555 HYSTERETIC OSCILLATOR formula for "f" incorrect

Discussion in 'Feedback and Suggestions' started by SgtWookie, Jul 22, 2011.

  1. SgtWookie

    Thread Starter Expert

    Jul 17, 2007
    under THEORY OF OPERATION, formula for f given as:
    f = 0.7 / RC
    which is not correct.

    The correct approximation formula is:
    f = 1 / 0.7(RC)
    which can also be expressed more simply; if harder to remember:
    f = 1.44 / (RC)

    I also suggest that the approximate formula for t (time) be added just prior to the calculation for frequency; as:
    t = 0.7(RC)

    ...mentioning that R is in Ohms, and C is in Farads.
    Last edited: Jul 22, 2011
  2. Wendy


    Mar 24, 2008
    Actually I disagree. My rebuttal is on the 555 Hysteretic Oscillator thread in detail.

    1 time constant is RC, or 1TC = 1RC. It takes 0.7 TC to charge the RC circuit, then another 0.7 TC to discharge the circuit, for a total of 1.4 TC (give or take a large variance).

    So the period is:

    P = 1.4 TC or 1.4 RC

    Since the frequency is inverse, and I choose to leave the number on the top (an arbitrary choice):

    F = 1 / (1.4 RC) or 0.7 / (R C)

    Had me going for a sec though, I've based an awful lot of work on this simple formula. Me heart skipped a thump it did.
    Last edited: Jul 22, 2011
  3. ErnieM

    AAC Fanatic!

    Apr 24, 2011
    Bill is correct here in his frequency and period calculations. If you want to do this exactly you can pull out this magic formula:

     t = RC \ast |ln \frac{(Vss - Vi)}{(Vss - Vf)}|

    Vss is the steady state voltage of the cap, or the voltage when time is infinity
    Vi is the initial voltage on the cap
    Vf if the final voltage on the cap
    t is the time it takes to charge from Vi to Vf

    So for the charge phase:
    Vi = 1/3 Vcc
    Vf = 2/3 Vcc

    The time to charge (t1) is just:

     t1 = RC \ast |ln \frac{(Vss - 1/3*Vcc)}{(Vss - 2/3*Vcc)}|

    Which reduces to:

     t1 = RC \ast |ln 2| \sim RC\ast .693

    To discharge:
    Vi = 2/3 Vcc
    Vf = 1/3 Vcc

    Also the time to discharge (t2) is just:

     t1 = RC \ast |ln \frac{(Vss - 2/3*Vcc)}{(Vss - 1/3*Vcc)}|

    Which reduces to:

     t1 = RC \ast |ln .5| \sim RC\ast |-.693| = RC\ast .693

    Total period is just the sum:

     Period = t1 + t2 \sim .693 + .693 = 1.386

    I call this a magic formula as I don't have a reference to it's derivation. If there is any interest I could work it out, it's just solving the general cap voltage equation of:

     Vc(t) = Vss * ( 1 - e^{\frac{-t}{RC}})

    One thing to note is the formula is correct for any Vi or Vf, even those outside of the range of zero to Vss.
    Last edited: Jul 25, 2011