In the following amplifier circuit, the op-amp is ideal. The volt

SOLUTION

Concept:

The voltage gain for inverting configuration of an Op-Amp is given as:

\(\frac{{{V_0}}}{{{V_i}}} = \frac{{ - {R_2}}}{{{R_1}}}\)

Also, the saturation voltage for an Op-Amp is the maximum voltage that the output can reach.

Application:

The output voltage for the given inverting configuration will be:

\({V_0} = - \left( {\frac{{{R_2}}}{{{R_1}}}} \right) \times {V_i}\)

\({V_0} = \left( { - 3} \right)\left( {\frac{{ - 6.2}}{1}} \right) = 18.6\;V\)

Since V₀ = 18.6 is greater than the saturation voltage of the Op-Amp (+10 V), the output voltage will saturate at + 10 V.

∴ With V₀ = +10 V and applying KCL at the inverting terminal, we get:

\(\frac{{{V^ - } - \left( { - 3} \right)}}{{1k}} + \frac{{{V^ - } - \left( {10} \right)}}{{6.2\;k}} = 0\)

(V^- + 3) (6.2 k) + (V^- - 10) 1k = 0

6.2 V^- + 18.6 + V^- - 10 = 0

7.2 V^- = - 8.6

\({V^ - } = \frac{{ - 8.6}}{{7.2}}\)

V^- = -1.2 V

Alternate Method:

We can also do KCL analysis to evaluate the output voltage i.e. Applying KCL at the inverting terminal, we get:

\(\frac{{0 - \left( { - 3} \right)}}{{1K}} + \frac{{0 - {V_0}}}{{6.2\;K}} = 0\)

\({V_0} = \left( { - 3} \right)\left( {\frac{{ - 6.2K}}{{1K}}} \right)\)

V₀ = 18.6 V

Since V₀ > 10 V, V₀ will saturate at V₀ = + 10V.

Now, Applying nodal analysis at the inverting terminal, we get:

\(\frac{{{V^ - } - \left( { - 3} \right)}}{{1K}} + \frac{{{V^ - } - 10}}{{6.2\;K}} = 0\)

V^- = -1.2 V