Electronics - Operational Amplifiers - Discussion
Discussion Forum : Operational Amplifiers - General Questions (Q.No. 29)
29.
What is the frequency of this 555 astable multivibrator?
Discussion:
11 comments Page 1 of 2.
Murugan said:
7 years ago
T(high) = 0.693(RA+RB)C.
T(low) = 0.693(RB)C,
T = T(high)+T(low),
F = 1/T.
F = 1.44/(RA+2RB)C.
T(low) = 0.693(RB)C,
T = T(high)+T(low),
F = 1/T.
F = 1.44/(RA+2RB)C.
(2)
Madhur said:
1 decade ago
Its coming around 241HZ.
(1)
Sunny said:
1 decade ago
@vid
Explanation which you had given is wrong.
For Astable multivibrator, f=(1/T)= 1/(charg. time + discharging time)
or, f = 1/[0.69(Ra+2Rb)c].
Explanation which you had given is wrong.
For Astable multivibrator, f=(1/T)= 1/(charg. time + discharging time)
or, f = 1/[0.69(Ra+2Rb)c].
(1)
Rakesh singh said:
1 decade ago
F = 1.44/(R1 + 2 R2)C
Buvana said:
1 decade ago
How is it can be correct? It is obscure for me.
Karuppasamy kpk said:
1 decade ago
According to the formula,
F= 1.44/((1000+2*6000000)10`9)
F=240.50 hz
Therefore,no one option is correct.
how can u say option B correct...?
F= 1.44/((1000+2*6000000)10`9)
F=240.50 hz
Therefore,no one option is correct.
how can u say option B correct...?
Vid said:
1 decade ago
For astable t=1.4rc
f=1/(1.4(ra+2rB)c
Then i am getting 119.0
f=1/(1.4(ra+2rB)c
Then i am getting 119.0
Kelly Jones said:
1 decade ago
f= 1.4427/[R1+(2*R2)]*C.
f= 1.4427/[1k+(2*3M)]*1.0 nano.
f= 1.4427/(1k+6M)*1.0 nano.
f= 1.4427/6.001M*1.0 nano.
f= 1.4427/6.001 milli.
f= 240.41 Hz.
f= 1.4427/[1k+(2*3M)]*1.0 nano.
f= 1.4427/(1k+6M)*1.0 nano.
f= 1.4427/6.001M*1.0 nano.
f= 1.4427/6.001 milli.
f= 240.41 Hz.
Abbasi said:
10 years ago
For free running multivibrator.
f = 1.44/(R1+2R2)C.
Hence f = 239.96 Hz.
f = 1.44/(R1+2R2)C.
Hence f = 239.96 Hz.
Sanjoy said:
10 years ago
Ton = (RA+RB) C1 = 3 ms.
Toff = RB*C1 = 3 ms.
T = Ton+Toff = 6 ms.
F = 1/T = 1/6 ms = 167 HZ.
Toff = RB*C1 = 3 ms.
T = Ton+Toff = 6 ms.
F = 1/T = 1/6 ms = 167 HZ.
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