The solar panel is 15.4V 200mA. I'm using a 6P3T rotary switch to select between AAA, AA, and 9v batteries, and a 3PDT toggle switch to select between 2 and 3 batteries in AAA or AA mode. Energizer 8.4V 175mAh batteries will charge at 1C, for either 66 or 90 minutes, I haven't decided yet. If I choose 66 minutes, I'll have to use the rotary switch pole that selects battery holder (sw1F in my schematic) and remember not to fill more than 1 battery holder at a time, since PGM 2 and 3 can no longer be tied together. AAA batteries (I have EBL 800mAh) will charge for 4hr, 24min. AA batteries will charge from a LM317 constant current source set to 200mA (max317 maxes out at 4hr, 24min, unless anyone knows a way to disable the timer?). in addition to all this, I have a LM7805 regulator going to a usb power output, always wired on. What will happen if I try to charge batteries and charge with usb at the same time? Also, should I have a diode going to the MAX713 power rail? there's already one after the 2n6109, so the batteries won't discharge through the panel, right? the LM317 and LM7805 sections already have a 1n4001 in their supply. Thanks everyone for the answers. Sorry for the crowded schematic, I just learned KiCad. any other suggestions welcome.

 

I think your using a sledge hammer to crack a walnut. But do be it. Before you do any more, prototype what you have. Breadboard, perfboard anything. You can then test all your scenarios. And BTW you don't need as many steering diodes as you've put in. The regulators are single quadrant ( look that up)

 

Pretty sure U2 won’t work with just one resistor and Pin3 doesn’t appear to be connected.
Overall, it is not easy to follow. Found myself redrawing parts of the circuit to see how it works. No engineering manager would accept this schematic. You probably know how it should work but anyone else would give up.
You should have rules about drawing:
Inputs to the left
Outputs to the right
Lines not crossing
All connections have a “dot”
Etc

 

To use the LM317 (U2) in constant current mode, you need to connect the batteries to be charged to the other side of R6. The LM317 holds a constant voltage (1.25V) between the OUT pin and the ADJ pin. By taking the output at the ADJ pin, you get a current of I = 1.25V/R6.