Question: [![enter image description here](https://i.stack.imgur.com/yfclx.jpg)](https://i.stack.imgur.com/yfclx.jpg)Hi
I have tried to extract the transfer function of the filter above, as you see it a high order filter.

I have combined the \$C\_1\$ in parallel with \$R\$ & \$C\_2\$ then make a voltage divider in order to add the inductor.
I always miss the capacitor or I don't get the same result. Where is my mistake?

[![enter image description here](https://i.stack.imgur.com/ZX4xA.png)](https://i.stack.imgur.com/ZX4xA.png)

Answer: 
![schematic](https://i.stack.imgur.com/UKoJ2.png)

[simulate this circuit](/plugins/schematics?image=http%3a%2f%2fi.stack.imgur.com%2fUKoJ2.png) – Schematic created using [CircuitLab](https://www.circuitlab.com/)

*Figure 1. Current limiting simplicity: add R2.*

Let's check your maths: from \$ P = I^2R \$ we can get \$ I\_{MAX} = \sqrt {\frac {P}{R}} = \sqrt {\frac {0.473}{10000}} = \sqrt {0.0000473} = 6.9~mA \$, as you correctly calculated. Let's be conservative and round down to 6 mA.

The maximum current for a visible LED will be in the case of a red one as red has the lowest forward voltage drop - typically about 1.8 V at a reasonable brightness. That leaves us with 1.5 V to drop across the series resistor at 6 mA. Mr. Ohm says \$ R = \frac {V}{I} = \frac {1.5}{6m} = 267~\Omega \$ which we will round up again to the nearest value of 270 Ω.

Simply pop this somewhere in the circuit to set the minimum resistance in the circuit to a safe value for the pot.

Thanks for the reference to my pot smoker.
You should place a series resistor to limit maximum current. Eventually could add an emitter follower to increase maximum current.