Question 64 CEL02 - Chief Engineer - Limited (Alt)

The Correct Answer is B ### Explanation for why Option B is correct: The smallest incremental step in voltage that a Digital-to-Analog Converter (DAC) can produce is known as the resolution or Least Significant Bit (LSB) voltage. This value is determined by the output voltage range and the number of discrete steps the DAC can generate. 1. **Determine the Number of Steps:** An $N$-bit DAC can resolve $2^N$ discrete levels (steps). * For an 8-bit DAC, the number of steps is $2^8 = 256$ levels. * The number of *intervals* between these levels is $2^N - 1$, but we calculate the step size by dividing the total range by the total number of levels available ($2^N$). 2. **Calculate the LSB Voltage (Resolution):** The LSB voltage is calculated using the formula: $$\text{LSB Voltage} = \frac{\text{Maximum Output Voltage Range}}{\text{Number of Steps} (2^N)}$$ * Range = 10 volts (0 to 9 volts, although standard convention for DAC calculation uses the maximum voltage output capability or full-scale range (FSR). In this specific problem context where the range is defined as 10 volts, we use 10 V). * $2^8 = 256$ $$\text{LSB Voltage} = \frac{10 \text{ V}}{256} \approx 0.0390625 \text{ V}$$ 3. **Compare to Options:** * $0.0390625$ volts is closest to $0.04$ volts. Therefore, the smallest incremental step is approximately $0.04$ volts. ### Explanation for why the other options are incorrect: **A) 0.03 volts:** While $0.039$ volts is close to $0.04$ volts, $0.03$ volts is too low and does not represent the precise calculated resolution of $10/256$. **C) 0.625 volts:** This value is far too large. A step size of $0.625$ volts would imply a much lower resolution (fewer bits). If the resolution were $0.625$ V, the DAC would only have $10 / 0.625 = 16$ levels, which corresponds to a 4-bit DAC ($2^4=16$), not an 8-bit DAC. **D) 1.25 volts:** This value is even larger than option C. A step size of $1.25$ volts would imply only $10 / 1.25 = 8$ levels, corresponding to a 3-bit DAC ($2^3=8$). This is incorrect for an 8-bit system.