I've got a few questions about a task from my exam preparation paper. The sketch below shows the circuit in question. It is given that U_E varies between 0V...3V. The operating voltage U_B is 10V. The op amp and diode are ideal at the beginning (it does not say at which point in time they must be not seen as ideal anymore). The four tasks are the following:

1. Give an algebraic expression for the differential amplification \$v_1=\frac{dU_A}{dU_E}\$

My solution: \$v_1=1+\frac{R_2}{R_1}\$ (non inverting op amp)

2. What is U_E's value such that D1 is conducting? What is U_A then? Give algebraic expressions for U_E and U_A.

My solution: \$U_E>U_B*\frac{R_3}{R_3+R_4}\$ thus \$U_A>U_B*\frac{R_3}{R_3+R_4}*(1+\frac{R2}{R1})\$

3. What is the differential amplification v_2=dU_A/dU_E if D1 is conducting? Give an algebraic expression.

My solution: \$\frac{dU_A}{dU_E}=\frac{R2}{R4}+\frac{R2}{R3}+1+\frac{R2}{R1}\$

4. U_B is 10V. At which value U_E does U_A saturate? Give an algebraic expression for U_E. What is the differential voltage u_d as a function of U_E if saturation occurs?

I don't really know how to solve this task so I'd be glad if anyone could help me out.

Moreover I am not quite sure about my solution to question 2 and 3.


simulate this circuit – Schematic created using CircuitLab

  • \$\begingroup\$ Please consider making your post more clear by writing the equations in MathJax. Note EE uses \$ to start and end inline Mathjax to avoid confusion when talking about dollar amounts. (We still use $$ for display equations) \$\endgroup\$ – The Photon Jul 25 '17 at 16:41
  • \$\begingroup\$ #2 is correct if your definition of an ideal diode is 0V forward voltage. You may want to clarify the forward voltage of an ideal diode with the teacher. For #3, if D1 is an ideal diode, just replace it with a short circuit. For #4 just set up an equation for U_A < 10V. Once the op amp saturates, it can no longer change the inverting input to match the non-inverting input. \$\endgroup\$ – DavidG25 Jun 19 '18 at 20:09

The Diode's Anode is the same node as the Inverting Input. The Voltage Divider at it's Cathode is 10*200/(200+800)=2-Volts. So the Maximum Inverting Input Voltage is 2.7V. Since the Feedback voltage divider provides 1/3 of the output to the Inverting input the Max Vout will be 2.7*3=8.1 as any Vout exceeding that will forward bias the Diode. So Your Non-Inverting input is actually limited to a 2.7-Volt Input as well (which will output Vout = 8.1-Volts. I don't know how you fit that in an algebra equation. But uhhh that should help, It answers #2.

edit: your title says "Non-Linear", so their is likely some "non-linearities" in there. Because of the Diode

Good Luck!

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