Hi. I'm curious if it would be proper to do something like in the attached pic. You'll notice that I have a AC wall wart basically in parallel with a battery (say six AA penlights), both nominally supplying 9V to V_Supply (minus a diode drop). If the wart is unplugged, then D1 should be back biased and no current should flow through D1 and the battery should provide 9V minus a diode drop to V_Supply. If both wart and battery are present and the battery is fully charged, then no current should flow, since D1 sees 9V on either side. If the battery has lost some charge and is less than 9V, then some current less (than 10mA) should flow into the battery from the wart, and the battery should provide V_Supply with some voltage minus a diode drop. So, the idea is to trickle charge the battery, which might or might not be a rechargeable type. I wouldn't think that a current less than 10mA even to a non-rechargeable battery would cause the battery to heat up, explode or leak and I've heard conflicting stories as to whether it's safe to charge a non-rechargeable. But assuming it's rechargeable(s), would that be a reasonable way to trickle charge the battery and a non-problematic circuit? Thanks.
1\$\begingroup\$ place D10 between the battery and R3 \$\endgroup\$– jsotolaSep 1, 2019 at 20:03
\$\begingroup\$ @jsotola I think your suggestion conflicts with intension of OP of how the circuit should work. Putting a diode there (depending on polarization) either prevents the battery to be recharged or it prevents to battery to source in case the wall wart is unplugged \$\endgroup\$– HuismanSep 1, 2019 at 21:15
With no more information than you have given I would have to say that trying to recharge your battery in this way is a bad idea. Recharging batteries is a tricky thing to do properly, and you need to know exactly what kind of battery you are charging.
The arrangement you propose would not supply more than about 8.3V to the battery, even at extremely small currents. A trickle charge of 1mA reduces the charging voltage to under 8V. Is that enough to charge it? No one can say.
How much current does your load require? Let's say the load needs 10mA. If the wall wart is providing that current then the drop across the resistor is 6V and the drop across the two resistors is about 1.4V. The voltage available at the load falls to less than 2V and the voltage at the battery falls under 3V. The wall wart isn't doing you any good at all in this scenario.
\$\begingroup\$ Ugh... My idea has been slain outright by a guru, a master of the art (which is always more difficult than it seems). Thanks for putting it to rest so soundly, Elliot. \$\endgroup\$ Sep 2, 2019 at 15:28