5
\$\begingroup\$

With parallel termination of a transmission line, the line characteristic impedance \$Z0\$ is matched with the pull-down resistance \$Rt\$. But, in that case, shouldn't the receiver see half of the transmitted voltage?

In my understanding, with no reflection, the whole situation can be brought back to a voltage divider with equal resistances.

Obviously, I made an error somewhere. So, how can the receiver see the full voltage with parallel termination?

\$\endgroup\$
  • \$\begingroup\$ For a long line, terminated in Zo at the receiving end, the signal sees Zo for its entire journey. It doesn't see a voltage divider when it arrives at the receiving end. \$\endgroup\$ – Chu Feb 2 '18 at 10:45
  • \$\begingroup\$ Intuitively i though that accross an impedance, voltage dropped linearly. Voltage drops "in chunks" then ? \$\endgroup\$ – Sachiko.Shinozaki Feb 2 '18 at 10:47
  • \$\begingroup\$ I don't get exactly what you mean, but the signal is always travelling into an impedance of Zo (normally a resistance). When it arrives at the end of the line it still sees Zo, even though it's physically not the line anymore, but a resistance (whatever that may be: resistor, antenna...), that dissipates the power in the signal, so nothing gets reflected back to the sending end of the line. \$\endgroup\$ – Chu Feb 2 '18 at 12:14
6
\$\begingroup\$

Line impedance is not the same as a series resistance which would cause a voltage drop. Line impedance, altough it happens to be measured in ohms, tells how the electric and magnetic fields are related outside the wires (=in the space where the wave actually travels, it's not in the metal). You can use the line impedance in calculations of what happens to a wave in line joints and terminations, but it has no use in ohms law.

In ordinary 2 wire cables, but not in waveguides, you can make most of the wave calculations by using the voltages and currents which they cause to the wires, but remember, that the wave and thus also the actual energy flow is outside the metal, it's only guided by the wires.

In your case the wave comes from the source, which obviously has very low internal series resistance. The wave meets the matched load, no reflection is caused. If the signal is DC, the rest can be calculated with ohms law. You get full voltage if the wire hasn't remarkable DC resistance.

\$\endgroup\$
2
\$\begingroup\$

The characteristic impedance of a cable is the value that should be taken into consideration when terminating to avoid reflections; it doesn't mean that it acts as that impedance in a potential divider sort of way.

Consider a really short 50 ohm coax terminated in 50 ohms driven by a zero ohm source at a low frequency - the full driving voltage will appear at the load both mathematically and intuitively (if you think about it).

\$\endgroup\$
2
\$\begingroup\$

I think you are confusing, case A:

schematic

simulate this circuit – Schematic created using CircuitLab

With, case B:

schematic

simulate this circuit

The difference between A and B is only at the source. The voltages I indicated is the amplitude of the signal at that point.

Indeed you're right that there is a 2:1 voltage division between the voltage at Vsource and the output voltage. That's because Rsource and Rload make a voltage divider.

However the voltage at Vsource isn't really the input voltage of the transmission line. That voltage is only present at the right side of Rsource.

In case B this is more obvious as I have replaces the voltage source and series resistor with their Thevenin equivalent, a current source and parallel resistor.

Note how the characteristic impedance of the transmission line is not actually present in the circuit. The characteristic impedance is the impedance with which you should feed in and extract power from the transmission line. If you don't do that you get signal reflections.

\$\endgroup\$
  • \$\begingroup\$ So basically the voltage across the transmission line is 0, right ? \$\endgroup\$ – Sachiko.Shinozaki Feb 2 '18 at 10:56
  • \$\begingroup\$ Yes, it is. At least when we use the T-line properly and it is also lossless. And that's good because then we're not losing any power. All the power going in comes out again at the other end. Which is nice. \$\endgroup\$ – Bimpelrekkie Feb 2 '18 at 10:57
  • \$\begingroup\$ The voltage across the transmission line is not zero. \$\endgroup\$ – Chu Feb 2 '18 at 12:33
  • \$\begingroup\$ @Chu Indeed it would be more precise to say: The amplitude of the signal at the input of the T-line is the same as the amplitude of the output. The voltage between input and output is indeed not zero as a wave travels through the T-line and depending on the wavelength (frequency) and length of the T-line to voltage across the T-line could be zero but does not have to be. \$\endgroup\$ – Bimpelrekkie Feb 2 '18 at 12:47
  • \$\begingroup\$ Probably my fault - I didn't read it as the voltage across the length of the line. \$\endgroup\$ – Chu Feb 2 '18 at 15:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.