help with op amp circuit

[formulajoe]

gain of the circuit is -50 V/V, open loop gain is 200 V/V. There are three resistors in the circuit. The standard R2 and R2 and there is an R3 resistor.
Vo is across R3. It is connected to the output of the output going to ground. I need to find a value for R1 to make the gains works. The value for R2 is 100K and the value for R3 is 10K.
I don't know how to this problem. I can do it just fine with a finite open loop gain if R3 is gone and I can do it just fine with R3 as long as there is an infinite open loop gain.

Heres what I have

G = Vo/Vi = -50 = (-R2/R1)/(1+1+R2/R1)/A
Using this I found R1 to be 1500 ohms. Is this right?

[Tom Mattson]

I don't know about anyone else, but I need a picture.

[formulajoe]

heres a crude picture, but it should work.

[Tom Mattson]

G = Vo/Vi = -50 = (-R2/R1)/(1+1+R2/R1)/A


OK, now what does that "A" stand for?

[mezarashi]

OK, now what does that "A" stand for?

It stands for the finite open loop gain, and the equation is nearly right I believe. The standard form is:

G = (-R2/R1)/(1+(1+R2/R1)/A)

where G is the closed loop gain of an op-amp with finite open loop gain. Of course, as A approaches infinity, notice that the gain will be equal to -R2/R1, which is what you would get if you analyze the op-amp under ideal assumptions.

[uart]

I don't know how to this problem. I can do it just fine with a finite open loop gain if R3 is gone and I can do it just fine with R3 as long as there is an infinite open loop gain.

Heres what I have

G = Vo/Vi = -50 = (-R2/R1)/(1+1+R2/R1)/A
Using this I found R1 to be 1500 ohms. Is this right?

The resistor R3 is irrelevent as long as you dont need to assume finite open loop output impedance. In other words just ignore R3.

Your formular for G is wrong. It could be made correct however by "fixing up the brackets" in the formula. (A is the open loop gain).

Your initial solution of 1500 ohms is obviously incorrect because even if the open loop gain was infinite the value required would be 10000/50 = 200 ohms. So clearly the answer will be less than 200 ohms.

[BobG]

gain of the circuit is -50 V/V, open loop gain is 200 V/V. There are three resistors in the circuit. The standard R2 and R2 and there is an R3 resistor.
Vo is across R3. It is connected to the output of the output going to ground. I need to find a value for R1 to make the gains works. The value for R2 is 100K and the value for R3 is 10K.
I don't know how to this problem. I can do it just fine with a finite open loop gain if R3 is gone and I can do it just fine with R3 as long as there is an infinite open loop gain.

Heres what I have

G = Vo/Vi = -50 = (-R2/R1)/(1+1+R2/R1)/A
Using this I found R1 to be 1500 ohms. Is this right?
I'm not sure what formula you're using. Your drawing shows an inverting op amp, so the gain is just G=-(\frac{R_2}{R_1})

As drawn, R3 isn't going to affect your gain. Here's an easy way to show that:

The non-inverting input of the op amp is connected to ground, so it's zero. For an ideal opamp, the voltage of both inputs will be equal to each other, so the inverting input is also at zero volts. What does that mean for the current through R1?

The current through both inputs of the op amp have to be equal and will be zero. So what does that mean for the current through R2?

How many volts were dropped through R1 and what direction was the current going? What direction was the current going through R2 and how many volts will be dropped through R2 (Ohm's law)?

The current through R3 may vary according to the value of R3, but the voltage at the node of R2, R3, and the output of the opamp is determined by R1 and R2. If R3 weren't there, and knowing Kirchoff's Current Law, what could you tell about the current at the output of the opamp? Knowing that, why do you think R3 would be added?

[Tom Mattson]

I'm not sure what formula you're using. Your drawing shows an inverting op amp, so the gain is just G=-(\frac{R_2}{R_1})


I got exactly that too Bob, but it seems that you aren't supposed to model the op amp as ideal. If it were ideal then you wouldn't be dealing with open loop gains in the first place.

[mezarashi]

I'd just like to add some more comments with regards to the last two posts less Tom's post now. When the op-amp is non-ideal, then using the "standard" analysis procedures is not adequate. Let me add some proof to where the equation comes from:

Assume that op-amp is not ideal, thus Vout = -AVin, where Vin is the voltage at the negative input of the op-amp and A is the finite open-loop gain.

Then the voltage at the negative input is: -Vout/A
Reuse this in your normal circuit analysis and you will get the same "messier" equation :P

[formulajoe]

G = Vo/Vi = -50 = (-R2/R1)/1+(1+R2/R1)/A

This is my equation, I used this when I solved for the 1500 ohms. When I put it in the first time it was a typing mistake. This is incorrect?

[mezarashi]

G = Vo/Vi = -50 = (-R2/R1)/1+(1+R2/R1)/A

This is my equation, I used this when I solved for the 1500 ohms. When I put it in the first time it was a typing mistake. This is incorrect?

That's correct.

[formulajoe]

The resistor R3 is irrelevent as long as you dont need to assume finite open loop output impedance. In other words just ignore R3.

Your formular for G is wrong. It could be made correct however by "fixing up the brackets" in the formula. (A is the open loop gain).

Your initial solution of 1500 ohms is obviously incorrect because even if the open loop gain was infinite the value required would be 10000/50 = 200 ohms. So clearly the answer will be less than 200 ohms.


I dont undertstand where to go?

[formulajoe]

I used the equation Vo = V0/A - I1*R2, where I1 = (Vi - -(Vo/A))/R1. Solved it for R1 and got a value of roughly 2500. Is this correct?