# Determining the gain of an op-amp with feedback?

1. Dec 3, 2012

### richyw

1. The problem statement, all variables and given/known data

2. Relevant equations

I have two "golden rules" I was given which are "no current into the op-amp" and $V_{-} = V_{+}$

and the open loop gain is infinite

Basically my notes and textbooks are leaving me with pretty much nothing though

3. The attempt at a solution

tried determining the currents like we did in other methods. tried figuring out the case when x=1 and x=0. I don't get how current can flow across the resistor if V_=V+. I'm basically completely lost. No Idea where to start.

2. Dec 3, 2012

### Staff: Mentor

Using your "golden rules", given a voltage V_in at the input, what will be the voltage at the + terminal of the op-amp? (Hint: you're looking at a simple voltage divider).

So, what then is the voltage at the "-" terminal?

3. Dec 3, 2012

### richyw

well I would say $$V_{+}=\frac{xR}{(1-x)R+xR}V_{in}$$$$V_{+}=x$$

4. Dec 3, 2012

### richyw

and my golden rule says that V_=V+

5. Dec 3, 2012

### Staff: Mentor

Well, $x\,V_{in}$, right?

Good. So what's the current through the input resistor, R?

6. Dec 3, 2012

### richyw

oops I meant V-=xV_in

7. Dec 3, 2012

### richyw

the current through the input resistor would be $$\frac{V_{in}-xV_{in}}{R}=\frac{(1-x)V_{in}}{R}$$so$$(1-x)V_{in}=xV_{in}-V_{out}$$$$1-x=x-\frac{V_{out}}{V_{in}}$$$$\frac{V_{out}}{V_{in}}=2x-1$$?

8. Dec 3, 2012

### richyw

if my work is hard to follow I just said that the current through the input resistor must equal the current through the feedback resistor after the "so".

9. Dec 3, 2012

### richyw

this makes sense to me now so I hope it's correct haha.

10. Dec 3, 2012

### Staff: Mentor

Looks good. For someone who started out with "no idea", you've carried it off nicely

11. Dec 3, 2012

### richyw

thanks a lot. I guess usually I just have "no idea where to start". sucks on exams!