BU-405: Charging with a Power Supply

Learn how to charge a battery without a designated charger.

Batteries can be charged manually with a power supply featuring user-adjustable voltage and current limiting. I stress manual because charging needs the know-how and can never be left unattended; charge termination is not automated. Because of difficulties in detecting full charge with nickel-based batteries, I recommend charging only lead and lithium-based batteries manually.

Lead Acid

Before connecting the battery, calculate the charge voltage according to the number of cells in series, and then set the desired voltage and current limit. To charge a 12-volt lead acid battery (six cells) to a voltage limit of 2.40V, set the voltage to 14.40V (6 x 2.40). Select the charge current according to battery size. For lead acid, this is between 10 and 30 percent of the rated capacity. A 10Ah battery at 30 percent charges at about 3A; the percentage can be lower. An 80Ah starter battery may charge at 8A. (A 10 percent charge rate is equal to 0.1C.)

Observe the battery temperature, voltage and current during charge. Charge only at ambient temperatures in a well-ventilated room. Once the battery is fully charged and the current has dropped to 3 percent of the rated Ah, the charge is completed. Disconnect the charge. Also disconnect the charge after 16–24 hours if the current has bottomed out and cannot go lower; high self-discharge (soft electrical short) can prevent the battery from reaching the low saturation level. If you need float charge for operational readiness, lower the charge voltage to about 2.25V/cell.

You can also use the power supply to equalize a lead acid battery by setting the charge voltage 10 percent higher than recommended. The time in overcharge is critical and must be carefully observed. (See BU-404: What is Equalizing Charge.)

A power supply can also reverse sulfation. Set the charge voltage above the recommended level, adjust the current limiting to the lowest practical value and observe the battery voltage. A totally sulfated lead acid may draw very little current at first and as the sulfation layer dissolves, the current will gradually increase. Elevating the temperature and placing the battery on an ultrasound vibrator may also help in the process. If the battery does not accept a charge after 24 hours, restoration is unlikely. (See BU-804b: Sulfation and How to Prevent it.)


Lithium-ion charges similarly to lead acid and you can also use the power supply but exercise extra caution. Check the full charge voltage, which is commonly 4.20V/cell, and set the threshold accordingly. Make certain that none of the cells connected in series exceeds this voltage. (The protection circuit in a commercial pack does this.) Full charge is reached when the cell(s) reach 4.20V/cell voltage and the current drops to 3 percent of the rated current, or has bottomed out and cannot go down further. Once fully charged, disconnect the battery. Never allow a cell to dwell at 4.20V for more than a few hours. (See BU-409: Charging Lithium-ion.)

Please note that not all Li-ion batteries charge to the voltage threshold of 4.20V/cell. Lithium iron phosphate typically charges to the cut-off voltage of 3.65V/cell and lithium-titanate to 2.85V/cell. Some Energy Cells may accept 4.30V/cell and higher. It is important to observe these voltage limits. (See BU-205: Types of Lithium-ion.)

NiCd and NiMH

Charging nickel-based batteries with a power supply is challenging because the full-charge detection is rooted in a voltage signature that varies with the applied charge current. If you must charge NiCd and NiMH with a regulated power supply, use the temperature rise on a 0.3–1C rapid charge as an indication of full charge. When charging at a low current, estimate the level of remaining charge and calculate the charge time. An empty 2Ah NiMH will charge in about 3 hours at 750–1,000mA. The trickle charge, also known as maintenance charge, must be reduced to 0.05C. (See BU-407: Charging Nickel-cadmium; BU-408: Charging Nickel-metal-hydride.)

Last updated 2016-02-27

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The Battery and You
Batteries as Power Source


On June 30, 2011 at 10:19am
Niels-Erik Jensen wrote:

It is a bit surprising that you can charge a 2Ah(=2000mAh) battery using a 500 mA and a charging time of 3 hours.. 3h * 500mA = 1500 mAh.

The battery is heating up during charging so one would expect e.g. a 500 mA connected in e.g. 5 hours.

Please clarify. One answer could be that you should never discharge a 2Ah to more than 50% of full charge.

Please clarify. I have tried for 3 months to get an answer to this question and also how much energy (in%) is lost (to heating the battery) during decharging the battery

On July 6, 2011 at 11:06pm
Paul Peter wrote:

Give a brief idea of charging Lithium Yitrium based batteries

On July 7, 2011 at 1:20pm
Fredrick Stanley wrote:

can you send me information on how long should any cell phone charge fo

On July 9, 2011 at 2:24pm
rohit khatri wrote:

hi thanks a lot for the information you are providing.
i would like to know about dead cell and cell reversal or polarity reversal.

On August 13, 2011 at 8:56am
joel wrote:

Can you explain rapid pulse charging? I want to know how i can charge cell phones quickly in 10 mins.

On August 29, 2011 at 3:28pm
BWMichael wrote:

Joel, the faster the charge, the more damage you are doing to the battery

On March 21, 2012 at 6:50am
jamiedentims wrote:

hello again   - re your ” sailer post” thursday evening pm me and we will definatley get back to you   our systems are available now
jamie d

On June 26, 2012 at 4:53pm
michaelnalsim wrote:

well gerry it took me ages to find it here is there link
and details,ring them for advice , mention micky n recommened you

On February 22, 2013 at 9:10am
Nikhil wrote:

I want to pose a question here….. Plz help..
If I have 2 lead acid battery packs of 24V 26Ah each, and I want to charge them. Which one of the following techniques will take lesser time????

1) I connect them in series to make a 48V 26Ah unit, and charge this pack using a suitable charger for 48Volts.

2) I charge the 2 packs of 24V 26Ah simultaneously using two suitable chargers of 24 volts.

On June 14, 2013 at 6:41am
lead acid battery wrote:


On June 25, 2013 at 6:13am
Tango wrote:

Ok, I’m confused!
My wall charger supplies 5V to the phone’s microUSB port.
Is it :
a) the charger studying the battery voltage and current to regulate the delivered voltage and current, Or
b) the phone’s circuitry and charging algorithm that is instantaneously telling the charger to deliver x Volts and y Amps?
c) the phone’s circuitry receiving constant 5V from the charger and regulating the voltage and current internally according to battery condition?

Does the phone play any role here? or is it just the battery and the charger doing their thing?
If it’s the phone, I feel safe. If it’s the charger (which i doubt), then any third party charger should damage my battery as the charger is not made with my phone’s battery specs in mind. But this does not happen, third party chargers also work great.
That said, if all Li-ion batteries charge the same way (4.2V, 3% thresholds) then it’s not an issue.

If it’s the charger doing the thinking, then it’s regulating it’s voltage from zero to 4.2V during the charge cycle, whereas it is rated at 5V DC. And so is the microUSB port - rated 5V!

Soooo confused!

On July 21, 2013 at 11:20pm
Vinay wrote:

can we use two batteries in parralell

On October 18, 2013 at 2:09am
abhishek wrote:

how much current does a 12v 26 Ah battery can provide? and please ddo provide a suitable way to recharge it..

On October 26, 2013 at 5:21am
Chris Flores wrote:

How long should I charge the battery? And how should I know when the charging is complete with sealed lead acid?

On November 5, 2013 at 12:49am
Sam wrote:

Thank you for taking the time to write this article.

On November 19, 2013 at 3:32am

What is the total heat load of 12V 26AH x 32 no’s VRLA batteries connected to a 10KVA 3 phase UPS.

On December 25, 2013 at 7:46am
Imran wrote:

Plz tell me how increase battery time

On March 21, 2014 at 4:15pm
Stephen wrote:

This article has helped me understand better the basics of what “smart” chargers are doing.  My question is how a “smart” charger might determine the correct full charge point when it “knows” battery type (I tell it that), applied voltage, charger output current and time.  Since the switch to float voltage is dependent on amp-hr capacity of the battery and charge current, can a charger reliably determine battery capacity, therefore charge current with this limited information?  Thanks in advance for your assistance. I am concerned I may have purchased the wrong charger.

On March 30, 2014 at 4:32am
manu wrote:

I have a Lithium-Ion Rechargeable Battery Pack 11.1V 1500mAh. It contains 3 X Li-ion 3.7V 1500mAh cells.  Can I recharge the battery by directly connecting it to DC Power Supply. How ?

