Now copy the printed circuit board manufacturing panel, more than for the copper wire, copper metal itself determines the physical characteristics of the process in conducting there must be some resistance, wire inductance components affect the transmission of voltage signals, resistance components will affect the current signal transmission lines in high-frequency inductance of particularly protruding.

In the printed circuit board can be seen as a certain wire is rectangular copper strips the rules, we a long 10cm, width 1. 5mm, a thickness of 50μm wire, for example, can be seen by calculating the size of its impedance .

Wire resistance can be calculated by the formula:
R = ρL / s (Ω)
Type of L is wire length (m), s for the wire section area (mm2), ρ the resistivity ρ = 0. 02. Through calculation, the wire resistance value of approximately 0. 026Ω.

When a conductor of the wire away from the other, much larger than the width of its length, the wire inductance volume of 0. 8μH / m, then the wire is 10cm long with 0. 08μH inductance. Then we can derive from the following formula presented by the copy sheet out of wire inductance:
XL = 2πfL
Type of π is a constant, f is the frequency of the signal wires through (Hz), L is the inductance per unit length wire volume (H). So that we can calculate the wires were in the low and high frequency inductance:
When f = 10KHz when, XL = 6. 28 × 10 × 103 × 0. 08 × 10-6 ≈ 0. 005Ω;
When f = 30MHz when, XL = 6. 28 × 30 × 106 × 0. 08 × 10-6 ≈ 16Ω

The above formula we can see that in low-frequency signal transmission wire resistance than wire inductance, and inductance in high frequency signal is much greater than the lead wire resistance.