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To calculate wire inductance, first decide what geometry you mean:
\[ \boxed{L \approx 1\ \text{nH/mm}} \]
for a thin straight lead, jumper, or PCB trace in a typical layout. This is only an approximation; the real value depends strongly on the return path.
For a straight round wire in free space, with length \(l\) and diameter \(d\):
\[ \boxed{ L_{\text{nH}} \approx 0.2\,l\left[\ln\left(\frac{4l}{d}\right)-K\right] } \]
where:
Example: for a 100 mm long, 1 mm diameter wire at high frequency:
\[ L_{\text{nH}} = 0.2 \cdot 100 \left[\ln\left(\frac{4 \cdot 100}{1}\right)-1\right] \]
\[ L_{\text{nH}} = 20[\ln(400)-1] \]
\[ L_{\text{nH}} \approx 20[5.991-1] \]
\[ \boxed{L \approx 100\ \text{nH}} \]
Strictly speaking, inductance is not a property of an isolated piece of wire alone. Current must flow in a closed path, so the inductance depends on:
A single-wire formula is therefore a partial inductance approximation. It is useful for estimating the parasitic inductance of:
But if you know the return path, calculate the loop inductance instead. That will usually be more physically meaningful.
For a long, straight, round wire in air:
\[ \boxed{ L = \frac{\mu_0 l}{2\pi} \left[ \ln\left(\frac{2l}{r}\right)-K \right] } \]
where:
Since \(d = 2r\), this is equivalent to:
\[ \boxed{ L{\text{nH}} \approx 0.2\,l{\text{mm}} \left[ \ln\left(\frac{4l{\text{mm}}}{d{\text{mm}}}\right)-K \right] } \]
This formula assumes:
At low frequency or DC, current is distributed throughout the conductor cross-section. The magnetic field exists both outside and inside the conductor. Therefore the total inductance includes:
At high frequency, due to skin effect, current flows mostly near the surface of the conductor. The internal magnetic field becomes small, so internal inductance approaches zero.
That is why the constant changes:
| Condition | Constant \(K\) | Meaning |
|---|---|---|
| Low frequency / DC | 0.75 | Includes internal inductance |
| High frequency / RF | 1.0 | Internal inductance mostly gone |
The difference is usually modest. For many practical estimates, the result is close to the common rule:
\[ \boxed{1\ \text{nH/mm}} \]
Suppose you have a copper wire:
Use:
\[ L_{\text{nH}} = 0.2l\left[\ln\left(\frac{4l}{d}\right)-1\right] \]
Substitute:
\[ L_{\text{nH}} = 0.2 \cdot 50 \left[ \ln\left(\frac{4 \cdot 50}{0.8}\right)-1 \right] \]
\[ L_{\text{nH}} = 10[\ln(250)-1] \]
\[ \ln(250) \approx 5.52 \]
\[ L_{\text{nH}} = 10[5.52-1] \]
\[ \boxed{L \approx 45.2\ \text{nH}} \]
So a 50 mm wire has roughly 45 nH of inductance, which is close to the rough estimate of 50 nH.
If you have a forward and return wire running in parallel, the loop inductance is usually more useful than the isolated-wire estimate.
For two parallel round wires carrying equal and opposite currents:
\[ \boxed{ L_{\text{loop,nH}} \approx 0.4\,l \left[ \ln\left(\frac{2s}{d}\right)+K_p \right] } \]
where:
This shows an important design rule:
\[ \boxed{\text{Smaller spacing between forward and return conductors gives lower inductance.}} \]
For example, if the return wire is close to the signal wire, the loop area is small, so the inductance is reduced.
This is why twisted pairs, coaxial cables, and PCB traces over ground planes have much lower loop inductance than a single long wire with a distant return path.
For a wire or round conductor running parallel to a ground plane, the return current flows in the plane beneath the wire. A useful high-frequency approximation is:
\[ \boxed{ L_{\text{nH}} \approx 0.2\,l \ln\left(\frac{4h}{d}\right) } \]
where:
This is highly relevant for:
The closer the conductor is to the ground plane, the lower the loop inductance.
For a single circular loop made from round wire:
\[ \boxed{ L_{\text{nH}} \approx 0.2\pi D \left[ \ln\left(\frac{8D}{d}\right)-2 \right] } \]
where:
This applies when the loop diameter is much larger than the wire diameter.
For example, if:
then:
\[ L_{\text{nH}} \approx 0.2\pi \cdot 100 \left[ \ln(800)-2 \right] \]
\[ \ln(800) \approx 6.68 \]
\[ L_{\text{nH}} \approx 62.8(4.68) \]
\[ \boxed{L \approx 294\ \text{nH}} \]
A loop of wire can therefore have much more inductance than a straight wire of similar length because the loop area is large.
Once you know the inductance, its impedance at frequency \(f\) is:
\[ \boxed{ X_L = 2\pi f L } \]
For example, a 100 nH wire at 100 MHz has:
\[ X_L = 2\pi \cdot 100 \times 10^6 \cdot 100 \times 10^{-9} \]
\[ \boxed{X_L \approx 62.8\ \Omega} \]
This is why a seemingly harmless 10 cm wire can be a serious problem in RF, switching converters, fast logic, and decoupling networks.
At 1 kHz, the same 100 nH wire has:
\[ X_L = 2\pi \cdot 1000 \cdot 100 \times 10^{-9} \]
\[ \boxed{X_L \approx 0.00063\ \Omega} \]
So at low frequency it is usually negligible.
If you just need an engineering estimate:
\[ \boxed{L \approx 1\ \text{nH/mm}} \]
\[ \boxed{ L_{\text{nH}} \approx 0.2\,l \left[ \ln\left(\frac{4l}{d}\right)-K \right] } \]
To reduce inductance:
The dominant factor is often not wire diameter but loop area.
For a straight round wire:
\[ \boxed{ L{\text{nH}} \approx 0.2\,l{\text{mm}} \left[ \ln\left(\frac{4l{\text{mm}}}{d{\text{mm}}}\right)-K \right] } \]
with:
As a rule of thumb:
\[ \boxed{L \approx 1\ \text{nH per mm}} \]
But in real circuits, the most accurate answer comes from calculating the inductance of the complete current loop, not just the isolated wire. The return path and loop area usually dominate the result.