Digital Electronics - Flip-Flops - Discussion

Discussion Forum : Flip-Flops - General Questions (Q.No. 2)
2.
Which statement BEST describes the operation of a negative-edge-triggered D flip-flop?
The logic level at the D input is transferred to Q on NGT of CLK.
The Q output is ALWAYS identical to the CLK input if the D input is HIGH.
The Q output is ALWAYS identical to the D input when CLK = PGT.
The Q output is ALWAYS identical to the D input.
Answer: Option
Explanation:
No answer description is available. Let's discuss.
Discussion:
12 comments Page 1 of 2.

Sandeep said:   8 years ago
Q(n + 1) = D normally but in case -ve edge trigger Q(n) = D here FF will take i/o value which is there before apply -ve edge i/o.

Monika said:   9 years ago
This works out for me to get a clear cut idea about how particular matter or topic.

Thank you all.

Majoni said:   10 years ago
An edge-triggered flip-flop changes states either at the positive edge (rising edge) or at the negative edge (falling edge) of the clock pulse on the control input. The three basic types are introduced here: S-R, J-K and D.

Dipanjan said:   1 decade ago
What was PGT?

Khushboo c sharma said:   1 decade ago
Can anyone tell me what's the relation between CLK triggering and output of the Off.

AKARSH said:   1 decade ago
D stand for direct whatever will be the data at f/f it will transfer to o/p and on next clock it will transfer to next f/f, data will transfer directly on changing clock, f/f will respond according to -ve edge triggered clk in the same manner.

Bhagwat said:   1 decade ago
Negative edge triggered is created with the help of D-flip flop and D depends on Q.

Dinesh said:   1 decade ago
The o/p changes only at the -ve edge of clk.

Omm said:   1 decade ago
Point1:

FF is -ve edge triggered means the o/p will follow i/p only on transition of clock from 1 to 0.

It means that the o/p D=Qn=Qn-1 when clock changes from 1 to 0.

Yasmin said:   1 decade ago
Q depends on the value of D.


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