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question about L6920 step-up operation

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Hi,

I have not seen how the reference voltage which affects the feedback voltage is created.
But for me it seems it has to be created by some kind of charge pump/ switched-capacitor topology.

greets
 
Hi,

it is a step up circuit.
So even if the input is 1V or less the internal circuit will work.


two possible solutions:
1) It may use an internal voltage divider to make 1.23V on the feedback pin to be lower than 0.6V .. for internal processing

2) it may start switching below 1V of input voltage -- this switching generates a higher ouput voltage .. and this ouput voltage may be used to drive the feedback circuit. (more likely to me)

Klaus
 
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