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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:

1. [math]R_1 = (1.01\pm0.01)\ k\Omega[/math]
2. [math]R_2 = (1.01\pm0.01)\ k\Omega[/math]
3. [math]R_3 = (5.10\pm0.01)\ k\Omega[/math]
4. [math]\mbox{OP}\ \mbox{AMP}\ 741[/math]
5. [math]V_{ref} = (+11.90\pm0.01)\ V[/math]
6. [math]V_{cc} = (+11.90\pm0.01)\ V[/math]
7. [math]V_{ee} = (-12.11\pm0.01)\ V[/math]

Graph [math]V_{out}[/math] as a function of [math]V_{in}[/math]. Is there a hysteresis loop?

If the input voltage goes up my measured output voltages are:

If the input voltage goes down my measured output voltages are:

In the plot below I overplay the output voltages vs. input voltages for both cases as input voltage goes up (black line) and down (red line):

As we can clearly see there is the hysteresis loop. The reason for this loop is because the threshold voltages depends from output voltage. If the output voltage is high the threshold voltage is higher than for the case if output voltage is low.

Compare the threshold values to what is expected.

The theory does say (ch. 10.19 The Schmitt Trigger R. Simpson "Introductory electronics for scientists and engineers"):

[math]V_2 = \frac{R_{123}}{R_2}V_{ref} + \frac{R_{123}}{R_3}V_{out}[/math]

1) if the output is high:

[math]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]V_2^' = \frac{R_{123}}{R_2}V_{ref} - \frac{R_{123}}{R_3}V_{out_1}[/math]

where

[math]R_{123} = (R_1 || R_2 || R_3) = \frac{R_1 R_2 R_3}{R_1+R_2+R_3}[/math]

and [math]V_{out_1}[/math] and [math]V_{out_2}[/math] approximately equal to supply voltage [math]V_{cc}[/math]

The actual measured values of high and low output voltages (they do not really equal to [math]\pm V_{cc}[/math]) are:

1. [math]V_{out_1} = (+11.06\pm0.01)\ V[/math]
2. [math]V_{out_2} = (-10.54\pm0.01)\ V[/math]

Substituting all quantities in the formulas above:

[math]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]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]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}(10.54\pm0.01)\ V[/math]

and doing math and propagating errors we end up with the following predicted threshold voltages:

1) If the output voltage is high the threshold voltage are:

[math]V_2 = (6.41\pm 0.07)\ V [/math]

1) If the output voltage is low the threshold voltage are:

[math]V_2 = (4.46\pm 0.06)\ V [/math]

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