I think this question almost answers itself, but I've learned that's not always the case. It appears from my somewhat brief googling and catalog-trawling that Microchip sells the 3008 and 3208 ADCs. Other than the resolution itself, are there any meaningful differences in these two parts? The price difference is $1 (in quantity) so I realize that if you're building something (or hundreds of them, really) where you only NEED 10-bits it might be worth saving a buck, but for general purpose / hobbyist type applications is there any reason NOT to just go with the 12-bit version? I assume that the interfacing code is all the same across these?


Cost is one factor, as you note. $1 is a big difference in price for a lot of people.

Also consider the complexity in interfacing with the ADC. If it has a parallel interface, you need two extra pins for the 12-bit ADC versus the 10-bit. If it has a serial interface, then you don't need extra pins, but you need extra time to transfer two more bits1. If the serial interface is slower than the ADC, this limits the sample rate, since you can't start reading a new sample until you are done reading the previous.

This is all assuming that everything else is equal. In reality, that's probably not true. A careful reading of the datasheets is necessary to understand all the differences.

1: assuming, as The Photon points out, that the serial protocol used by the ADC doesn't transmit the measurements in octet chunks, in which case either 10 bits or 12 bits requires two octects, or 16 bits.

  • 3
    \$\begingroup\$ For a parallel ADC if you buy the 12 bit part and only need 10 bits you can just not connect the two lsbs. For a serial part, many if not most types transmit in byte-sized transactions. So they send 16 bits whether the actual resolution is 10 or 12 bits, they just zero out either the unused lsb's or msb's. \$\endgroup\$ – The Photon Jan 28 '14 at 23:34
  • \$\begingroup\$ @ThePhoton sure, but whether you connect the extra pins or not, they take up board space. \$\endgroup\$ – Phil Frost Jan 28 '14 at 23:51
  • \$\begingroup\$ True; also I checked the datasheet and for the parts named by OP, they actually do transmit only the needed bits over the serial interface. I thought there might be a power advantage to using the 10-bit part, but actually it has a higher Icc spec than the 12-bit part! \$\endgroup\$ – The Photon Jan 28 '14 at 23:59
  • \$\begingroup\$ The 3008 can run at a faster clock, so ends up with twice the max sample rate. \$\endgroup\$ – Pete Kirkham Jan 29 '14 at 0:04
  • \$\begingroup\$ @PeteKirkham is that the 10 bit or the 12 bit variety? I must admit, I was too lazy to find and read the specific datasheets for these parts. General indemnifying language added to answer. \$\endgroup\$ – Phil Frost Jan 29 '14 at 0:10

In addition to Phil's excellent observations, more bits may also mean more iterations of the algorithm underlying the sampling. For example, successive approximation ADC's will need more iterations to yield more bits. This can decrease your effective sampling rate, even for a device with parallel communication. Similar with a delta-sigma, where more bits can mean slower rates. Probably not so with a dual-slope.


Looks like the 10 bit throughput is twice the 12 bit throughput; 200 vs 100 ksps. 12 bit conversion also takes two more clock cycles than 10 bit conversion. What's the more important aspect of your analog signal, precise amplitude, or precise frequency? Would aliasing be a problem for you at maximum throughput?


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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