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From what I understand, the voltage doubles since the reflected and incident pulses add up. But then why don't all open terminations have double the source voltage since we can consider the voltage there to be a pulse of infinite duration?

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Yes, the voltage doubles when the pulse hits the open end of the cable, but this change then propagates back toward the source end of the cable. What happens there depends on how well the impedance of the source is matched to the impedance of the cable.

If the source is matched to the cable, the initial pulse was only half of the nominal source voltage to begin with, because of the voltage divider action between the source impedance and the cable impedance. When this pulse gets to the open end of the cable, it doubles to match the orignal source voltage. Once this change propagates back to the source, then the entire cable is at the nominal source voltage, and no further reflections occur.

If the source has a very low impedance with respect to the cable, then the backward propagating pulse will be reflected there, too, but now it will be inverted, but slightly attenuated. It will propagate once again to the open end of the cable, where the whole thing repeats. You'll see a decaying oscillatory waveform on the cable until all of the reflections die out — which happens quite quickly on a short cable. Eventually, the cable settles down at the nominal source voltage.

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  • \$\begingroup\$ +1. This seems to most clearly address what I believe to be the OP's question. \$\endgroup\$ Sep 14 '14 at 22:35
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That's because a "pulse of infinite duration" is not at all similar to "a pulse of finite duration".

For the reflection, and all what comes from that, to be significant you have to be working with a time base that is somewhat similar to the propagation delay of the cable. Since the latter is generally in the 10ns ballpark you can easily see why that won't work.

When you turn on a lab PSU you probably have a spike at its terminals but the cables inside are so short that it is barely noticeable.

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  • \$\begingroup\$ In other words: that double voltage is indeed there at the end of your cable, but it is very short, and as soon as you start to measure it you are loading it, which makes it disappear. \$\endgroup\$ Sep 14 '14 at 8:56
  • \$\begingroup\$ Yeah @WoutervanOoijen my answer is not a high quality one huh? I thought: easy question, easy answer. If you have some suggestions please tell me, I'll edit the answer, or just provide yours that would probably be better. \$\endgroup\$ Sep 14 '14 at 9:06
  • \$\begingroup\$ I did not want to replace your answer at all, I just wanted to add a slightly different perspective that might make it easier for Phil to understand. \$\endgroup\$ Sep 14 '14 at 9:28
  • \$\begingroup\$ @WoutervanOoijen ok, thanks then. I wasn't implying anything, I am just more than happy when a question has three or four answers. \$\endgroup\$ Sep 14 '14 at 9:29

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