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Would the circuit above works?

I'm trying to build a self sustaining solar powered Arduino controlled plant watering system. Since I'm using an electronic valve and not a pump, I'll assume it wouldn't need a lot of power. Can the battery be used while being charged by the solar panel? Would the charger circuit use up the juice when it's not charging the battery? I will be using this charger.

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1 Answer 1

Can the battery be used while being charged by the solar panel?  

Yes, while some charging circuits that combine the batt. charger & power supply into one unit don't work to charge & use at the same time (but many also do), using a separate charger & power supply always does.

Would the charger circuit use up the juice when it's not charging the battery?  

The charger circuit will use up some power, even when not running. How much depends on the charger you use. Whether or not that's "too much" depends on:

  • how much your panel supplies (which also depends on how much light hits it)
  • the power usage of your boost converter to run the 'duino & the valve (even when neither of them are using much/any power)
  • how much power thw 'duino uses (which actually depends on the software code you use)
  • how much power is used by the valve, how often it's used, and how long it's "on" at one time.

All of those are things that you'll have to consjder when choosing your battery, PV panels, software & components.


Here's a quick diagram to help with your understanding of how the battery's internal resistance works in the circuit.
enter image description here

  • Whenever there's more current coming from the charger than what the boost converter uses, there's a "backup" of current, which raises the back-EMF above the battery's current internal voltage, so current goes through the load, and through the battery's internal resistance, in parallel.
  • Whenever there's not enough current from the charger, power/current comes out of the battery's internal voltage, through the battery's internal resistance, then through the load (boost converter). Thus putting internal resistance & load in series, and lowers the voltage the charger sees in the circuit, since it effectively hooks in between the two "voltage divider" resistances (ignoring the reactivity of the converter, for now). Picture the "voltage divider" property like this (same circuit, just drawn differently):
    enter image description here
  • \$\begingroup\$ I've just finished high school with minimal knowledge in electronics, so correct me if I'm wrong. Wouldn't the internal resistance of the batteries 'decrease' the voltage of it when being used, 'tricking' the charger that it has a lower voltage that it actually has, therefore overcharging the battery? \$\endgroup\$
    – Infrasonic
    Feb 25, 2016 at 12:35
  • \$\begingroup\$ @Infrasonic If enough current is being used to draw current out of the battery (which is the only time when the battery's internal resistance will be resucing the voltage), then the battery is being discharged, so overcharging is no concern. Basically, in that case the charger's not providing enough power for the immediate load, so the batt. helps out until the charger can 'catch up,' then the battery stops providing power & can start charging again. (sorry for any cell-phone typos) \$\endgroup\$ Feb 25, 2016 at 13:20
  • \$\begingroup\$ What if even with the current draw the charger is able to provide enough current into the battery (more than the current out)? My Arduino will be using a soil moisture sensor and water the plant when the soil is dry, I assume current will be constantly drawn from the battery to the Arduino as it's monitoring the soil non-stop. \$\endgroup\$
    – Infrasonic
    Feb 25, 2016 at 14:47
  • \$\begingroup\$ If the charger supplies more than the cjrrent being drawn, then the battery is being charged, so the battery's internal resistance shows up in your circuit as a series resistance with the battery's voltage source, with both kf those in parallel with the load...so the charger 'sees' the voltage of the total system as higher than the battery's current available output voltage. \$\endgroup\$ Feb 25, 2016 at 14:58
  • \$\begingroup\$ @Infrasonic see updated answer post for graphical explanarion (easier to describe with schematic than words) \$\endgroup\$ Feb 25, 2016 at 15:22

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