Impact of circuit length in high frequency circuit

Question

I learned that after the RF antenna, we need to determine the matching circuit and calculate the line impedance depending on the FR4 characteristics.

I also learned that after we got the surface characteristics that we need to have 50 Ω, the length of the circuit is not important.

I don't understand why the length doesn't matter though the equation is: R = ρ × length / surface.

Like in a coaxial cable, the impedance is constant even if we cut the cable in a different place.

Andy aka · Accepted Answer · 2022-09-30T09:41:48.797

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I don't understand why the length doesn't matter though the equation is: R = ρ × length / surface.

Ask yourself this very important question: when you activate a light switch on your wall (and knowing that electricity doesn't travel infinitely fast), how much current begins to flow towards the light and, how can the right amount of current flow towards the light before it has reached the light. What mechanism is at play that allows current to flow even though it hasn't reached the light?

Answer: the cable presents an impedance (\$Z_0\$) to the switch that permits current flow based on ohm's law: -

$$I_{CABLE} = \dfrac{V_{SUPPLY}}{Z_0}$$

The fact that the impedance is resistive doesn't mean it has any relationship with the resistivity of the copper wire or the PCB track. It's part of transmission line theory. And, it behaves like an impedance of \$Z_0\$ from the moment a signal is applied at the source end to the moment it leaves at the load end.

edited Sep 30 '22 at 09:41

answered Sep 30 '22 at 09:36

Andy aka

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Thanks for your answer, I understand that the transmission line has a default impedance, generate by the natural inductance of a cable and the capacitance, generate by the shielding of the cable plus the skin effect that affect the impedance at high frequency, but I think that I miss something when I try to calculate the Z0 to have a practical example: Z0 = sqrt (Re² + Im²) But how the impedance is determined practically? I don't visualize the phenome. – Valentin Crettez-Bouard Sep 30 '22 at 11:53
The characteristic impedance of a cable or transmission line is: - $$\sqrt{\dfrac{R+j\omega L}{G+j\omega C}}$$ Where R, L, G and C are the series resistance, series inductance, parallel conductance and parallel capacitance per length of the line. Did you mean "[phenome](https://en.wikipedia.org/wiki/Phenome#:~:text=A%20phenome%20is%20the%20set,sum%20of%20its%20phenotypic%20traits.)"? – Andy aka Sep 30 '22 at 12:06
Here is how it is derived: [Derivation of Characteristic Impedance](https://electronics.stackexchange.com/questions/83807/derivation-of-characteristic-impedance/83836#83836). – Andy aka Sep 30 '22 at 12:13

score 3 · Answer 2 · answered Sep 30 '22 at 09:17

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You are confusing resistance to the characteristic impedance.

A coaxial cable, or any other transmission line, ideally has zero resistance so there is no losses, but for a wave or impulse travelling along the transmission line, the wave sees a characteristic impedance of 50 or 75 or whatever ohms it is manufactured, and this depends on the capacitance and inductance for unit length of the transmission line.

answered Sep 30 '22 at 09:17

Justme

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So if I understand the end of the respond, this depends on the capacitance and inductance for unit length of the transmission line. So why the impedance of cable do not change when we cut it at difference place? – Valentin Crettez-Bouard Sep 30 '22 at 11:56
@ValentinCrettez-Bouard it's the inductance/capacitance *per unit length*. The same reason the resistivity (not resistance) of a wire doesn't change when you cut it shorter, it's an intensive quantity – llama Sep 30 '22 at 21:48
@ValentinCrettez-Bouard *Characteristic impedance* is what an infinite amount of that cable is equal to. So an infinitely long 50 ohm cable acts the same as a 50 ohm resistor. This also works in reverse: a 50 ohm resistor acts the same as an infinitely long 50 ohm cable. So if we connect a 50 ohm resistor to the end of a 10-meter 50 ohm cable, the combination acts like an infinitely long cable, which obviously means the wave never reflects off the end! – user253751 Sep 30 '22 at 22:37

score 3 · Answer 3 · answered Sep 30 '22 at 12:14

The formula you used is for the ohmic resistance. Typically, ohmic resistance is much lower than the characteristic impedance. And for long coax cables, where it would seem important, the losses due to ohmic resistance are minor. Most of the loss is due to dissipation in the dielectric and some is also due to leakage through insufficient shielding (on cheap cables).

Short, quality coax cables, terminated in connectors suitable for the application, are quite “ideal” in terms of losses, and their length truly doesn’t matter if both the source and the load are well matched to the cable.

If you were using the coax for DC or very low frequency transmission and the currents were large enough, then the ohmic losses would play a role. But that’s not a typical use for coax.

Impact of circuit length in high frequency circuit

3 Answers3