ADC throughput calculation

Question

I was studying ADC basics from this document, AVR127: Understanding ADC Parameters. I have a question regarding the throughput calculation. The document reads,

Consider the case of single-ended conversion where one conversion takes 13 ADC clock cycles. Assuming the ADC clock frequency to be 1MHz, then approximately 77k samples will be converted in one second. That means the sampling rate is 77k.

May I know how they reached this value? (I know they divide the clock frequency by 13, but don't know why.) I am not able to find the logic behind it. If you can explain the math, it would be appreciated.

Answer 1

If the ADC clock frequency is 1 MHz, each clock lasts for 1 μs. So, if the ADC needs 13 clock cycles to compute a sample then it takes 13 μs to do so. That, is an effective sample throughput rate of 1/(13 μs) or approximately 76.92 kHz.

Answer 2

Write it out with the units. Arrange the ratios to produce the desired unit and cancel out undesired units. (A unit in both the numerator and denominator cancels out).

In this case, you are given both ADC Clocks per second and ADC clocks per sample. You want to calculate samples per second and you want the ADC clocks unit to cancel out. When you write it out with the units you will see that you have to divide ADC clocks per second by samples per ADC clock to get ADC clocks to cancel out and leave you with samples per second.

1,000,000 ADC clocks      1 sample        76,923 samples
-------------------- * --------------- = ----------------
      1 second          13 ADC clocks        1 second