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Quick question regarding SLA battery charging

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jst3712

New Member
Hi. If I am charging an SLA battery using a constant voltage source and a resistor in series to limit the current (high wattage), how do I calculate the correct resistor value, assuming the battery voltage will not drop below 11.8 volts?
Somebody told me a while back that it's not R=E/I because the battery will never be charged from zero volts, but instead, more like > 11.5V. I need someone to confirm this so that the charging process doesn't take longer than it potentially should.

Let's just say I want to charge my 7.2Ah battery with 14.5V and maximum of 1A current (battery says 14.4-15.0V <2.1A initial current). R=E/I >>> 14.5/1A = 14.5 ohms (closest value 15 ohms). Is this correct or should I be using a slightly different formula?

Will switch go to float voltage (around 13.6V) automatically.

One more question: If battery says "Initial current < 2.1A", why do they call it "Initial current"? Why not "maximum charging current"?. Thanks.
 
Last edited:

ericgibbs

Well-Known Member
Most Helpful Member
hi,
The voltage difference between the charger voltage and the actual battery voltage would determine the series resistor value.

Say 14.5- 11.8v = 2.7V drop at 1 amp is 2.7R,, the wattage would be 2.7V *1A= 2.7W,,, say 5Watt rating.

The problem is that the 11.8V will rise as soon as the 1A flows into the battery, so the 1A will reduce.

Ideally you want an active current limiter.

The Initial current of 2.1A will not be the same as the maximum charge current,, you would not normally charge continuously at the maximum
 

jst3712

New Member
hi,
The voltage difference between the charger voltage and the actual battery voltage would determine the series resistor value.

Say 14.5- 11.8v = 2.7V drop at 1 amp is 2.7R,, the wattage would be 2.7V *1A= 2.7W,,, say 5Watt rating.

The problem is that the 11.8V will rise as soon as the 1A flows into the battery, so the 1A will reduce.

Ideally you want an active current limiter.

The Initial current of 2.1A will not be the same as the maximum charge current,, you would not normally charge continuously at the maximum

Hmm, interesting. Thanks for clarifying.
So my 15 ohm resistor is way too high. I have now connected a 5.6 ohm (the only other 5W resistor I have) and the voltage is rising a bit faster now. I assume current would be limited to around 0.5A with this value, but that's ok.

It's going to be really interesting though as to what will actually happen when the battery level falls that low (11.8-12.0V)... I'm scared the current will be too high when I first activate the charger :confused: but I trust you, haha. I'll just have to test it for myself.
 

MrAl

Well-Known Member
Most Helpful Member
Hi

The only way to do this with a resistor rather than an actual control circuit is to switch resistors as the charge progresses.

You would work the resistor value anyway you want but it would go something like this...

Lets say you have a 14.5v source and the battery voltage is 11.5 volts to start and it takes 600ma max.
The required resistor value is:
R=(Vs-Vb)/Imax=(14.5-11.5)/0.6=5 ohms
The power in the resistor is:
P=Imax*Imax*R=1.8 watts
so you need a 4 watt resistor (twice the power P).

So say you start with a 5 ohm 5 watt resistor. Now after a short while the voltage rises to 12.0 volts. The current is now 0.5 amps, so that's still close to 0.6 amps so maybe wait a while more. At some point the battery voltage rises to 12.5 volts, and that means now a current of 0.4 amps flows, so maybe you want to increase that again to 0.6 amps for the full max current. Now we have 14.5-12.5=2 volts so to get 0.6 amps again we need a new resistor values of:
R=2/0.6=3.33 ohms, and the power P is around 1.2 watts so you now need a resistor of value 3.33 ohms rated for about 2 watts or better.

Now disconnecting the old resistor and connecting the new resistor the current will jump up for a few seconds and then decrease back down to around 0.6 amps so now you allow it to charge for some time again and possibly later change the resistor again.

So that's the basic idea to using different resistor values to charge the battery. But there is one small catch. We assumed the source voltage was a constant 14.5 volts. If that is not the case then the charge current will still vary somewhat. It depends highly on how much the 14.5v source changes, and this can lead to dramatic changes in charge current once the resistor values get small.
For example, with a 14.5v source and 13.5v battery using a 1 ohm resistor we get 1 amp charge current, but if that 14.5v were to rise to 15.5 we'd get 2 amps charge current, which is much higher and could be too much for the battery. So some thought about how to regulate the source is a good idea in case it comes from an unregulated source (like a regular unregulated wall wart) to begin with.
 

