# Frequency compensation circuit

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}\$$?

• What part is constant current? – winny Apr 8 at 9:11
• 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). – Ray Athan Apr 8 at 9:26
• I can't see the the load and the current feedback from your constant current source. What you have depicted is a simple buffer. – Marko Buršič Apr 8 at 9:35
• the buffer is used to prevent loading of the sensor by the compensation circuit – Ray Athan Apr 8 at 9:38

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