How to lower output voltage of DI-158 charger?
I built 5W CCCV charger per DI-158, it works just fine. But to use it as intended (charge battery) I need to lower CV output voltage to 3.6V to charge a single LiFePO4 cell. How to modify this circuit properly?
I can sure modify ratio of resistor divider to the FB pin and set output to 3.6V, but the CC portion gets "loose" - it's much wider than 10% now, and output collapses at below 1.8V (too soon). In the circuit description it says R5/R6 ratio sets both CC (e.g.power) and CV, but it does not explain about CC settings. Like if I halve both resistors, ratio and output voltage will stay the same, but how the circuit (and CC) is affected?
Bottom line - what are new values of components on DI-158 schematic to achieve 3.6V output CCCV charger? BTW I use planar transformer around E22 core.
Thank you,
Victor
Comments
So, how do I choose actual values (not ratio) for R5 and R6 (DI-158)?
There is nothing about it in the LNK616 spec other than
one sentence saying these resistors set CC and CV parameters.
I realize R5/R6 ratio defines output voltage in CV mode, but
how absolute values of R5 and R6 affect CC mode?
I have no clue how to choose them so the circuit is stable,
5W output power is maintained and current regulation in CC
mode is still within 10% as DI-158 describes.
I will reduce primary inductance by 3.6/5 times, but
should I change pri/sec turns ratio as well?
(BTW, since output voltage is now lower, I was able to get
higher output current for the same 5W power).
Can someone at PI explain these two points please?
Thank you, Victor
Please use PI-Expert design software to determine appropriate transfortmer design. This software is available under Design Support
Primary inductance must also be reduced by a factor of (3.8/5)