2
\$\begingroup\$

I cannot se how the device is a buck charger when charging and a boost on the output. Am I stupid or is the datasheet wrong?

"5A Single Cell Li-Ion Switching Battery Charger with Direct Charge, Power Path Management and USB OTG Boost Mode " https://www.richtek.com/assets/product_file/RT9468/DS9468-01.pdf

enter image description here

enter image description here

enter image description here

\$\endgroup\$
3
\$\begingroup\$

Buck Regulator for Charging Battery.
= Stepdown from 5V to Battery (CC then 4.2V then cutoff)

Boost Regulator as OTG (on-the-go USB power)
= Step-Up from Battery to USB or 3~4V to 5V

It is consistent and correct.

I have redrawn the switches to show more clearly the topology used.

enter image description here

\$\endgroup\$
  • \$\begingroup\$ so the input("vbus") is the output in Boost OTG mode? \$\endgroup\$ – Anton Ingemarson Mar 27 at 16:32
  • \$\begingroup\$ Yes Vbus is the output in Boost mode. Thus the design needs a Full bridge to change Vbat current direction and a bridge to swap PWM from series (Buck) to shunt mode.(Boost) \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Mar 27 at 17:04
  • \$\begingroup\$ They don't really want you to understand how they designed it, just to use it. \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Mar 27 at 18:27
1
\$\begingroup\$

A synchronous buck converter, operating in continuous mode, will work just fine in reverse. In a lossless circuit, the PWM ratio plus the continuous conduction in the inductor enforces a voltage ratio from input to output, but lets current go either way. It's just a matter of how you arrange your control circuit.

\$\endgroup\$
  • \$\begingroup\$ So the input (Vbus) will be the output in boost mod? \$\endgroup\$ – Anton Ingemarson Mar 27 at 16:35
  • \$\begingroup\$ I assume so -- it'd pretty much have to be, and it would fit with the notion of having a box that either takes 5V in and charges a battery, or puts 5V out. The datasheet is a bit light on that sort of detail. \$\endgroup\$ – TimWescott Mar 27 at 18:25

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.