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 ________.


Discussion:
12 comments Page 1 of 2.
Varun said:
1 decade ago
W=0.69*r*c.
Calculate you get the answer.
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
=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.
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.
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.
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.
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.
Pranav said:
9 years ago
General formula for pulse width is k*R*C.
We have to choose 'k' accordingly,
For monostable K=1.1
For a stable K = 0.69
For 74122, it is said that we should use k= 0.32.
We have to choose 'k' accordingly,
For monostable K=1.1
For a stable K = 0.69
For 74122, it is said that we should use k= 0.32.
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