Mr RB

Well-Known Member
You can put a resistor before the voltage regulator (I just said that in another battery charger thread :)) which will limit the peak current during initital charge but won't affect the voltage regulation during the rest of the charging process.
 

jst3712

New Member
Thanks for the replies guys... appreciate it.

I am not using a voltage regulator as such, but instead, a boost-buck DC-DC converter, and their main power source is a switchmode regulated adapter. The boost-buck is great because even if I connect a big'ish load to the supply rail (causing a voltage drop), it does not affect the output voltage of the boost-buck at all :) Whereas in my experience, an adjustable voltage regulator caused issues with the voltage connecting to the battery - but I don't realy want to get into that as it's past tense now.

In fact, I am planning on having 2 of these modules - one for 14.5V boost charge (when required), and one for 13.6V float charging (each with their own resistor and diode in series with their outputs. I will have a relay switching between the 2 modules, controlled by a micro-controller, which will also perform other tasks. For over-voltage protection for the battery, I will attach a crowbar circuit to intentionally blow a fuse located at the boost-buck inputs (the battery will have its own fuse as well).

MrAl: I like your suggestion with the different resistor values, but I think I will stick to just 1 resistor (per module) due to limitations with space that I will have on my PCB.

Another question!!!!! Once the battery gets full and it's time to switch to Float, is it ok to immediately supply the battery with the 13.6V (from 14.5V) - the other module - , or should I let the battery's voltage fall naturally a bit first, THEN apply the float voltage?
 

Mosaic

Well-Known Member
Slight OT ,but last week I hacked a Nokia Cell phone charger to act as a float supply for lead acid batts.

Basically there is a 4V Zener feeding the low side of the opto which drives the high side switching. Causes about 5V regulated.
Swapped in a 12V zener and I get 12.9 V reg. Placed a 1ohm resistor in the output leg to do a bit of current limiting since the supply is rated at 1000mA. Will put a 250mA fuse instead later.
All the stock output parts eg. capacitor, are rated 16V.
Works just great.
 

WTP Pepper

Active Member
Slight OT ,but last week I hacked a Nokia Cell phone charger to act as a float supply for lead acid batts.

Basically there is a 4V Zener feeding the low side of the opto which drives the high side switching. Causes about 5V regulated.
Swapped in a 12V zener and I get 12.9 V reg. Placed a 1ohm resistor in the output leg to do a bit of current limiting since the supply is rated at 1000mA. Will put a 250mA fuse instead later.
All the stock output parts eg. capacitor, are rated 16V.
Works just great.

12.9V will not fully charge a lead acid battery. I assume the 12.9V is measured off load, but it will prevent the lead acid from dying as it's above about 10V which is the lower voltage before you permantly wreck such a battery.
 

tvtech

Well-Known Member
Most Helpful Member
12.9V will not fully charge a lead acid battery. I assume the 12.9V is measured off load, but it will prevent the lead acid from dying as it's above about 10V which is the lower voltage before you permantly wreck such a battery.

Excellent post.

+1

Regards,
tvtech
 

jst3712

New Member
How about we go back to where we left off at #6 :D
Anyone have any concerns or suggestions with the proposed charging methodology? Let me know if you can't visualize it and I will get a schematic together.
Cheers.
 

tvtech

Well-Known Member
Most Helpful Member
Hi jst3712

A schematic would be good. Just so we are all on the same page :)

Regards,
tvtech
 

jst3712

New Member
hey.jpg
 

Mosaic

Well-Known Member
12.9V will not fully charge a lead acid battery. I assume the 12.9V is measured off load, but it will prevent the lead acid from dying as it's above about 10V which is the lower voltage before you permantly wreck such a battery.

Actually at below 12.5V the battery will sulphate and be permanently damaged, not 10V. Below 10.5V starter batteries are considered over discharged. That is the purpose of the $3 nokia float maintainer. A flat 12.9V device is not a charger. Also, i am in the tropics...which reduces the Lead acid voltage requirements. In cold countries the voltage will be different.
 

jst3712

New Member
You are actually quite rude.

Just post the schematic requested.....and that's it. You need to learn to be nice/decent to people to get help...Not just expect it.

Good luck with your circuit. I have no interest in this anymore.

tvtech

I didn't mean to be rude. I had a family emergency just as I attached the schematic, so I didn't have time to type anything.
Actually, I am very appreciative when it comes to online forums when I need help, so me appearing to be rude is your opinion - it's just the way it happened.
I do thank you for your assistance so far with this, and please understand that sometimes sh*t happens - to all of us. Thank you.
 
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