Why the conversion time of an ADC is calculated as: number of bits in resolution+number of programmed sample clocks+2/(ADC clock frequency) ? As far as I know for the successive approximation A/D converter, a clock cycle is the time it takes for a bit so number of bits in resolution would give us n cycles, so why this number is added to 2 and the whole thing is divided to the clock frequency? The microcontroller is HCS 12 and uses successive approximation A/D converter

  • \$\begingroup\$ You should try looking at some really complex SPI ADCs! \$\endgroup\$
    – Andy aka
    Commented Oct 1, 2015 at 23:48

1 Answer 1


The initial 2 clocks are to sample the input signal onto a capacitor.

Then there is a programmable time (The programmed sample clocks) where the sample is transferred to the capacitors that form the A/D.

Lastly the successive approximation algorithm then takes one clock per bit as you say.

Total conversion time

The division by the clock frequency is to convert the result to time (in us if you use the frequency in MHz) rather than a number of clocks.

The following application note has the details: AN2428

  • \$\begingroup\$ I thought "number of programmed sample clocks" accounted for the sampling time. Or is that just an adjustable amount on top of a fixed base of two clocks for sampling? \$\endgroup\$
    – Dave Tweed
    Commented Oct 1, 2015 at 23:54
  • \$\begingroup\$ Can you give a link to the datasheet? \$\endgroup\$ Commented Oct 1, 2015 at 23:58
  • \$\begingroup\$ It is a book: hsc12/9s12 an introduction to software and hardware interfacing \$\endgroup\$
    – Jack
    Commented Oct 2, 2015 at 1:19

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.