50 ohm cable

Discussion in 'General Electronics Chat' started by TAKYMOUNIR, Mar 20, 2013.

1. TAKYMOUNIR Thread Starter Active Member

Jun 23, 2008
351
1
in 50Ω or 75 Ω cable this mean that if i have 10 ampere go through this cable i will have voltage drop 500v or 750v in this cable
thanks

2. MrChips Moderator

Oct 2, 2009
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No. 50Ω or 75Ω cable is the AC impedance, not the DC resistance.

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3. kubeek AAC Fanatic!

Sep 20, 2005
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A side question, how do I find the frequency where the almost infinite parallel DC resistance starts getting close to the rated impedance, i.e. some corner frequency? I mean calculating that point, not measuring.

4. MrChips Moderator

Oct 2, 2009
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Any combination of DC resistance has no AC impedance.
Hence there is no corner frequency.

5. kubeek AAC Fanatic!

Sep 20, 2005
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Actually, I meant as I increase the frequency for say 0.01Hz to hundreds of megahertz, where does the cable start act as it has some impedance? Or does it have 50ohms even at that 0.01Hz?

6. Papabravo Expert

Feb 24, 2006
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As I understand the physics, the characteristic impedance is not a function of frequency. It is a function of the geometry of the coaxial cable. As you increase the frequency the losses change and the electrical length of the transmission line gets shorter and shorter. As the frequency goes up the wavelength becomes shorter and shorter to the point where the dimensions of the cable are of the same order of magnitude as the wavelength of the signal and it stops being a transmission line and starts being a waveguide.

7. Sue_AF6LJ Member

Mar 16, 2013
45
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It always remains a transmission line, it will become so lossy it is not usable as a transmission line anymore.

If you have a long enough transmission line and an analogue ohmmeter you will see the needle kick down toward the low resistance end of the scale. it will take a very short time returning to infinite resistance. This is not only due to the capacitance between the conductors but it is also due to the distributed inductance in the transmission line.

8. The Electrician AAC Fanatic!

Oct 9, 2007
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Graph 1 shows the situation. On that graph the corner frequency you're asking about is plainly visible at about 20 kHz.

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9. thatoneguy AAC Fanatic!

Feb 19, 2009
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When dealing with long cable runs, there aren't many options for changing the characteristic impedance of the cable. Without fiber, if a long run is needed, the signal is amplified by the predicted dB loss of the run, so the other end will see the signal at the correct/usable level. For very, very long runs, amplifiers are needed in the middle of the run, such as trans-oceanic cables, so a usable signal is left after the loss from distance. These losses increase with frequency.

The "effective bandwidth" of coaxial cable is a decision point made where the amplification to overcome losses is a greater hassle than using fiber. This point is usually somewhere in the 500Mhz to 1Ghz range for runs of significant length, but can be higher for short runs. At very low frequencies, such as those you are curious about, coax acts like two wires, like those to a Multimeter probe.

It's about impedance matching, in short runs. I'll use oscilloscope probes as an example, they are all coaxial cable, rather than fiber for the probes, though some can read fiber as well.

At lower frequencies, (<300Mhz) the mis-match doesn't mean much/anything, the capacitance of the probe and scope are dominant. Passive/simple probes are typically x10 so the impedance appears to be 10MΩ, going through a 50Ω characteristic impedance coax, then into the 1MΩ Oscilloscope front end. When the frequency increases, that doesn't work out as well.

Oscilloscope probe cables for 1Ghz+ have 50 ohm input impedance to match the cable characteristic impedance. At this frequency and low impedance, passive probes can't be used due to loading on the circuit.

The solution is active probes. Differential/Active probes are required to allow high impedance measurements, then "translating" them to 50Ω to feed the scope. Active probes run from thousands to tens of thousands of dollars for each probe.

10. crutschow Expert

Mar 14, 2008
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True, it's not the DC resistance. But an infinite length of ideal coax will show its characteristic impedance for DC or AC. Thus if you apply a step DC voltage to an infinite length cable, the current will be equal to the voltage divided by the cable characteristic impedance. For a finite length of cable it will look like its characteristic impedance for the time it takes the step to travel to the end of the cable and back to the start. At the point the reflected wave will add or subtract from the input current depending upon whether the other end is open or shorted.

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