The MSB frequency of a 4-bit ripple counter with 16 MHz clock is
Show Hint
In asynchronous counters, the MSB output always represents the slowest rate of change.
The frequency scales down as a power of 2: \( f_{\text{out}} = \frac{f_{\text{in}}}{\text{Modulus}} \), where modulus is \( 2^N \).
Step 1: Understanding the Problem:
The question asks to find the frequency at the Most Significant Bit (MSB) output of a 4-bit ripple counter when the input clock frequency is 16 MHz.
A ripple counter is an asynchronous counter where the clock input of each subsequent flip-flop is triggered by the output of the preceding flip-flop. Step 2: Key Formula or Approach:
In any binary counter, each stage acts as a divide-by-2 frequency divider.
For an \( N \)-bit ripple counter, the frequency at the \( N \)-th flip-flop output (which corresponds to the MSB) is given by:
\[ f_{\text{MSB}} = \frac{f_{\text{clk}}}{2^N} \]
Where:
\( f_{\text{clk}} \) is the input clock frequency.
\( N \) is the number of bits (or number of flip-flops). Step 3: Detailed Explanation:
Let us perform the stage-by-stage calculations for the 4-bit ripple counter:
Given:
Number of bits, \( N = 4 \)
Input clock frequency, \( f_{\text{clk}} = 16\text{ MHz} \)
• First stage output (\( Q_0 \) - LSB): The first flip-flop divides the clock frequency by 2:
\[ f_0 = \frac{16\text{ MHz}}{2} = 8\text{ MHz} \]
• Second stage output (\( Q_1 \)): The second flip-flop divides \( f_0 \) by 2:
\[ f_1 = \frac{8\text{ MHz}}{2} = 4\text{ MHz} \]
• Third stage output (\( Q_2 \multi \)): The third flip-flop divides \( f_1 \) by 2:
\[ f_2 = \frac{4\text{ MHz}}{2} = 2\text{ MHz} \]
• Fourth stage output (\( Q_3 \) - MSB): The fourth flip-flop divides \( f_2 \) by 2:
\[ f_3 = \frac{2\text{ MHz}}{2} = 1\text{ MHz} \]
This can be directly validated using the master formula:
\[ f_{\text{MSB}} = \frac{16\text{ MHz}}{2^4} = \frac{16\text{ MHz}}{16} = 1\text{ MHz} \]
Step 4: Final Answer:
The MSB frequency of the 4-bit ripple counter is 1 MHz.