Electrical – Lead Acid Battery Charger with Load in Parallel

batterieschargerpower electronicspower supply

I have a circuit with a sealed lead acid backup battery, which it charges automatically with a constant current of 350mA at ~15V.

I wish to connect an additional module to the circuit (gsm gateway) which draws between 50mA and 150mA, with peaks of 2A when sending an SMS or communicating through Wifi (periods of seconds), at 5v. DC-DC step down will be used.

That is: the charger would be providing 350mA (~5W) to the battery, and some of this current (~1W) would be diverted to the gateway. I assume the battery would automagically be providing energy for the short peaks (~10-12W). DC-DC step down efficiency will be higher than 75%.

1) Considering that most of the time the gateway will be resting in low power mode, what is the short and long term effect of connecting the gateway in parallel with the battery?

2) Is this potentially safe for the battery and charger?

3) Without power failures, would the battery still be mostly at 100% charge? I assume the duty cycle must be analysed.

4) From a system engineering point of view, would it be substantially better to have a >16v (higher than charger) power supply in parallel with the battery, protected by schottky diodes, or a power multiplexer? Battery recovery time would be better, but the number of components, power dissipation and space required would also be higher.

Best Answer

Just connect your load directly to the battery. The charger will provide 350 mA when the battery voltage is low enough.

Battery current = charger current - load current

Yes, it's really that simple. If the load is drawing 50 mA and the charger is putting out 350 mA, then the battery will be charged with 250 mA. If the load draws 350 mA, then the battery current will be zero. If the load draws 1 A, then the battery current will be -650 mA, meaning it will put out that much current.

The battery will be drained some whenever the device draws 2 A, but will then be recharged over a longer time by the charger when the device only draws 50 mA. You need to make sure of two things for this to work:

  1. The maximum high-current drain times the maximum time it needs to be sustained is well within the battery's capability. This guarantees that the battery can still deliver power at the end of a long high-current discharge event.

  2. The average load current is less than the average charge current capability. This guarantees that in the long run the battery will be charged.

Both these seem to be true from the limited description you ahve provided.

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