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I thought that the characteristic impedance was a parameter defined only for transmission lines. But now I have this question: is it defined also for a port? And, eventually, what does it represent and which are the physical parameters that determine it?

This doubt came to me when I saw that for instance it is possible to evaluate the input reflection coefficient of a transistor, which is not a transmission line.

For instance, let's consider this scheme:

enter image description here

These blocks are connected by wires, and not by transmission lines. But there are some reflection coefficients, which are defined starting by characteristic impedances. What do they represent?

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  • \$\begingroup\$ Please clarify why the input of a transistor is not a transmission line. \$\endgroup\$ – Peter Smith Jun 27 at 14:20
  • \$\begingroup\$ The purpose of your education NOW is to learn HOW to LEARN. What methods do you use? en.wikipedia.org/wiki/Characteristic_impedance then how is impedance mismatch measured? \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Jun 27 at 14:24
  • \$\begingroup\$ @Sunnyskyguy EE75, if I asked a question, the reason is that I DO NOT UNDERSTAND what I find on the web, and I need some explanations \$\endgroup\$ – Kinka-Byo Jun 27 at 14:27
  • \$\begingroup\$ input reflection coefficient of a transistor, which is not a transmission line. That generally applies to a transistor circuit which is biased (so that it can amplify a signal for example) and it will then have a small signal input impedance. That impedance will need to be matched (for optimum power transfer) to the transmission line feeding the signal to the transistor. Go research "RF LNA" and you will see examples. \$\endgroup\$ – Bimpelrekkie Jun 27 at 14:28
  • \$\begingroup\$ @Kinka-Byo If you ask a question and cannot find the answer then you are asking the wrong question due to a false assumption, so then change your assumptions. The answers are all out there. but one cannot find them with the wrong question. If you dont know the right question re-read the fundamentals such as the WIKI link until you understand how to learn how to ask yourself a better question \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Jun 27 at 14:32
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A port does not have a characteristic impedance. It has an input impedance, or a port impedance, or just an impedance.

The characteristic impedance of a line is the ratio of V to I that will produce a travelling wave on the line. It doesn't tell us what the actual ratio of V to I at any point on the line until we also know the magnitudes and phases of the travelling waves in both directions are.

The input impedance of a port tells use the actual ratio of V to I at the port terminals due to an AC excitation.

We also sometimes talk about the characteristic impedance of a system. This just means we'll design our system with a default characteristic impedance for the transmission lines in the system. (We might also choose to use different characteristic impedances for different parts of our system)

Once we have defined a characteristic impedance for our system, we can determine what the reflection coefficient is for a port with a given port impedance if it is used in that system. If we actually used lines with different characteristic impedance than the system characteristic impedance, we'd have to do some additional math to calculate the actual reflection coefficient where the line connected to a port, since both the line and the port will have non-zero reflection coefficients in that system.

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  • \$\begingroup\$ therefore, if you consider the scheme in my question, are the reflection coefficients calculated with respect to the characteristic impedances of the transmission lines which connect the components? And if are the components simply connected by brief wires? \$\endgroup\$ – Kinka-Byo Jun 27 at 15:41
  • \$\begingroup\$ @Kinka-Byo, since there are no transmission lines in your example, they must be calculated relative to the system \$Z_0\$ as explained in my last paragraph. \$\endgroup\$ – The Photon Jun 27 at 15:58
  • \$\begingroup\$ but what does determine the system's characteristic impedance, if transmission lines are absent? \$\endgroup\$ – Kinka-Byo Jun 27 at 16:03
  • \$\begingroup\$ @Kinka-Byo, it's in principle an arbitrary choice of the system designer. In practice different application areas almost always use specific system impedances. For example, video systems very often use 75 ohms and other RF systems very often use 50 ohms. But in a classroom context, you could get a problem saying "this is a 27.5 ohm system" and you'd be expected to be able to calculate the reflection coefficients relative to that arbitrary defined value. \$\endgroup\$ – The Photon Jun 27 at 16:06
  • \$\begingroup\$ Normally you'd choose to make the system impedance equal to the characteristic impedance of the cables you think are most likely to be used to connect to the system. \$\endgroup\$ – The Photon Jun 27 at 16:07
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A reflection coefficient such as s11, s22 depends only on impedance ratios to measure mismatch, and does not need but may include a transmission line.

E.g. a 50 ohm sig-gen puts out twice the expected voltage until a matched load of 50 Ohm load is applied. 

Why? 

There is no effective transmission line. 

Why? 

Because the propagation delay is almost nothing or at least LESS THAN it's rise time.

So matched impedances divides the voltage in half and that gives to excellent Return Loss? meaning minimal signal returned as an echo? or **Maximum Loss of Returned signal** Yes.

Right

So unloaded it is double what you expect if the input or "dial" is calibrated by this method.

Yes...

So I don't need a transmission line or a delay much longer than the rise time. (10~90%)

Right, but you could.

ok..

So what does a unloaded step pulse look like on a transmission line with 0 rise time? ( infinite bandwidth)

enter image description here

Is Return Loss used with power supplies?

Yes, it's used on the grid. but not generally for lab supplies.

We expect the load R always to be much greater than the power source.

But when huge Caps with low ESR are placed far away, then you need to consider it as driving an AC short circuit transmission line. with a mismatch and echo causing ringing with the damping factor determined by how small the loop resistance was.

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The port impedance is simply a measure of how the port responds to external sources - regardless of it being a transmission line or not. For an infinite piece of transmission line, the port impedance will equal the characteristic impedance, but for shorter pieces, the two can be different.

Think of it this way: A characteristic impedance of 50 ohm just means that when you source a current of 1 mA, 50 mV of voltage will be generated, or, if you apply 50 mV of voltage, 1 mA will flow into the port. A complex impedance means there will be a phase shift in this current-voltage relationship - An impedance of 50*j Ohms will have the same 1 mA and 50 mV but with a 90 degrees phase shift.

Now how does that lead to reflections? Imagine the gate of your transistor is a capacitor, and presents a complex impedance. You send in a signal with a 50 ohm source. The capacitor will present a complex impedance, so it has to 'create' a opposite voltage wave to cancel out the incoming one exactly such that the voltage across it matches what it should be.

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  • \$\begingroup\$ Thank you very much, I have another question. Regarding the example of the capacitor, is its characteristic impedance equal to its "electrical" impedance? (so, 1/jomegaC)? \$\endgroup\$ – Kinka-Byo Jun 27 at 15:22
  • \$\begingroup\$ After reading the excellent reply by The Photon I realize I got characteristic impedance and port impedance mixed up. \$\endgroup\$ – Joren Vaes Jun 28 at 3:42

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