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How to choose the bias resistors in TL431 and Optocoupler of a switching power?

Other Parts Discussed in Thread: TL431

I saw an article wrote:

Vo=15V

Let   R6=10k,  R5=(Vo/Vf-1)*R6=(12/2.5-1)*10=50K

Let   R1=470Ω,  If=3mA,  Vr1=If* R1=0.003*470=1.41V;  Vr3=Vr1+Vf=1.41+1.2=2.61V; 

Let   Ika =20mA,  Ir3=Ika-If=20-3=17,  R3= Vr3/ Ir3=2.61/17=153Ω; 
so   Vka=Vo’-Vr3=15.2-2.61=12.59V 

However, how can the diode current If be 3mA?

Also,  why is Vka 12.95V when Ika is 20mA?

What is the relationship between Vka and Ika?    Thank you!

  • There is no reason for IKA = 20mA to chose R3.
    R3 only needs to provide the minimum current for TL431

    If the output goes too high then the LED should be off. However the TL431 cathode draws up to a maximum of 1mA even when it would ideally be no current. The TL431 will draw more current as needed (up to >100mA [limited by resistors] or until VKA drop to 2V) to keep the loop in regulation

    For much better ESD protection, move R1 to PC817A cathode. With it in the original location TL431 is at risk for damage when arcing breaches occur at the opto- coupler.
  • Thank you. Now I also know how to calculate the current Ika in the stable state. It's determined by the current flowing out of the controller, or U1 in the picture.