# Transmission line equation

• posted

L > Tr/2/Td

where L is the length of the line, Tr is the rise time, and Td is the propagation delay.

or

L > 1/20/Td/F

where F is the frequency and the rise time is 10% of that.

Example:

F = 10Mhz and Td ~= 1.5ns/ft then the L > 40in to be considered a transmission line.

This is a formula I found online. The real question is what happens when your lenght is around the threshold? Is the L given above some nominal length which has a certain amount of ringing that is considered acceptable or is it the minimal length where transmission line effects start to become significant but are still relatively insignificant?

I'm about 35in and not sure it is really necessary to terminate for decent signal integrity or if it's really just a waste.

• posted

Well, if you don't know, the easy way is to test. Put a scope on it and see how ugly (or not) it is. It's going to be more critical (lower frequencies / shorter lines) if step response is important (e.g. building a 'scope), and less critical if you're just poking around (e.g., 74LS logic).

The rules of thumb are to get you in the order of magnitude. If you're an order of magnitude above, yep better terminate. If you're an order of magnitude below, screw it. If you're inbetween, take a closer look and see how much it really matters.

Tim

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• posted

Measure transmission lines in nanoseconds and compare to risetime of signals in them..

• posted

of

Huh? This does not say anything about the formula. Is L the threshold where reflection issues start to occur or only when they become significant?

• posted

More the later. There are *always* reflections, but the idea is that if the "speed" of your signals (for sine waves, pretty much just the frequency, as the maximum slew rate of a sine wave is just 2*pi*f V/s -- for digital signals, the slew rate of the edges) is slow enough that there's time for a handful of reflections to have occurred (based on the length of the transmission medium), you (or, say, your oscilloscope probe or your circuit) won't "see" the reflections but rather just the final voltage that the reflections are building up to, and hence you can treat everything like lumped elements.

The "lumped element" approximation used in standard circuit analysis effectively assumes that the speed of light (in your medium) is fast relative to your signals or, equivalently, the physical size of your circuity is small relative to a wavelength at the frequencies of interest.

How much any of this matters depends a lot of the type of signal you're sending, e.g., narrow band analog signals modulated at 1GHz care a lot less about a bit of mis-termination than, say, low-amplitude 1GHz digital signals (e.g., LVDS).

---Joel

• posted

Td is more like 1ns/ft on FR4 or in wire. 1.5ns is the number for free space.

Your formula is whacked. The frequency of the signal has no bearing on the issue. The rise time determines the frequency *content*, which is the important issue.

Think about it for a moment. If the transitions die before the edge resolves you won't see the reflections, rather the effects will effect the transition but not be noticeable as reflections directly. Where you'll see it is where the length of the line is greater than or equal to edge rate. If the line length is equal to the edge time you'll see the reflection in the middle of the edge, adding an overshoot (bad) or step (worse), depending. You'll see the effects outside the edge when the round trip time ( 2xlength) is greater than the edge.

As you can see, the frequency of the signal doesn't matter.

• posted

You can also look at this in the frequency domain.

A perfect square wave consists of a fundamental and odd harmonics (each with the amplitude 1/n in which n is the harmonic order) with zero phase shift between them.

To get a waveform that quite reasonably resembles a square wave, at least the 3rd and 5th harmonic should be present at correct phase and reasonably correct amplitude. Thus to analyze how the 1 GHz square wave behaves, you also have to look at the propagation at 3 and 5 GHz.

Even in a matched practical (i.e. lossy) transmission line, the phase between the harmonics is maintained, but the higher harmonics are attenuated more than the fundamental, so the edges are rounded off more or less, depending on the total attenuation.

When feeding a sine wave into a mismatched transmission line and measuring the voltage along the transmission line with some RF probe, There are going to be steady peaks and dips every quarter wavelength along the length of the transmission line due to the combination of the forward and reflective wave. There is also a 180 degree phase shift between the peak and next dip

Feeding different frequencies into the mismatched line will show peaks and dips at different places along the line. Feeding 1, 3 and 5 GHz (i.e. a square approximation) will show peaks and dips 3 and 5 times more closely along the line at 3 resp. 5 GHz than at 1 GHz.

If you move a probe slowly along the line, the amplitude and phase of the fundamental and harmonics will change constantly, combining into all kind of waveforms, some of which even resemble a square wave (at every 1/2 wavelength of the fundamental).

So when transferring a square wave across a mismatched transmission line, the length should be a 1/2 wave multiple at the fundamental frequency in that medium (i.e. corrected by the velocity factor). This may be acceptable for a constant frequency clock, but any random digital signal will have a large variety of fundamental frequencies, so this does not help, so you really have to match the transmission line system.

When is a line "long" in order to be considered a transmission line and require proper impedance matching ?

A 1/4 wave transmission line shorted at the far end will show infinite impedance at the generator end and an open 1/4 wave transmission line will look like a short circuit at the source. So definitely 1/4 wavelength is a significant length.

With typical PCB materials (with v=200 mm/ns) at 1 GHz this would be

50 mm. For a square wave the fifth harmonic (5 GHz) 1/4 wavelength is 1/20 wavelength of the fundamental or 10 mm. When relaxing the fidelity requirements for the square wave and considering only the 3rd harmonic (3 GHz), the 1/4 wavelength is the 1/12 wavelength of the fundamental or about 17 mm.

The common rule that a line longer that 1/10 wavelength is good for sine waves, but IMO for square waves something like 1/15 to 1/25 wavelengths should be used.

Paul

• posted

A dielectric that speeds waves up. Neat trick that!

;-)

Usual rule of thumb is 1ns/foot in free space or air, 1.5ns/foot in polyethylene.

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• posted

Oh, crap! Nevermind.

• posted

Crap happens:-)

```--
"Electricity is of two kinds, positive and negative. The difference
is, I presume, that one comes a little more expensive, but is more```
• posted

The formulas you are using are rule of thumb. An easier rule of thumb is transmission lines longer than 1/6 of your signal rise (or fall) time should be terminated, or at least scrutinized carefully. You are concerned about the transition times of your edges. The following article discusses termination issues in easy to understand language.

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