I'm not familiar with that term. Could somebody explain it to me? Is it the frequency at which the phase of the AC current running through the impedance goes to zero?
Edit: Here is the context in which it is used.
The ‘characteristic frequency’ of a certain impedance
accharacteristic-impedancefrequencyimpedance
Related Solutions
This seems the simplest mathematical way to derive characteristic impedance. Consider a "lump" of transmission line connected to the continuation of that transmission line (\$Z_0\$): -
- R is series resistance of cable for a given length
- L is series inductance of cable for a given length
- G is parallel conductance of cable for a given length
- C is parallel capacitance of cable for a given length
- \$Z_0\$ to the right is the continuation of the cable
Therefore the impedance looking into the left is: -
$$Z_0 = R + j\omega L + Z_0||\dfrac{1}{G + j\omega C}$$
$$= R + j\omega L + \dfrac{\frac{Z_0}{G+j\omega C}}{Z_0 + \frac{1}{G+j\omega C}}$$
$$= R + j\omega L + \dfrac{Z_0}{1 + Z_0(G+j\omega C)}$$
$$Z_0[1 + Z_0(G+j\omega C)] = [R+j\omega L][1 + Z_0(G+j\omega C)]+Z_0$$
$$Z_0 + Z_0^2(G+j\omega C) = R+j\omega L + Z_0[(R+j\omega L)(G+j\omega C)]+Z_0$$
$$Z_0^2(G+j\omega C) = R+j\omega L + Z_0[(R+j\omega L)(G+j\omega C)]$$
The important thing next is to recognize that \$(R+j\omega L)(G+j\omega C)\$ is insignificant as the "lump" approaches zero length and we are left with: -
$$Z_0^2(G+j\omega C) = R+j\omega L$$
hence $$Z_0 = \sqrt{\dfrac{R+j\omega L}{G+j\omega C}}$$
The skin depth doesn't really have much, if anything, to do with impedance. It simply defines a loss mechanism. Note that with a thicker track, you will have marginally less skin effect losses due to a larger surface area for the track.
The impedance of microstrip is inversely proportional to the thickness of the signal track, but the difference for 0.5 oz copper and 1 oz copper is quite small, and definitely smaller than the 10% impedance variation most fabricators will guarantee.
I often use surface layers for some high speed signals as there is marginally less dielectric absorption loss as the velocity of the signal is a bit quicker (due to the effective dielectric constant being lower as half of the signal is in air, not FR-4).
Surface layers are normally plated up to 1 oz or sometimes 2 oz thicknesses. Talk to your PCB supplier to find out what they normally do.
Your stackup looks normal for 1 oz plating on the surface and 0.5 oz plane copper thickness.
Best Answer
The impedance of a parallel connected resistor and capacitor is
$$Z = R\,||\frac{1}{j\omega C} = \frac{1}{\frac{1}{R}+ j\omega C} = R \,\frac{1}{1+j \omega RC} $$
This is of the form of a resistance times a dimensionless, frequency dependent quantity.
There is a particular frequency of interest,
$$\omega_0 = \frac{1}{RC}$$
which is the characteristic frequency. At this frequency, the impedance is
$$Z_0 = R \,\frac{1}{1+j \omega_0 RC} = R \,\frac{1}{1+j} = \frac{R}{\sqrt{2}}e^{-j\frac{\pi}{4}}$$