Op-Amp analysis without using op-amp gain

Question

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). 

Answer 1

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

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.

Answer 2

You can find an expression for your circuit with this equation:

$$V_{out} = A \left( V_{in} - V_B \right)$$

Lets say the open loop gain is 120dBV/V is also 10^12 which isn't uncommon for a real opamp to have.

So lets say Vin is 4V and V_B is 2V:

$$V_{out} = A \left( 4 - 2 \right) = 2V*10^{12} = 2 \times 10^{12} \ \mathrm{V} $$

Or lets say Vin is 1V and V_B is 4V:

$$V_{out} = A \left( 1 - 4 \right) = -3V*10^{12} = -3 \times 10^{12} \ \mathrm{V} $$

That is an insane amount of voltage, for an ideal op amp that can supply voltages like that the voltage would go to either positive or negative infinity. For a normal opamp, the voltage would hit the rails of the op amp, and guess what you've built: A comparator. A comparator compares two voltages and goes low or high depending on which one is higher.