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I’m a little confused regarding impedance matching with resistors.

As far as I know, the impedance of a trace is a part of its characteristics and it can’t be measured with an ohmmeter. For example, if we measure a trace of 50 ohm to the ground, the resistance will be infinite, and it's totally correct because we're talking about reactance effect of L & C components of the line.

Why do we match the output/input impedance with a parallel/series resistor?

I know this resistor is for absorbing the wave energy, but how we can relate this resistor to the line impedance matching?

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4 Answers 4

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Imagine an infinite board with an infinite trace on it. Apply 1V to the trace. What happens is that an impulsive wave, a step, propagates down the trace. If the current needed to keep this wave going is 0.02A, we say that the "characteristic impedance" of the trace is 1V/0.02A = 50Ω. The infinite trace behaves like a 50Ω resistor.

But that means that a 50Ω resistor behaves like our infinite trace. An infinite trace would swallow any waves transmitted and never reflect them back. Since the 50Ω resistor is electrically the same, it does the same thing.

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I know this resistor is for absorbing the wave energy, but how we can relate this resistor to the line impedance matching?

The characteristic impedance of the line is the ratio of voltage to current that can be carried by a forward or reverse-travelling wave on the line.

The resistance of the termination is the ratio of voltage to current that it allows.

If you relate these two impedances by making them equal, then all the energy in the wave is absorbed by the termination. If they aren't equal, then a reflected (reverse-travelling) wave must be produced at the point of termination in order for all of the system constraints to be met (Kirchoff's current law, the V:I ratio of the forward wave, the V:I ratio of the termination, and the V:I ratio of the reverse wave)

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if we measure a trace of 50 ohm to the ground, the resistance will be infinite

For a very short length of time the resistance measured (if you have the right equipment) will be 50 Ω. Afterwards it will become infinite.

But, in between, there will be voltages and currents sloshing around from end to end but, these perturbations will gradually settle down in a few microseconds (for a short line or trace).

Why do we match the output/input impedance with a parallel/series resistor?

Basically to avoid the perturbations mentioned above. If we match, the perturbations are low to zero in amplitude and wouldn't (for example) disrupt high speed data passing down the line.

I know this resistor is for absorbing the wave energy, but how we can relate this resistor to the line impedance matching?

If you choose a terminating resistor whose value doesn't match the characteristic impedance of the trace, the "unsuspecting" voltage and current travelling down the trace will "hit" a load (at the end) that doesn't match their ratio. That ratio is determined by the trace's characteristic impedance and you potentially have a problem with ohm's law unless a reflection back to the source is created.

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As far as I know, the impedance of a trace is a part of its characteristics and it can’t be measured with an ohmmeter. For example, if we measure a trace of 50 ohm to the ground, the resistance will be infinite, and it's totally correct because we're talking about reactance effect of L & C components of the line.

It can be measured by an ohmmeter, but the ohmmeter needs to be an extremely fast instrument that reads current on the scale of 1 nanosecond. One can construct such a practical ohmmeter from a pulse generator, a low-inductance current shunt (or a current probe), and an oscilloscope. When you apply a Heaviside voltage step to a lossless transmission line, you can see that a current flows into the line, and its value is the ratio between the applied voltage and the line's characteristics impedance.

Paradoxically, voltage and current are in-phase with each other, as if it's a resistor, but the ideal transmission line only contains inductors and capacitors. Since energy-dissipation is not allowed, the circuit should not be resistive. The solution is that, this circuit only becomes reactive when the pulse reaches the end of the line and starts reflecting back. After many reflections, the line reaches steady-state and behaves like a reactive component, as seen by an ordinary ohmmeter or impedance bridge. But before that occurs, it behaves like a resistor. Another way is to say that information about what is located at the other end of the transmission line propagates no faster than the speed of light inside the dielectrics.

Other phenomena found in ordinary circuits also exist for transmission lines, momentarily. For example, if you insert a series resistor between an ideal voltage source and a transmission line, at the mid-point, it forms a voltage divider. When the voltage source is turned on, initially only half the voltage can be measured on the line itself. The half voltage is doubled when the pulse reflects back from the far end of the line. These facts highlight the physical meaning of the name characteristics impedance - this impedance can be very real (in both senses of the word) indeed. It's also why characteristics impedance is sometimes called surge impedance.

If the line is infinite, the reflection never occurs and the line is always resistive. To simulate a infinite line with a finite line, simply terminate its far end with a resistor set to the line's characteristic impedance. This is why parallel termination eliminates reflection.

Note that for a lossy RCLG transmission line that physically exists, its behavior is significantly more complex than the lossless line, and these statements are only approximately correct within a short time window for a short line.

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