Jun 23, 2019 #1 B Blue10 Junior Member level 3 Joined May 28, 2019 Messages 28 Helped 0 Reputation 0 Reaction score 0 Trophy points 1 Activity points 186 Hi, How do I calculate the frequency of this circuit? Thanks.
Jun 23, 2019 #2 FvM Super Moderator Staff member Joined Jan 22, 2008 Messages 52,404 Helped 14,749 Reputation 29,780 Reaction score 14,094 Trophy points 1,393 Location Bochum, Germany Activity points 298,026 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).
Jun 23, 2019 #3 B Blue10 Junior Member level 3 Joined May 28, 2019 Messages 28 Helped 0 Reputation 0 Reaction score 0 Trophy points 1 Activity points 186 FvM said: 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). Click to expand... Why is R2=5k?
FvM said: 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). Click to expand... Why is R2=5k?
Jun 23, 2019 #4 FvM Super Moderator Staff member Joined Jan 22, 2008 Messages 52,404 Helped 14,749 Reputation 29,780 Reaction score 14,094 Trophy points 1,393 Location Bochum, Germany Activity points 298,026 RV1 center position, 10k/2 is effective R2 value (ignoring the diode voltage drop).
Jun 23, 2019 #5 B Blue10 Junior Member level 3 Joined May 28, 2019 Messages 28 Helped 0 Reputation 0 Reaction score 0 Trophy points 1 Activity points 186 FvM said: RV1 center position, 10k/2 is effective R2 value (ignoring the diode voltage drop). Click to expand... Thank you for your answer! So the frequency is : f = 1.44 / ((R1 + 2*R2)*c) = 13KHz ?
FvM said: RV1 center position, 10k/2 is effective R2 value (ignoring the diode voltage drop). Click to expand... Thank you for your answer! So the frequency is : f = 1.44 / ((R1 + 2*R2)*c) = 13KHz ?
Jun 23, 2019 #6 S srizbf Advanced Member level 5 Joined Apr 3, 2010 Messages 1,992 Helped 417 Reputation 840 Reaction score 329 Trophy points 1,363 Activity points 10,985 frequency of your circuit is: 1/((R2+k*Rv1)+(1-k)*Rv1) where 'k' as per figure is 0.41 - - - Updated - - - srizbf said: frequency of your circuit is: 1/((R2+k*Rv1)+(1-k)*Rv1) where 'k' as per figure is 0.41 Click to expand... 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 - - - Updated - - - srizbf said: frequency of your circuit is: 1/((R2+k*Rv1)+(1-k)*Rv1) where 'k' as per figure is 0.41 Click to expand... add 1.44 in numerator.
Jun 23, 2019 #7 B Blue10 Junior Member level 3 Joined May 28, 2019 Messages 28 Helped 0 Reputation 0 Reaction score 0 Trophy points 1 Activity points 186 srizbf said: frequency of your circuit is: 1/((R2+k*Rv1)+(1-k)*Rv1) where 'k' as per figure is 0.41 Click to expand... But by using your equation the frequency will be : 9 * 10^(-5) Hz !!
srizbf said: frequency of your circuit is: 1/((R2+k*Rv1)+(1-k)*Rv1) where 'k' as per figure is 0.41 Click to expand... But by using your equation the frequency will be : 9 * 10^(-5) Hz !!
Jun 23, 2019 #8 S srizbf Advanced Member level 5 Joined Apr 3, 2010 Messages 1,992 Helped 417 Reputation 840 Reaction score 329 Trophy points 1,363 Activity points 10,985 Blue10 said: But by using your equation the frequency will be : 9 * 10^(-5) Hz !! Click to expand... 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
Blue10 said: But by using your equation the frequency will be : 9 * 10^(-5) Hz !! Click to expand... 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
Jun 23, 2019 #9 crutschow Advanced Member level 6 Joined Feb 22, 2012 Messages 4,459 Helped 999 Reputation 1,996 Reaction score 1,123 Trophy points 1,393 Location Colorado USA Zulu -7 Activity points 25,279 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.
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.
Jun 23, 2019 #10 W wwfeldman Advanced Member level 4 Joined Jan 25, 2019 Messages 1,101 Helped 231 Reputation 461 Reaction score 297 Trophy points 83 Activity points 8,554 @Blue10 where did you find that circuit?
Jun 23, 2019 #11 Audioguru Advanced Member level 7 Joined Jan 19, 2008 Messages 9,457 Helped 2,151 Reputation 4,302 Reaction score 2,008 Trophy points 1,393 Location Toronto area of Canada Activity points 59,719 That circuit is all over the internet. Hey Google: 555 oscillator circuit with duty-cycle adjustment