If Vsat =13.5 V, what are the trip points and hysteresis in Fig. 36-16?

Question

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

Accepted Answer

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

the feedback fraction is

[latex]B=\frac{1k\Omega }{48 k\Omega }=0.0208[/latex]

With Formulas (36-6) [latex]UTP = BV_{sat}[/latex] and (36-7), [latex]LTP = -BV_{sat} [/latex] 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),[latex]H = 2BV_{sat}[/latex] 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.

Q. 36.5