Digital Electronics - Flip-Flops - Discussion

Discussion Forum : Flip-Flops - General Questions (Q.No. 45)
45.
An RC circuit used in a 74122 retriggerable one-shot has an REXT of 100 k and a CEXT of 0.005 F. The pulse width is ________.
70 s
16 s
160 s
32 s
Answer: Option
Explanation:
No answer description is available. Let's discuss.
Discussion:
12 comments Page 1 of 2.

Ram said:   8 years ago
Pulse width (tp) of Mono-stable multi-vibrator= 1.1 RC.
Pulse width (tp) of Non-triggerable Mono-stable multi-vibrator= 0.69 RC.
Pulse width (tp) of Re-triggerable Mono-stable multi-vibrator= 0.32 RC.
(2)

Varun said:   1 decade ago
W=0.69*r*c.

Calculate you get the answer.

Shesh said:   1 decade ago
W=0.69*r*c
=0.69*100000*0.005*10^-6
=345*10^-6
Not getting

Krishna said:   1 decade ago
By using this W=0. 69*r*c.
I can't getting the answer can explain me the answer.

Venkat said:   1 decade ago
Actual formula for this is pulse width = 0.32RC(1+(0.7/R))
HERE R= is in Kilo Ohms
So answer is 160 micro sec.

Gopi-009 said:   1 decade ago
W=0.32RC.

R and C known
W=0.32*100*0.005

W=0.16ms.

W=160*10*-6.

Kiran said:   1 decade ago
Can any one explain where I have to use 1.1RC, 0.69Rc and this 0.32RC.

Rkumar said:   1 decade ago
Can anyone please explain where I have to use 1.1RC, 0.69Rc and this 0.32RC ?

PRIYANKA said:   1 decade ago
One shot is monostable. So we have to use 1.1 RC.

Shahin said:   10 years ago
As per data sheet of 74122, if Cext >1000pF,

Then, width of the pulse, tW = k*Cext*Rext (1+0.7/Rext) k = 0.32.

Here Cext = = 0.005 uF > 1000 pF.

= 0.32*0.005 uF*100 k (1+0.7/100 k) = 160 us.

Thank you.


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