If V_{sat} =13.5 V, what are the trip points and hysteresis in Fig. 36-16?
If V_{sat} =13.5 V, what are the trip points and hysteresis in Fig. 36-16?
With Formula (36-4),B=\frac{R_1}{R_1+R_2 }
the feedback fraction is
B=\frac{1k\Omega }{48 k\Omega }=0.0208With Formulas (36-6) UTP = BV_{sat} and (36-7), LTP = -BV_{sat} the trip points are
UTP = 0.0208(13.5 V) = 0.281 V
LTP = -0.0208(13.5 V) = -0.281 V
With Formula (36-9),H = 2BV_{sat} the hysteresis is
H = 2(0.0208 V)(13.5 V) = 0.562 V
This means that the Schmitt trigger of Fig. 36-16 can withstand a peak-to-peak noise voltage up to 0.562 V without false triggering.