Question

An 8-bit ADC converts analog voltage in the range of 0 to +5 V to the corresponding digital code as per the conversion characteristics shown in the figure. For \(V_{in}\) = 1.9922 volt, which of the following digital output, given in hex, is true?

• A. 64 H
• B. 65 H
• C. 66 H
• D. 67 H

Hint:

For a standard linear ADC, the output code is found by \(\text{floor}( (V_{in} / V_{ref}) \times 2^N ) \), where N is the number of bits. Here, \(\text{floor}( (1.9922 / 5) \times 256 ) = \text{floor}(101.99) = 101\). Let's re-check the first method. The quantization level is \(D = V_{in}/resolution\). Okay, my first calculation was correct. 102 in decimal is 66H.

Answer 1

The Correct Option is C

Note: The provided graph shows a non-linear conversion for very small voltages and is not applicable for the given input of 1.9922V. We must assume the ADC is linear over its full range, and the graph is illustrative of quantization steps only. Step 1: Determine the resolution of a linear 8-bit ADC. An 8-bit ADC has \(2^8 = 256\) possible output levels (from 0 to 255). The full-scale voltage range is 5V. The resolution, or step size, is the voltage change corresponding to one LSB (Least Significant Bit). \[ \text{Step Size} = \frac{\text{Full Scale Range}}{\text{Number of levels}} = \frac{5 V}{256} \approx 0.01953125 \text{ V} \]
Step 2: Calculate the decimal digital value for the given input voltage. The digital output is found by dividing the input voltage by the step size and taking the integer part. \[ \text{Decimal Value} = \text{floor}\left(\frac{V_{in}}{\text{Step Size}}\right) = \text{floor}\left(\frac{1.9922}{0.01953125}\right) \] \[ \text{Decimal Value} = \text{floor}(102.000...) = 102 \]
Step 3: Convert the decimal value to hexadecimal. To convert 102 from decimal to hexadecimal: Divide 102 by 16: \(102 \div 16 = 6\) with a remainder of \(6\). The hexadecimal representation is formed from the quotient and remainder: \(66_{16}\) or 66H.