Determining U_ref on Schmitt Trigger

Question

If I have only starting values a U_high=3V and U_low=2V (pulse signal on V3 on schematics), can I determine a U reference, that's given us on the inverting input of the op. amp.?

My calculations were: Uref = (U_high - U_low)/2 = 2.5V. Is that correct?

I had difficulties from here on, because I got different values for R2 (which aswell I have to calculate) when I calculated R2 based on Uhigh and different when based on Ulow. How is that even possible?

This is supposed to be an ideal operational amplifier.

This is not a homework, but preparations for an exam.

^{simulate this circuit – Schematic created using CircuitLab}

Answer 1

What have we got?

• Op-amp is ideal. It's a Schmitt trigger so output will be +15 V or -15 V.

Let's start with condition that output is low (-15 V) and U_in is on the point of switching the output high.

^{simulate this circuit – Schematic created using CircuitLab}

Figure 1. Output is about to switch high.

$$ V_+ = U_{out} + \frac {(U_{in} - U_{out}) R_2}{R_1+R_2} = -15 + \frac {(3 - (-15)) R_2}{1k+R_2} = -15 + \frac {18\cdot R_2}{1k+R_2} \tag 1$$

^{simulate this circuit}

Figure 2. Output is high but input is at the lower threshold.

$$ V_+ = U_{out} + \frac {(U_{in} - U_{out}) R_2}{R_1+R_2} = 15 + \frac {(2 - 15) R_2}{1k+R_2} = 15 - \frac {13 \cdot R_2}{1k+R2} \tag 2$$

Solving simultaneously:

$$-15 + \frac {18\cdot R_2}{1k+R_2} = 15 - \frac {13 \cdot R2}{1k+R2} \tag 3$$

$$ \frac {31 \cdot R_2}{1k+R_2} = 30 \tag 4 $$

$$ \frac {R_2}{1k+R_2} = \frac {30}{31} \tag 5 $$

$$ {R_2} = \frac {30}{31}k + \frac {30}{31}R_2 \tag 6 $$ $$ \frac {1}{31}R_2 = \frac {30}{31}k \tag 7 $$

You should be able to solve for R2 and thensolve for U_ref.

I haven't double-checked this or finished the calculations. Let me know if this works.