Kirchhoff's Law for High Frequency Circuits: Explained

[SMOF]

Hello,

I am hoping someone can break down an equation for me. I am used to Kirchhoff's law in the form of i₁ + i₂ + i₃ = 0 etc. But recently in a High Frequency class, we were told ...'Let us apply the Kirchhoff's law to the equivalent circuit of a transmission line segment of length \deltaz. Using the voltage law, we get

V(z,t) = R\deltaz * I(z,t) + L\deltaz * (\deltaI(z,t)/\deltat) + V(z + \deltaz,t)'.

If anyone could help me break this down, or explain who it relates to the general form of the equation, that would be amazing.

Many thanks in advance.

Seán

 

[BackEMF]

Hello,

I am hoping someone can break down an equation for me. I am used to Kirchhoff's law in the form of i₁ + i₂ + i₃ = 0 etc. But recently in a High Frequency class, we were told ...'Let us apply the Kirchhoff's law to the equivalent circuit of a transmission line segment of length \deltaz. Using the voltage law, we get

V(z,t) = R\deltaz * I(z,t) + L\deltaz * (\deltaI(z,t)/\deltat) + V(z + \deltaz,t)'.

If anyone could help me break this down, or explain who it relates to the general form of the equation, that would be amazing.

Many thanks in advance.

Seán

An infinitesimal piece of transmission line would look as follows:

Now say the current flowing into the left side of this circuit is I(z,t) and the voltage across the two input terminals is V(z,t). Likewise, say the current flowing out of the right side is I(z+\delta z,t), and the voltage across the output terminals is V(z+\delta z,t). Just as in the diagram (except it uses x as the distance variable, and upper case delta symbols - but hopefully you get the idea).

Now simply apply the usual KVL equation, Ohm's Law and the inductor equation: V = L \frac{\text{d}I}{\text{d}t}. This will yield your equation above.

 

[SMOF]

Hey,

That's great! Thanks for the reply and the information.

Seán

 

[BackEMF]

No problem Seán. If you need any more help just let me know.