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

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