Electronic Devices - Linear-Digital ICs - Discussion
Discussion Forum : Linear-Digital ICs - General Questions (Q.No. 12)
12.
Determine the free-running frequency when R3 is set to 2.5 k
.
. 
Discussion:
6 comments Page 1 of 1.
Gitanjali said:
9 years ago
Timer 555 working as astable multivibrator,
Frequency = 1.44/{(r1+r2)*2*c}.
Frequency = 1.44/{(r1+r2)*2*c}.
(1)
Mukunf said:
9 years ago
Can anyone solve it clearly?
Anomii said:
7 years ago
Can anyone solve it?
Gopi said:
7 years ago
Please explain in detail @Gitanjali.
Jihad said:
7 years ago
Can anyone solve this in detail?
Cyran said:
6 years ago
Vc=[(R3+R4)/(R2+R3+R4)]*V+
Vc=[((2.5x10^3)+(18x10^3)) / (510)+(5x10^3)+(18x10^3)] *12 = 10.46V
fo=(2/(R1C1))*[((V+)-Vc)/V+]
fo=(2/(10x10^3)(220x10^-12)) * ((12-10.46)/12)= 116.39kHz.
Vc=[((2.5x10^3)+(18x10^3)) / (510)+(5x10^3)+(18x10^3)] *12 = 10.46V
fo=(2/(R1C1))*[((V+)-Vc)/V+]
fo=(2/(10x10^3)(220x10^-12)) * ((12-10.46)/12)= 116.39kHz.
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