# Find the gain in a non-ideal op amp - check my logic

1. Apr 3, 2012

### jkuhling

To start I have attached an image of the circuit. We are to relate the $V_{0}$to $V_{s}$. I am not exactly sure where to start with it. I know that $V_{-}$ and the $V_{+}$ on the op amp have to have the same voltage. The resistor that bridges the two inputs is throwing me off.

Not sure if this is the right way to start but if I look at the current through the 400$\Omega$ resistor as $I_{0}$. Then using KCL:

$I_{0}$=$I_{1}$+$I_{2}$.

Sense there is no current running through the op amp, then what ever the current is in the first 1200$\Omega$ is the same current in the second 1200$\Omega$. By this the current at $V_{0}$ to be the current through the 600$\Omega$ resistor.

If I put in the voltage across the head minus the voltage in the tail stuff for the currents and then solve to the $V_{0}$/$V_{s}$ should be the right way.

I am curious if I am thinking about this right or I am completely off my rocker.
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

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2. Apr 4, 2012

### rude man

Just write the usual node or loop equations (I'm a node man myself).

I don't understand what you mean by "head" and "tail" stuff. Are we flipping coins here?

Why is your post titled "non-ideal" op amp?

3. Apr 6, 2012

### Staff: Mentor

Don't let it. The presence of that resistor should pique your interest!
Correct. So with that pair of 1200Ω resistors forming an unloaded voltage divider, you can say the voltage at the op-amp's ⊕ input is half that at the junction of the three resistors. And you know that the voltage at the op-amp's ⊕ input must equal Vo, so you can immediately mark the voltage at the junction of the three resistors as equal to .... ?

This simplifies the network you need to analyze. Remember, amplifiers are not all designed to give a gain > 1.0