# Why the capacitor doesn't see R1 and R2

Why does the time constant of the capacitor with ideal op amp, is independent from R1 and R2? If I put test voltage source, it clearly does depend on it,because V+ is function of R1 and R2. therfor, The output voltage is also function of R1 and R2, and therfore the current of the capacitor is function of them.

• Because the output of an ideal op-amp behaves just like a voltage-controlled voltage source.
– G36
May 25, 2020 at 21:53
• I understand this. But as much as I know, in cases like this u need to use test source instead the capacitor, and in this case Vt/It is not R May 25, 2020 at 21:58
• You put this test voltage where?
– G36
May 25, 2020 at 22:01
• Which time constant are you looking for, exactly, and for which transfer function? When I attempt to solve this using the Generalized Time- and Transfer Constant method of Hajimiri (2010) I obtain a first-order time constant associated with the capacitor, which is dependent on the ratio $\frac{R_1 + R_2}{R_1}$. Depending on where you put your input and output, the transfer constant associated with that capacitor might make it irrelevant, but you didn't specify what transfer function you're actually solving for. May 25, 2020 at 22:03
• The test voltage should replace the capacitor. May 25, 2020 at 22:06

## 1 Answer

The op amp together with R1 and R2 is an inverting Schmitt trigger. As long as V- is smaller than R1/(R1+R2) *VDC (VDC being the operating voltage), the output will be +VDC. As soon as this limit is reached, the output voltage will change to -VDC and will remain there until V- will drop below - R1/(R1+R2) *VDC.

In both states, the output of the op amp will be a nearly perfect voltage source and will charge or discharge the capacitor with a time constant of tau=RC.

The values of R1 and R2 determine the levels where the switching will occur. However, they have no influence on the time constant.

P. S.: The amplitude of the output signal does not depend on R1 or R2 but only on the operating voltage.