Question

The output voltage of a 6-bit ladder type DAC with a digital input of 111001 is (Assume maximum output voltage is 10 V):

• A. \(9.04\text{ V}\)
• B. \(9.20\text{ V}\)
• C. \(9.06\text{ V}\)
• D. \(8.90\text{ V}\)

Hint:

To quickly double-check your answer, convert the binary fraction directly:
\(111001_2 = \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + 0 + 0 + \frac{1}{64} = 0.5 + 0.25 + 0.125 + 0.015625 = 0.890625\). Multiplying by \(10\text{ V}\) gives \(\approx 8.90\text{ V}\).

Answer 1

The Correct Option is D

Concept: A Digital-to-Analog Converter (DAC) converts a discrete binary code into a continuous analog voltage proportional to the code's value. For an \(n\)-bit resolution architecture, the total number of distinct step combinations available is given by \(2^n\). The analog output voltage (\(V_{\text{out}}\)) can be calculated using the fractional weight of the binary input relative to the full scale range: \[ V_{\text{out}} = V_{\text{max}} \times \left( \frac{\text{Decimal Value of Binary Code}}{2^n} \right) \] Alternatively, for architectures defined by their full-scale resolution limits across individual intervals: \[ V_{\text{out}} = V_{\text{fs}} \times \sum_{i=1}^{n} \frac{b_{n-i}}{2^i} \]

Step 1: Extracting the given parameters from the problem statement.
From the problem description, we have the following specifications:
• Total number of resolution bits, \(n = 6\)
• Maximum reference voltage boundary, \(V_{\text{max}} = 10\text{ V}\)
• Input binary sequence code \(= 111001_2\)

Step 2: Converting the binary input string into its base-10 decimal equivalent.
We evaluate the 6-bit binary code by multiplying each bit by its corresponding power of 2, starting with the Most Significant Bit (MSB) on the left down to the Least Significant Bit (LSB) on the right: \[ \text{Decimal} = (1 \times 2^5) + (1 \times 2^4) + (1 \times 2^3) + (0 \times 2^2) + (0 \times 2^1) + (1 \times 2^0) \] Calculating the individual exponential powers of 2: \[ 2^5 = 32, \quad 2^4 = 16, \quad 2^3 = 8, \quad 2^2 = 4, \quad 2^1 = 2, \quad 2^0 = 1 \] Substituting these values back into the calculation: \[ \text{Decimal} = (1 \times 32) + (1 \times 16) + (1 \times 8) + (0 \times 4) + (0 \times 2) + (1 \times 1) \] \[ \text{Decimal} = 32 + 16 + 8 + 0 + 0 + 1 = 57 \]

Step 3: Calculating the total number of digital combinations.
For a 6-bit system, the total number of quantization levels is: \[ 2^n = 2^6 = 64 \]

Step 4: Computing the final analog output voltage.
Substitute the computed decimal value (\(57\)), the total levels (\(64\)), and the maximum reference voltage (\(10\text{ V}\)) into the transfer equation: \[ V_{\text{out}} = 10\text{ V} \times \left( \frac{57}{64} \right) \] Let us calculate the fraction \(\frac{57}{64}\) using long division: \[ \frac{57}{64} = 0.890625 \] Now multiply this ratio by the 10 V reference: \[ V_{\text{out}} = 10 \times 0.890625 = 8.90625\text{ V} \] Rounding this value to two decimal places to match the standard format of the options gives: \[ V_{\text{out}} \approx 8.90\text{ V} \] This aligns precisely with Option (D).