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
Best Answer
What have we got?
Let's start with condition that output is low (-15 V) and Uin 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 Uref.
I haven't double-checked this or finished the calculations. Let me know if this works.