Electronics and Communication Engineering - Digital Electronics - Discussion
Discussion Forum : Digital Electronics - Section 4 (Q.No. 26)
26.
For the ring oscillator shown in the figure, the propagation delay of each inverter is 100 pico sec. What is the fundamental frequency of the oscillator output __________


Answer: Option
Explanation:
T = 500 x 10-12 = 5 x 10-10
= .
Discussion:
5 comments Page 1 of 1.
Durgesh said:
7 years ago
A cycle has both positive and negative half cycles. A number of cycles passing through a point in a given time we call it as a frequency.
So one, not gate conducts and change its state from one-half cycle to another half cycle. It conducts again and changes its state.
If final not gate at output conducts two times then only it produces a complete full cycle.
So, option C is correct.
So one, not gate conducts and change its state from one-half cycle to another half cycle. It conducts again and changes its state.
If final not gate at output conducts two times then only it produces a complete full cycle.
So, option C is correct.
Ravi said:
8 years ago
Yes, C the is correct answer.
Danish said:
8 years ago
Frequency of Ring Oscillator or Astable multivibrator is the inverse of the 2*n*propagation delay of each inverter.
So, few calculations will give 1 GHz as the answer. C is the correct answer.
So, few calculations will give 1 GHz as the answer. C is the correct answer.
Deep shukla said:
9 years ago
2 GHz is right. Frequency is not time period.
Abhi said:
10 years ago
Wrong answer it should be 1 Ghz.
Time taken to 0 to 1 is 100 psec.
Total 5 transition is 5x100 = 500 psec.
But time period becomes = 500+500 = 1000 psec.
Time taken to 0 to 1 is 100 psec.
Total 5 transition is 5x100 = 500 psec.
But time period becomes = 500+500 = 1000 psec.
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