# Determine the operating point of the circuit in the figure. Voltage divider, op amp, diode

simulate this circuit – Schematic created using CircuitLab

Since the operating point of the circuit is determined by analysing it in DC, the capacitors behave as open circuits.
At this point, as far as I understand it, the voltage at node V+ is determined by the voltage divider consisting of resistors R2 and R1. $$V_+ = E \frac{R_1}{R_1 + R_2} = 30\text{ mV}$$ Since the operational amplifier is ideal, due to the virtual short-circuit principle, node V- will have the same voltage. The voltage drop at the ends of resistor R4 causes a current to flow towards ground. The same current will flow through R5, since it cannot pass through the input of the op amp. This way I can determine $$V_x = R_5\frac{V_-}{R_4} + V_-=0,3V$$ But at this point either I find a negative Vout or the diode remains off and Vout = 0.
What am I doing wrong?

• Where did this circuit come from? Why don't you just run the simulation if you want to know the output voltage? Jan 19, 2022 at 0:19
• @ElliotAlderson Actually I wanted to know if my reasoning was correct Jan 19, 2022 at 9:47
• The simulation would tell you immediately if your reasoning was correct. Jan 19, 2022 at 11:59

Your equation for $$\V_+\$$ is correct but you've calculated the value incorrectly -- it is 147 mV, not 30 mV.
$$\V_-\$$ has the same voltage, as you've stated, so the current flowing from that node to ground through $$\R_4\$$ is $$\147\text{ mV} / 10\text{ k}\Omega = 14.7\mu\text{A}\$$. That same current flows from $$\V_x\$$ to $$\V_-\$$ through $$\R_5\$$ (since, as you've stated, no current flows into or out of the op amp's input) so $$\V_x\$$ is $$\14.7\mu\text{A} \times 90\text{ k}\Omega = 1.323\text{ V}\$$ above $$\V_-\$$, which is therefore $$\V_x = 1.47\text{ V}\$$. $$\V_{\text{out}}\$$ is 700 mV less than $$\V_x\$$ since the diode is on.