# Electronic – Op-Amp analysis without using op-amp gain

analysisbatteriescircuit analysisoperational-amplifier

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).


And I know that : $$V_-=V_+$$
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}.
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.