MOP
Example:
With ADCCLK = 30 MHz and sampling time = 3 cycles:
Tconv = 3 + 12 = 15 cycles = 0.5 µs with APB2 at 60 MHz
在ADC Clock = 30Mhz、12-bits Resolution的情況下，最小的Sampling Time可以達成0.5us一個Sample
所以
Sampling Rate = 1 / (0.5 * 10^(-6)) = 2Msp/s
由取樣定理可以知道，這樣等效的BandWidth為Sampling Rate的一半，即1MHz
