Find number of bits in analogue to digital converter

Question

For display after signal conditioning, the output of a pressure transducer is digitized. The digitized value should have an accuracy of ±0.01 percent FS. When the maximum magnitude of the absolute error voltage is to be 0.5 milliVolt, find out the number of bits needed in the analogue to digital converter and the full scale output voltage of the transducer conditioning amplifier which can be accommodated.

Answer 1

The digitized value consisting of an error of ± ½ levels or ± ½ LSB

 Hence Δ = 0.015% FS or 1 part in 6,666,

Subsequently 1 level = 2Δ= 0.03% FS or 1 part within 3,333 (2parts within 6,666)

 It means that the full output signal range take up 3,333 parts and therefore the number of levels is specified as:

L = 3333 levels.

The number of bits is specified by the power of the radix 2 that provides a decimal value of greater than or equivalent to 3333. 

As 2¹² = 4096. Thus, the number of bits needed is:

N = 12 bits.

When the error is ± ½ levels = ± 0.5mV in that case the quantum step voltage is:

 V_Q = 1 mV

When there are 4096 levels in that case the full scale output voltage by the transducer is:

 V_FS = (L - 1) V_Q = 4095 x 1 mV = 4095 mV

  V_FS = 4.095 Volt.