I've been trying to understand the principle of trace impedance (single ended and differential) for a while, read several articles (all the best google has to offer), but I have still some serious issues finalizing my understanding of it.

First of all, all this talk of fields around conductors made perfect sense for me, completely understandable on intuitive level in terms of where fields come from and what things affect it (trace width, distance to ground path etc.), no problem understanding physics principles behind it there (trace as inductors, resistors and capacitors - totally reasonable).

The problem starts when I hear the word "ohm" when talking about this impedance.

Pardon my ignorance, but when I hear the word "ohm", I understand it as resistance (or impedance) between two points when there is voltage applied to them. So if I see "the trace has 50 ohm impedance", I just don't understand it. To me, "50 ohm" means that if I apply 1 V (DC or some specific frequency or something), I will have 20 mA current.

I understand all the talk about the fields, but I fail to connect this understanding of how fields interact to the word "ohm". For now, it looks like measuring length with kilograms to me.

While writing this question, I was proposed by ee stackexchange to take a look at this: questions similar to what I have on ee stackexchange answered

And it did answer some questions (most of them, actually, such as why the trace impedance characteristic is frequency-independent), so if I get it right, these "ohms" are simply a trace geometry characteristic and have nothing to do with actual voltage and current in the circuits (no 20 mA current when I apply 1 volt?). Then why are they even called Ohms...(rhetorical question, I do realize it's just a final unit that comes out after mathematical operations no matter how counterintuitive for understanding it is).

So, if anyone could confirm or debunk my way of understanding trace impedance as something that has nothing to do with volts over amps and is just a geometrical characteristic (basically, "ohms that are not really ohms?"), I would be very glad.

I would be even more grateful if anyone had any good article or a book chapter that covers exactly that, preferably something that explained this to you personally; maybe it will help me to wrap up my understanding of it. I feel like I kinda get it, but I need some final "aha, now I get it!", and it's best when someone with knowledge answers it, I don't trust my own interpretation of things without anyone's confirmation on this matter.

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    \$\begingroup\$ Useful search term : transmission line theory. But be warned, you have quite a bit of reading to do. \$\endgroup\$
    – user16324
    Aug 26, 2021 at 11:58
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    \$\begingroup\$ @user_1818839 reading is exactly what I wanna do. And it's good to have recommendations from actual fleshy humans. Thank you \$\endgroup\$
    – Ilya
    Aug 26, 2021 at 12:32
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    \$\begingroup\$ Two more old questions that might help you: How is xΩ impedance cable defined? and Why doesn't controlled impedance depend on track length? \$\endgroup\$
    – The Photon
    Aug 26, 2021 at 17:43
  • \$\begingroup\$ "50 ohm means that if I apply 1 V, I will have 20 mA current" - yes, that's what you will observe for a certain time. Or, if you connect an actual 50 Ohm resistor at the end of the trace, you'll observe that current indefinitely. So it' Ohms that really are Ohms. \$\endgroup\$ Aug 27, 2021 at 9:18
  • \$\begingroup\$ While the signal is in flight in the transmission line, and no reflection has come back from the load to the source, the transmission line looks and behaves (from the perspective of the source) as a resistor with Z0 Ohms, where Z0 is the characteristic impedance of the transmission line (often 50 or 75 Ohms, but not always). \$\endgroup\$
    – user57037
    Aug 27, 2021 at 17:03

5 Answers 5


Ohms has dimensions volts/amps.

When applied to a resistor, it's the ratio of applied voltage to current through the resistor.

When applied to the behaviour of transmission lines, it's the ratio of the voltage wave going along the line and the current wave travelling with it.

When applied to the geometry of transmission lines, it the square root of the inductance/capacitance ratio for any given length of line. Inductance has units Vs/A, capacitance has units As/V, so their ratio L/C has units ohms2.

It's not a coincidence that the behaviour of a line is linked to its geometry by exactly the same factor, as the inductance and capacitance control the voltage/current ratio of waves on the line.

Two waves can travel at the same time on the line, one wave travelling in each direction. At each point, the two voltage waves add up to give you what you would read on an oscilloscope probing that point. The two current waves are differenced (as one is going forward, one backwards) to give you the current that would be read on an ammeter in series with the line. To read each wave in isolation requires a more complicated probe called a directional coupler.

