The input to the circuit shown below is v i = 2 sin ωt mV. The cu

SOLUTION

Concept:

An ideal Op-Amp with negative feedback follows virtual short-circuit, i.e.

V⁺ = V^-

Also, an ideal op-amp has an infinite input impedance, i.e. no current flows inside of the Op-Amp.

Application:

The given circuit is redrawn as:

Applying KCL at node A, we get:

i₁ + i₂ = 0

\(\frac{{{V^ - } - {V_i}}}{{1\;k}} + \frac{{{V^ - } - {V_0}}}{{10\;k}} = 0\)

Since, V⁺ = V^- = 0 V, the above equation becomes:

\(\frac{{0 - {V_i}}}{{1k}} + \frac{{0 - {V_0}}}{{10k}} = 0\)

-10 V_i – V₀ = 0

V₀ = - 10 V_i

Given V_i = 2 sin ω t mV

∴ V₀ = -20 sin ωt mV

Similarly, Applying KCL at the output node, we get:

i₀ + i₂ = i_L

i₀ = i_L – i₂

\({i_0} = \frac{{{V_0} - 0}}{{4k}} - \frac{{\left( {0 - {V_0}} \right)}}{{10\;k}}\)

\({i_0} = \frac{{ - 20\sin \omega t}}{{4k}} - \frac{{20\sin \omega t\;}}{{10k}}\)

i₀ = -5 sin ωt μA – 2 sin ωt μA

i₀ = -7 sin ωt μA