# Op amp with feedback simulate this circuit – Schematic created using CircuitLab

How to calculate Iout(current flowing through RL) of this circuit? All Resistors are of same value except the load resistor. Let the rail voltages be +15v and -15v.

• You have deleted the component designators (R1, R2, etc.) making discussion of your circuit nearly impossible. Please edit the schematic, double-click each component and put the numbers back in. – Transistor Sep 28 '18 at 22:55
• Welcome to EE.SE. Seems like a homework question to me. What have you done to try and solve this problem? We will help you but not do your homework for you. – Sparky256 Sep 28 '18 at 22:55
• This was asked in my today's interview at x company. I answered them, as it has positive feedback the output voltage will saturate to the rails and then apply ohms law at the output to calculate Iout. Is this correct? – Ravi Teja Sep 28 '18 at 23:01
• Unless the R's are all the same value. And it does look like homework. Hints: how much current flows into the inputs of the (ideal) op-amp? If the voltage at the - input is V1, what must the voltage at the + input be? – George White Sep 28 '18 at 23:03
• Thank you for the reply guys. R's are of the same value. The ideal op-amp has an infinite input impedance, so no current flows through the inputs. Considering virtual ground voltages will be the same. for ideal opamp V1=V2(+ve terminal voltage)=0 – Ravi Teja Sep 28 '18 at 23:17

## 1 Answer simulate this circuit – Schematic created using CircuitLab

Figure 1. OP's schematic redrawn with component designators and voltage reference points added.

$$\ V_- = \frac {R_1}{R_1 + R_2}V_O \tag 1\$$.

Since R1 = R2 then $$\ V_O = 2 V_- \tag 2\$$.

If the op-amp is working in linear negative feedback mode then $$\ V_- = V_+ \$$ so $$\ V_O = 2 V_+ \tag 3 \$$.

Summing the currents into V+ we get

\begin{align} I_{RL} & = I_{R3} + I_{R4} \tag 4 \\ & = \frac {V_{IN} - V_{+}}{R3} + \frac{V_O - V_+}{R4} \tag 5 \\ & = \frac {V_{IN} - V_{+}}{R3} + \frac{2V_+ - V_+}{R4} \tag 6 \\ & = \frac {V_{IN} - V_{+}}{R} + \frac{V_{+}}{R} \tag 7\\ & = \frac{ V_{IN}}{R} \tag 8 \end{align}

Perhaps surprisingly the output current is independent of RL.

Note how difficult it would be to write these equations without the component designations requested in my comment to your question.

• Thank you for the detailed explanation. I'm a new contributor so I was a little confused with the site interface and your comment. – Ravi Teja Sep 29 '18 at 1:30