Note that the impedance ratio does not apply to static conditions, only to propagating waves. If you connect a transmission line to a source and load and wait until it has reached the steady state, the voltage and current will be entirely defined by the source and load, not the line.

If a transmission line has a 50 ohm impedance, then connecting it abruptly to a 1 V source will cause a 1 V voltage wave and a 20 mA current wave to start travelling along the line.

When these waves get to the end of the line, they may find a 50 ohm resistor. In which case the voltage and current are in exactly the right ratio for the resistor. Current will flow in the resistor, one volt will be developed across it, and things will stop changing.

Alternatively, when the waves get to the end of the line, they might find an open circuit. Current cannot flow out of the end of the line, so a 20 mA wave is generated to flow back down the line, to cancel the current at the open circuit. This is the reflection. This propagates with a 1 V reverse travelling wave back towards the source, doubling the voltage on the line from 1 V to 2 V as it goes.

The waves might find a short circuit, or a resistor with some other value, or a cable of another impedance connected to its end point. In each case, a reverse wave would be generated such the ratio of the voltage sum to the current difference was correct for the V/I ratio of the following section, whether it was a real resistor or another transmission line.

I like to think of the repeated reflection process as a negotiation (inappropriately anthropomorphic, but I like it), whereby the source and load at the far end gradually come to an agreement about how much current the source should supply to the load. Let's say you have a 12 V battery, a 50 ohm transmission line, and a 10 ohm load, which will draw 1.2 amp when the line has reached steady state. Closing a switch at the battery end will send a 12 V 240 mA wave along the line. The load does not match the line impedance, so will send a -8 V -160 mA wave (still the 50 ohm ratio in the line) which adds up to 4 V and 400 mA (a 10 ohm ratio in the resistive load), dropping the line voltage to 4 V as it goes. This reflection will reflect off the source, and will continue up and down the line, changing the current at each transit. The size of the successive reflections falls with time as the current in the line rises in discrete steps to its final value.

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    \$\begingroup\$ That last paragraph was really helpful! \$\endgroup\$
    – DamienD
    Aug 26, 2021 at 20:06
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    \$\begingroup\$ That is one of the best explanations I've ever heard, bravo! \$\endgroup\$
    – canton7
    Aug 27, 2021 at 13:09
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    \$\begingroup\$ I like to think of it as a negotiation also. And because of propagation speed and causality and conservation of charge and energy, the only way they can negotiate is by way of reflections. That is just how I think of it. Not claiming it represents the physical world accurately. \$\endgroup\$
    – user57037
    Aug 27, 2021 at 17:00
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    \$\begingroup\$ It is important to note though, that the transmission impedance does not behave like an ohm resistor (except for DC, but one usually doesn't call it transmission impedance for DC). If you double the length of the line, the transmission impedance stays the same. \$\endgroup\$
    – lalala
    Aug 28, 2021 at 17:42
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    \$\begingroup\$ @lalala it behaves exactly like an ohmic resistor if the line is infinitely long, or terminated in a matched load. \$\endgroup\$
    – Phil Frost
    Aug 28, 2021 at 18:18

Transmission lines have a characteristic impedance. In some languages it is also called the wave impedance, because it is high frequency signals that feel the impedance as they travel along the transmission line, as the speed they travel is somewhat slower than speed of light.

The transmission line can be thought as infinitely small capacitances between the wires that need to be charged, but also these infinitely small capacitances are connected with infinitely small pieces of wire which have inductance, so capacitor charging is limited by inductance. The ratio of unit inductance and capacitance basically defines the impedance.

Assuming you take an infinitely long piece of ideal coaxial cable which has a 50 ohm characteristic impedance, and apply 1V voltage to one end of it, the voltage wave starts travelling along the cable and it will draw 20mA current from the 1V supply.

It really means ohms, volts and amps.

Which incidentally means that if you have a finite piece of transmission line, what happens at the other end of it when a wave with 1V and 20mA hits it. If it has an open end, there's no place for the current to flow when voltage hits 1V at the end. If there is a short circuit, the voltage will stay at 0V but 20mA flows. Those are discontinuities and the only way to make the signal come out properly from the trasmission line is to terminate it with something that allows 20mA to flow at 1V - which is a matching 50 ohm resistor, and thus it makes a short piece of transmission line to look infinite in length and so it prevents reflections that happen due to impedance mismatch.

