Applications of Operational Amplifier: Relaxation Oscillator

Relaxation Oscillator:

With positive feedback it is also possible to build relaxation oscillator which produces rectangular wave. The circuit is shown in fig. 2.

Fig. 2

In this circuit a fraction R₂/ (R₁ +R₂) = b of the output is feedback to the non-inverting input terminal. The operation of the circuit can be explained as follows:

Assume that the output voltage is +V_sat. The capacitor will charge exponentially toward +V_sat. The feedback voltage is +bV_sat. When capacitor voltage exceeds +bV_sat the output switches from +V_sat to -V_sat. The feedback voltage becomes -V_sat and the output will remain –V_sat. Now the capacitor charges in the reverse direction. When capacitor voltage decreases below –bV_sat (more negative than –bV_sat ) the output again switches to +V_sat.This process continues and it produces a square wave. Under steady state conditions, the output voltage and capacitor voltage are shown in fig. 2. The frequency of the output can be obtained as follows:

The capacitor charges from -β V_sat to +β V_sat during time period T/2. The capacitor charging voltage expression is given by

This square wave generator is useful in the frequency range of 10Hz to 10KHz. At higher frequencies, the slew rate of the OPAMP limits the slope of the output square wave.

The duty cycle of the output wave can be changed replacing the resistance R by another circuit consisting of variable resistance and two diodes D₁ and D₂ as shown in fig. 3.

Fig. 3	Fig. 4

When the output is positive then D₂ conducts and D₁ is OFF. The total feedback resistance becomes R_b+R and charging voltage is reduced by V_D. During the interval when the output is negative then D₁ conducts and D₂ is OFF. The charging resistance becomes R+R_a. The total charging and reverse charging period is decided by total resistance (2 R + R_a + R_b) = constant. Therefore frequency will remain constant but duty cycle changes. The capacitor voltage and output voltage of the oscillator are shown in fig. 4.

The duty cycle is given by 

By varying R_a and R_b the duty cycle can be charged keeping frequency constant.

Example - 2

Characterize the astable mulitvibrator shown in fig. 5. Establish the frequency range.

Fig. 5

Solution:

By observing the way the output section is connected, we conclude that the output voltage oscillates between V_L = - 5V and V_H = + 5V.

To calculate the oscillation frequency range, two extreme values for R₁ and R₂ are used. Thus, when the wiper of the potentiometer is at its leftmost position, R₁ = 120 kΩ and R₂ = 10 kΩ. At the other extreme, R₁ = 20 kΩ and R₂ =110 kΩ.

Substituting the values of R₁ and R₂ in the expression of T, we obtain the two extreme values of T, i.e., T_min and T_max

.

Therefore,

f_min = 285 Hz and f_max = 4.6 kHz