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In reading the definition of impedance matched wiring a question popped into my head that I can’t seem to find the answer to. Assuming a 50 ohm impedance line that is infinity long, then in theory if I attached a voltmeter I am able to see 50 ohm resistance. However what keeps me bugged is what would be the current graph look like for a very long but not infinite transmission line that terminates in an actual load.

Suppose we have a very very long 50 ohm impedance matched transmission line that is connected in a 1k resistor (like a basic lamp circuit with very long wires). If I apply a 5 volt DC at the source, assuming an infinitely fast sampling ammeter I should first see the current of 0.1A before the current settles to 5 mA. Is this assumption correct or am I missing something. Thank!

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    \$\begingroup\$ Probably wrong, but I can't "decode" your setup. For example: "Suppose we have a very very long 50 ohm impedance matched transmission line that is connected in a 1k resistor" . Well then it is not matched, 50ohm line shall be terminated with 50 ohm, not 1k. \$\endgroup\$
    – Marko Buršič
    Commented Sep 5, 2019 at 8:57
  • \$\begingroup\$ You say 0.1A. For that to happen the 50 Ohm line should be grounded but in your case it is not, there is a 1k resistor in series. You are calculating 0.1A based on which electric law? (BTW 50 Ohm in transmission lines is not a usual resistance it is for RF) \$\endgroup\$
    – cm64
    Commented Sep 5, 2019 at 9:21
  • \$\begingroup\$ Sorry to clarifiy is meant that transmission line itself is a impedance matched to 50 ohms butt it is connected a load \$\endgroup\$
    – attle
    Commented Sep 5, 2019 at 10:07

1 Answer 1

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From your description, yes, initially you see the line-charging current of 5v/50_ohm or 0.1 amp current.

After the initial edge has propagated the entire length of the line and arrived at the (non-matched) 1,000 ohm lumped resistor load, the current into the resistor is only 5volt/1,000_ohm or 5milliamp. There will be a reflection, because the extra current, the 0.095amp, is still flowing; the differential-equation solution has boundary conditions on the energy and these boundary conditions require a large ( 95% ) overshoot on the voltage. However the reflected wave imposes a total current of 5v/1,000_ohm or 0.005 amp, and as that reflected wave propagates back toward your current-meter you STILL read 0.1 amp up until the reflected wave arrives back at the source.

Then your current meter changes readings, from 0.1 amp to 0.005 amp; that change occurs after a total delay of TWICE the electrical length of your cable.

Notice we've not discussed what happens when the reflected wave has arrived back at your 5 volt source. If that source has 50 ohm Rout, what happens? If that source has 0 ohm Rout, what happens?

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for other description, and perhaps better description than I've written, search for 'Bergeron model'; this model was invented by a French hydraulic engineer, seeking to understand water-hammer (water being a non-compressible low-loss medium of energy transmission).

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    \$\begingroup\$ I think you get a higher voltage at the resistor when the reflection starts. \$\endgroup\$
    – Jasen Слава Україні
    Commented Sep 5, 2019 at 9:26
  • \$\begingroup\$ So based on your logic a simple resistor circuit would get a huge current when turned on because it has a zero like resistor in series(wires) before the resistor and that doesn't happen. \$\endgroup\$
    – cm64
    Commented Sep 5, 2019 at 9:27
  • \$\begingroup\$ Jason ---I bet you are correct. Can the voltage become 0.1 amp * 1,000 ohms, which makes V_far_end_at_load be 100 volts? How do we write the boundary conditions? A better query ( to you ) is "What is conserved?" Are the electrons conserved? Thus the line current sets the wave-solution? \$\endgroup\$
    – analogsystemsrf
    Commented Sep 5, 2019 at 9:32
  • \$\begingroup\$ cm64 ---- we need to view the wires as combined inductances and capacitances, thus the impedance (of twisted-pair, maybe 100 ohms; or a precise coax of 50 ohms) defines the V/I ratio. \$\endgroup\$
    – analogsystemsrf
    Commented Sep 5, 2019 at 9:34
  • \$\begingroup\$ analogsystemsrf-- sorry to add but does this mean if a high voltage( say 1000v) power supply is connected to any transmission line that is impedance matched, it would see an initial very high abit very very short current spike? If it does, does these spike affect power supplies longevity at all? \$\endgroup\$
    – attle
    Commented Sep 5, 2019 at 10:10

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