1
\$\begingroup\$

I am designing a flyback converter based on the UC3844 chip. As a starting point, I went through an eval board that is given in the Microsemi website. The schematic of the design is as shown below -

flyback schematic.

The document is eval board doc Now, the isolation transformer has an auxillary winding, which can be used to power the IC.

If the auxxiliary winding s used to power the IC, why use R19 ? How is the auxilliary winding serving any purpose here ?

\$\endgroup\$
  • 1
    \$\begingroup\$ Easy: R19 provides power to start the IC up when it has not started switching yet. For the power from the aux. winding to be available the IC has to start switching first. Without R19 it is not getting any power ! You would then have a no chicken and no egg situation ;-) \$\endgroup\$ – Bimpelrekkie Mar 8 '16 at 17:48
  • \$\begingroup\$ and the diode D10 prevents the current flow from the main input into the auxilliary. \$\endgroup\$ – Board-Man Mar 8 '16 at 17:49
  • \$\begingroup\$ Indeed, without D10 the Vcc would be shorted to ground by the winding. Vcc would remain zero => IC not starting. But when the IC has started and the aux. winding can provide power, D10 also rectifies the AC coming out of the winding and makes DC from it, which is what the IC needs. \$\endgroup\$ – Bimpelrekkie Mar 8 '16 at 17:51
  • 2
    \$\begingroup\$ @FakeMoustache That looks very much like the the right answer in the wrong place to me! Care to copy it to the right place? \$\endgroup\$ – Neil_UK Mar 8 '16 at 18:02
  • \$\begingroup\$ But of course :-) \$\endgroup\$ – Bimpelrekkie Mar 8 '16 at 19:41
5
\$\begingroup\$

Easy:-) Resistor R19 provides power to start the IC up when it has not started switching yet. For the power from the auxiliary winding to be available the IC has to start switching first. Without R19 it is not getting any power ! You would then have a no chicken and no egg situation ;-)

Diode D10 prevents the Vcc (supplied by R19) to be shorted to ground by the auxiliary winding. If D10 was not there, Vcc would remain zero and the IC would not start.

But when the IC has started and the auxiliary winding can provide power, D10 also rectifies the AC coming out of the winding and makes DC from it, which is what the IC needs and full operation can commence.

\$\endgroup\$

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.