  • \$\begingroup\$ tbh the last paragraph looks like abracadabra for me. I understood it until "but 20ma flows". And then there are some discontinuities (wat?), signal's coming out and termination (sounds very homophobic), where I totally lost it, because to me it sounds like all single-ended signals now must be terminated to ground with 50Ohm resistor. Which is not true (?), but which in my incorrect understanding follows from your words \$\endgroup\$
    – Ilya
    Aug 26, 2021 at 10:52

Let's say your friend, who's a bit of a prankster, has somehow managed to pick a resistor of value unknown to you, and delivered it to Mars. Your friend has also somehow managed to obtain a very long length of zero-resistance speaker cable, and one end is connected to the resistor, and the other end is in your hands, still on earth.

Your friend hands you an ohmmeter and declares, "I've hidden your car keys. Tell me the value of the resistor at the other end of this cable, and I'll return them."

You connect the ohmmeter to the cable. What value does the ohmmeter read? It can't possibly read the value of the resistor: to Mars and back is 15 minutes at the speed of light. You can't know the value of the resistor instantly without violating causality.

So what does the ohmmeter read? It reads the characteristic impedance of the cable, at least for the first 15 minutes.

This works because the ohmmeter applies a known voltage to its terminals, measures the current, (or it might apply a known current and measure the voltage, either way works) and calculates the resistance with Ohm's law. In the minutes after you've connected the ohmmeter, the only thing determining the ratio of volts per amperes is the characteristic impedance, which does relate to real volts and amps.

Of course eventually the wave started by the ohmmeter reaches the resistor, at which point the volts per amperes might be too high, too low, or just right, depending on the value of the resistor. Unless it was just right, another wave propagates from the resistor back at the ohmmeter carrying this information. When that reflected wave reaches the ohmmeter it will have to adjust its current up or down to maintain its fixed voltage, thus making another "guess" at what the resistor's value may be. These waves propagate back and forth many times as the ohmmeter's display asymptotically approaches the actual value of the resistor.

In less absurd situations, this often all happens so fast it might as well be instant, and so there's no reason to think about characteristic impedance. But sometimes you are designing something very large (like a power grid) or very high frequency (like high-speed digital and RF circuits). In those applications the time it takes these waves to propagate is significant, and characteristic impedance becomes relevant.

  • \$\begingroup\$ interesting analogy. Unusual, yet quite intuitively-friendly. Thank you for your contribution \$\endgroup\$
    – Ilya
    Aug 28, 2021 at 8:33
  • \$\begingroup\$ But sometimes you are designing something very large (like a power grid) or very high frequency (like high-speed digital and RF circuits). I’ve always wondered why it’s an issue in HF designs, but I’ve never had it explained why. This answers that! \$\endgroup\$
    – Cole Tobin
    Aug 30, 2021 at 21:01

so if I get it right, these "ohms" are simply a trace geometry characteristic and have nothing to do with actual Volts and Amps in the circuits (no 20ma current when I apply 1 volt?).

There will be an initial current of 20 mA that flows when you apply 1 volt to a 50 Ω transmission line. It will look like pure resistance too. This is because you are actually transmitting power down the transmission line. Useful power can only be extracted from a source when the load (the transmission-line) looks like a resistor wholly or partly.

When I say "transmission line" I mean coax, twisted pairs or PCB traces. If the line is characterized as being 50 Ω then 20 mA of current will initially flow. But it won't flow for very long.

However, to understand it better why not assume a perfect loss-less transmission-line of infinite length. It's loss-less so there is no power dissipation in the cable/coax/twisted-pair/traces.

So you apply 1 volt at the end of the cable and 20 mA will flow indefinitely. Power is taken from the source by the infinitely long cable and, for a 50 Ω cable with 1 volt applied, 20 mW is consumed. The fact that it isn't converted to heat is because that power is still flowing at near light-speed down the infinitely long cable forever\$^1\$.

So, if anyone could confirm or debunk my way of understanding trace impedance as something that has nothing to do with volts over amps and is just a geometrical characteristic (basically, "Ohms that are not really Ohms?"), I would be very glad.

As far as the source is concerned, the cable/trace looks like a pure 50 Ω resistor. Of course we can't have an infinitely long cable so the really interesting stuff happens when that power (volts x amps) reaches the end of the cable and finds that there isn't a load connected that matches 50 Ω.

