The Schmitt trigger circuit shown in the figure below uses Zener
The Schmitt trigger circuit shown in the figure below uses Zener diode with Vd = 0.7 V. If the threshold voltage V1 is zero and hysteresis voltage VH = 0.2 V, then what is \(\frac{{{R_1}}}{{{R_2}}}\) and VR?
A. R<sub>1</sub>/R<sub>2</sub> = 6 and V<sub>R</sub> = 0.15 V
B. R<sub>1</sub>/R<sub>2</sub> = 67 and V<sub>R</sub> = -0.15 V
C. R<sub>1</sub>/R<sub>2</sub> = 66 and V<sub>R</sub> = -0.10 V
D. R<sub>1</sub>/R<sub>2</sub> = 66 and V<sub>R</sub> = -0.15 V
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Right Answer is: C
SOLUTION
Concept:
The input voltage Vi changes the state of output therefore it exceeds its voltage level above a certain value called upper and lower threshold voltages.
The upper threshold point:
\({V_{UTP}} = {V_R} \times \frac{{{R_1}}}{{{R_1} + {R_2}}} + {V_0}\frac{{{R_2}}}{{{R_1} + {R_2}}}\)
Lower threshold point:
\({V_{LTP}} = {V_R}\frac{{{R_1}}}{{{R_1} + {R_2}}} - {V_0}\;\frac{{{R_2}}}{{{R_1} + {R_2}}}\)
Now, VH = VUTP - VLTP (hysteresis voltage)
Calculation:
In Schmitt trigger, Zener diode voltage vd = 0.7 V
Threshold voltage V1 = 0
Hysteresis voltage VH = 0.2 V
\(\Rightarrow {V_H} = {V_R}\frac{{{R_1}}}{{{R_1} + {R_2}}} + {V_0}\frac{{{R_2}}}{{{R_1} + {R_2}}} - \left( {{V_R}\frac{{{R_1}}}{{{R_1} + {R_2}}} - {V_0}\frac{{{R_2}}}{{{R_1} + {R_2}}}} \right)\)
\(\Rightarrow {V_H} = 2\;{V_0}\frac{{{R_2}}}{{{R_1} + {R_2}}}\)
Now, V0 = VZ + VD = 0.7 + 6 = 6.7 V
\( \Rightarrow 0.2 = 2 \times 6.7 \times \frac{1}{{1 + \frac{{{R_1}}}{{{R_2}}}}}\)
\( \Rightarrow 1 + \frac{{{R_1}}}{{{R_2}}} = 67\)
\(\Rightarrow \frac{{{R_1}}}{{{R_2}}} = 66\)
Now, (threshold) \({V_1} = {V_R} \times \frac{{{R_1}}}{{{R_1} + {R_2}}} + {V_0} \times \frac{{{R_2}}}{{{R_1} + {R_2}}}\)
\(\Rightarrow 0 = {V_R}\frac{1}{{1 + \frac{{{R_2}}}{{{R_1}}}}} + {V_0}\frac{1}{{1 + \frac{{{R_1}}}{{{R_2}}}}}\)
\( \Rightarrow {V_2} \times \frac{1}{{1 + \frac{1}{{66}}}} = - 6.7 \times \frac{1}{{1 + 66}}\)
\(\Rightarrow {V_R} = \; - 6.7 \times \frac{1}{6} = - 0.10\;V\)