Digital Electronics - Flip-Flops - Discussion
Discussion Forum : Flip-Flops - General Questions (Q.No. 34)
34.
A 555 operating as a monostable multivibrator has a C1 = 0.01 
F. Determine R1 for a pulse width of 2 ms.
F. Determine R1 for a pulse width of 2 ms.200 k
182 k
91 k
182
Discussion:
7 comments Page 1 of 1.
BALAJI said:
4 years ago
For monostable T= 1/0.69RC formula is used for astable only T=1/2.2 RC is used.
(1)
ASHISH KUMAR said:
7 years ago
For monostable pulse width is given by (Tw) = 1.1RC.
Minal said:
7 years ago
Thanks @Sona.
Shaquib said:
8 years ago
Well explained, Thanks @Sona.
Sona said:
1 decade ago
Pulse width for Astable multivibrator is 0.69RC. For mono-stable it is 1.1RC. Just see the circuit of both, it'll be clear that why there is change in formula.
M.V.KRISHNA/PALVONCHA said:
1 decade ago
Pulse width=0.69RC.
why 1.1RC is taken here?
why 1.1RC is taken here?
Prince said:
1 decade ago
w=2ms c=.01uf
nw W=1.1 RC
evaluating we get r=181.8 kohm
nw W=1.1 RC
evaluating we get r=181.8 kohm
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