# How do microstrips actually represent lumped components?

I'm having trouble understand intutively how microstrips repersent lumped element components.

All we are told is that, high impedance sections on a microstrip line repersent inductors and low-impedance repersent capacitors. How though? How does this flat piece of conductive material with a dielectric below it actually repersent a capacitor/inductor.

We can picture all transmissions lines as having some inductance and capacitance per unit length.

If we consider a microstrip line; the lower impedance it is, the wider it will be. This results in greater capacitance because, as you said, there are two pieces of conductor with a dielectric between them which is exactly the structure of a capacitor. So a lower impedance means a larger capacitor area and therefore larger capacitance.

For higher impedance lines, the capacitance is negligible compared to the inductance so we can model it as a lumped inductor in some circumstances. I don't know off the top of my head what impact the dimensions of microstrip have on the inductance.

Many microwave texts, such as https://www.amazon.com/Microwave-Engineering-4th-David-Pozar-ebook/dp/B008ACCCHO, will discuss this more formally and actually show the equivalence but I think the intuition is sufficient here.

How do microstrips actually repersent lumped components?

It is actually the other way round!

A microstrip is a distributed component. It has a length and that length has (continous) capacitance and inductance all over its length/width/height

This is difficult to do calculations on as the number of elements (inducators, capacitors, resistors) is basically infinite.

To simplify things a lumped components model can be used, which simplifies the infinite number of components to a finite number.

A micro strip line is the physical implementation of a transmission line. Suppose you have a load and a transmission line:

The impedance at distance $$\l\$$ is:

$$Z(l) = Z_0 \frac{1 + \Gamma e^{-2\gamma l}}{1 - \Gamma e^{-2 \gamma l}},$$

where $$\\Gamma\$$ is the reflection coefficient and $$\\gamma\$$ is the propagation constant.

If you assume that the transmission line is lossless and by setting the load to either open or close you get a purely imaginary impedance. Since the impedance of an inductor and a capacitor is:

$$X_\mathrm{L} = j \omega L, \qquad X_\mathrm{C} = \frac{1}{j \omega C},$$

the transmission line impedance looks like a capacitor or an inductor at some specific frequency. For example:

$$Z(l_0) = j x = j \omega_0 L.$$

It is important to note that this simple implementation of a capacitor or inductor is only valid in the neighborhood of said specific frequency. There are methods to improve the bandwidth, but they are outside the scope of this answer.

A microstrip is a form of transmission line in that the conductor has series inductance and there is capacitance to ground. The ratio of inductance to capacitance determine the characteristic impedance. If you terminate a transmission line with a resistor equal to its impedance, the input impedance remains constant and resistive as the length is changes. If the strip is narrow, its inductance will dominate and the input will look inductive. Conversely a broad strip will add more capacitance than inductance.

The base model for realizing microstrip duals of capacitors and inductors is the 1/8 wavelength strip. A 1/8 wavelength transmission line will have a reactance of the line's characteristic impedance. So if you have a 1/8 (electrical wavelength) wavelength of RG-58/U it will measure 50 ohms on a bridge.

Measuring at one end with the other end open the reactance will be capacitive. With the end opposite of the measuring terminals shorted it will be inductive. Consider the open circuit model to be two parallel conductors like capacitor plates. Consider the short circuited model to be inductive with a distributed length of wire providing the inductance.

In the formula above you are calculating the length of a line (stub). For an 1/8 wavelength the cotangent of (Bl) is 1. If the line is 50 ohms then the reactance is 50 ohms and the 'j' operator provides an indicator the value is reactive. The minus sign indicates the reactance is capacitive.

The next formula is for the end of the line being shorted.

For a 1/8 wavelength line the tangent of (Bl) is 1, the 'j' operator is positive, thus the reactance value is positive. This indicates the line is inductive.

As you can probably see, the formulas allow for calculating the length of the line more than and less than 1/8 wavelength. The formula's final calculated value will tell you if your line is capacitive or inductive by the math sign. In general when using microstrip you try to use the shortest length as it takes up the least amount of real estate on your PCB.

Keep in mind this assumes lossless line and no propagation delay in the line. Losses will add equivalent series 'Resistance' and propagation delay will require physically shortening the strip length when compared to your calculated length.

Regards