On April 22, 2014 at 10:25pm
Ruchita wrote:

I have lead acid battery of 4 V.While charging,How can I know whether battery is fully charged or not? Please help me It’s very important for me to know.

On May 21, 2014 at 11:04am
Bill Geldy wrote:

Can I replace 3 12v 17ah batteries (connected in parallel) with a deep cycle marine battery. If so, what size marine battery would you recommend?

On May 31, 2014 at 11:39pm
Suwan Khwakhali Shrestha wrote:

why my MPS 12v series battery going heat during charging?

On June 11, 2014 at 1:03am
ajay wrote:

12v adapter with battery charger pin - 20no’s
plz contact 9949341135,9912132004

On June 24, 2014 at 7:07am
Colin Williamson wrote:

Please advise at what voltage should my charger be set in order to charge 2 x 12v batteries connected in series

On July 10, 2014 at 12:38am
Andre Van den Wyngaert wrote:

@Ruchita: 4V lead acid battery you can charge with 4.8V to do a quick charge. Float charge is 4.6V and you can apply that voltage forever. If you apply 4.6V and the current drops to a low value, the battery is full.

On July 10, 2014 at 12:42am
Andre Van den Wyngaert wrote:

@Colin Williamson:
I assume the 12V batteries are lead based batteries. Each battery takes 13.8 float charge, or 14.5 quick charge (need to terminate manually in that case). So for 2 batteries in series, you need respectively 27.6V or 29V.

On August 21, 2014 at 5:44am
Diogo Saraiva wrote:

I didn’t understand.
Can I use my radioamateur source (RPS1230SWD) to charge my 12V 1.2AH lead-acid battery?
I can only set voltage… (and noise offset)
Can you help me please?
Thank you very much

On October 19, 2014 at 10:48pm
hashan wrote:

Thanks for provinding hese information.i want to know if we want to give current and charge runng car battery(12V). what are the current and voltage we should supply.

On October 20, 2014 at 1:25am
Andre Van den Wyngaert wrote:

@ Diogo Saraiva:
RPS1230SWD is perfect for charging 12V lead acid. Set it to 13.8V (is very easy: set the slide switch o 13.8V…) and connect the battery. + of the battery to the red connector, of course. Caution: if your battery is very empty, it might draw more current the first minute(s) than is good for it. So just for being sure, you can put a resistor between the charger and the battery. I would take 10 ohm approximately, with a rating of 2 watts or higher. This will limit the current to a safe value. So the charger’s black connector goes to the - of the battery, the charger’s red goes to the resistor. The other side of the resistor goes to the + of the battery. After having it connected for, let’s keep it safe, 15 minutes, you can bypass the resistor. You can also keep it in place; charging will take some more time then. You can leave the battery connected forever to 13.8V. It will stay full and ready for use. If you switch off the power supply, better remove the battery, because it might discharge slowly through the power supply.

On October 20, 2014 at 1:28am
Andre Van den Wyngaert wrote:

@hashan: what kind of power supply will you use? Most of the time, the power supply is in danger, not the battery…

On November 23, 2014 at 3:26pm
rod wrote:

Is there a portable automotive 12VDC battery charger that will also power small 12VDC loads or a few watts like a small air pump - while NOT being hooked to a battery?

Seems using a battery charger to make DC is not possible on “automatic” battery chargers because they “sense” a dead battery?  Any work around for this?

Thanks - Rod

On November 29, 2014 at 2:30pm
1989 wrote:

Do you need to use DC current to charge batteries or is AC current ok??

On December 2, 2014 at 9:53pm
Ramesh Subramani wrote:

i have two doubts..
1) Is the charging circuit is necessary to charge a battery of 12V from a DC DC converter which gives a boosted 15V from 5V input..? will it charge if i directly connect the output of converter to the battery itself..?
2) i have a 12V battery of capacity 10Ah..can i charge it using a 250mA current which is provided by the DC DC booster circuit..?

On December 3, 2014 at 1:48am
Andre Van den Wyngaert wrote:

@ Ramesh Subramani:
yes, you can charge a 12V battery with a DC/DC converter that upconverts 5V to a higher voltage. Normally you need 13.8V. But you will need to limit the current in a certain way. If the DC/DC converter is limited, it will work without any extras. But 15V is too high to leave the battery connected. AS soon as it reaches 14.5V you need to disconnect manually. Can’t you reduce the output voltage to 13.8V? 2 1N4001 diodes in series could do the job. Then you can leave the battery connected.
From you questioning I understand that you DC/DC can deliver 250mA. That is enough to charge the battery. But it will take some time, of course…

On December 3, 2014 at 2:10am
Andre Van den Wyngaert wrote:

No, it is not allowed to connect an AC source like a mains transformer directly to the battery. Depending on the size of the transformer or adapter, you will damage it, and the battery too. If you really need to use the AC voltage, you need a bridge rectifier between the adapter and the battery. And some current limiting, depending on the voltage of the AC source. After the bridge rectifier is pulsing DC, which is no problem.

On December 3, 2014 at 4:23am
Ramesh Subramani wrote:

thank you very much for your reply sir..
i have another doubt, as i read it in an article, the battery draws more current from the supply when it is empty and if it is full the current input to the battery is zero..if so then if i connect the DC DC converter directly which can deliver only 250mA, whether it will damage the converter.? or what happen to the current output from the DC DC converter..?

On December 4, 2014 at 2:00am
Andre Van den Wyngaert wrote:

@Ramesh Subramani:
that is exactly what I mean when I said you need some current limiting circuit. If you upconverter has no current limiting, it can be overloaded.
So there are 2 possibilities:
1. your upconverter is current limited. Then there is no problem: you can connect it directly to the battery (eventuelly through 2 diodes to lower the 15V to 13.8V, but it is better if you change the circuit so that the output voltage becomes 13.8V. Most of the time only one resistor need to change value to do this)
2. your upconverter is not current limited. You could add a small circuit to limit the current. Example: http://www.eeweb.com/blog/circuit_projects/additional-current-limiter-for-power-supply   (the leftmost circuit will do. Use 2.7 ohm in the emitter of the transistor to limit to 250mA)
Good luck!

On December 6, 2014 at 7:47pm
Ramesh Subramani wrote:

thank you sir..

On December 18, 2014 at 5:19pm
John Creed wrote:

I have a three wheeled electric scooter witht a 36v electrical system powered by 3 -12v SLA batteriues. I would like to replace the SLA batteries by makeing a 36v Lithium-Ion battery by soldering 30 - AA 1.2v batteries in series. I see no problem making the battery, what I don’t know is how to charge it. Can I use the lead acid charger that came with the scooter or do I need a 36v charger designed for Lithium-Ion batteries?
Thanks for any input,

On January 6, 2015 at 3:30am
Andre Van den Wyngaert wrote:

@ John Creed:
Are you sure your cells are 1.2V Lithium-Ion? Some well-known Chinese websites e.g. Alibaba state “1.2V Lithium cell” but when you read on, it appears to be Ni-MH or even Ni-Cd. Imho, 1.2V Li-ion does not exist, unless maybe someone placed some circuitry inside the battery to reduce the output voltage.
Anyway, for charging li-ion, you cannot use the lead battery charger anymore. Li-ion requires a very accurately regulated output voltage and current limiting.
Li-ion “native” voltage is 3.6V…3.8V So you could use 10 of those in series to get 36V.

On January 20, 2015 at 12:24pm
Samuel wrote:

I have a macrotel rechargeable 9v battery @290mah, and my charger delivers 15mah, and the charge time is eternal, so here is the matter, i don’t know if you have heard of those power supplies for guitar pedals, well anyway, this power supply delivers 9v at 100mah, so calculating it would take less than 3 hours to charge, so i want to know if i can do it, and if there are risks on doing it, hope you can help me, regards!

On January 21, 2015 at 1:25am
Andre Van den Wyngaert wrote:

@ Samuel:
9V 290mAh - I assume it is NiMH since Lithium has higher capacity. 9V NiMH is actually 8.4V (7x 1.2V cells). You need at least 1.5V…1.6V per cell to charge, so 7x1.5V= 10.5V. If your guitar pedal supply delivers 9V exactly, you cannot charge the battery with it. If you connect it directly, you will see some current but quickly it will lower down to almost nothing.
A second issue would be the max current allowable. Those 9V cells, unless it is a fast charge type, prefer 1/10 of the capacity as charge current, so that would be 29mA, for 15 hours.
And a third one: you can never connect a standard power supply directly to a NiMH battery, unless it has a good current limitation circuit, and you can terminate the charge manually.

