Bond Wire Inductance
Understanding Bond Wire Inductance
The inductance of a single wire in free space depends on its length and diameter: L ≈ (μ0/2π)·l·[ln(2l/r) − 1], where l is length and r is wire radius. For 25 μm diameter gold wire, this gives approximately 0.7 nH/mm. The loop height also matters: higher loops increase inductance.
Parallel bond wires share current and reduce total inductance, but mutual coupling limits the reduction. Two wires with 100 μm spacing provide about 60% reduction (not 50%). The ground plane proximity further modifies the inductance through image currents.
L ≈ 0.2·l·[ln(2l/r) − 1] nH
l = length (mm), r = radius (mm)
25 μm Au, 1 mm: ~0.7 nH
Reactance at 10 GHz:
XL = 2πfL = 2π·10×109·0.7×10−9 = 44 Ω
Wire Configuration Inductance
| Configuration | L (nH, 1 mm) | Reduction | Use |
|---|---|---|---|
| Single wire | 0.7 | Reference | Signal |
| 2 parallel (100 μm) | 0.4 | 43% | Power |
| 4 parallel (100 μm) | 0.25 | 64% | Ground |
| Ribbon (250 μm) | 0.3 | 57% | High current |
Frequently Asked Questions
How much?
0.5-1.0 nH/mm single wire. 25 μm Au, 1 mm ≈ 0.7 nH. Parallel wires reduce but mutual coupling limits gain. At 10 GHz: 44 Ω.
RF impact?
Source wires: negative feedback (gain reduction). Gate/drain: matching element. ±100 μm length variation shifts match. Must model accurately.
Modeling?
Simple: 0.7 nH/mm lumped. Better: include mutual inductance. Best: 3D EM simulation. Above 20 GHz: transmission-line effects.