enter image description here

Hello All ,

In the Above circuit , TL431 is used to regulate the output voltage which is set by resistors RH1 and RL1. resistor RB is bias resistor to provide the current to regulate the output whenever output is not at rated value. the purpose of capacitor CF1 and RF1 is not getting understood. when Vout goes down will the CF1 provide the bias current to the TL431 ? whats the application of CF1 and RF1 here . kindly help me understand this .

  • 1
    \$\begingroup\$ CF1 is a capacitor so it blocks DC and therefore cannot supply bias (which is DC). My guess is that Cf1 and Rf1 are a frequency compensation network to provide a more direct path from output to input (of the TL431) to improve stability. It might be needed since R8 and the LED are in series with the output which is something the TL431 might not like. \$\endgroup\$ – Bimpelrekkie May 30 '18 at 13:57

Look at this simplified itnernal layout (from its datasheet):


You can see there's an opamp in there driving a transistor (NPN), whose collector is considered the output (the anode is grounded). This means that, by taking a quantity from the output (cathode), and bringing it to the input, you create a negative feedback (because the collector inverts the phase), so RF1 and CF1, together with RH1 and RL1, are there to provide a zero in the overall transfer function: \$f_Z=\frac{1}{2\pi C(R_{F1}+R_{H1}||R_{L1})}\$. That zero compensates for the loss of phase of the converter at around the switching frequency, ensuring it does not oscillate. Here's a quick proof:


The zero frequency is at \$\frac{1}{2\pi 1e-9[10000 + (1000||1000)]}\approx15.16\text{kHz}\$. Looking at the readings, you see the phase is 45o at around 13.2kHz, which is different than the math because of the gain-bandwidth product of the opamp, plus the parasitic capacitances of the transistor itself (both thrown in there for exemplification, only, no other reason), which influence the magnitude and the phase, both.

In short, the TL431, together with the feedback network, acts like an error amplifier, frequency compensated.

  • \$\begingroup\$ @a concerned citizen this is best explaination . thank you so much \$\endgroup\$ – Rohan May 31 '18 at 8:20
  • \$\begingroup\$ @Rohan You're welcome, but if you consider this to be the answer you're looking for, mark it down, so that others, in the future, might also know. In fact, that mark says "thank you", in itself. \$\endgroup\$ – a concerned citizen May 31 '18 at 8:45
  • \$\begingroup\$ Instead of a mark down, I think you mean upvote and/or "accept the answer" \$\endgroup\$ – Bimpelrekkie May 31 '18 at 17:40
  • \$\begingroup\$ @Bimpelrekkie Not upvote, "accept an answer", that's what I meant. Voting is only if you found it particularly useful, I think, so I do not mention it. \$\endgroup\$ – a concerned citizen May 31 '18 at 17:54
  • \$\begingroup\$ @ a concerned citizen where do i find that mark option? i am unable to locate it near the answer. \$\endgroup\$ – Rohan Jun 1 '18 at 8:10

The bigger picture is likely a flyback converter and regulation is done on the primary side but, output voltage sensing is done on the secondary side with the LED (part of an opto-isolator) passing control levels to the main switching converter circuit on the primary (unseen in the OP's picture).

Given the lack of a full schematic it might be reasonable to conclude that CF1 and RF1 help stabilize the control loop. They will tend to reduce the AC gain of the TL431 and this might be needed for stability in some circuits.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.