Editing Relaxation oscillator (section)

== Comparator&ndash;based relaxation oscillator ==
{{unreferenced section|date=November 2024}}

Alternatively, when the capacitor reaches each threshold, the charging source can be switched from the positive power supply to the negative power supply or vice versa. The earlier inverting [[Schmitt trigger]] animated example operates on the same principle (since the Schmitt trigger internally performs comparison). This section will analyze a similar implementation using a [[comparator]] as a discrete component.

[[Image:OpAmpHystereticOscillator.svg|thumb|A comparator-based hysteretic oscillator.]]

This relaxation oscillator is a hysteretic oscillator, named this way because of the [[hysteresis]] created by the [[positive feedback]] loop implemented with the [[comparator]] (similar to an [[operational amplifier]]). A circuit that implements this form of hysteretic switching is known as a [[Schmitt trigger]]. Alone, the trigger is a [[bistable multivibrator]]. However, the slow [[negative feedback]] added to the trigger by the RC circuit causes the circuit to oscillate automatically. That is, the addition of the RC circuit turns the hysteretic bistable [[multivibrator]] into an [[astable multivibrator]].

=== General concept ===
The system is in unstable equilibrium if both of the inputs and the output of the comparator are at zero volts. The moment any sort of noise, be it thermal or [[Electromagnetic radiation|electromagnetic]] [[noise]] brings the output of the comparator above zero (the case of the comparator output going below zero is also possible, and a similar argument to what follows applies), the positive feedback in the comparator results in the output of the comparator saturating at the positive rail.

In other words, because the output of the comparator is now positive, the non-inverting input to the comparator is also positive, and continues to increase as the output increases, due to the [[voltage divider]].  After a short time, the output of the comparator is the positive voltage rail, <math>V_{DD}</math>.

[[Image:Series-RC.svg|thumb|Series RC Circuit]]

The inverting input and the output of the comparator are linked by a [[Series and parallel circuits#Series circuits|series]] [[RC circuit]].  Because of this, the inverting input of the comparator asymptotically approaches the comparator output voltage with a [[time constant]] RC.  At the point where voltage at the inverting input is greater than the non-inverting input, the output of the comparator falls quickly due to positive feedback.

This is because the non-inverting input is less than the inverting input, and as the output continues to decrease, the difference between the inputs gets more and more negative. Again, the inverting input approaches the comparator's output voltage asymptotically, and the cycle repeats itself once the non-inverting input is greater than the inverting input, hence the system oscillates.

=== 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>.