2
\$\begingroup\$

Not sure how to tackle this issue. I have a 1.28 MHz digital clock signal with 0 = 0V, 1 = 1.5V. I need to buffer it and split it for distribution, some of the outputs will travel approximately 100 feet. I'm flexible with the output voltage but I was thinking 5 TTL. The high voltage requirements of 5V may slow it down.

Option 1) Find a clock buffer distributor chip that works, then use line driver chips

Option 2) Use a voltage comparator circuit to change levels (or a different technique), then buffer distribute and line drive

Option 3) ?????

\$\endgroup\$
2
  • 1
    \$\begingroup\$ I'd transmit differentially at a low level to avoid EMI to other stuff. \$\endgroup\$
    – Andy aka
    Commented Jul 29, 2020 at 19:21
  • \$\begingroup\$ You could use a plastic optical fibre - something like the HFBR-1521Z goes fast enough (5MB) and can transmit comfortably over that distance (100m+). It's not cheap, you will need a receiver at the other end, but you don't need to worry about common mode, ground or emissions - so at system level may end up being cost effective \$\endgroup\$ Commented Jul 29, 2020 at 21:39

1 Answer 1

5
\$\begingroup\$

If you are 100feet away then ...

  • You should probably be using a differential signal. Otherwise you will probably radiate a lot of RF energy and fail any kind of emissions requirements set by the FCC or other regulatory bodies.
  • You should use a controlled impedance transmission line (such as a twisted pair cable), otherwise you will get reflections that will ruin your signal quality. A CAT5 Ethernet cable or similar would be fine.
  • There may be significant ground offset at each end of the line, therefore using TTL or other ground referenced signaling may have problems.

The solution is to just use an RS485 driver and receiver.

RS485 is made for long distance, medium speed data like this.

Analog Devices / Linear Tech makes some transceivers that operate up to 20MBps and support very large ground offsets (20V or more).

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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