# Inverting Op Amp question

I'm having trouble with this particular inverting Op Amp question. I know that there are 2 nodes and I must apply KCL on the nodes. What I tried was $\frac{0-V_i}{49k\Omega}+\frac{0-V_o}{79k\Omega}=0$ and I'm not sure about the second KCL equation. and I'm pretty sure that R3 isn't parallel with R4. The answer is -68.8

• $V_-$ is virtual ground, so $R_2$ and $R_4$ are in parallel. With that knowledge you can simplify $R_3$ – jippie Apr 13 '13 at 7:51

There is an easier way to think about this problem, assuming ideal components. First, since the noninverting pin is grounded, the inverting pin is at virtual ground. This means:

$V_{R1}=V_{in}$.

$\therefore I_{R1}=I_{R2\parallel R4}=I_{R3}$

Calculate the voltage drop across $R_{2\parallel 4}$ in series with $R_3$, and you'll arrive at $V_o$ from there.

• Possible correction: R2 and R4 are in parallel to ground. This combo is in series with R3. – helloworld922 Apr 13 '13 at 6:47
• That's right, not sure how I missed that when I specifically pointed out virtual ground. – Matt Young Apr 13 '13 at 6:50
• i figured the equation is: -Vin (R2/R1) (1 + (R3/R2) + (R3/R4)) = Vout – George Randall Apr 13 '13 at 7:19

FWIW, here's another method; form the Thevenin equivalent circuit looking into R2 from the inverting input.

The equivalent circuit is, by inspection:

$V_{TH} = V_{OUT}\dfrac{R_4}{R_3+R_4}$

$R_{TH} = R_2 + R_3||R_4$

Now, there's just one node to consider. The KCL equation for the remaining node is, by inspection:

$\dfrac{V_{IN}}{R_1} + \dfrac{V_{TH}}{R_{TH}} = 0$

Thevenin Equivalent circuit comes in handy solving this T Negative feedback network. At the node between R4 and R2.You can apply Thevenin Equivalent such that Vout thevenin=( R4/(R3+R4)) and Requivalent2=(R3//R4). Now you can solve like normal inverting op amp configuration.

KCL at V-:

$$\\frac{V1 - V_{in}}{R1} + \frac{V1 - V2}{R2} = 0\$$

KCL at V2: (just above R4)

$$\\frac{V2-V1}{R2} + \frac{V2-0}{R4} + \frac{V2-V_{out}}{R3} = 0\$$

Vout should equal:

$$\V_{out}=-V_{in} \frac{R2}{R1} (1 + \frac{R3}{R2} + \frac{R3}{R4})\$$