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NE555 astable frquency

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Blue10

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Hi,
How do I calculate the frequency of this circuit?
NE555 frequency.png

Thanks.
 

For a rough estimation, assume RV1 in center position, ignore diode voltage drop, use formula for astable operation in data sheet (R1=1k, R2=5k).
 

For a rough estimation, assume RV1 in center position, ignore diode voltage drop, use formula for astable operation in data sheet (R1=1k, R2=5k).

Why is R2=5k?
 

RV1 center position, 10k/2 is effective R2 value (ignoring the diode voltage drop).
 

RV1 center position, 10k/2 is effective R2 value (ignoring the diode voltage drop).

Thank you for your answer!
So the frequency is : f = 1.44 / ((R1 + 2*R2)*c) = 13KHz ?
 

frequency of your circuit is:

1/((R2+k*Rv1)+(1-k)*Rv1)

where 'k' as per figure is 0.41

- - - Updated - - -

frequency of your circuit is:

1/((R2+k*Rv1)+(1-k)*Rv1)

where 'k' as per figure is 0.41


add 1.44 in numerator.
 

frequency of your circuit is:

1/((R2+k*Rv1)+(1-k)*Rv1)

where 'k' as per figure is 0.41

But by using your equation the frequency will be : 9 * 10^(-5) Hz !!
 

But by using your equation the frequency will be : 9 * 10^(-5) Hz !!

again a typo.

the term 'c' is left out in denominator.

so the freq is of the form:

1.44/((R2+kRv1)+(1-k)Rv1)C2
 

Since the frequency of the 555 doesn't vary much with the setting of the duty-cycle pot, the value you get using 1/2 the pot resistance should be good enough for a reasonable estimate of the frequency.
 

@Blue10
where did you find that circuit?
 

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