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Count-up ADC numerical (VFS = 16 V)

Short answer

PRF = 1/0.1 ms = 10 kHz. Resolution = 16/15 ≈ 1.067 V. For 6.25 V, floor(6.25/1.067)+1 = 6 gives output 0110, conversion time 0.6 ms.

A count up type ADC of 4 bit has full scale voltage (VFS) = 16.00V. The minimum conversion time is 0.1ms. Determine: a) PRF of clock used b) Resolution c) The Digital output and the conversion time for Vin = 6.25V

In a count-up (counter) type ADC, the counter increases by one per clock pulse from zero. The DAC output grows by one step each time. Conversion ends when the DAC output first exceeds Vin.

Given

  • n = 4 bits, VFS = 16.00 V (DAC output for 1111)
  • Minimum conversion time = 0.1 ms

a) PRF of the clock

The minimum conversion time is one clock period (the comparator trips after the first pulse):

T = Tc(min) = 0.1 ms
PRF = 1/T = 1/0.1 ms = 10 kHz

b) Resolution

Resolution = VFS / (2ⁿ − 1) = 16 / 15 = 1.067 V per step
%Resolution = 1/(2⁴ − 1) × 100 = 6.67 %

c) Digital output and conversion time for Vin = 6.25 V

The counter stops at the first count N where N × 1.067 V > 6.25 V:

CountDAC output (V)> 6.25 V?
5 (0101)5.33No
6 (0110)6.40Yes, stop
N = floor(6.25 / 1.0667) + 1 = 5 + 1 = 6
Tc = N × T = 6 × 0.1 ms = 0.6 ms

Answer:

  • PRF = 10 kHz
  • Resolution = 1.067 V (6.67 %)
  • Digital output = 0110 (decimal 6), conversion time = 0.6 ms

Maximum conversion time for reference: 15 × 0.1 = 1.5 ms.

(If a book defines resolution as VFS/2ⁿ = 1 V, the counter stops at count 7 (7 V > 6.25 V), giving 0111 and 0.7 ms. The method is the same.)

✓ Answer checked by Claude Opus 5.5 (high effort)

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