Solve Ideal Op Amp Question: KCL & KVL

[jackstock]

Homework Statement

Homework Equations

KCL & KVL.

The Attempt at a Solution

Using KVL:

-9 + 10kI₁ + 6 = 0
-10 + 20kI₂ + 6 = 0
12 + 30kI₃ + 6 = 0
40kI₄ + 6 = 0

:. I₁ = 0.3mA
I₂ = 0.2mA
I₃ = -0.6mA
I₄ = -0.15mA

Now here's where the problem is. When I take KCL at the bottom node:

I₁ + I₂ + I₃ + I₄ = I₆ [the op amp is ideal so the current through the 6V source is zero]
I₆ = -0.25mA

:. V_o = -0.25m*20k = -5V [which is apparently incorrect]

However, when I take KCL at the inverting input to the op amp:

I₁ + I₂ + I₃ + I₄ = I₅
I₅ = -0.25mA

Using KVL again:
V_o - 6 - 0.25m*100k = 0
V_o = 31V [which is apparently correct]

So my question is why do I get different answers depending on which node I take KCL at? Is there something obvious which I'm missing? Thank you in advance for your help.

 

[The Electrician]

It's because I1, I2, I3, I4 and I6 are not the only currents into the bottom node. The opamp's power supply also provides a current into the bottom node. You will notice that the opamp's output absorbs (or supplies) the sum of I5 and I6, which ultimately finds its way to the power supply. So there is a current into the bottom node not shown on the schematic.

But, summing currents into the inverting input works because there are no hidden currents into that node.

 

[jackstock]

Thank you, that makes sense. For future reference, is there a way to identify where these currents from the power supply will turn up so this doesn't happen again?

 

[The Electrician]

The power supply will be connected to the opamp V+ and V- pins (or just V+ if a single supply voltage is used) and the "hidden" currents will be injected into the circuit wherever the supply return is connected. Don't use that node (probably the reference or "ground" node) as a node for summing currents.

 

[DeeNos]

I would first suggest double-checking your calculations to ensure they are correct. It is possible that there was a mistake made in one of the equations, resulting in the discrepancy in the answers.

Additionally, it is important to note that when taking KCL at the bottom node, the current through the 6V source should not be assumed to be zero. In an ideal op amp, the input currents are assumed to be zero, but the output current can still be non-zero. So, in this case, the current through the 6V source should be included in the KCL equation.

In general, when solving circuit problems involving ideal op amps, it is important to consider the assumptions and limitations of the ideal op amp model. If there are any doubts or discrepancies in the answers, it may be helpful to simulate the circuit using a circuit simulator software to verify the results.