On January 23, 2015 at 3:12pm
Samuel wrote:

Hey Andre, Thank you for your answer i really aprecciate it. So if you say that my option isn’t as much good, where can i get a charger that is more efficient?. I’m not from USA so i would really thank if you can give a web page or some dealers that ship worldwide and be trustful obviously, thank you again Andrew, Regards smile

On January 23, 2015 at 5:21pm
Glen Kinder wrote:

Hi - I have a 36v 1.6 amp.  scooter battery (3-12’s)  Would it do harm to charge them with a 36v -1.8 amp charger?

On January 26, 2015 at 2:12am
Andre Van den Wyngaert wrote:

@ Samuel:
you can find lots of chargers at Ebay but they are slow (13hours for 175mA so over 20 hours for your battery). One that might solve your problem is called “Intelligent Battery Charger for 1-4 PP3 8.4V (9V) NiCd/NiMH batteries. UK seller” It states it fills a 280mAh battery in 3 hours…

On January 26, 2015 at 2:23am
Andre Van den Wyngaert wrote:

@Glen Kinder:
in order to answer that question, some more detail would be welcome… e.g. battery type number, and charger type number. But I assume it is a scooter battery charger for the right battery chemistry. So imho the 12.5% more current will normally do no harm. But that’s only my opinion.

On March 25, 2015 at 10:21pm
zam wrote:

It helped me a lot but my question is that while we are discharging the lead acid battery we usually concern only with DC voltage level. So is it that AC voltage is not recommended?

On April 24, 2015 at 6:21am
Ian wrote:

Do 9v battery phone chargers work?

On April 24, 2015 at 11:19pm
khan g wrote:

I have a question. I have 2 x 12v both 175AH lead acid bettries connected in series. I want to charg them fully in one hour. Can any one help me to know what current and voltage will be appropiet for fast charging.

On July 23, 2015 at 8:16am
Shree wrote:

Dear sir
    I want to use a old HCL laptopbattrey in to an emergency light .
But how can i charge it with the help of my laptop charger rating 19 volt output ?..battery rating is 11.1 volt 4.4 ah ..
How can i reduced charger current value with help of rasistance or divode wich sufficient for charging the above battery..
Please help me sir ..

On July 24, 2015 at 12:10am
Andre Van den Wyngaert wrote:

to khan g:
So you have 24V 175Ah in total. Theoretically you need something like 28…30V 175Amps to charge it in one hour. (because 175Ah theoretically means it can deliver 175A one hour long, so you charge with the same) But before you even try that, I suggest to check carefully if it is allowed without cooking the batteries…
29 or 30V 175A means 5250 watts. That is quite a power supply!

On July 24, 2015 at 12:22am
Andre Van den Wyngaert wrote:

to Shree:
I’d say I need more info regarding the battery. Since it is 11.1V I guess it is Lithium-ion. And these are quite critical regarding charge voltage. If the voltage is too high, they might overheat and even explode.
Anyway, you will need to build a circuit that reduces and stabilizes the 19V input to the desired value, and that limits the current to a certain value, also depending on the battery type.
If your battery is 3x Li-ion “standard” cells in series, you will need 12.3V to be on the safe side and yet charge it sufficiently. Next question is: how much current can your charger deliver? Do you want slow and easy charge, or as fast as possible?
Best regards, Andre

On July 24, 2015 at 1:24am
shreekant wrote:

dear sir
    it is lithium ion battery having six cell ..and i want slow and easy charge .. what should i do ? how can i make this circuit to reduce voltage ..

thanks sir

On July 24, 2015 at 6:42am
Andre Van den Wyngaert wrote:

Hi shreekant,
I cannot think of a more simple circuit than this one:
LM317 is a very common and cheap linear regulator.
You will have to build 2 stages. First stage is the current limiter stage, see figure 26 in the datasheet. For R1 I would take 2.2ohm 2watt. That will give you 1.25/2.2=0.568A charge current. Input of this circuit is your 19V adapter.
Behind this stage comes the voltage regulator stage from figure 1. R1 is 240ohm 1/4w. For 12.3V output you need R2 to be 2120ohm. You can use a small trimmer resistor there, say 5K. Now you can trim the output voltage to exactly 12.3V (without battery connected) using a multimeter. That means 4.1V per cell and is safe. Now you can connect the battery. When empty, you voltage will drop a bit, since the current limiter stage lowers its output voltage to keep the current at 568mA. After a while, the voltage will be 12.3V and the current will become lower than 568mA, until it is near zero after some hours. You can keep the batteries connected. When you keep them connected with the input 19V disconnected, there will be a certain small discharge. But I guess this is not important, since you plan to use it as emergency supply and the load current will be a lot higher than this leakage.
One remark: since the regulators are linear (to keep things simple), they will become warm, certainly when the batteries are empty. So you need to screw the regulators onto a small cooling profile. It won’t be more than 5W in total.
Good luck!

On July 24, 2015 at 8:19am
Shreekant wrote:

Thank u very much sir ...
A last question i have also a solar penal out voltage is 12 volt wich is used in solar lantern .. can i used this power source to charge the same battery mentioned above .i .e 11.1 volt lithium ion battery 6 cell ,4.4 ah ...

Thanking u ..
You are so helpfull and awesome person sir
Thanks a lot .

On July 27, 2015 at 1:26am
Andre Van den Wyngaert wrote:

Hi Shreekant,
it won’t do any harm if you connect it instead of the 19V laptop supply. The regulators will limit the current and voltage. If there is something to limit, of course. I don’t know how much current your panel can deliver. And the linear regulators have some voltage drop, so the 12V might be a bit short. Also the efficiency of the circuit I described is not very high. There are switchmode circuits with chips that are specifically intended to find the optimal load point (max output power) for the solar panel. Of course these are a bit more complex to build.
If you have a small solar panel, you could try to limit the current to a much lower value, like 100mA or so. Charging will be slower of course, but it’s for free…

On August 12, 2015 at 9:16pm
Anthony Adverse wrote:

I to have a scooter battery pack at 24v, its 2x 12v 4.5AH SLAs, if these are connected in parallel, effectively 12v 9AH is it relatively safe to hook this up to a lead acid car charger, the one I have is fairly dumb.  I guess the question really is, will a lead acid charger work ok for sla batteries?



On August 13, 2015 at 2:28am
Andre Van den Wyngaert wrote:

Hi Anthony Adverse,
not knowing your charger I’d say it is a risk. You can connect the batteries, and they will charge for sure, but carefully measure the current and voltage. If the initial current is above 3Amp: stop immediately. Also measure no-load (“open”) voltage. 13…14V: OK. 14…15V: you will need to terminate manually once the batteries are full (meaning: charge current has dropped to less than 100mA or so). When higher than 15V: risk for damage. But this also depends on the impedance of the charger; if it drops voltage to a lot lower with just a small load, it could be usable.
Possible solutions to use this charger:
- If open voltage is too high: put diodes in series, like 1N4500…1N4508. Cost nothing. These are 3Amp diodes and drop 0.6V each (without current; drop is 0.8V with 3A current). Say your open voltage is 16V, then you can put 3 diodes in series. Gives you approx 14.2 open voltage, and will reduce the current.
- If open voltage is OK but initial charge current is too high: add a resistor in series. One ohm or a few ohm could do. Take 10W resistors so you won’t burn them. Or a 55W halogen car front light bulb could be used as series resistor as well.
- If both are too high: add resistor AND diodes in series until open voltage is near 14V and initial charge current is below 3A.
This will require some experimenting. But measuring the initial charge current is essential. Once it has dropped to a safe value, keep on monitoring the battery voltage. Once is has raised to 14.8…15V: stop charging. Good luck!

