From what I understand, since Vs=1, the node connected to the (-) terminal of the opamp has to be at 1V as well. I also know that no current flows through th e opamp. I'm really in the dark on this, as all of my professor's material is of the basic principles and has nothing on the level of this homework. Any help would be appreciated...
Correct. Ideally, the summing junction (inverting input or '-' pin) will equal the voltage presented at the non-inverting input or '+' pin by virtue of negative feedback. If Vs = 1V then your gain should be no more than 2 if you can only source 200mA.
Using KCL at the (-) terminal, I got this: 1/R1 + (1-Vo)/R2 =0 ...Rewrote it to this... Vo = (R2/R1) +1 So I basically want to maximize R2 and minimize R1... But the output can only be 200mA, so (Vo-Vs)/R2 = 200? Subbing in for Vo and Vs gives... (((R2/R1) +1)-1)/R2 = 200 (R2/R1)/R2 = 200 1/R1=200 R1=.005 kOhm Therefore, since R1 + R2 = 40, R2 =39.995 So... Vo = (R2/R1) +1 = 8000 But the question asks for the gain... Is that just the ratio between Vo and Vs? So the answer would be 8000/1? I'd appreciate any further commenting or pointing out of errors in my work.
You can only source 200 mA through the Load resistor. That will tell you the value of Vo. You already know what Vs is and Vo can be calculated. Draw the circuit ... leave the values for R1 and R2 blank. Annotate on your drawing the known values ... Vs ... I load. You know the total for R1 and R2. You should be able to work this out. The gain is the ratio of Vo and Vs.
Ah, you're right. What in the world was I thinking? My mind was on a given input voltage and a given gain that would allow 200 mA through a 20 ohm resistor. Would have made sense if Vs = 2V. Let's see if I can redeem myself. Av = 4 if Vs = 1V Rg = 10k Rf = 30k