I identified the basic op amps in the figure, but don't know what type of op amp is the one circled in green. Could someone please help me finish this problem?
You may have mislabeled some components. The two components circled in green are resistors. The triangle is the op amp. It is probably an ideal one, so it has no real-world counterpart.
Yes, it's an ideal op amp. But how would I go about solving the one circled in green according to these equations?
The resistors circled in green form a simple voltage divider. So the voltage presented to the +ve op-amp terminal V+, is related to Vi by V+=(1/3)*Vi. You can assume this is the case because an ideal op-amp would have zero bias current into the input terminals. From the +ve terminal to the op-amp output terminal the circuit is a non-inverting stage with the output voltage given by (1+4/2)*V+. Between the op-amp output terminal and the output point (node Vo) there is another voltage divider with divider factor 5/10=0.5 This may help you to find the overall gain Vo/Vi.
You have three circuits there. V1 goes into voltage divider, result is V1'. V1' goes into non-inverting amplifier, result is Vo'. Vo' goes into voltage divider, result Vo. Do the math. I get ½.
I'm still lost on how 1/2 is the answer: For the first voltage divider on the left, I obtained: V1 = (1/3) Vi Then for the voltage output of the non-inverting op-amp, I got: V2=(1+ (4/2) ) * V1 = (3)*(1/3) Vi=Vi Then I obtained Vo with the following voltage division: Vo = [ 5 / (5+5) ] *V1 = 5/10 * Vi = (1/2) Vi So I divided Vo/V1 = [ (1/2)Vi] / [(1/3)Vi] = 3/2 What did I do wrong?
It's not clear what your terms Vi and V1 indicate. In any case I think you really want the relationship Vo/Vi where Vi is the left most input node. You already have Vo = [ 5 / (5+5) ] *V1 = 5/10 * Vi = (1/2) Vi So Vo/Vi=1/2
I want to give some general notices/ideas about op-amp without going in deep details in equations. So op-amp always tries to bring the voltage difference in its inputs to 0: prov: As ideal opamp has infinite gain, hence out =A(inpls - inmns), so as A is infinite and out is constant => (inpls - inmns) should equal to 0 => inpls = inmns Based on this principle you can understand all the above brought schemes. BTW: GOOD Question!!