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
B0 (LSB)
B1
B2
B3 (MSB)
ANSWER
The stuck-at-1 bit is B2, option (C).
Official answer: CSource: ISRO's revised and final answer key of 16 February 2024
For a 4-bit unipolar DAC with a reference, , where is the input code. One LSB is .
The comparator threshold corresponds to , which is . A healthy converter first reaches the threshold at count 12.
At count 8 () the correct output is , which is below the threshold. The comparator still tripped, so the faulty input code must be at least 12 at that count. The stuck bit must therefore turn into or higher.
If B2 is stuck at 1, then becomes , giving , so the comparator trips. For counts 0 to 7, the largest possible code is , giving , so there is no crossing before count 8. This matches what was observed.
If B3 is stuck at 1, count 4 () would read as and the crossing would already appear at count 4, not at 8.
If B1 is stuck at 1, count 8 reads as , which is . If B0 is stuck at 1, count 8 reads as , which is . Both are below , so neither produces a false crossing at count 8.
Hence the bit stuck at 1 is B2.
KEY IDEA
Convert the threshold to a DAC code ( is code 12, ). Then find which stuck bit changes code 8 () to a code of at least 12 without causing an earlier crossing.
WATCH OUT FOR
Choosing B3 because it carries the most weight. A stuck MSB would trip the comparator at count 4, not at 8. Checking every count before 8 for a premature crossing is what separates B2 from B3.
From the ISRO EC 2023 paper. Every question of this paper · The full paper (PDF, ISRO) · All 4 ISRO ADC & DAC questions, with notes
