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I calculated Vo for this ideal op amp circuit using the formula: \$R_2\$/\$R_1\$(\$V_2\$-\$V_1\$) = \$V_o\$
I believe the correct \$V_o\$ = -14V I did not get this result using the above formula. It seems the formula is not correct for this circuit but I don't understand why. Would someone please enlighten me?

schematic

simulate this circuit – Schematic created using CircuitLab

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    \$\begingroup\$ (1) You are using a wrong equation for the output voltage. Notice that that this is not a differential OpAmp topology. To derive a correct formula, use the Kirchhoff's Current Law. (2) The rest of the enlightenment is in your own hands. After all, this is a homework. \$\endgroup\$ Mar 12 '16 at 5:13
  • \$\begingroup\$ Daveythewavey, ask yourself for which reason you wanted to use the mentioned formula? Did you derive it? No. So - where does it come from? \$\endgroup\$
    – LvW
    Mar 12 '16 at 9:39
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It seems the formula is not correct for this circuit but I don't understand why

That's not a differential amplifier.

To solve that circuit you can proceed in at least two ways (assuming ideal op-amp). Here are a few hints (you won't get the full solution):

  1. Superposition. Null the two independent sources \$V_1\$ and \$V_2\$ one at a time, and then sum the two partial responses. By nulling \$V_2\$ you're left with an inverting amplifier; by nulling \$V_1\$, with a non-inverting amplifier.
  2. Impose ideal op-amp conditions. Recall that an ideal op amp, when there is a feedback path from the output to the input, adjusts its output to have $$v_\mathrm{n}=v_\mathrm{p},$$ where \$v_\mathrm{n}\$ is the potential of the inverting input and \$v_\mathrm{p}\$ that of the non-inverting one. In your circuit, this implies \$v_\mathrm{n} = V_2\$. Knowing this, you can easily apply Millman's theorem at the inverting input (or Kirchhoff's laws, if you feel more comfortable), considering the circuit containing the sources \$V_1\$ and \$v_\mathrm{o}\$. Then solve for \$v_\mathrm{o}\$.
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schematic

simulate this circuit – Schematic created using CircuitLab

Figure 1. Inverting op-amp with offset on non-inverting input.

  • The trick here is to remember that the op-amp doesn't know where your GROUND or 0 V is. This should be obvious from the schematic as there is no 0 V connection to the op-amp.
  • In the inverting amplifier configuration the op-amp will try to adjust the output until the inverting input voltage is the same as the non-inverting input. This will happen when both are at +2 V above ground and your effective input is 4 V relative to the non-inverting input. -The gain of the circuit will be \$-\frac {R_F}{R_I} = -4\$ relative to the non-inverting input so

$$ V_{OUT} = -(V_1 - V_2)\frac{R_F}{R_I} + V_2 = (6-2)\frac{-4}{1} + 2 = -14 V$$

I did not get this result using the above formula.

In a practical experiment you might not get 14 V out. If this is the case you are probably driving the op-amp too close to the negative power rail. You have to either increase the negative power-rail voltage (watch the specifications) or decrease your 6 V input signal.

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  • \$\begingroup\$ transistor: you speak about "the gain". Which gain are you referring to? Vout/V1 or Vout/V2? I rather think, it is a simple case for superposition of TWO output values caused by TWO gain values. The formula in the question is NOT correct because it gives 4*(2-6)=-16V. \$\endgroup\$
    – LvW
    Mar 12 '16 at 12:38
  • \$\begingroup\$ You're right. I misunderstood some of his statements. Does it look right now? \$\endgroup\$
    – Transistor
    Mar 12 '16 at 12:47
  • \$\begingroup\$ The result is OK, of course. However, where is the logic behind the calculation? Here is my approach: Vout1=-4*6=-24V and Vout2=2*(1+4)=+10V. Hence, Vout=-24+10=-14V. \$\endgroup\$
    – LvW
    Mar 12 '16 at 16:30
  • \$\begingroup\$ Yes, considering the output as the sum of an inverting amplifier and a non-inverting amplifier is a good method. The logic of my method is to take the non-inverting input as reference, calculate the output relative to that and then add in the non-inverting input voltage as an output offset. \$\endgroup\$
    – Transistor
    Mar 12 '16 at 17:17
  • \$\begingroup\$ I like your method: it's elegant and fast, and would be my favourite in many situations: however, I suspect it's not easily understood by students because it requires a deeper level of understanding about the op amp (frequently, students have difficulties in understanding that the op amp doesn't know where the ground is). \$\endgroup\$ Mar 12 '16 at 19:41
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From TI/NSC:

enter image description here

Or...

  1. because of the way opamps work, the voltage on OA1- must be equal to, and the same sign as, the voltage on OA1+
  2. If OA1+ is at +2V, then OA1- must also be at +2V.
  3. If OA1- is at +2V then R1 has 4 volts across it.
  4. If R1 has 4V across it then the current through it must be \$ I =\frac{4V}{1000\Omega} = 0.004 A \$
  5. Since R1 and R2 are in series, the current through R2 is the same current through R1.
  6. Since R2 is 4000 ohms and there's 0.004 A through it, it must be dropping \$ 0/004A \times 4000 ohms = 16 \text{ volts.}\$
  7. Since OA1 is an inverting amplifier and OA1- is positive, OA1 OUT must be negative.
  8. Since the voltage on the left-hand side of R1 is +2 volts, the total drop across R2 is 16 volts, and OA1 OUT must be negative, OA1 OUT must be at -14 volts.
  9. R3 doesn't matter since it's not in the loop.
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