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Chapter 36

Q. 36.5

If V_{sat} =13.5 V, what are the trip points and hysteresis in Fig. 36-16?

Step-by-Step

Verified Solution

With Formula (36-4),B=\frac{R_1}{R_1+R_2 }

the feedback fraction is

B=\frac{1k\Omega }{48 k\Omega }=0.0208

With 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.