---
title: "Count-up ADC numerical (VFS = 16 V)"
subject: "Advanced Electronics"
chapter: 3
key: "semester_4/advanced_electronics:count-up-adc-16v-numerical"
source: https://www.bctengineer.com/q/semester_4/advanced_electronics/count-up-adc-16v-numerical
site: BCT Engineering
---

# Count-up ADC numerical (VFS = 16 V)

Advanced Electronics · Chapter 3: Digital-to-Analog and Analog-to-Digital Conversion

- Asked in: 2072 Chaitra, 2+3+3 marks
- Video: [Counter Type ADC Explained](https://www.youtube.com/watch?v=4zwtA2pG6Vc) — ALL ABOUT ELECTRONICS
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/advanced_electronics/count-up-adc-16v-numerical
- Whole chapter: https://www.bctengineer.com/md/semester_4/advanced_electronics/chapter-3

## Question

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

## 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.

## Answer

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):

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

#### b) Resolution
```text
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:

| Count | DAC output (V) | > 6.25 V? |
|---|---|---|
| 5 (0101) | 5.33 | No |
| 6 (0110) | 6.40 | Yes, stop |

```text
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.)

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. The minimum conversion time is one clock period. For 0.1 ms the PRF is:
   A. 10 kHz
   B. 1 kHz
   C. 100 kHz
   D. 0.1 kHz
   Answer: A. PRF = 1/0.1 ms = 10 kHz. (Key point: PRF)

2. Resolution with VFS = 16 V and 4 bits (using 2^n - 1) is:
   A. 1 V
   B. 1.067 V
   C. 0.8 V
   D. 4 V
   Answer: B. 16/15 = 1.067 V per step. (Key point: Resolution)

3. Digital output for Vin = 6.25 V:
   A. 0101
   B. 0111
   C. 0110
   D. 1000
   Answer: C. 5 x 1.067 = 5.33 V < 6.25 V, 6 x 1.067 = 6.40 V > 6.25 V. (Key point: Output count)

4. Conversion time for Vin = 6.25 V:
   A. 0.5 ms
   B. 1.5 ms
   C. 0.1 ms
   D. 0.6 ms
   Answer: D. 6 pulses x 0.1 ms = 0.6 ms. (Key point: Conversion time)

## Related in Advanced Electronics

- [Counter-type ADC numerical (VFS = 12 V)](https://www.bctengineer.com/question.md?key=semester_4%2Fadvanced_electronics%3Acounter-adc-12v-numerical) — Advanced Electronics, chapter 3
- [Dual slope ADC numerical (12 kHz clock)](https://www.bctengineer.com/question.md?key=semester_4%2Fadvanced_electronics%3Adual-slope-numerical-12khz) — Advanced Electronics, chapter 3
- [Count up/down and tracking type ADC](https://www.bctengineer.com/question.md?key=semester_4%2Fadvanced_electronics%3Acount-up-down-tracking-adc) — Advanced Electronics, chapter 3

## Related in other subjects

- [8-bit DAC LSB, MSB, resolution, full scale](https://www.bctengineer.com/question.md?key=semester_5%2Finstrumentation_ii%3Adac-lsb-msb-resolution) — Instrumentation II, chapter 4