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, illustrating non-linear conversion at very low voltages, is not applicable to the 1.9922V input. We assume a linear ADC response across its full range; the graph serves only to demonstrate quantization. Step 1: Determine the resolution of the 8-bit ADC, assuming linearity.An 8-bit ADC yields \(2^8 = 256\) discrete output levels (ranging from 0 to 255), given its digital nature.With a 5V full-scale range, the resolution (voltage increment per LSB) is:\[ \text{Step Size} = \frac{\text{Full Scale Range}}{\text{Number of levels}} = \frac{5 V}{256} \approx 0.01953125 \text{ V} \]
Step 2: Calculate the digital representation of the 1.9922V input.The ADC's digital output is derived by dividing the input voltage by the step size and truncating to the nearest integer:\[ \text{Decimal Value} = \text{floor}\left(\frac{V_{in}}{\text{Step Size}}\right) = \text{floor}\left(\frac{1.9922}{0.01953125}\right) \]\[ \text{Decimal Value} = \text{floor}(102.000...) = 102 \]
Step 3: Convert the resulting decimal value (102) to its hexadecimal equivalent.Divide 102 by 16: \(102 \div 16 = 6\) with a remainder of \(6\).The hexadecimal representation is thus \(66_{16}\), commonly written as 66H.