An ideal op-amp circuit and its input waveform are shown in the f

SOLUTION

Since we have a feedback from the output to the non-inverting terminal, we have a positive feedback circuit.

The given circuit is a Schmitt trigger circuit and the output can be any of the two steady-state saturation values.

Case I:

For higher saturation output V_OH = 6V, the given circuit is redrawn as:

Applying KCL the non-inverting terminal, we get

i₁ + i₂ = 0

\(\frac{{{V_{TH}} - 6}}{{2k}} + \frac{{{V_{TH}} - 0}}{{1k}} = 0\)

V_TH – 6 + 2V_TH = 0

V_TH = 2 V

(V_TH can also be obtained by using voltage division)

∴ The upper threshold voltage for the op-amp will be 2 V, i.e when V_in exceeds 2V, the voltage at the inverting terminal will exceed the voltage at the non-inverting terminal and the output will saturate at – 3V.

From the input waveform, this transition will occur at t = t₂

Case 2:

Similarly, if the output is at the lower saturation level of V_OL = -3 V, the equivalent circuit is drawn as:

Applying KCL at the non-inverting terminal, we get:

i₁ + i₂ = 0

\(\frac{{{V_{TL}} - \left( { - 3} \right)}}{2} + \frac{{{V_{TL}}}}{1} = 0\)

V_TL + 3 + 2 V_TL = 0

V_TL = -1V

(Here also, we can use voltage division to evaluate V_TL)

∴ The lower threshold voltage will be -1 V, i.e. when the input voltage (V_i) is less than V_TL, the voltage at the non-inverting terminal will exceed the voltage at the inverting terminal and the output will saturate at + 6V again.

From the input waveform, this transition will occur at t = t₄

∴ For various inputs, we have the following outputs:

V_in < 2, V₀ = + 6 V

V_in > 2, V₀ = 3 V

V_in < -1V, V₀ = +6 V

V_in > -1V, V₀ = -3 V

∴ The output waveform will be: