Editing Multivibrator (section)

=== Astable multivibrator using an op-amp ===
Assume all the capacitors to be discharged at first. The output of the [[Operational amplifier|op-amp]] V<sub>o</sub> at node '''''c''''' is +V<sub>sat</sub> initially. At node '''''a''''', a voltage of +β V<sub>sat </sub> is formed due to voltage division where <math>\beta=\left [ \frac{R2}{R1+R2} \right ]</math>. The current that flows from nodes '''''c''''' and '''''b''''' to ground charges the capacitor C towards +V<sub>sat</sub>. During this charging period, the voltage at '''''b''''' becomes greater than +β V<sub>sat </sub>at some point. The voltage at inverting terminal will be greater than the voltage at the non-inverting terminal of the op-amp. This is a comparator circuit and hence, the output becomes -V<sub>sat</sub>. The voltage at node '''''a''''' becomes -βV<sub>sat</sub> due to voltage division. Now the capacitor discharges towards -V<sub>sat</sub>. At some point, the voltage at '''''b''''' becomes less than -β V<sub>sat</sub>. The voltage at the non-inverting terminal will be greater than the voltage at the inverting terminal of the op-amp. So, the output of the op-amp is +V<sub>sat</sub>. This repeats and forms a free-running oscillator or an astable multivibrator.

If V<sub>C</sub> is the voltage across the capacitor and from the graph, the time period of the wave formed at capacitor and the output would match, then the time period could be calculated in this way:[[File:Astable Multivibrator using Op-Amp Graph.jpg|thumb|Graph showing the output waveform of the op-amp and the waveform formed across the capacitor C.]]<math>V_c = V_c(\infty)+[V_c(0)-V_c(\infty)]e^\tfrac{-t}{RC}
</math>

<math>V_c(t)= V_{sat} + [-\beta V_{sat} - V_{sat} ] e^{\left ( \frac{-t}{RC} \right )}</math>

At ''t'' =''T1'',

<math>\beta V_{sat} = V_{sat} (1-[\beta +1]e^{\tfrac{-T1}{RC}})</math>

Upon solving, we get:

<math>T1 = RC\ln\left [ \frac{1+\beta}{1-\beta} \right ]</math>

We are taking values of R, C and β such that we get a symmetrical square wave. Thus, we get ''T1'' = ''T2'' and total time period ''T'' = ''T1'' + ''T2''. So, the time period of the square wave generated at the output is:

<math>T=2RC\ln\left [\frac{1+\beta}{1-\beta} \right]</math>