I'm trying to understand this material from C.Basso regarding TL431 compensation (Designing Compensators for the Control of Switching Power Supplies

If you refer to pg.110, there is one example regarding TL431 based on type 2 compensator.


From the example, it stated that it require to provide 15dB at 5kHz with 50 degree phase boost. Basically, I understand all the calculation & phase boost operation using type 2 compensator.

However when it comes to pg.111, it mentioned that with Cpole = 2nF (this is based on pole value in optocoupler @ 4kHz) & C2 = 581pF, the bandwidth cannot be reached, therefore crossover frequency need to be reduced.



  1. From the slide above, how they know that the bandwidth cannot be reach with these Cpole (2nF) & C2(581pF) value? Any calculation example would help.
  2. If bandwidth cannot be reached, what will happen to the system? Any example case will be help, because it’s hard for me to imagine the impact of bandwidth to power supply system.
  • 1
    \$\begingroup\$ @VerbalKint might be able to help (for reasons that may become obvious). \$\endgroup\$
    – Andy aka
    Mar 13, 2020 at 10:34

1 Answer 1

  1. The output pole of this compensation network (type 2) is set by the output capacitor of the optocopler + the pullup resistor of the primary's side controller. First he calculates the output capacitor in order to set a pole @ fp, and for this case it would require a capacitor C_calc = 581pF connected to the output of the optocopler. However, the optocopler already has an output capacitor (2nF), meaning that he would have to add a "negative" capacitor (Cout = C_calc - 2nF) in order to make it work. As you know it is not possible, therefore, he selects a (smaller) frequency (C ~ 1/(2pi R f_p)) which leads to a required output capacitance bigger or equal the already existent 2nF.

  2. The bandwidth determines how fast your circuit responds to a change in the output. Upon modifying the output voltage, the feedback and the controller need some time to read it out and check what to do in order to achieve proper regulation. How fast your circuit process it, is determined by the bandwidth. Failing to reach a reasonable bandwidth can lead to instability problems. e.g.:

Too small bandwidth (slow feedback): let's say you connect a highly capacitive load. The output voltage will drop a lot, but will take time for the circuit to respond and increase the output voltage. No

Too big bandwidth (fast feedback): You circuit might want to regulate very small variations of the output voltage caused by noise. This can also lead to instability, as the loop control will find itself trapped in this always out-of-regulation loop.

  • 1
    \$\begingroup\$ Oui, being the author of the seminar, I agree with vtolentino explanations. Either you select a crossover frequency from which you have a certain (positive) value for the added capacitor at the feedback pin or the optocoupler pole is already too low and there is not much you can do. In this case, you work backwards: with that opto capacitance (augmented by 100 pF for noise purposes) what crossover frequency is possible? Another trick with the TL431 and its fast lane, there is a minimum gain the system delivers, naturally narrowing the choice for the crossover point. \$\endgroup\$ Mar 13, 2020 at 11:59
  • \$\begingroup\$ Hi @vtolentino, thanks for your clear explanation! Now I'm clear about that. \$\endgroup\$
    – Lutz Fi
    Mar 17, 2020 at 2:31

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