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I have seen that most of RF printed board circuits have a GND plane which is connected to the power supply ground terminal and serves as a return path for current from different components on the board.

Something like this:

enter image description here

What I have seen in many datasheet is also that this plane is assumed as reference potential for the output ports and indicated with the symbol:

enter image description here

like the classical GND terminal of an Op - Amp or other simple circuit.

What I do not understand is: How can be this plane represented by a single node? How can it be a reference voltage?

At high frequencies it is a transmission line, so its voltage will have some spatial oscillation, so it will not have a constant voltage. How can it be considered a reference potential for output ports?

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  • \$\begingroup\$ captcha makes the anwers uselessly slow At high frequencies, the return currents either remianunder th etraces, or very close. \$\endgroup\$ Commented Jul 3, 2020 at 6:27
  • \$\begingroup\$ An op-amp doesn't have a GND terminal. \$\endgroup\$
    – Andy aka
    Commented Jul 3, 2020 at 8:23
  • \$\begingroup\$ The transmission line is usually not represented on the schematic at all. \$\endgroup\$
    – user20574
    Commented Jul 3, 2020 at 9:57
  • \$\begingroup\$ @analogsystemsrf, how does the return currents under the traces lead to the ground plane having a uniform potential at all locations? \$\endgroup\$
    – The Photon
    Commented Jul 3, 2020 at 16:10

2 Answers 2

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Everything about transmission line theory is a useful approximation, not a detailed description of physical reality.

When transmission line effects are important, it's only an approximation to assign a scalar potential to any point in the circuit. Whether on the transmission line, on the ground plane, or at the terminations. Nonetheless we approximately define the scalar potential, and we find this is useful for predicting the circuit behavior at a high level.

In the microstrip transmission line, the actual behavior comes from the electromagnetic wave travelling between, and (approximately) confined by the boundary structure formed by the ground plane and the trace. We can define an (approximate) potential at each axial location along the line by the integral between the ground plane and the line (along the shortest possible path) at that axial location.

It isn't important whether the potential varies between different points on the ground plane or whether the potential varies between different points along the trace. It's only important that the (approximately defined) potential difference between the trace and the ground plane varies along the length of the line. Since this is all that matters, it's useful to assume that the ground plane remains an equipotential and that all the potential changes happen on the trace.

From experience we know that this approximation is close enough to reality to allow us to predict the circuit behavior as observed at the terminations of the transmission line reasonably accurately.

In summary: You are correct that in a high-speed circuit there will be potential variation between points on the ground plane. Nonetheless we make a useful approximation and assume the ground plane is at a uniform potential and we find this approximation is useful for predicting circuit behavior.

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It's all dependent on how the ground plane is constructed, any material has resistance, which means any current flowing in a ground plane will create voltage. For most designs this can be approximated to zero volts. For example 1oz copper has 5mohms of resistance per square inch. 10mA of current would produce at most 5uV of voltage, for most designs this could be considered zero. this means the ground plane can be represented by a single node at DC. High-speed signals need a transmission line to propagate. However for a microstrip line we simply need a reference plane not a ground plane, in the example above it doesn't matter what plane we use, as long as there plane is capacitively coupled to the source a load of the transmission line.

In fact at high speeds just a copper plane cannot carry high frequency signals because it has too much inductance.

For high speed signals the current follows the path of lowest impedance so to create low impedance it is necessary to create the right amount of capacitance between the microstrip line and the ground plane. When the micro strip line is sized correctly with the right amount of capacitance per unit length versus the right amount of inductance per unit length the wave can continue with very little loss. Most of the energy of the wave is not carried by the electrons but in the magnetic and electric fields created by source and carried by the transmission line.

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