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The image of the circuit is shown below and it required to find \$V_0\$,

enter image description here

My first attempt at solving this problem is by changing the current source into a voltage source with 1-V and 2k\$\Omega\$ resistance. The fact that the inverting and non-inverting terminals aren't grounded make this problem look difficult.To the point which, I don't know how to proceed with this question or where to start. I would appreciate any help.

Following some thought and another schematic from a hint suggested by Alfred, I produced a schematic representing our work.

enter image description here

And my solution for the problem is below,

Using node equation at nodes A and B we have,

$$\frac{V_A-V_B}{1k}=-0.5 \text{mA}$$ $$\frac{V_B-(2+V_A)}{1k}=-0.5-x$$ where x is the current that is sent in the output of the op-amp. Using KCL, at the bottom node near the current source we see that the same current that goes through the op-amp also goes through the \$2\text{k}\Omega\$ resistor. Hence, we have,

$$\frac{-V_A}{2k}=x$$

Replacing this in the second equation,

$$\frac{V_B-(2+V_A)}{1k}=-0.5+\frac{V_A}{2k}$$

And and solving the equations yields \$V_A=-2 \text{-V}\$ and \$V_B=-1.5 \text{-V}\$

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  • \$\begingroup\$ Nodal analysis is still useful. 'Ground' is just referencing a point to be 0V. Instead of calling it 'ground', you could call it V1 V2 and it is still the same. You can start by making an equation for each node, then continue from there. \$\endgroup\$ – Pyxzure Jan 31 '14 at 13:57
  • \$\begingroup\$ Regarding your 2nd schematic: \$v_-\$ is not equal to \$V_O\$. By KVL, \$v_- = V_O + i_x \cdot 1k\Omega\$ \$\endgroup\$ – Alfred Centauri Jan 31 '14 at 19:55
  • \$\begingroup\$ @AlfredCentauri Isn't \$v_+=v_o=v_-\$ \$\endgroup\$ – user29568 Jan 31 '14 at 19:58
  • \$\begingroup\$ @user29568, no, why do you think that? \$\endgroup\$ – Alfred Centauri Jan 31 '14 at 20:13
  • \$\begingroup\$ Is \$v_0\$ representing the voltage across the resistor connected to the ground? If yes, then the node I have labeled as \$V_0\$ should have voltage \$V_o\$ no? \$\endgroup\$ – user29568 Jan 31 '14 at 20:16
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I don't know how to proceed with this question or where to start.

If there is (net) negative feedback, then you proceed by setting the voltage across the op-amp input terminals equal to zero:

$$v_+ = v_-$$

Note that with zero volts across the input terminals, the 2k resistor in parallel with the current source is irrelevant; there is zero volts across it so there is zero current through it. You may remove it from the circuit without changing the solution.

This should get you started.


@AlfredCentauri I still don't see the bottom loop, do you mean the loop v+ connected to VB then connected to the voltage source and then the resistor and finally VA. Is that considered a loop even with the op-amp? And when I do I still don't get your equation.

schematic

simulate this circuit – Schematic created using CircuitLab

This is the bottom-most loop and KVL clock-wise 'round the loop starting with the voltage across the 1k resistor is:

$$i_1 \cdot 1k\Omega -2V + V_B - V_A = 0 $$

rearranging yields

$$V_B = V_A - i_1 \cdot 1k\Omega + 2V$$

If the presence of the voltage source above is puzzling, recall that the output of the ideal op-amp is an ideal (controlled) voltage voltage source referenced to ground which I've shown explicitly here.

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  • \$\begingroup\$ When you refer to voltage, do you mean voltage difference or voltage potential at a point? If one of the terminals aren't connected to a ground how can we say that the voltage at the terminals is zero. The difference is zero as expected from any ideal op amp, but the voltage at terminals could be any value but equal. Right? Further, if there is no change of voltage across the resistor, how would that affect the circuit? Wouldn't that make it a short? \$\endgroup\$ – user29568 Jan 31 '14 at 18:24
  • \$\begingroup\$ I was wondering if by any chance you knew any good textbooks that cover Op-amps, filters, convolution, and Fourier, and other related material. \$\endgroup\$ – user29568 Jan 31 '14 at 19:49
  • \$\begingroup\$ @user29568, \$v_+\$ is a node voltage which is, by definition, the voltage difference between a circuit node and the 'ground' node. So, for example, to measure \$v_+\$, place the red lead of your voltmeter on the non-inverting terminal of the op-amp and the black lead on the ground terminal. \$\endgroup\$ – Alfred Centauri Jan 31 '14 at 19:50
  • \$\begingroup\$ I understand the definition of the node voltage, but I seem to be interpreting your answer "across the op-amp input terminals equal to zero" as \$v_+=v_-=0\$. I don't think that is what you meant, perhaps you where referring the voltage difference between the two input terminals. \$\endgroup\$ – user29568 Jan 31 '14 at 20:13
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    \$\begingroup\$ @user29568, that's what voltage across means; the voltage across two terminals is the voltage difference between the terminals. \$\endgroup\$ – Alfred Centauri Jan 31 '14 at 20:19

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