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## Q. 36.5

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

## 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.