Consider an op-amp circuit and input to the circuit as shown in f

SOLUTION

Concept:

V^- = V⁺ = 0 V (Virtual Ground concept)

\(i = \frac{{{V_{in}}}}{R}\)   ---(1)

Since no current flows inside the Op-Am, we have:

i = i_c

The equation of current across the capacitor is given as:

\({i_c} = C\frac{{d{V_c}}}{{dt}}\)

\(\frac{{d{V_c}}}{{dt}} = \frac{{{i_c}}}{C}\)

Using equation (1), we can write:

\(\frac{{d{V_c}}}{{dt}} = \frac{{{V_{in}}}}{{RC}}\)

\({V_c}\left( t \right) = \frac{1}{{RC}}\mathop \smallint \limits_0^t {V_{in}}dt + {V_c}\left( 0 \right)\)      ---(2)

Where V_c(0) is the initial voltage across the capacitor.

Applying KVL from output to inverting terminal, we can write:

V₀ + V_c (t) = 0

V₀ = -V_c(t)

Using equation (2), we get:

\({V_0} = - \frac{1}{{RC}}\mathop \smallint \limits_0^t {V_{in}}\left( t \right)dt + {V_c}\left( 0 \right)\)

Application (Conceptual Method):

Since the given circuit is an integrator, the circuit will integrate the input waveform and produce the output.

Also, since the input is applied at the inverting terminal, the output will be invertible.

∴ For the given input waveform, the integrating waveform will be:

Now, Due to the inverting terminal, 180° phase shift is added, and the actual output waveform will be:

Conventional Method:

For 0 ≤ t ≤ 1 ms, the output voltage will be:

\({V_0} = 0 - \frac{1}{{RC}}\mathop \smallint \limits_0^t 1.dt\)

With R = 10 k and C = 0.1 μF, the above equation becomes:

\({V_0} = - \frac{1}{{10 \times {{10}^3} \times 0.1 \times {{10}^{ - 6}}}}\mathop \smallint \limits_0^t dt\)

\(= \frac{{ - 1}}{{{{10}^{ - 3}}}}t\)

At t = 1 ms, the output voltage will be:

\({V_0} = \frac{{ - 1}}{{{{10}^{ - 3}}}} \times {10^{ - 3}} = \; - 1\;V\)

Similarly,

For 1 ≤ t ≤ 2 ms, the output voltage will be:

\({V_0} = - 1 + \frac{1}{{RC}}\left( {t - 1} \right)\)

Where -1 V is the initial voltage at t = 1 msec.

\({V_0} = - 1 + \frac{1}{{{{10}^{ - 3}}}}\left( {t - 1} \right)\)

At t = 2 m sec, we get:

V0 = -1 + (2 - 1) = 0 V

∴ The graph represented in Option (1) is correct.