Question 15.7: Objective: Design a Schmitt trigger circuit for the photodet...

The Schmitt trigger circuit is the inverting type, for which the voltage transfer characteristics are shown in Figure 15.30(b). From Equations  (15.77(a)) and (15.77(b)), the hysteresis width is

V_{T H} = V_{S} + \left(\frac{R_{1}}{R_{1}  +  R_{2}} \right) V_{H}             (15.77(a))
V_{T L} = V_{S} + \left(\frac{R_{1}}{R_{1}  +  R_{2}} \right) V_{L}            (15.77(b))

so
0.060 = \left(\frac{R_{1}}{R_{1}  +  R_{2}} \right) [5  −  (−5)] = 10 \left(\frac{R_{1}}{R_{1}  +  R_{2}} \right)

which yields R_{2}/R_{1} = 165.7. We can find the reference voltage from Equation (15.76), which can be rewritten to obtain
V_{REF} = \left(1 + \frac{R_{1}}{R_{2}} \right) V_{S} = \left(1 + \frac{1}{165.7} \right) (2) = 2.012  V
Resistor values of R_{1} = 100  \Omega and R_{2} = 16.57  k\Omega will satisfy the requirements. The crossover voltages are thus V_{T H} = 2.03  V and V_{T L} = 1.97  V.

Trade-offs: If we use standard-valued resistors R_{1} = 120  \Omega and R_{2} = 20  k\Omega, the hysteresis width is
V_{T H}  −  V_{T L} = \left(\frac{R_{1}}{R_{1}  +  R_{2}} \right) (V_{H}  −  V_{L} )

= \left(\frac{0.12}{0.12  +  20} \right)  [5  −  (−5)] → 59.6  mV
If we are able to use a reference voltage of 2.012 V, then the switching voltage is
V_{S} = \left(\frac{R_{2}}{R_{1}  +  R_{2}} \right) V_{REF} = \left( \frac{20}{0.12  +  20} \right) (2.012) = 2.0  V
Resistor tolerances will also affect the results, but will not be considered here.
Comment: In this case, the output chatter effect is eliminated for noise signals with amplitudes lower than 30 mV. The hysteresis width can be adjusted up or down to fit specific application requirements in which the noise signal is larger or smaller than that given in this example