Example-1:

1. Consider the pulse generator shown in fig. 3. In the quiescent state (before a trigger pulse is applied), find V₂, V_O and V₁.
2. At t = 0, a narrow, positive triggering pulse v whose magnitude exceeds V_R is applied. At t = 0+, find V_O and V₁.
3. Verify that the pulse width T = RC ln (2 V_O) / V_R.

Fig. 3

Solution:

(a). Before a trigger pulse is applied, the circuit is in stable stage with the output at v_O = +V_O ( ≈ V_Z + 0.7). The capacitor C is charged with the polarity shown in fig. 3.

Thus, v₁ ≈ 0.7V

and v₂ = -V_R

(b). At t = 0, a narrow positive triggering pulse of higher magnitude is applied. The capacitor C voltage can not charge instantaneously. Therefore, v₂ becomes positive and greater than v₁ (≈ 0.7 V). The comparator output changes.

Thus, v_O= - (V_Z + 0.7V) = -V_O

Since capacitor C voltage can not change instantaneously, therefore,

v₁ = 2 V_O

(c). The input trigger pulse is of very short duration therefore, after the short duration pulse the voltage v₂ returns to (-V_R). But the output remains –V_O because v₁ is at – 2V_O.

The capacitor now starts charging exponentially with a time constant t = RC through R towards –V_O, because diode is reverse biased.

V_C = (-V_O – V_O) ( 1 – e^{-t / RC} ) - V_O

The voltage at point v₁ is, thus, given by

V₁ = - V_O – v_c

= - V_O + 2 V_O (1 – e^{-t / RC} ) - V_O

= - 2 V_O e^{-t / RC} – t / RC

When v1 voltage becomes more than V_R, the comparator output switches back to +V_O. Let at t = T, the voltage v₁ becomes – V_R

The capacitor now starts charging towards +V_O through R until v_c reaches +V_O and v₁ becomes 0.7 V. The waveforms at different points are shown in fig. 4.

Fig. 4

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