GradFinalLab RS
Construct a Schmitt Trigger using the 741 Op Amp
Draw the Schmitt Trigger circuit you constructed. Identify the values of all components
To construct the circuit above I am going to use the following components and voltages:
- [math]\displaystyle{ R_1 = (1.01\pm0.01)\ k\Omega }[/math]
- [math]\displaystyle{ R_2 = (1.01\pm0.01)\ k\Omega }[/math]
- [math]\displaystyle{ R_3 = (5.10\pm0.01)\ k\Omega }[/math]
- [math]\displaystyle{ \mbox{OP}\ \mbox{AMP}\ 741 }[/math]
- [math]\displaystyle{ V_{ref} = (+11.90\pm0.01)\ V }[/math]
- [math]\displaystyle{ V_{cc} = (+11.90\pm0.01)\ V }[/math]
- [math]\displaystyle{ V_{ee} = (-12.11\pm0.01)\ V }[/math]
Graph [math]\displaystyle{ V_{out} }[/math] as a function of [math]\displaystyle{ V_{in} }[/math]. Is there a hysteresis loop?
Identify the input voltage threshold levels at which a [math]\displaystyle{ V_{in} }[/math] will produce [math]\displaystyle{ V_{out} \approx V_{cc} }[/math]
The theory does say (ch. 10.19 The Schmitt Trigger R. Simpson "Introductory electronics for scientists and engineers"):
[math]\displaystyle{ V_2 = \frac{R_{123}}{R_2}V_{ref} + \frac{R_{123}}{R_3}V_{out} }[/math]
1) if the output is high:
[math]\displaystyle{ V_2 = \frac{R_{123}}{R_2}V_{ref} + \frac{R_{123}}{R_3}V_{out_1} }[/math]
2) if the output is low:
[math]\displaystyle{ V_2^' = \frac{R_{123}}{R_2}V_{ref} - \frac{R_{123}}{R_3}V_{out_1} }[/math]
where
[math]\displaystyle{ R_{123} = (R_1 || R_2 || R_3) = \frac{R_1 R_2 R_3}{R_1+R_2+R_3} }[/math]
The actual measured values of high and low output voltages are (they do not really equal to [math]\displaystyle{ \pm V_{cc} }[/math]):
- [math]\displaystyle{ V_{out_1} = (+11.06\pm0.01)\ V }[/math]
- [math]\displaystyle{ V_{out_2} = (-10.54\pm0.01)\ V }[/math]
Substituting all values in the formulas above:
[math]\displaystyle{ R_{123} = (1.01\pm 0.01)\ k\Omega || (1.01\pm 0.01)\ k\Omega || (5.10\pm 0.05)\ k\Omega = (0.459\pm 0.003)\ k\Omega }[/math]
[math]\displaystyle{ V_2 = \frac{(0.459\pm 0.003)\ k\Omega}{(1.01\pm 0.01)\ k\Omega}(11.90\pm 0.01)\ V + \frac{(0.459\pm 0.003)\ k\Omega}{(5.10\pm 0.05)\ k\Omega}(11.06\ pm 0.01)\ V }[/math]
[math]\displaystyle{ V_2^' = \frac{R_{123}}{R_2}V_{ref} - \frac{R_{123}}{R_3}V_{cc} }[/math]
and doing math and handling the error propagation we end up with the following threshold voltages:
[math]\displaystyle{ V_2 = }[/math]
[math]\displaystyle{ V_2^' = }[/math]
Compare the threshold values to what is expected.
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