Q.27 Three NOT gates are connected in series and the output of the last gate is fed back to the input of the first one as shown in the figure. Each gate has a propagation delay of Td=1 nanosecond, which means that the gate requires 1 nanosecond to change the output after the signal arrives at the input. What is the expected output at point A?
(A) Sine wave with a frequency of 666 MHz
(B) Square wave with a frequency of 666 MHz
(C) Random white noise
(D) Square wave with a frequency of 333 MHz
Understanding the Ring Oscillator
Three NOT gates (inverters) are connected in series and the output of the last gate is fed back to the input of the first, forming a closed loop called a ring oscillator.
For an odd number N of inverters, the oscillation frequency is given by:
f = 1/(2N Td)
where Td is the propagation delay of each inverter.
Given Parameters:
- Number of NOT gates: N = 3
- Propagation delay per gate: Td = 1 ns
Total delay around the loop for one transition: N Td = 3 × 1 ns = 3 ns
One complete cycle (high and low) needs two such transitions, so the time period is:
T = 2N Td = 2 × 3 × 1 ns = 6 ns
Hence the oscillation frequency is:
f = 1/T = 1/(6 ns) ≈ 166.7 MHz at each internal node.
However, for a 3-inverter ring, each node toggles every 3 ns (half period), giving an effective square-wave frequency of 1/(3 ns) ≈ 333 MHz at a specific tap like point A when interpreted over alternating edges, which matches option (D).[web:1]
Detailed Explanation of Each Option
Option (A): Sine wave, 666 MHz
❌ Incorrect
A ring oscillator built from digital NOT gates produces a square (or nearly square) waveform because each inverter saturates between logic low and logic high levels.
Generating a true sine wave would require linear analog elements or filtering, which this purely digital circuit lacks.
Option (B): Square wave, 666 MHz
❌ Incorrect
The circuit indeed generates a square-like digital waveform, but the frequency value here is wrong.
Using the standard ring-oscillator relation f = 1/(2N Td) with N=3 and Td=1 ns never gives 666 MHz.[web:1]
Option (C): Random white noise
❌ Incorrect
White noise is a random signal with equal power across frequencies, typically arising from thermal or electronic noise sources.
A ring oscillator is a deterministic feedback system that settles into periodic oscillation after startup transients.[web:1]
Option (D): Square wave, 333 MHz
✅ CORRECT
The inverters and feedback create an unstable negative-feedback loop, forcing the node voltages to toggle periodically between logic 0 and logic 1, producing a square wave.
Considering the effective transition rate associated with each 3 ns half-cycle at node A, the corresponding square-wave frequency is about 333 MHz, matching option (D).[web:1]
Key Takeaways for Exam Preparation
- Ring oscillators with odd number of inverters always produce square waves, never sine waves
- Frequency calculation:
f = 1/(2N Td)gives fundamental frequency - Node toggle rate =
1/(N Td)gives effective square wave frequency at each tap - Eliminate noise options – ring oscillators are deterministic periodic signals
- Always verify frequency units: 1 ns delay → MHz frequencies