On September 9, 2015 at 7:30am
gerry wrote:

my union 12V1.2Ah sealed rechargeable mx12012 battery is dead, used for my intruder alarm system backup in case of power cuts.

please can you advise cheapest battery charger that couls recharge this battery

On September 10, 2015 at 1:24am
Andre Van den Wyngaert wrote:

Hi Gerry,
If “dead” means your voltage is very low, like below 9V, you need to throw away the battery and buy a new one. You cannot completely recover its capacity by any means.
If you want to try to get it alive anyway, connect a DC power supply to it. Before connecting + to + and - to -, set it to 13.8V and limit the current to 100mA approx. Leave it connected for a few days. But it will probably not draw any current at all.
I hope this answers your question. If your question is about replacing the intruder alarm charging circuit, that is a more difficult one. I looked a few minutes but couldn’t find any suitable chargers on the internet. You need 13.8V since the battery is connected to it all the time. And you need a current limiting of approx 200mA. Please see my reply to shreekant a few posts up. It describes a simple 2-stage charging circuit.
Good luck!

On September 11, 2015 at 12:36am
gerry mc wrote:

Thanks you Andre.

Yes the battery has just discharged due to a few mains failures over the last 10 years!
I did buy a new battery £23 but I thought if I could re-charhge the old one I would have one on stby. I will try what you recommend “Before connecting + to + and - to -, set it to 13.8V and limit the current to 100mA approx. Leave it connected for a few days”- one again thanks for your time and info.

Kind regards

On September 22, 2015 at 8:30pm

I need to know what voltage and amps rating does the charger need to be to charge the battery safely on My Son’s Samsung Galaxy Nexus mobile phone the battery is A Samsung Near Field Communication 3.7 V Li-ion 1850 mAh BATTERY?
these are the extra numbers in letters that are on the battery they may help you.
EB-L1D71VZ 1850 mAh VZW : SAM1(or it’s an I)515BATS DPP DC120725 PLZ reply back as soon as possible because tomorrow’s my son’s birthday he’s going to be 7 years old and I want to make sure he has a charger for it I picked up the phone at a pawn shop For him.Thank you for your time.

On September 23, 2015 at 1:05am
Andre Van den Wyngaert wrote:

@Dick Murray:
I quickly googled the Samsung galaxy nexus, and, like most phones, it has a micro-USB port to charge it. So it is easy: you can use any micro-USB phone charger. You can even connect it to a PC with a USB-to-micro-USB cable, and it will charge. The phone itself takes care of the charging process; you don’t have to worry about it. I would go for a 5V 1000mA charger. All micro-USB chargers are 5V. PC USB is 5V too. Finally there is some standardization! Some chargers are only 500mA, but most are 1000mA and even higher. Should cost less than 10 dollar…

On October 1, 2015 at 10:21am
Iyala Ore Nigel wrote:

Hi Andre,
I have a 48volts dc motor, which I operate with a (4 x 12volts 7.5Ah sealed lead battery) connected in series.

Can I use a battery eliminator of equivalent capacitor to charge the battery, and at the same time run the dc motor ?

What am trying to achieve is to keep the motor running when there is power outage and the battery eliminator to take over when power is restored.

Are there risks involved ? Do I stand to fry my batteries ? Current limiting resistor specification is welcomed since I can find it here in my locality.

Integrated circuits are scarce here and 48volts battery chargers are so expensive, which is why I am working with a battery eliminator.

Please advice me in an email. Thanks a million

On October 2, 2015 at 1:24am
Andre Van den Wyngaert wrote:

@ lyala Ore Nigel,
some questions first: what is the power of the motor (how much current)?
Which type of Battery eliminator (voltage and max current)
your email address if you want me to mail you grin
best regards,

On October 16, 2015 at 12:14pm
Iyala Ore Nigel wrote:

Hi Andre

The power of the motor is not specific, I bought it from a scrap store, the spec label was removed even before I purchased it from the scrap store.

The only thing I am sure about the dc motor is it can only operate with a 12 volts battery from UPS backups, which I connected 4 of same battery in series to achieve 48 volts at 7.5 Ah. And the desired speed was achieved

As for the battery eliminator, here is the link http://www.learningaboutelectronics.com/Articles/How-to-build-a-DC-power-supply.php to the site from where I got the schematic on how to build it to achieve about 53 volts dc.

My question is, can I charge the batteries and run the Motor from the output of the battery eliminator ? Will the batteries get over-charged ?

Any solution or idea will be appreciated.
E-mail is (phoniflake@yahoo.com)

On October 17, 2015 at 1:05am
Farrukh wrote:

I have a 12v and 35 amp lead acid generator’s battery. I want to charge it. Which charger is best and what its output(amperes an volts) values should be. May i use a 19V 5amp Laptop adapter to rcharge it?plz help

On October 20, 2015 at 2:07am
Andre Van den Wyngaert wrote:

@ lyala Ore Nigel:
theoretically yes, you can do that. You can connect the batteries to a voltage of 4 x 13.8V = 55.2V continuously. If that is no problem for the motor, go ahead! Problems that can occur are:
1. maybe the power supply cannot deliver enough current to charge the batteries + power the motor. Certainly when the batteries are empty that can be a problem. You can put a series resistor towards the batteries, so the motor will always run and less current goes to the batteries. Or you can build a small current limiter circuit (see one of my previous posts). Such a circuit limits current in one direction only; no current passes in the opposite direction. But you can add a diode. Charging goes through the limiter circuit; discharging (to the motor) goes through the diode. 
2. Your link shows a voltage regulator with LM317. LM317 cannot handle 55.2 volts; it will blow. You will need to find an alternative with higher max voltage.
3. If 55.2 it too high for the motor, you need to add some more circuitry to reduce that voltage to 48V. The batteries will keep their charge with 55.2V, but once loaded they will drop to 52V initially, and then go to 48V when almost empty.
Best regards,

On October 20, 2015 at 2:18am
Andre Van den Wyngaert wrote:

you can use the laptop supply, but you will need some circuitry to reduce the current and voltage. When you connect the power supply straight to the battery, it will draw over 5A, and the laptop supply will limit the current (hopefully - if it doesn’t, it will be damaged). At a certain point, the battery will be almost full, and the voltage goes above 14.5V. At that point you must take away the charger to avoid damage to the battery.
To make things more safe: see my post of july24. You can build a simple current limiter + voltage regulator.
If that is not possible, you can add a series resistor 2.2 ohm with at least 25W max dissipation. It will reduce the current. But the voltage restriction stays.

On October 22, 2015 at 3:12am
Thilak wrote:

I would like to charge my 12V Lead acid battery from AC-DC Converter.The output of AC-DC converter is adjustable anywhere between 13 and 16 V. Since my battery is used as backup in case of power failure,I am wondering whether I can charge with a float voltage of 13.6 V. I do not want to use any specific charging circuit. If I apply 14.4 Charging voltage directly to the battery continuosly,what will happen.


On October 23, 2015 at 12:44am
Andre Van den Wyngaert wrote:

Yes, you charge with a float voltage of 13.6V or a bit higher, 13.8V. If you are sure the voltage stays there and does not rise, you can keep it connected forever. No need to go to a higher voltage. 14.4…14.5V is for quicker charge, and can reduce sulfatation. But you cannot keep it connected; you must manually terminate the charge as soon as the current has dropped to a certain value. From that point on you can set you supply to 13.8V and keep it connected. If you keep it at 14.5V, the battery will get damaged.
The only concern in this story is the current. If your AC/DC converter is current limited, there won’t be any problem. It will limit the charge current at any time to the value you set. If your supply is not current limited, you will need to watch carefully what happens when you connect the battery (measure the current). You can set a lower initial voltage to reduce the current to a safe value. Or add some series resistor to reduce the current until the battery voltage is near 13.6V and you can bypass the resistor.
Best regards, Andre

On October 26, 2015 at 12:32am
Thilak wrote:

Thank you very much for the response. I have another question.If I charge at float voltage,will it reach 100% charge.I understand it will take a long time.But I am worrying about the self discharging rate.

On October 26, 2015 at 1:23am
Andre Van den Wyngaert wrote:

I suggest you read the article on this site about charging lead acid, BU-403. After many charge cycles, the battery could become less performing if you never apply the top charge at higher voltage…
In my opinion, it all depends on the use. If your battery is always (almost) full and never really discharged, the continuous 13.8V will be fine. If it is often discharged quite deep, you better use some charging circuit that takes care of the topping charge.

