Editing Relaxation oscillator (section)

=== Example: Differential equation analysis of a comparator-based relaxation oscillator ===

[[Image:opamprelaxationoscillator.svg|thumb|300px|Transient analysis of a comparator-based relaxation oscillator.]]

<math>\, \! V_+</math> is set by <math>\, \! V_{\rm out}</math> across a resistive [[voltage divider]]:

:<math>V_+ = \frac{V_{\rm out}}{2}</math>

<math>\, \! V_-</math> is obtained using [[Ohm's law]] and the [[capacitor]] [[differential equation]]:

:<math>\frac{V_{\rm out}-V_-}{R}=C\frac{dV_-}{dt}</math>

Rearranging the <math>\, \! V_-</math> differential equation into standard form results in the following:

:<math>\frac{dV_-}{dt}+\frac{V_-}{RC}=\frac{V_{\rm out}}{RC}</math>

Notice there are two solutions to the differential equation, the driven or particular solution and the homogeneous solution.  Solving for the driven solution, observe that for this particular form, the solution is a constant.  In other words, <math>\, \! V_-=A</math> where A is a constant and <math>\frac{dV_-}{dt}=0</math>.

:<math>\frac{A}{RC}=\frac{V_{\rm out}}{RC}</math>

:<math>\, \! A=V_{\rm out}</math>

Using the [[Laplace transform]] to solve the [[Homogeneous polynomial|homogeneous equation]] <math>\frac{dV_-}{dt}+\frac{V_-}{RC}=0</math> results in

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

<math>\, \!  V_-</math> is the sum of the particular and homogeneous solution.

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

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

Solving for B requires evaluation of the initial conditions.  At time 0, <math>V_{\rm out}=V_{dd}</math> and <math>\, \! V_-=0</math>. Substituting into our previous equation,

:<math>\, \! 0=V_{dd}+B</math>

:<math>\, \! B=-V_{dd}</math>

==== 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>.