resistance R1. And that's connected to this terminal, as

you will, the inverting terminal. So, you got the feedback resistor and the

input resistor both connected to this terminal.

And the non-inverting terminal of the up end is grounded.

So, we want to analyze this circuit and compute V out as a function of the

signal, the signal voltage. So, we're going to start by looking at

this node that we've labeled A, and apply Kirchhoff's current law to that node.

Now, Kirchhoff's current law says that the algebraic sum of all the currents

flowing into that node is 0. So, i1 is flowing in, i2 is flowing in,

so they're both positive. And iN is flowing out, so that's

negative. So, those three have to sum up to zero.

Now, [COUGH] I can then write an expression for i1 and i2 in terms of the

voltages. Now i1, the current through this

resistor, is just going to be the difference of Vs minus Vn.

So, Vs is the voltage on this side of the resistor.

The voltage on this side of the resistor is Vn.

And so, the current i1 is just that voltage difference, Vs minus Vn over R1.

Then, the second current, i2, the voltage difference across that is V out over

here. The m over here, and so the current, i2,

is V out minus VN over R2. So, that's the second term.

And then in is leave it as it is. Now, we're going to now use the

assumptions that are part of the ideal op amp model.

the first one is that Vp and VN are equal.

And the second one is that there's no current flowing into either of these.

the non-inverting or the inverting input. So, these are both zero.

So, that's the ideal op amp assumption. Okay, so with those assumptions and that

equation, remember that we're looking back up here.

We made Vp equal to 0 and Vp equals VN, so VN has to be zero.

So, I can go back in this equation and set VN equal to zero everywhere I see it.

And the equation at node then reduces to this simple equation, Vs over R1 plus V

out over R2 equals 0. And so, I can solve that for V out as a

function of Vs, the source of voltage, and it's just minus R2 over R1.

And so, the voltage gain of this, the output voltage divided by the input

voltage is negative. The ratio of the feedback resistor to the

input resistor. So, if I put in a small sinusoidal signal

what's going to come out of this is a sinusoidal signal that is amplified by a

factor of R2 over R1, and it's going to be inverted.

So, I get a larger sine wave that is also inverted.