Question

For the given digital circuit, $A=B=1$. Assume that AND, OR, and NOT gates have propagation delays of $10$ ns, $10$ ns, and $5$ ns respectively (wires have zero delay). Given that $C=1$ at turn–on, the frequency of steady–state oscillation of the output $Y$ is _____.

• A. 20 MHz
• B. 15 MHz
• C. 40 MHz
• D. 50 MHz

Hint:

For a single-loop ring oscillator, $f=\dfrac{1}{2\sum\text{(gate delays)}}$ when there is an odd number of inversions in the loop.

Answer 1

The Correct Option is A

With $A=B=1$, the feedback path that determines the oscillation contains one NOT gate and the OR and AND gates, i.e. an odd number of inversions (one) around the loop, so the circuit behaves like a ring oscillator.
Loop delay $T_{\ell}=\underbrace{5}_{\text{NOT}}+\underbrace{10}_{\text{OR}}+\underbrace{10}_{\text{AND}}=25\text{ ns}$.
The oscillation period is twice the loop delay, $T=2T_{\ell}=50\text{ ns}$, hence the frequency \[ f=\frac{1}{T}=\frac{1}{50\text{ ns}}=20\ \text{MHz}. \] Therefore, \fbox{20 MHz}.