Editing Relaxation oscillator (section)

==== Frequency of oscillation ====
First let's assume that <math>V_{dd} = -V_{ss}</math> for ease of calculation.  Ignoring the initial charge up of the capacitor, which is irrelevant for calculations of the frequency, note that charges and discharges oscillate between <math>\frac{V_{dd}}{2}</math> and <math>\frac{V_{ss}}{2}</math>.  For the circuit above, V<sub>ss</sub> must be less than 0. Half of the period (T) is the same as time that <math>V_{\rm out}</math> switches from V<sub>dd</sub>. This occurs when V<sub>−</sub> charges up from <math>-\frac{V_{dd}}{2}</math> to <math>\frac{V_{dd}}{2}</math>.

:<math>V_-=A+Be^{\frac{-1}{RC}t}</math>

:<math>\frac{V_{dd}}{2}=V_{dd}\left(1-\frac{3}{2}e^{\frac{-1}{RC}\frac{T}{2}}\right)</math>

:<math>\frac{1}{3}=e^{\frac{-1}{RC}\frac{T}{2}}</math>

:<math>\ln\left(\frac{1}{3}\right)=\frac{-1}{RC}\frac{T}{2}</math>

:<math>\, \! T=2\ln(3)RC</math>

:<math>\, \! f=\frac{1}{2\ln(3)RC}</math>

When V<sub>ss</sub> is not the inverse of V<sub>dd</sub> we need to worry about asymmetric charge up and discharge times. Taking this into account we end up with a formula of the form:

:<math>T = (RC)  \left[\ln\left( \frac{2V_{ss}-V_{dd}}{V_{ss}}\right) + \ln\left( \frac{2V_{dd}-V_{ss}}{V_{dd}} \right)  \right]</math>

Which reduces to the above result in the case that <math>V_{dd} = -V_{ss}</math>.