An n-bit DAC has a full-scale (reference) of 5 V and a resolution (LSB size) of 5 mV. What is the value of n?
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A8
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B9
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C10
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D11
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E12
Answer
Correct Answer: 10
Explanation
Introduction / Context:DAC resolution links the number of bits to the smallest analog step size. Correctly relating LSB size to full-scale range is fundamental in data conversion design.
Given Data / Assumptions:
- Vref (full-scale) = 5 V.
- Resolution (LSB) = 5 mV = 0.005 V.
- Assume ideal straight-binary coding.
Concept / Approach:
Two common definitions exist: LSB = Vref / (2^n) or LSB = Vref / (2^n - 1). For ideal sizing and most quick estimates, LSB ≈ Vref / 2^n suffices. Using this yields a clean integer n.
Step-by-Step Solution:
Set 0.005 = 5 / 2^n → 2^n = 1000.n = log2(1000) ≈ 9.966 ≈ 10 bits.Using Vref / (2^n - 1): 5 / 0.005 = 1000 ≈ 2^n - 1 → 2^n ≈ 1001 → still n ≈ 10.Verification / Alternative check:
A 10-bit DAC has 1024 steps; LSB ≈ 5 / 1024 ≈ 4.88 mV, very close to 5 mV; the stated resolution is consistent with 10 bits.
Why Other Options Are Wrong:
- 8, 9 bits: yield LSBs significantly larger than 5 mV.
- 11 bits: LSB would be roughly 2.44 mV, smaller than specified.
Common Pitfalls:
Confusing Vref/(2^n) and Vref/(2^n - 1); both give n ≈ 10 here, but always check the convention used in the specific DAC's datasheet.
Final Answer:
10