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----- 2013/11/08 modify my question

I have an 'A' source has 5 MHz signal (0-5 V square wave) without knowing input impedance and a receiver 'B' has 75 ohm input impedance.

As I know I have to match the impedance match to avoid the reflection effect, but I will have the divider effect if I parallel a resistor to the 'B' input end or series connect a resistor to the 'A' output (means not 0-5 V input for 'B' anymore).

Should I do a 2x amplifier circuit to make it 0-10 V first to avoid the divider effect and use this op as a buffer and series connected to a 75 ohm to make the output impedance as nearly 75 ohm?

If there is any information I can study, please let me know.

Thanks for the help.

-----2013/11/08 upload the 'A' spec enter image description here

-----2013/11/08 upload the 'A' output via Oscilloscope

  1. sync output with 50 ohm parallel enter image description here

  2. sync output without 50 ohm parallel enter image description here

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  • \$\begingroup\$ I found that in different format, some (ex. TTL) can accept as high level even the source is divide by 2 (ex. 5V to 2.5V). Does it means the signal wont be affect even I match the impedance condition while I ignore the divider effect? interfacebus.com/voltage_threshold.html \$\endgroup\$
    – ccmkn
    Nov 8, 2013 at 17:35
  • \$\begingroup\$ What are the specs on B that make you think it might be damaged? \$\endgroup\$
    – The Photon
    Nov 8, 2013 at 17:35
  • \$\begingroup\$ How does your linked image apply? Are A and B digital logic chips? \$\endgroup\$
    – The Photon
    Nov 8, 2013 at 17:41
  • \$\begingroup\$ it is 75 ohm terminator and accept TTL signal. so I think it is better not to exceed 5V. \$\endgroup\$
    – ccmkn
    Nov 8, 2013 at 17:42
  • \$\begingroup\$ 'A' is laser sync output and 'B' is a frame grabber card. I connect them by a BNC cable. Just wondering if I have to do anything about the impedance and divider effect. \$\endgroup\$
    – ccmkn
    Nov 8, 2013 at 17:44

2 Answers 2

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Your oscilloscope measurements show that your source is not impedance-matched.

To avoid reflections due to impedance mismatch at the load, I recommend to simply design this as a 75-Ohm system:

schematic

simulate this circuit – Schematic created using CircuitLab

If no (or very little) reflection comes back from the destination end, it won't cause any issues to have the source not impedance-matched, and you will get your full 5-V signal at the destination end (provided the source is able to drive a 75-Ohm load, which your oscilloscope measurements show it is).

You will need choose a coaxial cable with 75-Ohm characteristic impedance, and it will be connectorized with 75 Ohm BNC connectors. According to Wikipedia these connectors "can be made to" intermate with 50-Ohm BNC, but to avoid damaging things you might prefer to use an adapter at the source end. Since your load has a 75-Ohm termination, its connector ought to be the 75-Ohm type also, but it would be wise to double-check.

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  • \$\begingroup\$ great drawing! so you think it is ok for the input voltage is divided by 2 in order for impedance match condition? I am wondering if this will affect the high voltage level judgement by destination. \$\endgroup\$
    – ccmkn
    Nov 8, 2013 at 17:55
  • \$\begingroup\$ If your "laser sync" driver is specified to generate a 5 V signal with 50-Ohm matched output, then the "SRC" element should actually be producing 10 V p-p, meaning they've already accounted for the divider effect. But if you can link to the manual for the device or scan the relevent specs and post an image it would help to be sure. \$\endgroup\$
    – The Photon
    Nov 8, 2013 at 18:14
  • \$\begingroup\$ I attach the spec as reference. How do I know the "SRC" is producing 10V p-p? I try to connect the "Source" directly to oscilloscope with 50 ohm parallel (for impedance match), and it shows the 2.5V p-p. It means the "SRC" is providing 5V p-p, right? or I miss something? \$\endgroup\$
    – ccmkn
    Nov 8, 2013 at 18:29
  • \$\begingroup\$ How do you know it has 50 Ohm output impedance? Is there another part of the spec you didn't share? \$\endgroup\$
    – The Photon
    Nov 8, 2013 at 18:33
  • \$\begingroup\$ It doesnt mention 50 ohm output impedance in spec. Im sorry if this is a misleading. I attach the oscilloscope test for reference. So I cannot say it is 50 ohm output impedance when I see the reflection effect is disappeared with 50 ohm in parallel? \$\endgroup\$
    – ccmkn
    Nov 8, 2013 at 18:47
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At 5 MHz, you might not even have to worry about matching at all if the cable is short. How long is the cable you're using? If it's less than around 3 meters, you will probably be all right.

If you do need to match it over a narrow bandwidth, then you can build a matching network with capacitors and inductors. However, I'm not sure if this would work very well for your signal - square waves have a lot of harmonics and many of those harmonics would not be matched correctly if you build a narrow band matching network.

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  • \$\begingroup\$ Without knowing rise and fall time, I don't see how you can say he'll be okay at 3 m. If you need more information to answer a question, it's better to ask for clarification in a comment than to post an answer based on wild guesses about OP's situation. \$\endgroup\$
    – The Photon
    Nov 8, 2013 at 17:40

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