On October 29, 2015 at 11:39pm

i want to charge 12v /7ah lead acid battery. the output configuration is 54v dc/84v dc/94 vdc. Please suggest the charging technique for it

On October 30, 2015 at 2:37am
Andre Van den Wyngaert wrote:

@Richa Mishra: So you have 54V DC available, and want to charge one 12V lead battery?
I guess best way is to step down the voltage to a usable voltage. But your input voltage is quite high; most commercial regulators don’t like that. An old, good and easy one is LT1076HV (Farnell 2102591); needs only 8 external parts and you have a good buck converter with low power losses (80% efficiency). You can set the output voltage to e.g. 18V and then add the circuit like I described july 24. That circuit is linear, but since it only gets 18V input, it won’t dissipate too much heat.
Assuming you will limit the current to 1Amp with the described circuit (july24), you will draw 1A from the 18V stepped down voltage. The LT1076HV is a step-down switching regulator, so you will draw approx 400mA from the 54V source. The LT1076HV only accepts 60V at its input, so you cannot use the 84V or 94V. And the 60V max is for the HV version only; don’t use LT1076 but LT1076HV.
Easier solution? You can find DC/DC converters; input voltage max e.g. 140V, out 24V. Will work when you add the linear limiting circuit I described july24. But you will buy like a 100USD for that DC/DC converter… LT1076 + surrounding parts will be around 15 USD.
greetings, Andre

On October 30, 2015 at 2:54am

Thank you Andre.

i have only 230 volts ac supply. and i have to charge 12volts/ 7ah batteries which are in configuration of 54v/84v/94v dc.

So my idea was 230 volts———-step down transfromer(18 volts)—-rectification——-dc-dc converter(push-pull)(step up to give 94v dc)

please see if it is viable.Basically i need to charge 94 v dc maximum.(each battery of 12 v )


On October 30, 2015 at 3:17am
Andre Van den Wyngaert wrote:

Hey Risha,
your idea can work. But a step up from 18V to 94V that can deliver 1A (to get a reasonable charge time) is not cheap; it is a 100W you have to build. LT3796 (Linear technologies) could be a good solution. It is actually a high voltage LED driver, but since there are many parameters to set (voltage and current limiting) you can use it as a battery charger (LT also says this in the description). And you can come in with a higher voltage, like 48V.
Good luck, and be careful with those voltages!

On October 30, 2015 at 3:23am

thank you Andre,

i am planning to make push pull dc-dc converter on my own. do u have any suggestions on which would be other better option to choose from dc-dc converter to step up

On October 31, 2015 at 2:08pm
Thilak wrote:

Dear Andre,
I am float charging my 12V Lead Acid battery(security alarm system) with AC-DC Converter(13.8V) which has a current limitation of 1 ampere. In response to your answer ,I need some suggestion on selecting series resistor to limit current.I am planning to use 10 Ohm resistor by considering the battery voltage does not drop below 12V(at the worst case-10.8V ) at any condition.
Please correct me if I am wrong.

On November 4, 2015 at 3:19am
Andre Van den Wyngaert wrote:

@Thilak: If you use a 6…7Ah or comparable sealed lead acid accu, you don’t have to limit the 1Amp at all. That is, if you AC/DC converter has current limiting. If it hasn’t, you better limit the current indeed. An “empty” accu will show 12V approx. So when you apply 13.8V, the current is (13.8-12)/R where R is your resistor. With 10 ohm, your current will never be higher than 0.18A. If the battery drops way below 12V, current will be higher; but I assume you never want that to happen; it will damage the battery. But assume you allowed it to drop to 11V, which is the absolute lowest you can allow. If you connect it to 13.8V, I would aim for 1 amp to flow. That means you need 2.8 ohm; say 3.3 ohm. It could at that moment dissipate (2.8 x 2.8)/3.3 = 2.37W, so you pick a 3W or 5W resistor. but within a minute or so, you voltage will be above 12, and the current will drop to a lot lower value; at 12V battery voltage it will be 0.55 amp; at 13V voltage it becomes 242mA. Sounds perfect to me.
Best regards,

On November 8, 2015 at 12:53am
thilak wrote:

@Andy:Thank you very much for your detailed explanation.I am using 2.1Ah sealed Lead acid battery.I tested with 10 Ohms and applied the voltage of 13.8V after the fully charged battery discharges for an hour.The charging current measured was 166mA .After few hours ,I noticed that it was not at all charging and still consumes the same current.Then I reduced it to 5 Ohm resistor and now it looks like charging.(atleast there is a reduction in the charging current).Tthe datasheet mentions that the floating voltage appied would be 13.6(+/- 1%).If I use 10 ohm resistor,almost 1.7V drops across the resistor and you will have only 12.1V applied to the battery.So I reduced a resistor value to 5 ohm ,now it consumes 135mA and available voltage across the battery is 13.125.
I would like to know whether this applied voltage would be sufficient to charge the battery ?Else I prefer to go without resitor as suggested.Thanks once again.

On November 9, 2015 at 1:30am
Andre Van den Wyngaert wrote:

@thilak: A question: where exactly did you put the resistor? In series with the battery, or in series with the AC/DC adapter? If it is in series with the AC/DC adapter, the consumption of the alarm system itself will cause the voltage drop over the resistor. In that case, your alarm system is drawing the current and not the battery. If the voltage over the battery is only 13.125, that is quite low and it won’t fully charge. So maybe you better use no series resistor at all. Only when the battery is very empty, it could draw over 1 amp for a short time, which could damage your AC/DC adapter. But if it is an electronic adapter (not a transformer but and electronic switching circuit), chances are good that it is current limited and won’t get damaged.
As an alternative, you could use a resistor in series with the battery only. But at power failure, the battery won’t be able to power the alarm system then. So you should add a diode over the resistor. Charging is through the resistor only, discharging through diode + resistor in parallel. Diode 1N4001 will do.
Best regards,

On November 9, 2015 at 4:39am
Thilak wrote:

@Andy: I put the resistor in series with battery(not with adapter) and also added diode in parallel to create low impedance path when operated at battery power(connecting battery+resistor in parallel with load).
I have only tested with this combination and sees a drop of around 1V across the resistor and now the battery voltage is 12.8.In that case,either I have to get rid of resistor or slightly increasing the DC Output of the converter.

On November 9, 2015 at 5:17am
Andre Van den Wyngaert wrote:

if there is 1V across resistor, current is 1V/5R=0.2Amp. So the battery charges. If you give it enough time, I think the voltage will rise until finally it will reach 13.8V and current is zero…
This will fully charge the battery. But it will take time.

On November 13, 2015 at 9:51pm
K B D Prasadrao wrote:

can i charge Li ion 2500mah battery with cell phone charger with 6v 800mah output

On December 7, 2015 at 12:04pm
Edwin wrote:

Hi Andre, thanks for a very informative site!

I have 2 sealed lead acid battery (70AH each) linked by a VSR charged by the alternator when the engine is running. I want to install a 100w solar panel with either an mppt or solar charge controller.

Question - can I leave the solar to charge the batteries at the same time as the alternator is running? Do I have to turn one off?

Thanks for your response!
regards. Edwin

On December 9, 2015 at 2:42am
Andre Van den Wyngaert wrote:

Hi Edwin,
I did not make this site; I am only like everybody reading the posts and trying to answer where possible… I live in Belgium so even my English is not so good!
Your question is clear but I don’t have the answer; I’m sorry. I guess you want to keep the solar system connected to the 2nd battery? When the generator stops, the batteries are automatically disconnected from each other by the VSR. But you want the second battery to keep on charging from the solar panel. I think most solar panel regulators are one-direction so it is always impossible to feed energy from the battery or generator back into the solar regulator. So I think there is a good chance you can connect the solar to battery 2. To be sure, add an amp meter in series. If you have a meter that can only measure small currents, you can put a resistor in series, like 0.1 ohm 10 watts.
10 amp through it will give 1V across it, so you can measure with a simple multimeter in the DC volts setting. 1V means 10 amps. When it is dark and the generator runs, no current should flow towards the solar panel. When it is daylight and the generator is off, no current should go to the generator either.
Best regards,

On December 9, 2015 at 2:55am
Edwin wrote:

Hi Andre,
Thanks for your response! Will there be any conflict if the alternator is supplying 12v at the same time as the solar panel to battery 1 and 2?

You English is perfect! smile


On December 9, 2015 at 3:49am
Andre Van den Wyngaert wrote:

Hi Edwin,
I cannot answer your question without having all the details e.g. which regulator/charger, which solar charger, which VSR, ... That is why I suggested to measure the currents so you see what is happening. But if both the regulator and the solar charger do their job, and don’t accept current to flow back from the battery, I think there will be no conflicts. Depending on the internal circuitry, it is of course possible that e.g. the generator regulator thinks that the battery is full and stops charging, because the solar charger is charging the same battery. It all depends on the circuit. But I think it is quite harmless to test and check currents.

