0
\$\begingroup\$

I see reference design of Texas Instruments PMP40766 Universal input, 500-W CC/CV e-Bike charger reference design , In the design I don't understand the reason of diodes at the feedback circuit I share below from the Schematic — PMP40766 . And the 1uF capacitors at between shunt resistors. Can anyone explain the circuit. I guess the purpose of complete network is compensation for get the desired output feedback. But diodes? Thanks. enter image description here

\$\endgroup\$

1 Answer 1

0
\$\begingroup\$

In the design I don't understand the reason of diodes at the feedback circuit

They seem to be there for a softer transition from voltage regulation (CV) to current regulation (CC). They also play a role in a softer startup although the controller IC (UCC256403) has its dedicated softstart function.

The R87-C42 pair is normally a compensation network for the voltage loop. Likewise, C40 is a compensation component for the current loop. They will appear in the transfer functions for both loops so they will be effective on overall stability and dynamic behaviour (response) of the circuit.

And the 1uF capacitors at between shunt resistors.

C44 is there for high-frequency filtering: C39 and 41 are electrolytic capacitors therefore, they are not as effective as an MLCC in filtering out the higher-frequency noise components. So is C46. But I wouldn't place C46 there as the noise current will flow through the shunt resistor even though the sensed voltage is fed to a differential filter formed by R67-R70-C45 (fc~150 kHz).

\$\endgroup\$
2
  • \$\begingroup\$ I understand. I met with this design because of my digital controlled llc project I am currently at feedback design stage so I think it would be helpful for researching diffirent designs. \$\endgroup\$
    – MSB
    Commented May 5 at 15:25
  • \$\begingroup\$ Since this circuit control CC/CV stage with analog. I am planning to control it with MCU via Current and Battery Voltage feedback. \$\endgroup\$
    – MSB
    Commented May 5 at 15:27

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.