# 555 HYSTERETIC OSCILLATOR formula for "f" incorrect

#### SgtWookie

Joined Jul 17, 2007
22,210
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.

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#### Wendy

Joined Mar 24, 2008
22,155
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.
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.

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#### ErnieM

Joined Apr 24, 2011
8,053
Bill is correct here in his frequency and period calculations. If you want to do this exactly you can pull out this magic formula:

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

Where:
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.

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