On December 9, 2015 at 3:11pm
Edwin Ling wrote:

Thanks Andre! I’ll check the current as advised.

On December 26, 2015 at 1:02am
Hassan wale m. wrote:

Pls I want to build charger for ( 4) 12volt battery. So I want to know the recommended diode that can be use that will charge all this battery very vast…  Mail me (hassanwale12@gmail.com)
  I will also like it if the picture for the diode is attached

On January 1, 2016 at 1:44pm
tahir wrote:

I have 12 v 17 amp max power supply output.  The problem is this power supply does not charge as my other charger 12 v 5 amp ( made by transformer ) . this transformer charger charged to bettery 3 hour but power supply charger not charged 15 hours too. Any body plz tell

On January 4, 2016 at 2:37am
Andre Van den Wyngaert wrote:

the problem is probably the voltage. Your other charger with transformer will have a higher output voltage. Althought it says “12V 5A” , the voltage will be higher when the load current is lower than 5A. Your lead battery needs 13.8V for a full charge. So you cannot charge it with 12V. Your new supply is 12V 17A, so the transformer is bigger and more stabilized. This means the voltage is not high enough…
Can you adjust the output voltage higher in some way? Is it a regulated power supply, or just a transformer with a rectifier?

On January 8, 2016 at 12:21am
phil byrne wrote:

I have a 36v Li-ion battery rated at 28aH. I am trying to design a simple system using a solar panel, controller etc to fit on a Bob trailer behind my bicycle so that I can charge a spare battery as I am riding. Can I use a 12v solar panel with a booster controller?
What precautions should I take to ensure the battery doesnt overcharge?
The mains charger I use has output of 36v at 4 amps
Thank you

On January 11, 2016 at 3:24am
Humbert .s. Dkhar wrote:

Which device can recharge the 12V7.5AH sealed rechargeable battery

On March 7, 2016 at 9:11am
Paul Hilling wrote:

Where do the charging control circuits reside for Li-Ion powered devices (e.g., cell phones, laptops, etc.)? Do they reside in the device itself or in the charger? Everything I have found thru searches discusses charging circuitry but never explains where same resides. I have read that one should be careful about using chargers that are not approved by the device mfgr but if the limiting circuitry resides in the device (i.e., that which prevents overcharging), my thought is that a charger with higher output capacity (e.g., 500ma vs. 3a @5v) might be the only consideration since a fast charge could be detrimental to some batteries.
What am I missing, if anything?

On March 8, 2016 at 5:15am
Andre Van den Wyngaert wrote:

@Paul Hilling:
the charging circuit is in the phone or laptop. Theoretically, you can connect any 5V power supply and the equipment will take care that the batteries are charged.
BUT… how can the charging circuit know how much current the 5V source (adapter, USB, car charger, ...) can supply?
A smartphone e.g. can draw over 1 amp from 5V to do its quick charge thing. But if you connected a 5V 0.5 amp max. charger, it will be overloaded. Also many USB ports will be overloaded with 1 amp.
So the manufacturers build in some recognition system over the USB lines. If the smartphone recognizes its charger, it will draw max current and charge the quickest way. But if it just gets 5V without the recognition, it will draw a lot less, a “safe” current.
You can find more info on http://www.obddiag.net/usb-power.html and other websites…
So actually you are free to connect a 10amp 5V charger - the smartphone won’t recognize it and will only draw the current it considers “safe” - which is often near 300mA. “10A” on a charger means it can maximum deliver 10A. If you draw more, it gets damaged. But your smartphone will never draw that much, so it’s safe to connect it.

On March 20, 2016 at 8:28pm
Zulkefli Aris wrote:

My 12v 7Ah battery current voltage is 2.7v. Can it still be charged?

On March 21, 2016 at 1:31am
Andre Van den Wyngaert wrote:

@Zulkefli Aris:
A 12V battery that has now 2.7V will probably be permanently damaged. What I do in such cases is to keep the battery connected to a power supply at 15V approx., current limit 1A. It probably won’t draw much current; maybe only a few mA. But when the battery is still usable, it will start increasing to higher values.
Some people say you can help the battery by giving it a high current pulse. You can try that if the current described above stays low. Put 30V DC or even more from your DC power supply onto the battery; current of the supply should be limited or you will damage the supply. First set the power supply, then connect it for a few seconds to the battery. Repeat that a few times. Then go back to step 1: 15V, limited to 1 amp. Once current is above a few 100 mA, lower your voltage to 14.5V and keep it there until the current is back below 100mA again, and the battery is full. But I’m afraid it will never reach its original capacity again.

On March 22, 2016 at 10:18am
Adam Mahmud wrote:

l’m given a question as : a circuit arrangement of 5 ohm standard resistor ,5 dry cells,key and voltmeter a ammeter. no rheostat, how will take frequents readings in order to plot a graph of I against V?

On March 23, 2016 at 1:30am
Andre Van den Wyngaert wrote:

@Adam Mahmud:
is this a quiz question or so grin ?
If this is all you have, you can measure at least 6 points of the V/I curve, and then connect the points with a line to get the complete V against I curve.
I assume all cells are identical.
I assume your resistor is exactly 5ohm. In that case you only have to measure voltage, not current. I = V / R and R is fixed 5 ohm.
Below I describe how to estimate the 6 points of the curve:
Point 1: all 5 cells in series; measure open voltage. Current = zero
Point 2: One cell only; connect voltmeter. Connect the resistor and read voltage. Current = read voltage / 5 ohm. Your point is this current vs the read voltage x 5 (since you actually want the result for 5 cells).
Point 3: take now 2 cells in series and connect the voltmeter. Connect the resistor and read the voltage. Current is this read voltage / 5 ohm. Point on the curve is this current vs the (readout voltage / 2) x5.
Point 4: same with 3 cells; of course V = (readout/3)x5
Point 5: same with 4 cells; V = (readout/4)x5
Point 6: same with all 5 cells. V = readout.
Connect the resistor as short time as possible; otherwise the test is not correct since the batteries will discharge too much.

On May 7, 2016 at 8:16am
stuart wrote:

I have charged batteries since the 1960’s. first were the ni-cads. In all charging, one cannot charge any battery with a charger of the same voltage. For example, a 12 volt auto battery; this will not charge properly or fully with a 12 volt charger! Look for a load voltage on a battery charger of 14—15 volts or higher. voltage greater than this is alright. Current is not!. Use a resistor that limits the current to a max of amp/hr rating of the cell or less. When is a a battery fully charged? When you notice the battery temperature is greater than ambient. But don’t get stupid in letting the temp. soar. 100 degress F is plenty. Remember, any battery is no more than a resistor and capacitor across the terminals. All you are really doing is a chemical reaction and when its over, heat is all that is left

On May 9, 2016 at 9:42am
Glenda E. Sia wrote:

May I use 19v 2.1ah (laptop) power supply to charge 12v 45ah car battery?

On May 10, 2016 at 12:49am
Andre Van den Wyngaert wrote:

@ Glenda E.Sia:
don’t connect it directly to the battery. The voltage is too high for the battery, and you will probably overload the laptop power supply. (When the power supply is current limited, it could work though; the power supply would limit its current and reduce its output voltage. But I would not count on that.)
1: If your plan is to give the battery some extra charge and then disconnect the charger, then you can connect a resistor in series with the battery and the charger. This will limit the current. Resistor value should be (19V-14V)/2Amp = 2.5 ohm. You can use a 3.3 ohm to be on the safe side for the power supply warm-up. Power of the resistor must be 10 watts at least and it will become very hot. With a 20W version it will be cooler. Or you can use 2 pcs 1.5 ohm in series, each 10W. As soon as the battery voltage reaches 14.5…14.8V, disconnect the charger.
2: if your plan is to keep the charger connected ‘forever’ to keep the battery full, you need some simple voltage regulator otherwise you will damage your battery over time; see one of my previous answers.

