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Transmission line termination?

  1. Mar 27, 2008 #1
    Hello All,

    I have two PCBs, board A and board B. Board A has a 50MHz Crystal that I am making available through a high speed digital isolator (IL711-2E) to a chip that is on board B.

    Here's the datasheet on the isolator IC http://www.nve.com/Downloads/il711-2.pdf".

    Now, I was thinking about using an SMA cable to go between the boards but given the high frequency, I'm thinking that it might be a good idea to treat the cable like a transmission line and properly terminate the ends. But I don't know much about matching the ends.

    I know that my isolator IC can source a maximum of 10mA at 5V and that the other end of the line will goto a CMOS digital input on the destination board.

    I spoke to a friend about this and he said that I should put a capacitor and termination resistor in parallel to ground with the CMOS input on board B, and that I should put a resistor in series with the isolator on the board A side. But I'm not sure what values to use for the resistor. Just from looking at the specs for the output drive of the isolator, it can source 10 mA @ 5V so should I assume that it's output impedance is 500 Ohms? If so, then should I use 500 Ohm resistors for the beginning and end resistors mentioned above?

    Any help greatly appreciated.

    Jason O

    P.S. Forgot to mention but I don't expect the cable to be any longer than 6 feet.
    Last edited by a moderator: Apr 23, 2017
  2. jcsd
  3. Mar 27, 2008 #2


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    Staff: Mentor

    That's a pretty fast signal to be routing 6 feet, but certainly can be done with the proper driver, transmission line, and termination.

    The best way to do it is to use 50 Ohm coax for the cable (BNC or SMA ends), a 50 Ohm line driver IC, and forward terminate at the receive end with a 50 Ohm (moderate power) resistor. The receiver will then only get a 0V to 2.5V signal, if the 50 Ohm source had an open-circuit output voltage of 5V, so you will need to do some level translation, or other handling of the signal if you want to get back to a 5V clock at the receive end. You don't need a capacitor in series with the 50 Ohm forward termination resistor, as long as you use a real 50 Ohm line driver at the TX end.

    If you don't want to invest in the power of a real 50 Ohm line driver IC, then the next best way to do this is with a "back termination" resistor, placed in series with the output of the TX device. There is no forward termination used at the RX end, when you use the back termination technique. You choose the value of the back termination resistor to be the Zo of the cable (50 Ohms), minus the output impedance of the line driver IC. This is done so that there is a minimal re-reflection of the signal back off of the TX end. So, when the TX output transitions, that propagates down the 50 Ohm coax to the RX IC input, where a potitive reflection is generated (because the RX input Z is > 50 Ohms). That positive reflection propagates back up the coax, but is damped out by the series combination of the back termination resistance and the output Z of the driver IC (which total 50 Ohms).

    Back terminations are more commonly used in lower-power circuits, because standard logic families do not have enough output power to support a full 50 Ohm (or even 100 Ohm) forward termination.
  4. Mar 27, 2008 #3
    Hi Berkeman,

    Thanks for the great tips. Just one question though, how does the impedance matching work if I don't want to use a line driver and the output impedance of the isolator IC is 500 Ohms? You said that using the back termination method, you calculate the R value by subtracting the impedance of the cable from the output impedance of the driver IC. So in that case, that would be 50Ohms - 500Ohms, which tells me something is not right here.

    Does the resulting negative number mean my IC us too weak to drive the line? Or is there something else I'm missing?

    Jason O
  5. Mar 27, 2008 #4
    You can always whip out a Smith Chart and calculate line length and stub length of a single stub tuner.
  6. Mar 27, 2008 #5
    For most efficient power transfer your source and load impedances should be matched to the transmission line.

    If you are going to use a 50 ohm cable, then you need to put matching networks to make the source and load look like 50 ohms to the transmission line.

    Generally, the impedance of CMOS, and other digital IC's is very high. There is alot of matching network designs out there, like the L or T or Pi networks. However these are essentially low or high pass filters as well.

    If your 50 MHz crystal is a self contained oscillator then its output is a square wave, which is made up of odd harmonics. If you construct a matching pad like that it will filter your clock signal into a sine wave.

    If so then another viable option is go with a transformer design whose response is broadband. I think 10:1 transformers would do the trick ballpark.
  7. Mar 28, 2008 #6
    OK, so if I understand generally what you are all saying, unless I can find some coax cable with 500 Ohms of impedance, I'll need to use a line driver or a transformer that matches the line impedance?
  8. Mar 28, 2008 #7


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    What kind of wimpy driver are you using? A 74AC244 will have a Zout of 10-20 Ohms: Zout ~ Vol/Iout ~ 0.36V/24mA = 15 Ohms.

    It's true that an HC244 gate has a higher Zout ~ Vol/Iout ~ 0.33V/6mA = 55 Ohms, but still nowhere near 500 Ohms.
  9. Mar 28, 2008 #8
    Well, at the moment, I don't actually have a line driver, just the output from the digital isolator IC (the IL711-2E) which is just for galvanic isolation. I don't believe that it was specifically designed for driving cables which is why I was wondering if something with that high an output impedance could do it without some other interface in between.
  10. Mar 31, 2008 #9


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    No, you will need some sort of buffering. Either a gate like the 74AC244 with back-termination, or a real line driver IC. That's a requirement when you start working with moderate frequencies in the 10's of MHz and moderate distance like several meters....
  11. May 27, 2008 #10
    Hey I just thought of something. If I use the back termination method with a line driver like the 74HC125 and I'm driving a line that has a lower inpedance (100 Ohms for instance), and the output impedance of the IC is 142 Ohms, should I still add a 100 Ohm resistor in series with the line?

    Jason O
  12. May 27, 2008 #11
    Ok, Nevermind the above question. I just reread what berkeman wrote about calculating the output impedance and I realized that I calculated it wrong (I used the supply voltage / max output current)... Also realized that the 74HC244 IC is better than the HC125 since the enable input triggers all of gates rather than each one individually.
  13. May 27, 2008 #12


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    Staff: Mentor

    Yeah, and HC is generally too wimpy to be driving a transmission line with. The HC244/245 parts may be okay, as long as the Zo numbers are in line with what you are doing.
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