BQ25505: Expected output

Dean,

You can set the desired output voltage with the VBAT_OV resistors. If your question is how much output current (ISTOR) can I get with VBAT_OV=VSTOR=4.2V and the inputs above, you can use a power balance to estimate: eff=Po/Pi=(VSTOR*ISTOR)/(0.8V*1.2mA) where eff is pulled from the datasheet efficiency curves 2 and 3 (~82%).

Dean,

The  battery charger uses a dc/dc boost converter to boost its input voltage to a higher output voltage.  Wikipedia has a good explanation of boost converters.  All dc/dc converters use input power, voltage and current, (VIN*IIN) to provide some level of output power (VOUT*IOUT), with some losses in the circuitry. Even with 100% efficiency (no losses), 100%=VOUT*IOUT/(VIN*IIN) you can see that if VOUT is higher than VIN, IOUT has to be smaller than IIN.  It is easier to estimate the output power using known efficiency than try to compute all the losses in the components, especially for a boost converter running in discontinuous mode (inductor current goes to zero at the end of each switching cycle).   For the bq25505, the output voltage is set using the VBAT_OV, since it is intended to charge a battery.  For a source with high output impedance, the charger has the MPPT feature, which limits the amount of current being pulled from the source in order to prevent collapsing the voltage it sees all the way to zero.  So, you can use the VOC_SAMP resistors to set the MPP point for your HiZ source (e.g. solar panel).  The internal PFET in between the boost converter VOUT (VSTOR) and VBAT is used to prevent over discharge of the battery, assuming the battery doesn't have an internal protector FET.