Question
Design and draw the circuit of an inverting Schmitt trigger using op-amp with hysteresis of 40% of ±VSAT-
Answer :
Word Count : 660
We want an inverting Schmitt trigger whose switching thresholds are symmetric at ±V\_TH with V\_TH equal to 0.40·V\_SAT (i.e. the non-inverting threshold node must be ±0.4·V\_SAT). I will assume the op-amp saturates to ±V\_SAT and use that symbolically, then pick convenient resistor values that realize the required divider ratio. For the standard inverting Schmitt configuration (Vin → R\_IN → (–) input, (+) input fed by a resistor divider between V\_OUT and ground), the noninverting voltage V+ equals the fraction β of the output: $$ V_+ = \beta\;V_{out},\qquad \beta=\frac{R_2}{R_1+R_2} $$ with R1 between Vout and V+ and R2 between V+ and ground. When Vout = +V\_{SAT} the threshold seen by the (−) input is V\_{TH+} = β·(+V\_{SAT}). When Vout = −V\_{SAT} the threshold is V\_{TH-} = β·(−V\_{SAT}). Thus the thresholds are symmetric and ________ _____ __________ _________ _______ ________ ____ _________.
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We want an inverting Schmitt trigger whose switching thresholds are symmetric at ±V\_TH with V\_TH equal to 0.40·V\_SAT (i.e. the non-inverting threshold node must be ±0.4·V\_SAT). I will assume the op-amp saturates to ±V\_SAT and use that symbolically, then pick convenient resistor values that realize the required divider ratio. For the standard inverting Schmitt configuration (Vin → R\_IN → (–) input, (+) input fed by a resistor divider between V\_OUT and ground), the noninverting voltage V+ equals the fraction β of the output: $$ V_+ = \beta\;V_{out},\qquad \beta=\frac{R_2}{R_1+R_2} $$ with R1 between Vout and V+ and R2 between V+ and ground. When Vout = +V\_{SAT} the threshold seen by the (−) input is V\_{TH+} = β·(+V\_{SAT}). When Vout = −V\_{SAT} the threshold is V\_{TH-} = β·(−V\_{SAT}). Thus the thresholds are symmetric and ________ _____ __________ _________ _______ ________ ____ _________.
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