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You are comparing different A/D converters that will have 3.2 V and -3.2 V as their positive and negative reference voltages, respectively. Out of the following, converter _ can resolve the smallest change in voltage, which is approximately_ . (m = milli and u = micro)

(A) 16-bit, 100 kHz

(B) 8-bit, 1 MHz

(C) 15-bit, 512 kHz

(D) 19-bit, 256 kHz

(E) 17-bit, 330 kHz

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    $\begingroup$ Are you testing us? This is not a homework completion site. Show which you think is correct and explain why you think that - then you may get some feedback. $\endgroup$ – Solar Mike Apr 19 '19 at 6:47
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    $\begingroup$ Please show your current work? What help do you need? $\endgroup$ – Mahendra Gunawardena Apr 19 '19 at 10:37
  • $\begingroup$ Two words: Keithley Instruments. $\endgroup$ – William Hird May 20 at 16:53
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I order to solve this problem there one needs to understand the following

  • ADC range which is the maximum and minimum ADC input. For this problem the range is 3.2V, -3.2V.
  • ADC resolution which is the smallest distinguishable change in input. For a 12 bit converter this would be 6.4V/4096 = 0.0015625V

Below is an example for range of 3.3V based on a 12 bit converter.

enter image description here

Base on the above detail it should be possible to find the solution to above problem. In order to learn details I suggest checking out the following references.

Reference:

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