I have a question concerning this op-amp problem for which I need to find an expression for \$v_{out}(t)\$, where \$V_B\$ has a constant value:
simulate this circuit – Schematic created using CircuitLab
So far, I've found that we can describe \$v_{out}(t)\$ by the following equation :
$$v_{out}(t) = R_2 \times I_{R_2} $$
And I know that :
$$V^{-} = V^{+}$$.
Now, I have some problems to use \$R_1\$ in the analysis, since I know that the current passing through it is \$0 A \$.
Therefore, I tried to use another equation with the gain of the op-amp :
$$V_{out} = A \left( V_{in} – V_B \right)$$
But I don't think I can find an expression. Therefore, I wanted to know if it's possible to find an expression without using this last equation ?
Edit : Added the original problem. The text is the following :
Pour ce circuit VB est constante. Déterminer vout(t) en fonction de
vin(t) et des autres paramètres. Calculer sa transformée de Laplace
en fonction de Vin(s).
Best Answer
This is not true for your circuit.
This rule becomes true when you set up a circuit with negative feedback. Your circuit does not use negative feedback, so you can't use this rule.
In your circuit, \$v_{out}\$ will be approximately equal to the positive supply rail when \$v_{in}>V_B\$. \$v_{out}\$ will be approximately equal to the negative supply rail when \$v_{in}<V_B\$.
To know the output when \$v_{in}\$ is very close to \$V_B\$, you'd need to know the op-amp's gain and offset voltage. To analyze it for a quickly varying \$v_{in}(t)\$ you'd need to know more about the internal workings of the op-amp, its slew rate limit, etc.