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schematic

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

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

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

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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.

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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})\$

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