Lumped Discipline – Why It Fails as Dimension Approaches Wavelength

lumpedtheorywavewavelength

I'm studying some basic introduction to the lumped matter discipline on my own. I completely understand where the first two constraints come from (i.e. lumped matter can only interact through their terminals, which is consistent with KVL and KCL), however, while I do understand what is meant by the last constraint, I don't feel like I get the full picture!

Here is a quote from the book I'm reading:

The signal timescales must be much larger than the propagation delay of electromagnetic waves through the circuit.

This part is pretty trivial, since the propagation of em waves is finite we have to agree to choose a large enough timescale to allow propagation to happen. (right?)

Then the author states this rule in another way. This is where I loose it:

Put another way, the size of our lumped elements must be much smaller than the wavelength associated with the V and I signals.

My question is, if all electromagnetic wave travel at the same speed, then why does wavelength matter? Said another way, why is it that as circuit or component dimension approaches wavelength this rule does not hold? (I'm really keen on knowing the why, the underlying logic, and physics behind it.)

My best guess at the moment is that because if the circuit is smaller than wavelength it will be in the first cycle of the wave and thus can be treated as equal along all point on that circuit. Am I correct?

Best Answer

The signal timescales must be much larger than the propagation delay of electromagnetic waves through the circuit.

This part is pretty trivial, since the propagation of em waves is finite we have to agree to choose a large enough timescale to allow propagation to happen. (right?)

We're not choosing a time scale for analysis; both of these are properties of the system we're analyzing. By "the signal timescales" they mean the time scales like the periods of the highest-frequency components of the signals; the time scale over which a signal changes significantly. By "the propagation delay … through the circuit" they mean the time for an EM wave to propagate across the physical size of the circuit (or other effects such as a delay line component).


if all electromagnetic wave travel at the same speed, then why does wavelength matter? … My best guess at the moment is that because if the circuit is smaller than wavelength it will be in the first cycle of the wave and thus can be treated as equal along all point on that circuit. Am I correct?

Yes, basically correct.

Imagine the input signal as an electromagnetic wave propagating across the circuit. As a specific example, imagine that a single input signal (say, a clock or carrier signal) is being distributed to many points on the circuit through a branching wire or PCB trace.

  • If the wavelength is large compared to the circuit board (or other form of construction) then the voltage at any point in this distribution system is nearly equal.
  • But if the wavelength is small, then the voltage any given point in the circuit will see for that same distributed signal is different depending on where that point is located (or rather, how much distance of trace or transmission line the signal passed through to reach it, which is not necessarily a straight line). The signal distribution exhibits a propagation delay and corresponding phase shift.

"Within the first cycle of the wave" is not a sufficient condition, because that includes points at which the voltage is equal and opposite (180° out of phase) to that at other points, and many interesting effects occur with merely 90° of phase shift. A rule often quoted in amateur RF work is that you want to stay within the first tenth of the cycle — or in the usual form described in terms of wavelengths, that the length of the component you wish to consider as a lumped element must be less than 1/10 of the wavelength of the signal.