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schematic

simulate this circuit – Schematic created using CircuitLab

I am attempting to solve the above circuit for V0. Current and Voltage variables are from me. Thus far I've used KCL to say that $$i_3 = i_2 + i_1$$ and by ideal op amp function, $$ V_1 = 1V $$ $$V_3 = 2V $$

From there I say that $$ \frac{-1}{10k} + i_2 = \frac{2-V_0}{40k} $$

But that doesn't really get me any closer to finding V0, what am I missing?

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    \$\begingroup\$ Actually I think I see it I should be able to find V2 by using the voltage drop over R2, since the current should be the same as over R1. \$\endgroup\$ – Daniel B. Jan 26 '14 at 4:22
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    \$\begingroup\$ Do not re-use the symbols V1, V2 and V3 for different purposes in the same schematic!!! \$\endgroup\$ – Kaz Jan 27 '14 at 22:52
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Try solving for V2 first, since it does not depend on the right-hand amplifier.

Once you have that, you can easily solve for Vo.

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Hard way
Write KCL for the points marked V1 and V3. Solve the resulting system of equations for V2 and Vo.

Easy way
OP1 is wired as a non-inverting amplifier without any quirks or caveats. You can use the appropriate formula V2. Then do a KCL for the point, which is marked V3 on the schematic.

P.S.
I'm deliberately not giving more details, because this is a training exercise for the O.P.

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Correcting the names of the node voltage: I will name the voltage at the negative terminal of the second amp \$v_2\$ not \$v_3\$, also name the output voltage of the first amp by \$v_3\$ not \$v_2\$, and name the current passing through \$R_3\$ as \$I_2\$:

$$v_1=1V$$

Gain of first amp $$A1 = \frac{20}{10} +1 = 3$$

so \$v_3 = 3 \times 1=3\$V. $$I_1=\frac{v_1-v_3}{20k\Omega} = 0.1 mA$$ $$I_2= \frac{v3-v2}{30k\Omega} = 1/30 mA$$ $$v_2-v_o = 1/30 mA \times 40 k\Omega = 4/3 V$$

so \$V_o = v_2-4/3 = 2 -4/3 = 2/3\$V

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This cascade of two inverting stages can be solved mentally.

The first op-amp's + input is at 1V, therefore so is the - input. This is the inverting stage's reference voltage. The input to the stage is 0V, which means that is sitting -1V below ref. This offset from ref is what is amplified. The gain is -2X (-R2/R1), and so the output must be +2V above ref, or 3V.

Next, the 3V output of stage 1 is 1V above the reference voltage (2V) of the second inverting stage. The gain is -4/3 = 1.33, and so the output is 1V x -1.33 = 1.33V below the ref voltage: 0.66V.

The currents can be worked out using Ohm's Law. For instance we know that node V1 is at 1V, node V2 is at 3V. So the voltage drop across the resistor in the i1 direction is -2V, and I = V/R. Yes, the current is negative: conventional current flows opposite to the arrow drawn for i1.

I suggest working with a hard-copy of the schematic and adding the information to it as it becomes clear, rather than solving equations on a separate sheet. Any time you know a voltage at a node, or any other fact, label it right there on the schematic. Then dependent facts will "pop out".

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In general, the process of working these is pretty straightforward. Start with what you know, solve for what you don't. You know that an op-amp with negative feedback (a path from the output to the negative input) tries to make its inputs equal. (In the real world there would be concerns about the voltage rails of the op amp, but this is obviously a textbook problem so we won't worry about such.) The positive input voltages are given to you, fixed by the voltage sources. From this fact, we also know the negative input voltages.

So that means we know the voltage on both sides of the 10k resistor: 1V and 0V. There's 1V across this resistor, and it's 10k, so the current through it is 100 uA, flowing left. We have current leaving a circuit node. It has to also enter that node somewhere, and there are only two possible paths: through the 20k resistor, or through the input of the op-amp.

Another assumption we make about op-amps is that current can't flow into or out of their input terminals. In the real world, that's not true, but for simple analysis of DC circuits with resistances in this range it's probably close enough.

So we know there's 100 uA going through the 20k as well, flowing left. We know the left side of the 20k is at 1V, and it's got 100 uA x 20k = 2V drop across it, so we get a voltage of 3V on the right side of the 20k.

Continue the process. The left side of the 30k resistor has 3V on it, and the right side has 2V. That means there's one volt across it, giving 33 uA through it, flowing right.

That 33 uA can't flow into the input of the op amp, so it has to go through the 40k. 33 uA through a 40k resistor gives a voltage drop of 1.33 volts. The left side of the 40k resistor is at 2V, and the right side is 1.33V lower, giving .66 volts.

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  • \$\begingroup\$ Sorry, who said the voltages at the noninverting inputs were -1 V and -2 V? I'm not seeing that anywhere. \$\endgroup\$ – Joe Hass Jan 28 '14 at 0:49
  • \$\begingroup\$ @JoeHass Sorry, copypasta. :) \$\endgroup\$ – Stephen Collings Jan 28 '14 at 1:09
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This is a cascaded opamp circuit. We can solve this by dividing it into two stage. First opamp act as non-inverting amplifier. Second amplifier act as a differential amplifier. Non-inverting amplifier is fed as one of input of differential amplifier. This is the key point we need to understand.

Now consider the first stage. Since it is a non-inverting amplifier, its gain is (1+R1/R2)=(1+20k/10k)=3. So the output of first stage is Vout=3V1.

Now consider the second stage. This is actually a differential amplifier stage, since it has input in both terminals. So in order to analyze it, we need to apply SUPER POSITION PRINCIPLE. According to the principle, we need to find the output produced by each input by short circuiting all other inputs and then add up all output values.

So short circuit non-inverting input and find the output cause by inverting input . We can see that, now it become an inverting amplifier. Hence output will be Vo1=(-R4/R3) x Vin1 where (-R4/R3) is gain. So V1 = (-40k/30k) x Vin1 = -4V1 (since Vin1 = 3V1). Now ground inverting input. Now it is a non-inverting amplifier. So Vo2 = (1 + R4/R3) x Vin2 = (1 + 40k/30k) x V2 = 7V2/3. So according to superposition priciple, output will be Vo = (Vo1 + Vo2) = (7V2/3 - 4V1)(General Case). Here V1=1v and V2=2v. Hence Vo = (7 x 1)/3 - 4 x 2 = (2/3)V

I think this will be the answer. If there is any error in it, please inform me.

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