$\begingroup\$ You need to know the separation of the wires. \$\endgroup\$ – analogsystemsrf May 9 '17 at 3:39
  • \$\begingroup\$ I'll update my question to include the fact that I know the separation of my wires \$\endgroup\$ – Guy Lee May 9 '17 at 3:49
1
\$\begingroup\$

If you know the separation of the wires

$$ L = (\frac{\mu_o * Length}{\pi}) * \ln(\frac{separation}{radius} + SDfactor)$$

where separation is the center_to_center distance, to be greater than 2*radius

https://en.wikipedia.org/wiki/Inductance shows derivation of these formula

As the link explains, there is a fudge factor to model skin effect. For low frequencies, where current uses the entire cross-section, add 1/4 inside the "ln" arqument. At high frequencies, add nothing.

Now lets put this equation to use. Consider a MCU SPI dataline, 1mm from the sensitive signal trace, and parallel for 40mm. What is the mutual inductance? We need to know the radius, so assume its 1/2 * 0.25mm (10 mils = 0.25mm). Yes, I'm using a flat trace circuit, in an equation for "round wires".

$$L_{Ind} = \left(\frac{4*\pi*10^-7 \frac{H}{m} * 0.04m) }{ pi}\right)* \ln(\frac{1mm}{ 0.125mm})$$

$$L_{Ind} = 4*0.04 * 10^-7 * ln(8) $$ {ignoring the skin-effect factor} $$L_{Ind} = 0.16 * 10^-7 * 2.08 = 0.32 * 10*-7 = 32 * 10^-9 = 32 nH$$

Suppose the dataline has 100pF Cload, with 2.5 volts/2.5nanoSeconds slewrate; the charging current \$I = C * \frac{dV}{dT} = 0.1nF * 1\frac{V}{ns} = 0.1 A\$. We'll assume the dataline current rises to 0.1 amps in 1nanoSecond, remains there for 0.5nS, and decays back to zero in 1nS.

What voltage is induced by the mutual inductance, from the dataline into signal trace?

V = L * dI/dT = 32nH * 0.1 amps/1nanoSec [knowing the nano cancel] = 3.2 volts.

NOW.....what is the benefit of separation? If we just have wires in air (no underlying planes), we see the mutual inductance (the coupling) drops very slowly because of ln(separation/radius).

Fortunately there can be underlying planes.

\$\endgroup\$
  • \$\begingroup\$ What constitutes as low frequencies and high frequencies? My signals are around the audible range ~1kHz - 18kHz so my thinking is that this constitutes as low frequency? \$\endgroup\$ – Guy Lee May 9 '17 at 4:18
  • \$\begingroup\$ High frequency is where the skin depth is much less than the wire radius. You can find a calculator here: rfcafe.com/references/calculators/skin-depth-calculator.htm . For 1-18 kHz, the skin depth varies from 2 mm to 0.5 mm. \$\endgroup\$ – Evan May 9 '17 at 4:30
  • \$\begingroup\$ For 1mm wires, does this SkinEffect lead to pulse distortion of the audio signals? Or are you signals purely sins? \$\endgroup\$ – analogsystemsrf May 9 '17 at 5:29
  • \$\begingroup\$ The skin effect will have some effect on the audio at 18 kHz (that 0.5mm is RADIUS effect, and 1mm DIAMETER means the most you ought to expect is a decibel, when the speaker wire is long (about 4 ohms), and feeds a 4 ohm speaker. Most of the cross sectional area is closer to the surface than 0.5mm even for thicker 2mm wires. \$\endgroup\$ – Whit3rd May 9 '17 at 6:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.