The Lower Threshold Point (LTP) voltage of the Schmidt trigger with an ideal op.amp shown in Figure is _______.

Question

Question

This question was previously asked in

ISRO Scientist Electronics 2024 Official Paper

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Answer 1

Explanation:

Schmitt trigger circuit:

F2 U.B Deepak 23.02.2020 D 5

V0 = + Vsat (or) -Vsat

\({V_{ref}} = \frac{{{R_2}}}{{{R_1} + {R_2}}}{V_0}\)

\({R_{01}} = {R_1}||{R_2} = \frac{{{R_1}{R_2}}}{{{R_1} + {R_2}}}\)

VUT is the upper threshold point if the output is +ve reference

VLT = Lower threshold point if the output voltage is –ve reference

\({V_{UT}} = \frac{{{R_2}}}{{{R_1} + {R_2}}}\left( { + {V_{sat}}} \right)\)

\({V_{LT}} = \frac{{{R_2}}}{{{R_1} + {R_2}}}\left( { - {V_{sat}}} \right)\)

F2 U.B Deepak 23.02.2020 D 6

Calculation:

On comparing this circuit with the question provided, we get:

In the Schmitt trigger

Vi = Vp-p, R1 = 10 kΩ, R2 = 5 kΩ , Vref = 0 V

Sinusoidal saturation voltage = ± 5 V (Vsat)

\({V_{LT}} = \frac{{{R_2}}}{{{R_1} + {R_2}}}\left( { - {V_{sat}}} \right)\) = 5 x (-5)/15 = -1.667 V