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schematic

simulate this circuit – Schematic created using CircuitLab

The diagram above is part of my constant current circuit which function as frequency compensation circuit. I was given the time constant from square wave test which is \$7.5\mathrm{ms}\$. How do I find the values of \$R\$ and \$C\$ if I want to achieve time constant of \$0.05\mathrm{ms}\$?

the linear differential equation of the circuit would be: $$ RC\cdot\frac{\mathrm{d}(E_{in}-E_{out})}{\mathrm{d}t} - (E_{in}-E_{out}) = E_{out}\cdot\frac{R}{R_c} $$ Note: None of the voltage input or output is given.

So far I have tried letting \$E_{in}=0\$ and get the equation by integrating both sides, then do the same thing by letting \$E_{out}=0\$. So I am stuck with two exponential equations. I am lost as to where do I start with, do I just assign a value for \$E_{in}\$?

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  • \$\begingroup\$ What part is constant current? \$\endgroup\$ – winny Apr 8 at 9:11
  • \$\begingroup\$ Ein is connected to another part of the circuit which has resistor that act like sensor. I have updated the diagram but the part that matters is the frequency compensation circuit (RC). \$\endgroup\$ – Ray Athan Apr 8 at 9:26
  • \$\begingroup\$ I can't see the the load and the current feedback from your constant current source. What you have depicted is a simple buffer. \$\endgroup\$ – Marko Buršič Apr 8 at 9:35
  • \$\begingroup\$ the buffer is used to prevent loading of the sensor by the compensation circuit \$\endgroup\$ – Ray Athan Apr 8 at 9:38
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the buffer is used to prevent loading of the sensor by the compensation circuit

The simplest LPF is a RC, output of opamp goes through the resistor and capacitor. $$\tau=RC$$

schematic

simulate this circuit – Schematic created using CircuitLab

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