Hi Papabravo, is the SD6201 Boost Converter, datasheet: https://datasheet.lcsc.com/lcsc/1804250821_SHOUDING-SD6201-AF_C171633.pdfIt is how a boost converter works. The diode blocks the voltage on COUT from returning to the Lithium Cell so it continues to build up when the switch is open, and the inductor charges the capacitor. When the switch is closed, the capacitor can only discharge into the load. This process repeats each switching cycle.
You didn't say which part you were using in your example. The steady state behavior may be different from the transient behavior.
I believe that data sheet figure is in error and the diode should be there, since it is shown in two other circuit examples.In the Datasheet, you can see that the first diagram says typical aplication circuit and in that diagram there is not diode
The second paragraph makes a note about what I presume to be the diode in question. What's that about? Seems to be broken English..I believe that data sheet figure is in error and the diode should be there, since it is shown in two other circuit examples.
Why add a diode if the FETs do the work?If the part is indeed a synchronous boost regulator, then operation without the diode is possible. In such a regulator it is not uncommon to add a diode in parallel with the synchronous switch and its associated body diode. Nothing wrong with that
To reduce the switching losses.Why add a diode if the FETs do the work?
It is possible that with the external diode you have more range on the possible duty cycle values. Since the output voltage is inversely proportional to (1-D), a larger duty cycle results in a larger boost ratio. Without some kind of an internal block diagram or a simulation model it will be hard to tell for certain.So the diode is apparently needed only for an output greater than 4.3V.
It's not clear to me how the diode helps raise the output voltage.
This is achieved because the DC path formed by the diode reduces power dissipation of the flyback FET during it's OFF state? As well as some noise reduction.To reduce the switching losses.