If I place a rechargeable battery in series with an active power supply to boost the voltage, will the battery ever need to be replaced?
\$\begingroup\$
\$\endgroup\$
8
-
\$\begingroup\$ I doubt the opamp provides 5 A directly, so there is a significant part to this circuit I know nothing about. \$\endgroup\$– polwelMay 26, 2022 at 17:58
-
1\$\begingroup\$ It absolutely is important. I am 100% certain your problem can be solved in a simpler way. Not my -1 btw. \$\endgroup\$– polwelMay 26, 2022 at 18:10
-
3\$\begingroup\$ But yes, you'll eventually drain the battery. It will carry a current at least as large as the opamp's quiescent current, which is listed in the datasheet. \$\endgroup\$– polwelMay 26, 2022 at 18:12
-
1\$\begingroup\$ you haven't provided a schematic of your circuit, but since the purpose of the battery is to extend your available voltage, I have difficulty believing that "it is constantly being charged via the active supply". Nevertheless, whether it is constantly being charged, or more likely _dis_charged, your circuit will not work for long. Either the battery will be drained, or it will be over-charged, and be damaged, possibly starting a fire. \$\endgroup\$– Math Keeps Me BusyMay 26, 2022 at 18:25
-
1\$\begingroup\$ Charging the battery requires you to generate a voltage across the battery to force current into it. If you don't already have power supplies generating the voltage you need, how would you charge the battery? Or if you do have a power supply generating the voltage you need, why use the battery? Drawing a circuit diagram would probably help a lot here. \$\endgroup\$– user1850479May 26, 2022 at 18:28
|
Show 3 more comments
1 Answer
\$\begingroup\$
\$\endgroup\$
Here's the schematic.
The rechargeable battery is in 'series addition' with the power supply.
It is bound to get discharged.
Please note the polarities.
There is no possibility of the battery getting charged in this configuration.
