I was wondering… I have read about ideal op amp, that we could change this circuit :
Into this one (with the ideal model), where the input resistance Ri is infinite, and the output resistance negligible (so Ro = 0 Ohm) :
simulate this circuit – Schematic created using CircuitLab
We then told me that, since the Ri is infinite, the voltage at node A is zero (since Vi = 0), so Va = 0 (node voltage).
But what happens if we have that?
Will the voltage at A still be zero? Does it changes absolutely nothing to have Vs2 there? I'm confused on this one. I've seen only problems with the + gate of the op amp at ground.
Thanks!
Best Answer
To find the answer, write an equation for the node voltage \$v_A\$ using superposition. For simplicity, assume the input resistance \$R_I\$ is infinite.
$$v_A = v_S \frac{R_2}{R_1 + R_2} + v_O \frac{R_1}{R_1 + R_2}$$
We also have
$$v_I = v_{S2}-v_A $$
$$v_O = Av_I $$
Thus, the first equation becomes
$$v_A = v_S \frac{R_2}{R_1 + R_2} + A(v_{S2}-v_A) \frac{R_1}{R_1 + R_2}$$
Rearranging yields
$$v_A = \frac{v_SR_2 + Av_{S2}R_1}{(1 + A)R_1 + R_2}$$
This is the general expression for \$v_A\$. As the gain \$A\$ gets very large,
$$v_A \approx \frac{Av_{S2}R_1}{AR_1} = v_{S2}$$
In the ideal case where the gain is 'infinite', this expression is exact. This is why we say there is a virtual short between the two input terminals of an ideal op-amp with negative feedback - the negative feedback ensures the inverting input voltage is identical to the non-inverting input voltage.