I have a board layed like the in the image below. I have a microcontroller that generated clk1, which feeds into U1. After 8 clock cycles, the clock passes through to U2 and does this 30 times. It's a shift register.

The microcontroller generates clk2, which is running in parallel across 30 some components.

I've included a place holder for possible termination resistor at the source pins itself (R1 and R2) so I can maybe experiment with different values to get my signals looking correct, however I do not know if this will work.

So my question is, if adding those place holder resistors is not enough to correct for SI issues, what can I add to the board that can help me correct it after ?


simulate this circuit – Schematic created using CircuitLab

  • \$\begingroup\$ That's a huge fanout on clk2. \$\endgroup\$ Commented Apr 3, 2015 at 22:38
  • \$\begingroup\$ @ScottSeidman ya. I checked the input/output current capabilities and it doesn't seem like it would be an issue. \$\endgroup\$
    – efox29
    Commented Apr 3, 2015 at 23:01
  • \$\begingroup\$ It's not current that will do you in, it's gate capacitance \$\endgroup\$ Commented Apr 3, 2015 at 23:03
  • \$\begingroup\$ @ScottSeidman Recommendations ? Add buffers along the way ? \$\endgroup\$
    – efox29
    Commented Apr 3, 2015 at 23:14
  • \$\begingroup\$ Not much experience, but maybehttp://www.digikey.com/product-highlights/en/lmk00725-clock-fanout-buffer/51437?WT.srch=1&mkwid=sbZCtG9zd&pcrid=62953614075&pkw=_cat:digikey.com&pmt=b&pdv=t \$\endgroup\$ Commented Apr 3, 2015 at 23:43

1 Answer 1


Add a pair of resistors to the end of the lines. One up to VCC and the other to ground. This will allow you to terminate the line at the end in addition to (or instead of) source termination.

The problem with source termination is it works best with only one endpoint and you have many. The intermediate points will see a step on the edges, the placement of which will depend on the relative impedance of the line to the source resistor.

Parallel termination will allow you to reduce the reflections which works better with lots of loads. Problem is you need the pair of resistors because the driver may not be able drive the desired resistance to ground or VCC. So the pair of resistors can be chosen to provide the desired impedance pulled to some intermediate voltage which the driver can handle.


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