my future fellow Electrical engineers.
I can't figure out how one gets -500(b) as the Open-loop gain.
Using node analysis:
\$i_1 = \frac{v_- – v_{in}}{100 Ohms}\$
\$i_2 = \frac{v_- – v_{out}}{100k Ohms}\$
and then substituting in \$v_- = v_+ – \frac{v_{out}}{A}\$. As the open loop gain of an op amp is not infinite. However, using this method, I got a wildly different answer (assuming I solved for \$\frac{v_{out}}{v_{in}}\$ correctly).
Best Answer
The question is not asking for the open loop gain. The question is telling you the open loop gain is 1000. You are supposed to calculate the closed loop gain, given that the open loop gain is 1000.
Let's assume Vout is 1V. Then V- must be -0.001V (because of open-loop gain). Then the current through the 100k will be 1.001V/100k = 10.01uA. The same current flows through the 100 Ohm resistor. So Vin is -0.001V - 10.01uA * 100 Ohms = -0.002001V.
So the closed loop gain is 1 /(-0.002001), which is about -500.
In an ideal op-amp, the gain for this inverting configuration would be Gideal = -R2/R1 = -100k/100 = -1000.
There is also a general formula for op-amps when open-loop gain is not infinite. The formula is:
Gain, G = Gideal * ( A / (A + 1 + R2/R1))
Where R2 is the feedback resistor, R1 is the other resistor, A is the open-loop gain. This also holds true for non-inverting op-amps.
If that formula looks familiar, maybe you were supposed to memorize it for this class.