PCB board conductor anti-interference formation principle

Why PCB board wires will form anti-interference?

Because in the printed circuit board copy board manufacturing, its wires are generally copper wire, and copper, the physical properties of this metal determines its conductive process there must be a certain impedance, the inductive component of the wire will affect the transmission of voltage signals, resistive components will affect the transmission of current signals, in high-frequency lines in the inductance of the impact of the line is particularly prominent.

In the printed circuit board on a section of the wire can be seen as a very regular rectangular copper strip, we take a section of 10cm long, 1.5mm wide, the thickness of 50μm wire as an example, through the calculation of the size of the impedance can be seen.

Wire resistance can be calculated by the formula:

R = ρL/s (Ω)

Where L is the length of the wire (metres), s for the cross-sectional area of the wire (square millimetres), ρ for the resistivity ρ = 0. 02. Through the calculation of the wire resistance value of about 0. 026Ω.

When a section of wire away from other conductors, its length is much greater than the width of the wire, the self-inductance of the wire is 0. 8μH/m, then 10cm long wire has an inductance of 0. 08μH. Then we can find out the inductance presented by the copyboard wire by the following formula:

XL=2πfL

where π is a constant, f is the frequency of the signal passing through the wire (Hz), and L is the self-inductance per unit length of the wire (H). In this way we can calculate the inductive resistance of the wire at low and high frequencies respectively:

When f=10KHz, XL=6. 28×10×103×0. 08×10-6≈0. 005Ω;

When f=30MHz, XL=6. 28×30×106×0. 08×10-6≈16Ω

Through the above formula, we can calculate that the wire resistance is larger than the wire inductive resistance in low frequency signal transmission, while the wire inductive resistance is much larger than the wire resistance in high frequency signal.