The circuit shown represents

SOLUTION

Analysis:

Given: 

V⁺ = V_A and V^- = V_Z

When V_i acts as a trigger to change the output state, the capacitor C₁ changes until the output state returns to the original state.

Initially, the capacitor is uncharged.

Let V₀ = + V_sat

Now, the voltage across capacitor is given by:

V_l(t) = V_f + (V_i - V_f) e^{-t / τ}

With V_i = 0 V and V_f = +V_sat, we can write:

\({V_c}\left( t \right) = {V_{sat}} + \left( {0 - {V_{sat}}} \right){e^{ - t/{R_1}{C_1}}}\) 

\({V_c}\left( t \right) = {V_{sat}}\left( {1 - {e^{ - \frac{t}{{{R_1}{C_1}}}}}} \right)\) 

\({V_c}\left( t \right) = {V_{sat}} - {V_{sat}}{e^{ - t/{R_1}{C_1}}}\) 

\({V_{sat}} - {V_c}\left( t \right) = {V_{sat}}{e^{ - t/{R_1}{C_1}}}\)    ---(1)

From the given figure:

V₀ – V_c(t) – V_A = 0 and V₀ = V_sat

V_sat – V_c(t) – V_A = 0

V_sat – V_c(t) = V_A  ---(2)

From (1) & (2), we get:

\({V_A} = {V_{sat}}\;{e^{ - t/{R_1}{C_1}}}\) 

Let the trigger input applied, be in the triangular, i.e.

With \(\;{V_Z} = \frac{{d{V_i}}}{{dt}}:\)

Now, 

Hence a pulse of short duration is generated with the help of a trigger input.

∴  It is a monostable multivibrator.