buck-boostswitchesvoltage
I am working with a power source of a range from 0-20V and I need to regulate the voltage to 12V. I plan to use a buck boost converter to stabilize the voltage, the problem is the input range is 5-32V. I need the power to not be lost when it's generating less than 5V.
What I am thinking of is multiplying the voltage if it is less than 5V but I don't know how to do that. I have looked into MOSFET switches but I don't understand how those work.
Allow me to make this circuit suggestion:
simulate this circuit – Schematic created using CircuitLab
For 1 V < Vin < 6 V the boost converter converts to a 6 V internal voltage and the LDO regulates that 6 V down to 5 V
For 6 V < Vin < 12 V the voltage at the Boost converter's output will "follow" the input voltage with some voltage drop due to a (Schottky) diode, so Vin = 7 V => Vmid = 6.5 V and Vin = 12 V => Vmid = 11.5 V.
Remember that boost converters have this basic circuit:
So when Vin is higher than the configured (regulated) output voltage, the output will follow the input voltage with a voltage drop from the coil's resistance and the diode.
The bicycle "dynamo" you have shown in the schematic is an alternator. (A dynamo is an electrical generator that creates direct current using a commutator.) This means that the voltage on the input to your buck-boost converter will be seeing half-wave rectified AC unless the capacitor is big enough to maintain voltage during the valleys. (Why is there no C value shown on the schematic?)
If you edit your question to show an accurate schematic of your setup we may be able to help further.
Best Answer
Assuming you want to get 620 mA @ 12 V as an end result ...that means approximately 7.5 W
At 1 V input and assuming ~80 % efficiency you need to supply 9.5 A from your source.
Not an impossible scenario but quite a challenge.
Re-visit your specs and work out exactly what you can provide from your source and you'll be able to figure out what can be done in the convertor and what power you might deliver.