On May 23, 2016 at 12:02am
Ajinkya wrote:

Hi Andre,
My Electric vehicle is 48Vdc battery powered,
when running it is taking 6 Amp current,I have connected four (04) nos. of 12 V /50Ah batteries in series to get 48Vdc.
When found dischrged ,i checked the voltge of individual battery & found that first battery dischrged to almost 6 volts ,because it is drawing more current,
when connected to charge again ,it was not fully with my automatic charger,,,
Please help me in solving above problem,,,



On May 25, 2016 at 10:28pm
premanand wrote:

I live in a remote area in india,I charge my 48 v battery bank using a 24 v peddle generator,which gives 12-24 Ac current which is rectified to Dc and then converted to 64 v dc ,there is a mechanism to control the current ,the problem is

please help me

On May 26, 2016 at 8:49pm
sujan paudel wrote:

we can use two 12v battery at a parrel combination it can charge or not

On May 30, 2016 at 2:02am
Andre Van den Wyngaert wrote:

I think you experience a typical problem with series batteries. Actually all 12V batteries are 6 cells of 2V in series. So you have 24 cells of 2V in series. No problem. But when discharging too deep, the cell or battery that has the lowest voltage gets in trouble. It might even get reverse polarized during the discharge. When one of your 4 batteries behaves bad now, I think it is defective and should be replaced. To avoid this in the future, don’t discharge too deep. NONE of the batteries should go below 12V, to be on the safe side. Measuring a total of 48V is no certainty: the 3 stronger batteries can still have e.g. 12.5V, while the weakest one already has 10.5V… So better don’t keep on driving until you ‘feel’ the batteries er empty… Charge them as often as possible.
Kind regards, Andre

On May 30, 2016 at 2:08am
Andre Van den Wyngaert wrote:

1: since you have a current control mechanism, I think this is normal. The power supply automatically drops its output voltage to reduce the current to the set value.
2: I have no experience with this. Sounds like the load is suddenly increasing. You could monitor the load current of the alternator to make sure it is not overloaded. Are you sure it happens at 50V exactly? Could it be that the batteries are full and the control circuit decides to lower the current? Do you have details of the control circuit?
Good luck!

On May 30, 2016 at 2:12am
Andre Van den Wyngaert wrote:

@sujan paudel: you can always connect 2 identical batteries in parallel (+ to + and - to -). Voltage stays 12V. It will charge perfectly, but it will take 2x the time to charge if you use a current limited charger, since each battery gets only half the current.

On July 10, 2016 at 11:53pm
munawer wrote:

can i charge a 175 ampair battery with a 12 v DC supply

On August 6, 2016 at 7:39pm
Santosh wrote:

Can I charge 40Ah battery using 15Vdc/1.1A source. Is it recommended? What would be the charging time?

On August 7, 2016 at 1:46pm
sadiq Ali wrote:

I have power supply 13.8v 15amp additional recharges battery& back up battery.

We need to use that 1 To 2 hours with back up battery.
How many battery AMP we can use with our power supply?

On August 8, 2016 at 11:54pm
Andre Van den Wyngaert wrote:

@ munawer:
You need more than 12V to charge the battery. If you can put the power supply to 13.8V that would be perfect. It must be current limited, otherwise you can destroy the power supply. If you don’t have electronic current limitation in the power supply, you can put it to e.g. 15V and add a resistor in series. The value depends on the maximum current that your supply can deliver.

On August 9, 2016 at 12:03am
Andre Van den Wyngaert wrote:

Yes, you can grin
It will be a slow charge, but that won’t harm the battery.
Assuming this is a mains adapter you are talking about, it won’t have current limiting. So you will overload the adapter if you connect it directly to the battery. You need to put a resistor in series, meaning: negative - of power supply to - of battery. + of power supply to the resistor, and other side of the resistor to + of battery. Value of the resistor will be 3.3 ohm, 5 watts (or higher watts is ok too). Resistor will heat up but that is no problem.
Charging time: if you start with 12V on the battery (=as good as empty) it will take approx. 80 hours. But that is a quite wild guess. The current will decrease during the charge. When your battery voltage reaches 13.8V, you can consider it full.

On August 9, 2016 at 12:06am
Andre Van den Wyngaert wrote:

@sadiq Ali: sorry but that question is not completely clear; you will need to give some more detail: how is the setup? Do you have 2 batteries in parallel, connected to one charger? You want to know max load current with the charger connected?

On August 19, 2016 at 1:48pm
Luke wrote:

Hi guys I am looking to make a mobile sound system that runs an amplifier and speakers at 24 volts. So I am going to run two 12 volt batteries in series. My questions are if I wire in a 5.5 mm female connection could I charge the batteries with a 24 volt power cord that comes with the amplifier. Also should I only charge the batteries when they are disconnected from the circut. Finally the amplifier I dought runs at 19 to 24 volts if the power cable is 19 volts will that not be powerful enough to charge the batteries?

On August 24, 2016 at 12:22am
Andre Van den Wyngaert wrote:

you will need 2x 13.5 ... 13.8 = 27V ... 27.6V to charge the batteries. You can put that voltage onto the batteries and they will charge until full. That voltage won’t damage them, so you can keep them connected.
If you want your amplifier connected all the time, that is possible. The amplifier will work and in the mean time the batteries will charge. When there is a power cut, the amplifier will keep on working on the batteries. BUT your amplifier must allow the 27.6V, otherwise you will damage it.
If you put 19V on 2 batteries in series, they won’t charge.
If you put 24V on the 2 batteries in series, they won’t charge either, or just a little bit. The voltage is not high enough.
Possible solution: you keep your 2 batteries connected to a 27.6V power supply. You also connect your amplifier at the same time, but through a voltage reduction. How that will look will depend on your amplifier. Might simply be some diodes in series; 0.7V loss for each diode.
When you don’t use the charger but work from battery only, you can bypass or partially bypass the reduction circuit.
Keep in mind that a fully charged lead battery delivers 13V approx, so you have 26V when they are full. Can your amplifier work on 26V?  Make sure before connecting it!
If it doesn’t, you can keep the reduction in place, or partially (reduce the number of diodes).

On September 5, 2016 at 7:04am
A.Chandran wrote:

I had purchased one SLIME12VoltDC15Amps180Watts car/bike tyre inflator for using with 12 Volt Lead Acid battery of 6cells withan adaptablr usable thru car cigarette lither socketI want to use it indoor using domestic line voltage of 220volt 0f 3 -5 Amps for filling air in Bike or schooter etc without using 12 volt lead acid car battery.I have got separate Battery charger of 12 volt Dc for charging 12 volt car or bike Lead acid battery for emergency purposes.How can I use this SLIME inflator for inflating Bike&Cycle; tyres etc from inside my home.What will be the Amps of this BATTERY charger>To increase the Battery charger Amps   to 15 Amps or so .kindly advice simple procedure to follow easily.Plz send reply to my above E mail.Thankx(A.Chandran( what i have to add to my present battery charger(A,Chandran.Engineer)5-9-2016

On September 7, 2016 at 7:10am
A.Chandran wrote:

Regret to not that u have not sent me ther equired reply for my quiry dated Septr 5th 2016
  Plz answer correctly this time that I want to run&use; this SLIME 12 voltDc tyre inflator by my domestic line 230 Volts.It is mentioned in that slime instruction manual the specification as 12Volt 15AMps or 180 watts like that.Plz suggest what type of line adaptor I have to use for using this SLIME inflator-pump for inflating my Bike tube from home insted of connecting with 12 volt Lead acid Battery inside my car.Plz suggest directly instead of telling round about way as yu r claiming experts in Battery,Transformer,Charger etc usage experts by email reply as given early at the earliest. Thanks(A.Chandran)7-9-2016

On September 8, 2016 at 12:40pm
yomi wrote:

I don’t know more about batteries, I do have li-ion batteries I collect together 3.7v I want to make an external battery for my laptop so I arrange 11.1v,  3x in series and 4 more in parallel to increase the amperage,  the problem am facing here is the charging, I have AC/DC adaptor that read output 18.5v 3ah, how will I use this adaptor to charge this battery? and is their a way I can monitor the charge to avoid over charge or explosion

On September 12, 2016 at 2:21am
Andre Van den Wyngaert wrote:

to A.Chandran:
First of all, the replies here are from other users of the website, not from batteryuniversity.
So I am just a user, trying to give an answer. And I was on vacation last week grin.
If your inflator says 12V 15A 180W, you will need an adapter that can deliver 12V 15A. It is as simple as that indeed.
If you really want to do this, the best solution is to buy a power supply. Cheapest solution could be a Tracopower TOP200-112, or something comparable. It can deliver 12V 200Watts, and accepts anything between 85…264VAC at its input. Cost at Farnell approx 100 euro, but you will probably find it for cheaper. This is a PCB board with an input and output connector, nothing else. So for safety, you will have to build it into some enclosure. It won’t get very warm, certainly not for your short and intermittent use.

