- 2072 Kartik · 2+2+3 marks
Counter-type ADC numerical (VFS = 12 V)
Short answer
Resolution = 12/(2^4−1)=0.8 V. Max time = 15×0.1 ms=1.5 ms, min=0.1 ms. For 4.25 V, floor(4.25/0.8)+1=6 → output 0110, conversion time 0.6 ms.
For the ADC shown below, the full scale output voltage (VFS) = 12.00V. The clock PRF is 10KHz. Determine: i) Maximum and minimum conversion time ii) Resolution iii) The output count and the conversion time for Vin = 4.25V. [Figure: 4-bit counter (count-up) type ADC: a comparator (+V supply, GND) compares Vin with the DAC output; the comparator output and the CLOCK feed an AND gate that clocks a 4-bit UP COUNTER (outputs D0 (LSB), D1, D2, D3 (MSB), cleared by START); the counter outputs drive a 4-bit DAC whose output returns to the comparator; the comparator output is also the EOC signal.]
In a counter (count-up) type ADC, the counter starts from zero at START and counts clock pulses. The DAC output rises one step per pulse. The counter stops when the DAC output just exceeds Vin, because the comparator output goes LOW, which disables the AND gate and signals EOC.
Given
- n = 4 bits, VFS = 12.00 V (DAC output for 1111)
- Clock PRF = 10 kHz, so T = 1/10 kHz = 0.1 ms
i) Maximum and minimum conversion time
- Maximum: the counter may have to go through all 2ⁿ − 1 = 15 counts.
Tc(max) = (2ⁿ − 1) × T = 15 × 0.1 ms = 1.5 ms
- Minimum: one clock pulse (for a very small input).
Tc(min) = 1 × T = 0.1 ms
(Some books count 2ⁿ pulses for the maximum, giving 1.6 ms.)
ii) Resolution
With VFS taken as the output for the full code 1111:
Resolution = VFS / (2ⁿ − 1) = 12 / 15 = 0.8 V per step
In percentage terms: 1/(2⁴ − 1) = 6.67 % of full scale.
iii) Output for Vin = 4.25 V
The counter stops at the first count whose DAC voltage is greater than Vin:
| Count | DAC output (V) | DAC > 4.25 V? |
|---|---|---|
| 4 (0100) | 3.2 | No |
| 5 (0101) | 4.0 | No |
| 6 (0110) | 4.8 | Yes, so the comparator switches |
N = floor(4.25 / 0.8) + 1 = 5 + 1 = 6
Tc = N × T = 6 × 0.1 ms = 0.6 ms
Answer:
- Tc(max) = 1.5 ms, Tc(min) = 0.1 ms
- Resolution = 0.8 V
- For Vin = 4.25 V: output count = 6 = 0110 (D3 D2 D1 D0), conversion time = 0.6 ms
(If resolution is instead taken as VFS/2ⁿ = 0.75 V, the count is still 6, so the output and the 0.6 ms time do not change.)
✓ Answer checked by Claude Opus 5.5 (high effort)
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