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    Coaxial cable capacitance

    Dear All,

    I'm planning on using a transistor to drive a 75 Ohm coaxial line (RG6) with just a digital signal. From collector to VDD there will be a 75 Ohm resistor. The signal at the collector is than AC-coupled onto a RG6 coaxial cable. At the end of the long cable (100m, neglect the attenuation for now) there is a 75 Ohm termination.

    So if my transistor is off, the 75 Ohm pull-up resistor needs to pull the coaxial line to VDD, but there is also the 75Ohm termination where the current flows through. I was wondering how fast this would be as there is 5*R*C needed for proper settling. I was wondering what the capacitance of a coaxial cable is. In the datasheet I find 20.6pF/ft or 67.59 pF/m. Does this simply scale? So does my pull-up resistor see 6.759nF for a 100m cable? This would mean that 5*R*C= 2.5µs and that the fastest I can switch is around 400kHz?

    Does this make any sense? Or am I missing some transmission line magic because these cables are typically used up to 1GHz?

    Thanks

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    Re: Coaxial cable capacitance

    If you've got a 75 ohm termination, you're only going to be able to pull to VDD/2.



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    Re: Coaxial cable capacitance

    Hi Barry thanks for the reply. Yes I understand because of the voltage divider. But my concern is more the capacitance of the coaxial cable. At what speed will I be able to switch? Is this coaxial transmission line really 6.7nF of capacitance I need to consider, or don't I need to consider this because the capacitance of a coaxial tranmission line is of a distributed nature and the pull-up resistor won't "see" this?



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    Re: Coaxial cable capacitance

    Hi,

    Yes you miss the characteristic impedance.

    You just need to drive the 75 Ohms of the cable impedance, that's it. This already includes the cable capacitance.
    And it's independent of cable length.

    Let's say you have a high to low transition. And it takes 2ns. Then this transition takes place in a limited length of cable.
    I don't know the cable parameter, but let's assume the signal speed in the cable is about 70% of c.
    70% × 300,000km/s × 2ns = 0.42m
    0.42m x 67.6pF/m = 28.4pF
    Now let's say tau = R ×C then R = tau / C = 2ns / 28.4pF = 71.4 Ohms.

    Oh, not far away from 75 Ohms.

    ****
    Now take a slower signal, let's say 10ns.....do the same calculation and get the same result: 71.4 Ohms.
    Independent of frequency and independent of cable length...

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    Re: Coaxial cable capacitance

    https://www.rfcafe.com/references/el...coax-chart.htm

    Coax specs. If you are worried 'bout capacitance.
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    Re: Coaxial cable capacitance

    Quote Originally Posted by The_Dutchman View Post
    or don't I need to consider this because the capacitance of a coaxial tranmission line is of a distributed nature
    Yes, the L' and C' of the cable are distributed and do NOT create a low pass behaviour.


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    Re: Coaxial cable capacitance

    The input impedance of an ideal transmission line terminated with its characteristic impedance is just this impedance, in your example equivalent to a 75 ohm resistor.

    Switching speed will be mainly limited by the transistor parameters. In addition, frequency dependent transmission line attenuation is causing a waveform distortion, 100 m RG6 will give the below shown pulse form for square wave input.

    Click image for larger version. 

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    Please notice that your driver circuit is not terminating the transmission line correctly in on state. Possibly reflections due to improper terminated far end will be reflected back in this state.


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    Re: Coaxial cable capacitance

    Thank you all for giving me a more clear sight on how this works as I was confused with the capacitance stated in the datasheet, now it is more clear to me. I also looked up some video's on youtube on this topic and basically I learned that you can drive a coax with whatever impedance you want, as long as the far end is terminated exactly in the characteristic impedance there will be no/verry little reflection distorting the wave form (at any point on the line). If this is not the case and the far end is not exactly matched, there will be a reflected wave arriving back at the source, and then you have a problem when this impedance is not matched because it will reflect again and so on.. to me this is basically the reason why they use the source impedance matching to "absorb" any reflections coming back from the far end.

    As FvM mentioned, I understand that in the on-state the transistor will be on and will have a very low on resistance, maybe a few ohms, and therefore the line is not matched which is important for a improper terminated far end as this will cause a reflected wave travelling back towards the source and this will neither be terminated at the source because of the low on impedance of the switch. Possibly a push-pull output stage with a 75 Ohm series resistor might provide a better driver circuit? It should be better matched in the on & off state and will provide equal rise & fall times?



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    Re: Coaxial cable capacitance

    Possibly a push-pull output stage with a 75 Ohm series resistor might provide a better driver circuit? It should be better matched in the on & off state and will provide equal rise & fall times?
    A high speed CMOS buffer, e.g. 74AHC series works perfectly. The series resistance should be adjusted for the buffer output resistance.



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