I was reviewing a buck regulator circuit. Circuit was designed for 3.3 V output (Vout1)
simulate this circuit – Schematic created using CircuitLab
From this buck regulator output, a coupled inductor(L1), diode (D) capacitor (C) combination is used to generate an output Vout2. This Vout2 is used to power an LDO which will generate a 5 V output. Can anyone suggest how this L1-C-D combination works?
The chip is from TI and its data sheet can be found here: LMR23630
I am sure that it is a boosting circuit as the LDO output of Vout2 driven circuit is 5V.
But further how this is working, what is the design and what is the value of Vout2 is unknown for me.
Kindly advice.
Best Answer
It's a TI flybuck, a combination of buck and flyback converters by using the same core for both the buck and flyback.
Pros: simple way to get one more output and also isolation on the flynack output if needed (in this case it's not used due to shared ground).
Cons: big limitations on duty-cycle and poor cross-regulation, hence why there is an LDO downstream of the flyback output.
Vout2 will most likley vary with input voltage and to some extent the load on the buck output (3.3 V), but probably designed to stay above 5 V + the dropput of the LDO over the entire specified input voltage and output current range.
For details on how to design a flybuck converter, I would refer you to TI's guide here: AN-2292 Designing an Isolated Buck (Fly-Buck) Converter (Rev. C)