I'm trying to figure out how the voltage shunt regulator for my case is working. Below an extract of the electronic schematic of an isolated flyback board Ref NCP11187A65P45WGEVB from OnSemi which has two outputs voltage 12V and 16V. enter image description here

And it uses as shown a voltage shunt regulator NCP431BC. So I want to understand as the Vout(voltage output) is the 12V why the Vref (2.5V) for the NCP431 is picked from the 16V and the 12V ? Is there any explanation for that ? Below the complete schematic for the entire board: enter image description here

Thanks for your support.


1 Answer 1


The large resistor R209 to the "16V" rail makes this connection almost irrelevant. That means the "16V" rail is not regulated by the 431, its voltage could be much higher or lower without intervention.

The main voltage divider to set the output is R206 and R211+R212, which sets the 12V rail.

  • \$\begingroup\$ Thanks for your reply. However what is the purpose of connecting the 16V to the Vref. Otherwise if I disconnect the R209 resistor, is there any impact ? \$\endgroup\$
    – geek225
    Commented Sep 1, 2022 at 9:30
  • \$\begingroup\$ Very strange. 1.2M+680k will set the 16 V rail to 127 V for a 2.5 V reference voltage. \$\endgroup\$
    – winny
    Commented Sep 1, 2022 at 9:42
  • \$\begingroup\$ @geek225 It is indeed strange as winny writes. The capacitors on the 16V would give way long before the regulator would attempt to reduce the voltage. If the values are indeed as shown in the schematic, there will be basically no impact when disconnecting the 16V rail from the regulator. \$\endgroup\$
    – tobalt
    Commented Sep 1, 2022 at 10:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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