Question

For a clock frequency of 20 MHz, the output frequency of a flip-flop is:

• A. \(5\text{ MHz}\)
• B. \(20\text{ MHz}\)
• C. \(10\text{ MHz}\)
• D. \(40\text{ MHz}\)

Hint:

A single flip-flop operating in toggle mode always divides the input frequency by 2 (\(f_{\text{out}} = \frac{f_{\text{clk}}}{2}\)). Connecting \(N\) flip-flops in a cascade chain creates a ripple counter that divides the frequency by \(2^N\).

Answer 1

The Correct Option is C

Concept: Flip-flops (such as JK or T flip-flops) configured in a toggle state alter their output logic level on each active transition of the clock signal (either on the rising edge or the falling edge).
• Let us look at a rising-edge triggered toggle flip-flop. When a clock pulse arrives, the output toggles from \(0\) to \(1\).
• When the next consecutive clock pulse arrives, the output toggles back from \(1\) to \(0\). This means it takes exactly two complete cycles of the input clock signal to generate one single complete cycle (one high interval and one low interval) at the flip-flop's output. Therefore, a basic toggle flip-flop acts as a fundamental divide-by-2 (\(\div 2\)) frequency scaler.

Step 1: Finding the mathematical expression for frequency conversion.
The frequency of a periodic signal is inversely proportional to its time period (\(f = \frac{1}{T}\)). Let \(T_{\text{clock}}\) represent the time period of the incoming master clock signal, and \(T_{\text{out}}\) represent the time period of the resulting flip-flop output waveform. From the physical behavior described above: \[ T_{\text{out}} = 2 \cdot T_{\text{clock}} \] Expressing this relation in terms of frequency parameters: \[ \frac{1}{f_{\text{out}}} = 2 \cdot \left(\frac{1}{f_{\text{clock}}}\right) \] Inverting both sides gives the standard frequency division formula: \[ f_{\text{out}} = \frac{f_{\text{clock}}}{2} \]

Step 2: Calculating with the given frequency value.
We are given that the incoming system master clock frequency is: \[ f_{\text{clock}} = 20\text{ MHz} = 20 \times 10^6\text{ Hz} \] Substituting this value into our division equation: \[ f_{\text{out}} = \frac{20\text{ MHz}}{2} \] Dividing the values yields: \[ f_{\text{out}} = 10\text{ MHz} \] The output waveform frequency is exactly \(10\text{ MHz}\), which matches Option (C).