Question

The minimum number of comparators required in a \(6\)-bit flash analog-to-digital converter is ________. (answer in integer)

Hint:

A flash ADC places one comparator at each boundary between adjacent quantization levels; an n-bit converter has 2^n levels and therefore 2^n - 1 boundaries.

Answer 1

Correct Answer: 63

Step 1: Recall what a flash ADC is.
A flash analog-to-digital converter, also called a parallel ADC, is the fastest type of ADC. It compares the input analog voltage simultaneously against a set of reference voltage levels using a bank of comparators, and the pattern of comparator outputs is then decoded into a binary code.

Step 2: Recall how many output codes an n-bit converter must produce.
An \(n\)-bit ADC must be able to represent
\[ 2^{n} \]
distinct digital output codes, ranging from \(0\) up to \(2^{n}-1\).

Step 3: Understand why comparators sit between levels, not at each code.
The full input range is divided into \(2^{n}\) equal steps, or quantization levels, by a resistor ladder of reference voltages. A comparator is placed at each boundary between two adjacent quantization levels, so that the comparator output flips as the input voltage crosses that boundary. Since there are \(2^{n}\) levels arranged in a line, there are only
\[ 2^{n} - 1 \]
boundaries between them, so exactly \(2^{n}-1\) comparators are needed, not \(2^n\). The two outer ends of the range do not need boundary comparators of their own.

Step 4: Apply the formula for a 6-bit flash ADC.
Here \(n = 6\), so
\[ \text{Number of comparators} = 2^{6} - 1 \]

Step 5: Evaluate.
\[ 2^{6} = 64 \]
so
\[ 2^{6} - 1 = 63 \]

Final Answer:
The minimum number of comparators required is
\[ \boxed{63} \]