\$^1\$ If you lit a lamp in the middle of the universe, the lamp still emits power into free-space. It's the same mechanism for a cable but, a cable constrains the voltage and current just as a fibre-optic cable would for the lamp.

  • \$\begingroup\$ I have to admit, your reply sounds very smart. So smart, that I'm not sure I understand all of it. Maybe I do. Maybe, not everything. Basically, you mean 50Ohm is the resulting reactance for the front of the signal? For the moment when the voltage actually jumps from 0 to 1V in my case? Btw, I think you should write scripts for the movies. " so the really interesting stuff happens when that power (volts x amps) reaches the end of the cable and finds that there isn't a load connected that matches 50 Ω. " Haven't seen cliffhanger this good for a while. So what happens then? :D \$\endgroup\$
    – Ilya
    Aug 26, 2021 at 10:20
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    \$\begingroup\$ @Ilya it gets very complicated and lengthy to explain what happens next but basically, if the volts and amps cannot find a suitable value load resistor to dissipate the power they implicitly carry, then that power (some or all of it) gets reflected back to the source. And then it gets more complex and complexity builds upon complexity. \$\endgroup\$
    – Andy aka
    Aug 26, 2021 at 10:23
  • \$\begingroup\$ Yeah I've read about reflections; but this seems like it should happen to ALL single-ended signals then? I mean, they all have some trace impedance, and then when the front of the signal reaches the input of some chip, which is usually megaohms or something, then it gets reflected? So extra noise on single-ended signal like that? Apologize for keeping this going; also, you didn't answer the question in the first comment, and I think it's essential for me to make sure I understand what's going on (also, not marking your answer as correct yet in case someone else says something) \$\endgroup\$
    – Ilya
    Aug 26, 2021 at 10:27
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    \$\begingroup\$ You shouldn't accept any answer for at least a day to give others the opportunity to make answers. However, you have the reputation to upvote any helpful answer @Ilya \$\endgroup\$
    – Andy aka
    Aug 26, 2021 at 10:28
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    \$\begingroup\$ Basically, you mean 50Ohm is the resulting reactance for the front of the signal? - yes, for any length of t-line, the initial reactance presented by it is defined by its characteristic impedance (50 ohm, 75 ohm etc..) \$\endgroup\$
    – Andy aka
    Aug 26, 2021 at 10:33

All circuits have resistance (R), inductance (L) and capacitance (C). Taken together, these form the overall impedance of a circuit.

Regardless, the unit of impedance is still Ohms. Expressed another way, DC resistance is a special case of impedance where there is only the resistive element and no reactive elements (L and C) in play.

Transmission lines have an additional element: signal flight time. They are modeled as a series of L-C unit impedances which determine the net impedance of the line. The trick is, with a transmission line the signal flies down the wire as a wave. But the driver only ‘sees’ a net unit impedance at the driven end, not the whole lumped L and C of the line.

Here's a rather fanciful animation of transmission lines that may help you visualize their behavior (youtube link)

And a Falstad simulation

  • \$\begingroup\$ All circuits? What's the resistance of a diode? \$\endgroup\$
    – Phil Frost
    Aug 28, 2021 at 0:09
  • \$\begingroup\$ A function of its applied voltage. \$\endgroup\$ Aug 28, 2021 at 0:41
  • \$\begingroup\$ If you're comfortable saying resistance can be a function, then you don't need inductance or capacitance. You can say for example the resistance of an inductor is a function of a voltage integral. Which really, is just saying that all circuits have some functions that define the current and voltage, and you can take the ratio of the two. This is correct, and useful in some small-signal models, but not what most people mean by "resistance" without further context. Only linear circuits have resistance, by the more usual definition. \$\endgroup\$
    – Phil Frost
    Aug 28, 2021 at 18:14
  • \$\begingroup\$ Exactly. A pure 'resistance' has a simple relationship to applied voltage. It's the only thing that does. Inductance, capacitance, transmission line, diode junction, memristor... have more complicated models, with time-varying elements (some historical). OP is trying to grasp the difference between a transmission line and a pure resistance, which is impossible to explain except as the transmission line's reaction to time-varying inputs and the resultant launched waves. That said, all real circuits (even diodes) have R/L/C components that must be taken into account. \$\endgroup\$ Aug 28, 2021 at 18:31

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