# Relaxation oscillator

Relaxation oscillator

A relaxation oscillator is an oscillator in which a capacitor is charged gradually and then discharged rapidly. It is usually implemented with a resistor or current source, a capacitor, and a "threshold" device such as a neon lamp, diac, unijunction transistor, or Gunn diode. For simplification below, a single "threshold" device will be replaced by a set of comparators and a SR Latch.

The capacitor is charged through the resistor, causing the voltage across the capacitor to approach the charging voltage on an exponential curve. In parallel with the capacitor is the threshold device. Such devices don't conduct at all until the voltage across them reaches some threshold (trigger) voltage. They then conduct heavily, quickly discharging the capacitor. When the voltage across the capacitor drops to some lower threshold voltage, the device stops conducting and the capacitor can begin charging again, repeating the cycle. If the threshold element is a neon lamp, the circuit also provides a flash of light with each discharge of the capacitor.

The electrical output of a relaxation oscillator is usually a sawtooth wave. If only a small portion of the exponential ramp is used (that is, if the triggering voltage of the threshold device is much lower than the charging voltage source), the ramp will approximate a linear ramp but if a truly linear sawtooth is required, then the charging resistor should be replaced by some sort of constant current source.

Op Amp-Based Relaxation Oscillator

One example of a relaxation oscillator is a hysteretic oscillator, named this way because of the positive feedback loop on the operational amplifier, which exhibits hysteresis.

General Concept

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

In other words, because the output of the op amp is now positive, the non-inverting input to the op amp is also positive, and continues to increase as the output increases, due to the voltage divider. After a short time, the output of the op amp is the positive voltage rail, $V_\left\{DD\right\}$.

The inverting input and the output of the op amp are linked by a series RC circuit. Because of this, the inverting input of the op amp asymptotically approaches the op amp 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 op amp 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 op amp'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 Op-Amp based Relaxation Oscillator

$, ! V_+$ is set by $, ! V_\left\{out\right\}$ across a resistive voltage divider:

:$V_+ = frac\left\{V_\left\{out\left\{2\right\}$

$, ! V_-$ is obtained using Ohm's law and the capacitor differential equation:

:$frac\left\{V_\left\{out\right\}-V_-\right\}\left\{R\right\}=Cfrac\left\{dV_-\right\}\left\{dt\right\}$

Rearranging the $, ! V_-$ differential equation into standard form results in the following:

:$frac\left\{dV_-\right\}\left\{dt\right\}+frac\left\{V_-\right\}\left\{RC\right\}=frac\left\{V_\left\{out\left\{RC\right\}$

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, $, ! V_-=A$ where A is a constant and $frac\left\{dV_-\right\}\left\{dt\right\}=0$.

:$frac\left\{A\right\}\left\{RC\right\}=frac\left\{V_\left\{out\left\{RC\right\}$

:$, ! A=V_\left\{out\right\}$

Using the Laplace transform to solve the homogeneous equation $frac\left\{dV_-\right\}\left\{dt\right\}+frac\left\{V_-\right\}\left\{RC\right\}=0$ results in

:$V_-=Be^\left\{frac\left\{-1\right\}\left\{RC\right\}t\right\}$

$, ! V_-$ is the sum of the particular and homogeneous solution.

:$V_-=A+Be^\left\{frac\left\{-1\right\}\left\{RC\right\}t\right\}$

:$V_-=V_\left\{out\right\}+Be^\left\{frac\left\{-1\right\}\left\{RC\right\}t\right\}$

Solving for B requires evaluation of the initial conditions. At time 0, $V_\left\{out\right\}=V_\left\{dd\right\}$ and $, ! V_-=0$. Substituting into our previous equation,

:$, ! 0=V_\left\{dd\right\}+B$

:$, ! B=-V_\left\{dd\right\}$

Frequency of Oscillation

Note that half of the period (T) is the same as the first time that $V_\left\{out\right\}$ switches. This occurs when $V_-=frac\left\{V_\left\{dd\left\{2\right\}$.

:$V_-=A+Be^\left\{frac\left\{-1\right\}\left\{RC\right\}t\right\}$

:$frac\left\{V_\left\{dd\left\{2\right\}=V_\left\{dd\right\}-V_\left\{dd\right\}e^\left\{frac\left\{-1\right\}\left\{RC\right\}frac\left\{T\right\}\left\{2$

:$frac\left\{1\right\}\left\{2\right\}=e^\left\{frac\left\{-1\right\}\left\{RC\right\}frac\left\{T\right\}\left\{2$

:$lnleft\left(frac\left\{1\right\}\left\{2\right\} ight\right)=frac\left\{-1\right\}\left\{RC\right\}frac\left\{T\right\}\left\{2\right\}$

:$, ! T=2ln\left(2\right)RC$

:$f=frac\left\{1\right\}\left\{2ln\left(2\right)RC\right\}$

This analysis is only partially correct. The derived frequency of oscillation is based on the time period of the initial charge, which is not the same as subsequent charges/discharges and therefore incorrect.

This can be visually verified by examining the figure of simulation results that accompanies this article. Notice the initial charge time is shorter than all subsequent periods, implying that the derived frequency (based on this initial time period) would be higher than its true frequency of operation.

The initial conditions were that V0 = 0. This is true when the system is at rest and sufficient for the initial charge. But after the first and every subsequent transition (when Vt = Vdd/2 or when Vt = Vss/2) this initial condition is no longer true.

For subsequent transitions in which the capacitor is charging the initial condition would be that V0 = Vss/2 and not V0 = 0. This effectively changes the value of B in the derivation above:

At time 0, Vout = Vdd and V_ = Vss/2.
Vss/2 = Vdd + B.
B = Vss/2 - Vdd. (B does not equal -Vdd).

So, the charging equation becomes (for all charging periods except the initial charge):

$V_\left\{-\right\} = A + Be^\left\{-frac\left\{t\right\}\left\{RC$

$frac\left\{1\right\}\left\{2\right\}V_\left\{dd\right\} = V_\left\{dd\right\} - \left(frac\left\{1\right\}\left\{2\right\}V_\left\{ss\right\} - V_\left\{dd\right\}\right)e^\left\{-frac\left\{T\right\}\left\{2RC$

$\left(frac\left\{1\right\}\left\{2\right\}V_\left\{ss\right\} - V_\left\{dd\right\}\right)e^\left\{-frac\left\{T\right\}\left\{2RC = frac\left\{1\right\}\left\{2\right\}V_\left\{dd\right\}$

$e^\left\{-frac\left\{T\right\}\left\{2RC = frac\left\{V_\left\{dd\left\{V_\left\{ss\right\} - 2V_\left\{dd$

$-frac\left\{T\right\}\left\{2RC\right\} = ln\left(V_\left\{dd\right\}\right) - ln\left(V_\left\{ss\right\} - 2V_\left\{dd\right\}\right)$

$T = \left(2RC\right) \left[ln\left(V_\left\{ss\right\} - 2V_\left\{dd\right\}\right) - ln\left(V_\left\{dd\right\}\right)\right]$

$T = \left(2RC\right) \left[frac\left\{ ln\left(V_\left\{ss\right\}\right) \right\}\left\{ ln\left(V_\left\{dd\right\}\right)+ln\left(2\right) \right\} - ln\left(V_\left\{dd\right\}\right)\right]$

* Multivibrator
* RC oscillator
* Solar Engine
* FitzHugh-Nagumo model

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