Need a Tiny Bit of Help with 555 Oscillator Frequency Formula

[mmmboh]

With regards to this http://www.electronics-tutorials.ws/waveforms/555_oscillator.html, I figured out how to get t₂ fine, but I have a problem figuring out how to get t₁...I know when the negative-going waveform at pin 2 crosses Vcc/3, pin 3 goes to Vcc, and pin 7 gets internally disconnected, and the capacitor charges to 2Vcc/3.
Now using the formula V_capacitor=V₀(1-e^-t/RC), what I think "should" be right is to set it up like this: 2Vcc/3=(Vcc/3)*((1-e^{-t/(R_A+R_B)C})...but of course this doesn't make sense, the voltage across the capacitor can't go above the voltage that's charging it. The final capacitor voltage should be half the initial for the equation to work I believe. Can anyone clarify how I am suppose to set this equation up please?

Thanks!

 

[vk6kro]

There are many calculators available for doing this, but these are the usual formulas used:

T1 (on-time) = 0.693 * (R1 + R2) * C1
T2 (off-time) = 0.693 * R2 * C1
T (period) = 0.693*( (R1 + 2R2) * C1 )
F (frequency) = 1/ T
= 1.44 / ( (R1 + 2R2) * C1 )
D (duty cycle) = (T1 /(T1+T2))*100

Note that capacitor values are in Farads. R1 is the top resistor and R2 is the lower one.

This website seems OK:
http://www.horrorseek.com/home/halloween/wolfstone/TechBase/com555_555TimerCalc.html

What would be better is one that gave the resistor sizes given the frequency. Usually you are given the frequency and need to know the resistor sizes, not the other way around.
Like this:
F (frequency) = 1 / T
= 1.443 / ( (R1 + 2R2) * C1 )

So (R1 + 2R2) =1.443 / ( F * C1) ...C in Farads, again.

eg if F = 500 Hz and C1 = 0.1uF then (R1 + 2R2) = 28860 ohms.
If R1 = 1000 ohms then R2 = 13930 ohms

 

[mmmboh]

Right, but the problem is I need to actually derive the formula, for t2 I figured it out, but I can't figure out how to derive it for t1.

 

[vk6kro]

Suppose you had a power supply of 9 volts.

The capacitor will charge from 3 volts to 6 volts from a supply of 9 volts.

So, initially, there is 6 volts across the series resistance and the capacitor charges until there is 3 volts across the series resistance and an extra 3 volts across the capacitor.

So, you ignore the initial 3 volts charge on the capacitor and just regard it as a 6 volt supply charging a capacitor up to half the supply voltage of 6 volts.

 

[mmmboh]

Oh I see. Thanks!

 

[staticd]

This might help...