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I would like to determine the transfer function of the following circuit made of TL431. According to the different AN that I have read (and about this subject I really really thank Christophe BASSO, its work is so nice to read and the whole power electronic field become so simple to learn), the current mode controller should contain a pull resistor at the input of the compensation pin. I drawn the pull up resistor on the following schematic. Nevertheless for determining the transfer function, I need to know its value... Does anyone know how to determine its value ?

enter image description here

Here is the datasheet :

http://www.farnell.com/datasheets/1782047.pdf

Have a nice day !

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I need to know its value

You can estimate it from the data sheet: -

enter image description here

So, it has an open voltage of typically 5.5 volts and it has a shorting current of 1.2 mA. This means it typically has an internal pull-up resistor of 5.5/0.0012 = 4583 ohms.

This is a typical value and as you can see it might be 5.5/0.0022 (which equals 2500 ohms) or 6/0.0012 = 5000 ohms.

I'd probably assume it was around 3500 ohms.

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