Losses in Transmission Line

Ideally, a transmission line presents itself as a lossless medium for propagation of the high frequency signals. In most cases this is true for small lengths of transmission lines. However, if the length of the transmission line is large, or the geometry of the transmission line is such that it contains too little copper, or if the frequency of the impressed signal is very high, there is attenuation in the signal as it propagates from one end to the other. As PCB designers we may ignore the losses in transmission line for very small length of the transmission line. If the length of the transmission line is larger (say, greater than 10 inches) we must be aware of the transmission line losses and the ways to estimate it. There are three types of Transmission line losses Resistive loss, Dielectric loss, and Radiation loss.

Resistive Losses

Simply put, a resistive loss is the loss due to the finite series resistance the transmission line. The finite resistance of a transmission line gives rise to two undesirable effects. The first obvious result is the attenuation of the signal as it propagates down the transmission line. The second undesirable effect is that it distorts the signal by way of unequal phase shift of different frequency components of the signal.

Estimation of Resistive loss for High Frequency Signals

Let us assume the R is the series resistance per unit length, L is the inductance per unit length and C is the capacitance per unit length of a transmission line. At frequency w above R/L, the resistive loss for X inches is given by

Attenuation = e^{-[RX/[2*v(L/C)]]}



In terms of dB the loss per inch is given by
Loss per inch = 4.34 [R/v&sqrt;(L/C)] dB

We can use the following formula for estimating the resistance per inch of a PCB trace

R = [6.787 x 10^-7/(WT)] (1 + (Thermal_Coeff*(temp - 20)) Ohms per inch

Where R = Resistance per inch
W = Width of the PCB trace
T = thickness of the PCB trace
Thermal_Coeff = .0039 ohm / degC

temp = temperature at which the resistance is calculated.
Note that as the temperature increases, the resistance of the PCB trace increases. We may like to keep the formula simple by calculating the resistance per unit length at 20 deg C using

R = [6.787 x 10^-7/(WT)] Ohms per inch

You can use the online calculator to estimate the resistance of a given length of wire at http://referencedesigner.com/cal/cal_05.php

Example - A transmission line formed of microstrip of trace width 5 mils and width 1 mils spaced 5 mils above the ground has a characteristic impedance of 66.9 Ohm for Er = 4.0. Find the attenuation for 10 inch of such trace at 20 deg C. Solution – The resistance of 10 inch of the trace is given by

R = [6.787 x 10^-7/(WT)] x 10 inches
= 1.357 Ohms
The attenuation in dB for length L is given by



Loss for length X is = 4.34 [RX/v(L/C)] dB



Here RX = 1.357 Ohms



So,

Loss = 0.08 dB

You will often find the use of dB in literatures. It is time you learn what it means and how it is calculated. If you are already aware of the power ratio calculations in dB you may skip these sections.


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