Kitz Forum
Broadband Related => Broadband Technology => Topic started by: Weaver on November 29, 2020, 03:37:41 AM
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This seems like a stupid question but I realise that I have never known the answer. In DSL, if you measure the potential differences between the A conductor and true earth (the planet) and the B conductor and earth, what will those potential differences (voltages) be? I’m wondering if A = X Volts and B = -X Volts ? Or whether one side is 0 V or what ?
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With PERFECT balance, one side is at V volts wrt earth and the other is -V. In practice they wont be exact, With care the out of balance component (ie same on both sides) is typically - 30 dB probably better is possible at audio frequencies.
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Assuming that you're talking about the on-hook DC voltages, then the A leg is at ~0V relative to earth, and the B leg is at -50V.
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And the signal voltages, on top of dc? Are at +V and -V ?
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Out of interest I've just measured mine with respect to mains earth (which is connected to a long copper rod in the ground at the side of my house).
I got A = +3.5V, B = -53.3V.
:)
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Sand the signal voltages, on top of dc? Are at +V and -V ?
With xTU transceivers connected (and operational) at either end of the link the RF voltage present is miniscule. It is impossible to measure with the average DVM. Perhaps with a high quality, high bandwidth, oscilloscope something might be seen . . .
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@tubaman: thank you very much!
@burakkucat: Agreed. :-) I’m not sure what the signal voltage at the ATU-C end is in my case but downgrading that by 64 dB is going to have an impressive effect, 1V reduced down to 0.631 mV, according to my calculator.
I was wondering if the signal voltages were antisymmetrical.