So I need help with the following exercise on op-amps (picture). I know how basic op-amps work but here I dont know how they got to the formula (circled in red). Do you need to use KVL or KCL ? I know that the gain on a normale inverting op-amp is V0=-Zf/Zi*Vi.
Ok so now I applied KCL at the two nodes and this is what I got (picture 2). But after rewriting it and replacing Vs by its value, it still isn't the same equation as the answer given, what did I do wrong ?
Electronic – Op-amp exercise: Two stage inverter
operational-amplifier
Related Solutions
There is an easier way to think about this problem, assuming ideal components. First, since the noninverting pin is grounded, the inverting pin is at virtual ground. This means:
\$V_{R1}=V_{in}\$.
\$\therefore I_{R1}=I_{R2\parallel R4}=I_{R3}\$
Calculate the voltage drop across \$R_{2\parallel 4}\$ in series with \$R_3\$, and you'll arrive at \$V_o\$ from there.
I don't know how to proceed with this question or where to start.
If there is (net) negative feedback, then you proceed by setting the voltage across the op-amp input terminals equal to zero:
$$v_+ = v_-$$
Note that with zero volts across the input terminals, the 2k resistor in parallel with the current source is irrelevant; there is zero volts across it so there is zero current through it. You may remove it from the circuit without changing the solution.
This should get you started.
@AlfredCentauri I still don't see the bottom loop, do you mean the loop v+ connected to VB then connected to the voltage source and then the resistor and finally VA. Is that considered a loop even with the op-amp? And when I do I still don't get your equation.
simulate this circuit – Schematic created using CircuitLab
This is the bottom-most loop and KVL clock-wise 'round the loop starting with the voltage across the 1k resistor is:
$$i_1 \cdot 1k\Omega -2V + V_B - V_A = 0 $$
rearranging yields
$$V_B = V_A - i_1 \cdot 1k\Omega + 2V$$
If the presence of the voltage source above is puzzling, recall that the output of the ideal op-amp is an ideal (controlled) voltage voltage source referenced to ground which I've shown explicitly here.
Best Answer
As Tony Stewart commented, the circuit as shown will not work. Vo will be saturated, and V1 will be 0.4 the level of V2, but with the opposite polarity. Two inverters in a row produce positive feedback, not negative.
As a starting point, let's say Vs is set at zero, and Vo is 1 volt. Then V1 will be be -.4, and Vo will integrate even higher. The process only stops when Vo saturates. This is classic positive feedback, and there will be no reliable response to an AC Vs.
In order to get the circuit to work, the + and - inputs of one of the op amps (preferably the second) must be swapped.