PCB Tech - What loss will occur in the circuit board transmission line

The PCB transmission line contains at least two wires-one for the signal and the other for the return path. The complex circuit board net is a combination of this simpler transmission line structure. From the perspective of PCB design, understanding these structures (microstrip, stripline, and coplanar) is beneficial to designers and manufacturers.

What is the loss of the transmission line?

The transmission line structure has different loss mechanisms. The total loss of the PCB transmission line is called the insertion loss (αt). It is the sum of conductor loss (αc), dielectric loss (αd), radiation loss (αr) and leakage loss (αl).

αt = αc + αd + αr + αl

The effect of leakage is negligible because PCB has a very high volume resistance. Radiation loss is the energy lost from the circuit due to radio frequency radiation. This loss depends on frequency, dielectric constant (Dk) and thickness. For a particular transmission line, the loss will be much higher at higher frequencies. For the same circuit, when a thinner substrate and a higher Dk value are used, the radiation loss will be smaller.

In this article, we will only discuss the transmission line loss related to the conductor loss (αc) caused by the signal trace resistance and the dielectric loss (αd) caused by the PCB dielectric, the latter with the loss tangent/dissipation factor to measure.

αt = αc + αd

Characteristic impedance and loss mechanism

In our previous PCB transmission line series, we gave you the characteristic impedance of a transmission line (it is the impedance seen by the signal and has nothing to do with frequency):

R = line conductor resistance per unit length (pul)

L = the inductance of the line conductor loop pul

G = conductance between signal path and return path (due to dielectric) pul

C = the capacitance pul between the signal path and the return path (it increases with the Dk of the dielectric)

For a uniform transmission line, R, L, G, and C are the same at every point, so Zc has the same value at every point on the transmission line.

For a sinusoidal signal with a frequency of f (ω = 2πf) propagating along the line, the voltage and current expressions at different points and times are given by:

Where α and β are the real and imaginary parts of the PCB transmission line loss, given by the following formula:

At the frequencies we are interested in, R << ωL and G << ωC, therefore:

And: the loss of the PCB transmission line to:

This means that a wave propagates the loss of a PCB transmission line with a propagation delay per unit length, and attenuates as it propagates along the line.

The signal attenuation coefficient of a transmission line of length l is:

The attenuation or signal loss factor is usually expressed in dB.

Therefore, the dB loss is proportional to the line length. Therefore, we can express the above as dB loss per unit length:

We usually omit the minus sign and remember that it is a dB loss-always subtracted from the signal strength in dB.

The above is also called the total insertion loss per unit length of the transmission line, written as:

Now, the R/Z0 component of loss is proportional to R (resistance per unit length), which is called conductor loss and is caused by the resistance of the conductor that forms the transmission line. It is represented by'alfa'C. The loss of the GZ0 part is proportional to G-the conductance of the dielectric material, called the dielectric loss-denoted by "alfa" d.

Where R is the resistance of the conductor per inch.

Now, there are two conductors in the PCB transmission line-the signal trace and the return path.

Usually, the return path is a flat surface, however, the return current is not evenly distributed on the flat surface-we can prove that most of the current is concentrated on a strip that is three times the width of the signal trace, just below the signal trace.

Signal trace resistance in PCB transmission lines

Does the entire cross-sectional area of the signal trace participate in the signal current equally? The answer is: not always the case-it depends on the frequency of the signal.

At very low frequencies-up to about 1MHz, we can assume that the entire conductor participates in the signal current, so Rsig is the same as the "alpha" C resistance of the signal trace, namely:

ρ = Copper resistivity in ohms-inch Loss of PCB transmission line

W = trace width in inches (for example: 5 mils, or 0.005" trace 50 ohms)

T = trace thickness in inches (usually ½oz to 10oz, that is, 0.0007" to 0.0014")

For example, for a 5 mil wide trace:

For our purposes, we are interested in AC resistance at frequency f. Here, the skin effect enters the picture. According to the skin effect, the current with frequency f only spreads to a certain depth, which is called the skin depth of the conductor

We can see from the above that at 4MHz, the skin depth is equal to 1oz of copper thickness, and at 15MHz, it is equal to ½oz of copper thickness. Above 15MHz, the depth of the signal current is only less than 0.7 mils, and it keeps decreasing as the frequency increases.

Since we are focusing on high frequency behavior here, we can safely assume that T is greater than the skin depth at the frequency of interest, so we will use skin depth instead of T in the signal resistance formula. So we now have:

We use 2δ instead of δ, because the current uses all the periphery of the conductor-technically speaking, 2W can be replaced by 2(W+T).

The return signal propagates only with a thickness δ along the surface closest to the signal trace, and its resistance can be approximated as:

Increased conductor loss due to copper surface roughness at the conductor-dielectric interface

It is important to know that in the circuit board, the "copper conductor-dielectric interface" is never smooth (if it is smooth, the copper conductor is easily peeled from the dielectric surface); it is roughened into a tooth-like structure to increase the circuit The peel strength of the conductor on the board.