Where am I going wrong in analyzing this op-amp circuit?

I'm pretty new to circuit analysis and even newer to analyzing op-amp circuits. I'm trying to analyze the below circuit and find VA, but I don't think I'm setting it up correctly. I've tried to label the currents below:

I think that would give me the following equations:

1. i2 = VA/3
2. i3 = (VA - V_)/2
3. i3 = (V_ - V0)/3

However, I'm not sure that is correct and I'm also not sure how to handle VA - V+. I may be way off in how I'm approaching this, so any help would be greatly appreciated. Just having a lot of trouble understanding how to analyze op-amp circuits (particularly with respect to how current works in op-amp circuits).

• Do you know what a virtual ground is on an op-amp? Do you know how that shapes the input impedance? If you research op-amp virtual ground it will immediately tell you what V_ is. Commented Jan 20, 2023 at 18:02
• To really understand what is going on in this circuit, you need to draw the complete current paths (current loops). To do this, you must first draw the power sources connected to the operational amplifier. Commented Jan 20, 2023 at 19:24

Your circuit analysis thus far is spot-on.

The next step is to apply the defining relations for the op-amp.

The first defining relation, open-loop voltage gain $$\A_{VOL}\$$ is very high usually considered infinite.$$V_+-V_-=\frac{V_O}{A_{VOL}}.$$So with $$\A_{VOL}\$$ very high the difference between the inputs is considered as zero volts:$$V_-=V_+$$$$\V_+\$$ is connected to ground so therefore:$$V_-=V_+=0V$$ So this defining relationship places the 2-ohm resistor in parallel with the input 3-ohm resistor because they share the same voltage.

They therefore form a current divider so that $$\i_2\$$ and $$\i_3\$$ can easily be calculated.

The second defining relation says that the currents into $$\i_+\$$ and $$\i_-\$$ equal zero. So where does the current $$\i_3\$$ go? The output voltage of the op-amp decreases until all of $$\i_3\$$ flows through the feedback 3-ohm resistor.

Using this sequence in applying the op-amp defining relations, all the currents and voltages are easily calculated.

• Thanks. I think V+ being connected to ground is the key part I was missing. I wasn't sure if that meant it would be 0 in this particular case because it looks like it is also connected to the current source. Commented Jan 21, 2023 at 16:18

Because an op amp has very high gain, if it is working at all the input differential voltage must be almost zero. So you should assume that the voltage at the - input is zero.(It's not - quite - but it's very close.) Then you can calculate VA by treating the two resistors as being in parallel. Knowing VA, you can then calculate i3.

• To clarify: high gain and, in this implementation, feedback to the inverting input, are what cause the input differential voltage to be zero volts. If the op amp is connected as a comparator or some other configuration without negative feedback then it cannot be analyzed as having the same voltage.
– vir
Commented Jan 20, 2023 at 18:38
• Ok that's helpful. So then the equivalent resistor would be 6/5. And so then i1 = 0 so i2 is just -5, so that would mean VA is -6? Commented Jan 20, 2023 at 23:19
• @Coder1913 - Yup. Then you find for the output voltage. But THEN you take a look at overall circuit. For instance, if the op amp power supplies are +/- 5 volts, the output cannot be +9, and the assumption that the op amp "is working at all" isn't true. And keep in mind that, in addition to high gain, an op amp has high input impedances, so i1 is going to be zero anyways. And if i4 is the current through the feedback resistor, then i3 = i4. Commented Jan 22, 2023 at 17:01

As others have suggested, start by assuming that the voltage at the two input terminals is the same; this is a good starting approximation for this type of amplification circuit.

1. From there, the voltage across the 2 Ω resistor and the 3 Ω resistor that is connected in parallel with the 5 A current source needs to be determined (consider where the current of the 5 A source will flow).
2. Using these two voltages and currents, along with the approximation that the current flowing through (into or out of) the terminals of the op-amp is, in this case, negligible. The current flowing through, and the voltage across the 3 Ω resistor connected to the op-amp's output can also be determined.

This should provide you with a full analysis of the circuit. All voltages and currents are known.