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PMP10834: PMP10834

Luca Testa1
Prodigy 45 points
Part Number: PMP10834
Other Parts Discussed in Thread: PMP10833, PMP20122

Hi,

I'm thinking about using the same circuit as the one presented in your board PMP10834 for my design that needs an AC/DC with 5V output (max 0.5A).

To save space and costs, since I don't need a high precision on the 5V (il will just be used as input for simple DC/DC), I'm thinking about using just a resistor divider for the feedback signal (similar to PMP10833).

My question is: the GND of component UCC28C45D is the voltage between the sense resistor and the inductor. Shall this same voltage be the "GND" for the resistor divider for the feedback voltage (like your design PMP10833) ?

If so, I don't understand how UCC28C45D can sense the correct voltage, since (at DC) the voltage at the output of the inductor is the same as the voltage at the input of the inductor (the GND of component UCC28C45D). The output of the voltage divider would give 0V instead of the correct voltage for UCC28C45D to work properly.

Where am I wrong ?

Thanks a lot for your help !

Luca.

  • 0
    TI__Expert 3640 points

    Hi Luca,

    Since the GND of UCC28C45D is at the switch node, the output voltage is only feeding to the feedback resistor divider when the freewheeling diode (D5) is conducting current. That voltage is holding up by the capacitor on the feedback resistor divider (C2). Therefore, it is a "filtered" output voltage that being sensed. So the voltage regulation won't be great.

    There is a similar way you can sense output voltage like I did in PMP10833 design but with better output regulation is to use UCC287xx PSR controller like the PMP20122 design. UCC287xx PSR controller has sample and hold circuit in the IC and only sense output voltage when freewheeling diode is conducting. Therefore, you don't need to add a filter capacitor with large capacitance at voltage feedback resistor divider like C2 in PMP10833 design and can still have better output voltage regulation.

    However, since you output voltage is low (5V), you will need to find a way to supply IC VDD. One option is using a coupled inductor like I did in PMP10834 design. 

    Regards,

    Sheng-Yang Yu