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

Question

^{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?

Answer 1

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.