Timeline for time domain reflectometer determining impedance
Current License: CC BY-SA 4.0
8 events
when toggle format | what | by | license | comment | |
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Dec 26, 2020 at 14:08 | vote | accept | Yoomo | ||
Dec 20, 2020 at 12:55 | comment | added | Andy aka | At the instant the pulse is applied, the potential divider formed by the source impedance and the cable's characteristic impedance is created. It's instant so, if you measure after 1 ns or 10 ns or 100 ns or 1 us, you will get the same result (providing the cable is long enough so that far end reflections are avoided during the measurement period). | |
Dec 20, 2020 at 12:40 | comment | added | Yoomo | I am a bit confused. When I use a long square wave pulse like above the impedance will be a real value and can be calculated like this Vcable=1 volt⋅(100/150)? | |
Dec 20, 2020 at 12:29 | comment | added | Andy aka | @Yoomo use a pulse and pretty much after applying that pulse (via a resistor) look at the magnitude at the cable end where you apply the pulse. If the cable is short then you might get a reflection coming back too quickly and it screws up the measurement of course. | |
Dec 20, 2020 at 12:26 | comment | added | Yoomo | So with your formula I can calculate the impedance of the cable, when I am using either a pulse or a DC voltage? | |
Dec 20, 2020 at 12:24 | comment | added | Andy aka | @Yoomo If you are using pulses then this is going to be the case. The characteristic impedance will be \$\sqrt{L/C}\$ and that will be resistive above 1 Mohm. | |
Dec 20, 2020 at 12:22 | comment | added | Yoomo | But this assumes that the impedance of the cable is a real value? | |
Dec 20, 2020 at 10:53 | history | answered | Andy aka | CC BY-SA 4.0 |