Limit current drain from 12v7Ah Sealed Lead Acid Battery

I have a 12V 7Ah SLA battery which i use with my DC Fan motor (not sure about its voltage and current requirements may be between 12v to 18v). I use it with both battery and wall charger (non branded). The charger has 18v 7A~ output. While running on battery, the Fan runs for about 50 minutes on fully charged battery before becoming dead slow. I think the motor draws full 7Amps from battery constantly.

I want to extend that time because of power outage in my area by limiting the current drain from battery. Is it possible without power wastage / dissipation? Also if i use two 12v 7Ah in parallel to increase capacity what will happen.

• We are taking a shot in the dark. You need to find the data on the fan, is this a small boxer type fan, a car blower fan, ???
– Gil
May 31, 2021 at 14:57
• its almost identical with this one daraz.pk/products/… Jun 1, 2021 at 7:56

Based on the information, I would guess the fan drains the battery at ~ 0.8C = 0.8 x 7 = 5~6 Amperes. The following graph was extracted and adapted from Yuasa, a traditional SLA battery manufacturer.

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The added line in orange would be the initial discharge, more like a plateau. Bellow this “12V” region, the battery voltage drops more steeply, marked in red.

The recommended cutting voltage for such discharge rates 0.6C to 1C (4A to 7A) is 9.5V, for a 100% Depth of Discharge (DoD). Going lower than that wears even more the battery, more than the stress imposed by 100% DoD, as it is being over discharged.

Using 2 batteries in parallel has 2 advantages:

1. Running time can be slightly better than double = expect close to 2hours.
2. Battery Life is dependent also of how severe (current intensity in %C) it is charged and discharged. It is expected that you will have longer life of the units, not only their running time. Obviously you have here 2x the original investment, but it seems to be better sized for the intended UPS expected.

Battery Protection, could be addressed briefly too:

1. Discharge - Using or creating a simple cut-off device (or relay) to stop discharging as 1C beyond 9.5V is highly recommended; for lower discharge rates the cutoff voltage should be higher.

2. Charge - Limiting your charging current can be made with sophisticated modules as CV/CC, a must for LEDs, but an overkill for a SLA Floating Charger. I think that a simpler current limiting device as an incandescent lamp is enough, if you limit the Floating Voltage to 13.8V. This cutoff can be made with a relay and a TL431-based circuit to sample battery voltage and switch the relay. Somehow the used of a lamp in series is aligned with other contributor here, @Kartman. I suggest you see this post where I did some experiments and plotted the resistance of a filament lamp (12V x 5W) there. But if you would be going to buy a dedicated Buck power supply module as those based on XL4015 or LM2596, buy it with CV and CC trimpot adjustments, adjust the Voltage to 13.8V and limit the current between 0.1 and 0.2C (0.7A to 1.4A).

Reducing Fan current consumption - Applicable for most Fans, Blowers and Centrifugal Pumps:

Imagine you could reduce the motor Speed (and Voltage) to 80%, the Flow would also reduce to 80% of nominal (unregulated).

Torque (and Current) required by the Fan would then be (RPM80%/RPM100%)^2 = 64% of “nominal”, here about 3~4A.

Power needed would be even less, just 0.8^3 = 51% of nominal.

This means that if someone reduces the Speed slightly to 80%, it would be the same as Doubling the battery runtime, as the power is just half of nominal. In this case you might not even need to use Two Batteries. The following graph exemplifies with actual Fan numbers what would happen if one reduces the power speed to this 80% of nominal (as 1700 rpm there). Compare arrows of Power (Bold Green) with Speed (Bold Blue):

So, if you can use a Switched Mode CV module (as that “5A” rated module XL4015, but with extra Heatsink glued and near the Fan), you could adjust the Fan’s voltage to about 9V~10V and reduce the current consumption, doubling (or more) the running time.

Don’t limit directly the motor current, just the voltage: Startup and Acceleration torques demand higher currents than steady-state speeds.

When the Powerline returns, this reduced-power mode could be bypassed (manually, by relay, etc.) to resume nominal fan speed, while recharging the battery without overcurrent and preserving your Power-supply/charger from overload.

• Thank you so much. You answered so many of my questions in single post. clearly you are a pro. I do have a XL4015 module for charging the battery. I am using main power supply (18v 7A) to run the fan and charge the battery at same time. XL4015 output voltage is set at 13.6V (no pot to adjust current). I was just cautious to protect my battery from damage by high current charge. Can you recommend me a circuit to cutoff battery supply at low voltage? Jun 25, 2021 at 4:37
• @Omaid_it, there are some “modules” on those famous sales sites. However, you could do a simple cutoff relay, using a conventional a “12V auxiliary relay” for car. They are robust, support up to 20A~30A, and could be driven by most bipolar transistor. Use a diode (in anti-parallel) to protect the transistor from inductive spikes. HOW: use a 5~8V Zener diode with 1K resistor on the base of NPN (BC337 or 2N5551 are OK), emitter to ground, collector to relay; zener side connected to a 1K Trimpot: extremes connected to Ground and Bat+, central wiper to the Zener adjusting voltage to cutoff.
– EJE
Aug 27, 2021 at 0:43

Your SLA battery probably wont like being charged from 18V at 7A.

If you use two batteries in parallel, then it will run for twice as long.

You could use a pwm motor speed controller between your battery and fan. If you don’t require the fan to run full speed, then this might be a good solution for you. Pwm doesn’t waste the energy. If you run at 50%, then your fan will run twice as long.

• i am using XL4015 PWM buck (step-down) DC/DC converter to adjust charging voltage for battery which is currently set at 13.8V. Will that PWM Controller works with two 7ah batteries connected in parallel? May 31, 2021 at 8:31
• The XL4015 power supply will need to be adjusted to 13.8V and the current limited by using a 20W car light bulb in series. Otherwise you might overcharge the batteries and they will die. Two 7Ah batteries in parallel should be ok. The pwm controller doesn't care how many batteries are in parallel. May 31, 2021 at 9:15
• I am using a battery cutoff circuit to protect the battery from over charging which is connected on positive wire between power supply and XL4015. When reaching on specific voltage, the battery cutoff terminates the connection to the XL4015 which as a result stop battery from overcharging. While voltage out of XL 4015 is set at 13.8V, not sure about the current output of XL4015 and how to limit it. BTW, i am thinking to replace XL4015 with this circuit i2.wp.com/www.hackatronic.com/wp-content/uploads/2020/12/… May 31, 2021 at 9:36
• That circuit doesn't do anything that you don't already have. As I suggested, 20W 12V light bulb in series. That will limit the current. Cheap and easy. May 31, 2021 at 11:19
• Thank you so much. one more question, can i combine all the negatives in a circuit together? and PWM is best solution for limiting current drain from battery? May 31, 2021 at 11:30