Topic: Digital Circuits
A 4 bit unipolar DAC with 10V reference is fed to the input of a comparator, whose threshold is set as 7.5V. When the DAC counts from 0 and reaches 8, the comparator showed threshold crossing wrongly. It is found that one of the DAC input bit is stuck at '1'. Which is the stuck at '1' bit
- A. BO (LSB)
- B. B1
- C. B2
- D. B3 (MSB)
Answer: C. B2
Explanation: For a 4-bit unipolar DAC with a $10V$ reference, the step size is $10V / 2^4 = 0.625V$.
The threshold is $7.5V$. The DAC count needed to reach this threshold is $7.5 / 0.625 = 12$ ($1100_2$).
The problem states that the comparator tripped early at a count of 8 ($1000_2$). This implies the actual DAC output was evaluating to at least 12.
If input is $1000_2$ and one bit is stuck at 1:
- If B0 is stuck at 1, the value is $1001_2$ (9) -> $5.625V$ (No crossing).
- If B1 is stuck at 1, the value is $1010_2$ (10) -> $6.25V$ (No crossing).
- If B2 is stuck at 1, the value is $1100_2$ (12) -> $7.5V$ (Crosses threshold!).
Therefore, bit B2 is stuck at '1'.