An alternative could be that you use your car battery charger. But it must be able to deliver 15A. If it can only do 5A or so, you cannot easily make it to deliver 15A.
Since it is a charger, it will deliver 14V or even some higher. Your compressor will get this voltage instead of 12V, so it will consume more power and run slightly faster. Also, many chargers don’t deliver pure DC but a pulsing (rectified) DC. A compressor motor won’t get harmed, but might behave a bit different and produce a different noise too. You could add an output capacitor in parallel with the voltage, to flatten the pulses a bit. 4700µF 16V will do. 

On September 12, 2016 at 2:41am
Andre Van den Wyngaert wrote:

to Yomi:
be careful with Li-ion batteries! Carging of a 3.7V cell is normally with 4.1V and you have to limit the charge current to a safe value. If you give more voltage per cell, or allow a lot higher current, they can explode.
I charge a 3.7V cell with 4.1V…4.15V at 1Amp max. The current stays a while at 1A, and then lowers. Once low enough, they are full.
With 3 cells in series, make sure they are identical cells. Otherwise they won’t get the same voltage during charge. But as long as you limit the charge current, there is no real danger.

This means you cannot connect your adapter directly to the batteries. Li-ion cells require a specific charger. In your case, you need 3x 4.1V = 12.3V. Must be quite exact.  And the current must be limited to the sum of all parallel branches. But check that the current goes to all branches equally, otherwise one might overheat while the others don’t charge at all.

On September 12, 2016 at 9:40pm
Yomi wrote:

thank u, but if I build 5 in series to make 18.5v and 3 parallel to make 7.8A can I charge it with a adaptor of 18.5v 3A? also if the batteries read below 2.5 can they still charge thank u.

On September 26, 2016 at 11:21am
Amal wrote:

sir I hav a 6v 400mah servo battery and is dead.when I checked with micrometer it showed 5.3v.if I want to charge it what type of charger I should buy.

1) a charger of 6v above and 400mah above
2) a charger below 6v and less than 400mah
3) a charger about 6v and below 400mah

also I have a doubt that is the battery charging pushing for voltage and pulling for current. so if I use a high mah charger will it damage the battery .also will the current of 400mah battery will increase beyond 400 mah if it is overcharged

On September 27, 2016 at 1:02am
Andre Van den Wyngaert wrote:

to Yomi:
5 in series need 20.5V to charge correctly. It won’t work with 18.5V adapter: they will not get a full charge.
4 in series need 16.4V so that could work. Put a resistor between your charger’s + lead and the + of the battery pack. Your charger can give 3A, am I right? Then put (18.5-16.4)/2A (2A instead of 3A to be safe) = 1 ohm. A 1 ohm 5W resistor will be fine.
While charging, monitor the voltage of each cell individually. It should not go above 4.15V per cell. And carefully check the cell temperatures. With so many batteries you are talking about, and only 2 amp of charging current, they should not warm up.
Measuring of total charging current: simply measure the voltage over the 1 ohm resistor. Since it is 1 ohm, the voltage over the resistor = the current through it. If you measure 2V, you have 2 amp. The 2 amp divides in the number of branches of batteries. 4 branches means each branch gets 0.5 amp. Should be safe for most Li cells.

On September 27, 2016 at 1:14am
Andre Van den Wyngaert wrote:

to Amal:
what you should buy depends on the battery type. 6V 400mAh is probably a small NiMH 5-cell battery pack.
A suitable charger will give a higher voltage than 6V so it can push current into the batteries. But that current must be limited, otherwise the batteries will get destroyed.
When they are empty, you actually should charge with 1/10C = 40mA, during 14 hours. That is the universal rule.
Best solution is to find a charger for 6V NiMH 400mAh. But that will be hard, maybe impossible. So you need another solution.
What I would do: use a simple 12V DC adapter; very cheap. Connect it to the battery pack: - of adapter to - of battery; + of adapter to a resistor; other side of the resistor to + of battery pack. So the resistor is in series with the battery pack. Resistor value is 120 ohm, 0,5W. Very cheap too. Plug the adapter in. Start counting 14 hours. Disconnect everything. Battery should be full!

On September 27, 2016 at 11:26pm
Andre Van den Wyngaert wrote:

to Yomi:
I forgot an important thing. The circuit I described (just a series resistor) can overcharge the cells and get them damaged. So, once the cell voltages reach 4.1…4.15V, you need to stop charging manually.
A dedicated Li-ion charger will do that for you.

On September 29, 2016 at 11:01am
Amal wrote:

sir what Is the difference between mah , kwh and mj rating. also why batteries are specified by mah not kwh. pls tell the difference between amps and mah. will a 3000mah battery provides 3000mA per hour.

I have a 14.7v 4600mah 64wh battery. when I charged it with a rps(regulated power supply) with 17v why it is not recharging? is there any charger to charge it

On October 12, 2016 at 8:40pm
ibrahim isam wrote:

can you send to me the electronic circuit diagram for rechargeable battery

On October 16, 2016 at 11:05pm
Mark jones wrote:

Hello, I am seeking a bulk charger for a 20s large capacity (64ah) ev lipo pack, I’m looking at an 80v 12.5a switching power supply to stay under the max 84v for the pack and get reasonable quick charging. I’m just not sure that the power supply is correct, will it stop charging once the packs reach 80v? Is there a chance it could charge to an over voltage point? I know the packs can handle the current, and I do have a dedicated 40a lipo charger for routine charging but I need to remove batteries from vehicle, and to be able to charge the batteries onboard all connected to reach a sub 4.2v per cell point would be a huge benefit… cheers

On October 17, 2016 at 1:17am
Andre Van den Wyngaert wrote:

to Amal:
mAh, read “miliamperehours” is mA x hours. If your battery is 4600mAh it means it can deliver e.g. 460mA for 10 hours, or 46mA for 100 hours, or 92mA for 50 hours, and so on. This is theory. For very high currents, the losses increase and the time will be lower than expected.
kWh is kW x hours. 1 kWh = 1000 Wh. It is a different way of expressing the capacity of the battery. 64Wh means it can deliver 64 watts for 1 hour. Watts is directly related to mA, since 1 watt = 1A x 1V. So for your battery, 64Wh / 14V = 4.57Ah = 4570mAh. I used 14V instead of 14.7 since the voltage will drop gradually during the discharge.
mj is millijoule, also a measure of the energy stored. 1 J (joule) is 1 Watt x 1 second. You can find direct conversion tables between kWh and mJ on the internet.

14.7V battery: what type of battery is this? Lead? Lithium? It should charge with 17V in both cases.

On October 17, 2016 at 1:24am
Andre Van den Wyngaert wrote:

to Ibrahim Isam: there are plenty of circuits on the internet; the actual circuit will depend on many parameters like battery type (PB, Li, NiMH, ...) and voltage, current allowed vs charge time, and so on.
One website is http://www.linear.com/products/Battery_Charger_IC
Another one is https://www.maximintegrated.com/en/products/power/battery-management.html
They both propose possible solutions based on your input. When you pick a solution with a certain part, they give reference designs.

On October 18, 2016 at 1:00am
Andre Van den Wyngaert wrote:

to Mark Jones:
I think 80V should be safe in your case. That means 4V per cell, so they will never get completely full, but they will get 75% full. This is actually better to prolong battery lifetime. See article http://batteryuniversity.com/learn/article/charging_lithium_ion_batteries
The 80V supply must have current limiting, of course.
While charging, the current limiter will limit the current to 12.5A or whatever you choose. When the pack reaches 80V, the current will start dropping gradually. The article I mentioned says: terminate the charge when the current has dropped to 3% of 64 in your case = 2Amp approx. You can test and see if this means you have to get up at night or leave it on until the morning…
I am only a reader of these posts and this is my personal opinion as electronics designer. Always be careful with Lithium packs! To be very safe, check temperature rise of the pack.