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

Explanation:

Analysis of the 4-bit Unipolar DAC and Comparator System:

The given problem involves a 4-bit unipolar DAC (Digital-to-Analog Converter) with a 10V reference voltage and a comparator whose threshold voltage is set at 7.5V. The DAC counts from 0 to 15 (since it is a 4-bit DAC, it can represent 2⁴ = 16 levels). The comparator monitors the output of the DAC and determines when it crosses the threshold voltage.

It is observed that when the DAC count reaches 8, the comparator erroneously indicates threshold crossing. Upon analysis, it is found that one of the input bits of the DAC is stuck at '1'. We need to determine which bit is stuck at '1'.

Step-by-Step Solution:

1. Understanding the DAC Operation:

A 4-bit unipolar DAC converts a digital value (binary) into an analog voltage. The output voltage is given by:

V_out = (Digital Value / Maximum Digital Value) × Reference Voltage

Since the DAC is 4-bit, the maximum digital value is 15 (binary 1111). The reference voltage is 10V. Therefore:

V_out = (Digital Value / 15) × 10

For each binary value, the corresponding analog voltage is calculated as follows:

2. Comparator Threshold Analysis:

The comparator is set to a threshold voltage of 7.5V. It should ideally indicate threshold crossing when the DAC output voltage exceeds 7.5V. From the table above, this happens when the binary value exceeds 1011 (decimal 11), as for binary 1100 (decimal 12), V_out becomes 8V.

However, the problem states that the comparator erroneously indicates threshold crossing at binary 1000 (decimal 8). This suggests that the DAC is producing an incorrect output voltage due to one of its input bits being stuck at '1'.

3. Identifying the Stuck-at-'1' Bit:

To determine which bit is stuck at '1', we analyze the effect of each bit being stuck at '1' on the output voltage:

• B₃ (MSB): If the most significant bit (B₃) is stuck at '1', the output voltage would be significantly higher than expected for lower binary values, as B₃ contributes the largest weight (8 × Reference Voltage / 15). For binary 1000 (decimal 8), the output voltage would be correct, so B₃ is not the stuck bit.
• B₂: If B₂ is stuck at '1', it adds an additional weight (4 × Reference Voltage / 15) to the output voltage. For binary 1000 (decimal 8), the output voltage would incorrectly include this additional weight, resulting in an erroneous threshold crossing at the comparator. This matches the problem description, so B₂ is the stuck bit.
• B₁: If B₁ is stuck at '1', it adds a smaller weight (2 × Reference Voltage / 15). For binary 1000 (decimal 8), the output voltage would not exceed the threshold, so B₁ is not the stuck bit.
• B₀ (LSB): If the least significant bit (B₀) is stuck at '1', it adds the smallest weight (Reference Voltage / 15). This would not cause the erroneous threshold crossing at binary 1000 (decimal 8), so B₀ is not the stuck bit.

Correct Answer: The stuck-at-'1' bit is B₂.

Important Information

To further understand the analysis, let’s evaluate the other options:

• B₀ (LSB): The least significant bit contributes the smallest weight to the output voltage. If it is stuck at '1', the error in output voltage would be minimal and insufficient to cause the threshold crossing at binary 1000 (decimal 8).
• B₁: The second least significant bit contributes a slightly larger weight than B₀. If it is stuck at '1', the output voltage would still not exceed the threshold at binary 1000 (decimal 8).
• B₃ (MSB): The most significant bit contributes the largest weight to the output voltage. If it is stuck at '1', the output voltage would be significantly higher than expected for lower binary values, but it would not match the observed erroneous behavior at binary 1000 (decimal 8).

Conclusion:

By analyzing the effect of each bit being stuck at '1', we identify that B₂ is the stuck bit, as its incorrect contribution to the output voltage causes the comparator to erroneously indicate threshold crossing at binary 1000 (decimal 8). This analysis highlights the importance of understanding DAC operation and bit weighting in digital systems.