I'm working on a high speed ADC/DAC board. I saw some reference designs for the device, there are matching series resistors connected between output digital pins & controller.

Though, resistor network's value is given in the schematic. I'm curious to know, How do you calculate What resistors to be used for impedance matching.

What parameters of the input pins should be taken into account. Is there any formula for calculating the same ?


1 Answer 1


There are matching series resistors connected between output digital pins & controller

In these circumstances (fast digital signals crossing a significant distance on a PCB) you choose matching resistors to suit the impedance of the copper interconnection. The copper interconnection, just like coax cable or twisted pair cable exhibits characteristics that if not properly matched, can cause reflections when the fast edges of the signal reach the receiving device. This will inevitably lead to data corruptions.

On the other hand, if the tracks are not too long (compared to the wavelength of the fastest parts of the signal), maybe the resistors are just pull-ups. There are decent rules of thumb that govern the need to apply terminations of course.

If you need a more definitive answer, a section of the circuit would help as well as a picture of the copper tracks on the circuit board.

  • \$\begingroup\$ Though Im using high speed ADC/DACs my operating range is low near about 10Msps, Will wavelength matter in this case ? Or impedence matching is reuired ? \$\endgroup\$
    – rahulb
    Mar 27, 2014 at 8:35
  • 3
    \$\begingroup\$ The edges of your digital signal could be 10ns and it's always the edges in digital data that cause the reflection problems. Imagine a 2ns rise followed by a 2ns high followed by 2ns fall and 2ns low. That's a 8ns period and even if it were all smoothed out into a sine wave, it'd be a frequency of 125MHz - remember square waves carry infinite spectral content in harmonics. \$\endgroup\$
    – Andy aka
    Mar 27, 2014 at 8:39

Your Answer

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

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