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). 


  • 3
    \$\begingroup\$ Give there is no feedback, the opamp will drive to rails. \$\endgroup\$ – analogsystemsrf Apr 26 '19 at 4:59
  • \$\begingroup\$ Related: Op Amp operating conditions \$\endgroup\$ – The Photon Apr 26 '19 at 5:01
  • \$\begingroup\$ Are you really sure that the common node of R1 and R2 is to be grounded? I think, it makes more sense to ground the common node of Vin and R1. \$\endgroup\$ – LvW Apr 26 '19 at 8:12
  • \$\begingroup\$ Matt...don `t you think that moving the ground from the middle to the left node would enable negative feedback (voltage divider R1-R2) with an enormous influence on the gain properties? As shown, we have no feedback at all - and R2 acts as a simple load resistor without any influence!! \$\endgroup\$ – LvW Apr 26 '19 at 13:35
  • 1
    \$\begingroup\$ The problem description is very poor. The timely response is requested for an unknown function Vin(t). More than that, the frequency response Vout(s) is requested for an unknown gain function for the opamp A(s). Simply crazy! \$\endgroup\$ – LvW Apr 26 '19 at 19:03

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.

  • \$\begingroup\$ Ok, I see, but I know nothing about the opamp, not even the voltage inputs. \$\endgroup\$ – D. LaRocque Apr 26 '19 at 5:01
  • 2
    \$\begingroup\$ Then you don't have enough information to find \$v_{out}\$. \$\endgroup\$ – The Photon Apr 26 '19 at 5:01
  • \$\begingroup\$ I am afraid, the questioner has grounded the wrong node, dont you think so? \$\endgroup\$ – LvW Apr 26 '19 at 8:13
  • \$\begingroup\$ To me, the circuit as shown, makes absolutely no sense, Neither R1 nor R2 have any remarkable influence on the output-to-input characteristics. \$\endgroup\$ – LvW Apr 26 '19 at 13:38
  • 1
    \$\begingroup\$ @LvW, I agree the circuit given is probably not what OP intended. But it would make sense if they wanted a comparator, and wanted to include the output load and source resistance of the signal being tested in their model. \$\endgroup\$ – The Photon Apr 26 '19 at 15:53

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.


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