USB 3.7V Li-Ion Battery Charger

In this article we study a simple usb 3.7V li-ion battery charger circuit with auto-cut off, current control features.

How it Works

The circuit can be understood with the help of the following description:

The IC LM358 is configured as a comparator. The IC LM741 is not used since it is not specified to work with voltages lower than 4.5V.

Pin#2 which is the inverting input of the IC is used as the sensing pin and is attached with a preset for the required adjustments and setting.

Pin#3 which is the non-inverting input of the opamps is reference at 3V by clamping it with a 3V zener diode.

A couple of LEDs can be seen wired across the output pin of the opamp, for detecting and indicating the charging condition of the circuit. Green LED indicates the battery is being charged while the red illuminates as soon as the battery is fully charged, and supply is cut off to the battery.

How to Charge using USB Port

Please remember that the charging process can be quite slow and may take many hours, because the current from USB of a computer is normally very low and may range between 200mA to 500mA depending on which number port is used for the purpose.

Once the circuit is assembled and set up, the below shown design can be used for charging any spare Li-Ion Battery through the USB port.

First connect the battery across the indicated points, and then plug in the USB connector with your computer's USB socket. The green LED should instant become ON indicating the battery is being charged.

You can attach a voltmeter across the battery to monitor its charging, and check whether the circuit cuts off the supply correctly or not at the specified limit.


USB 3.7V Li-IOn Charger Circuit



Video Clip showing the automatic cut off action, when the Li-Ion cell is charged upto 4.11V:



Please note that the circuit will not initiate charging unless a battery is connected prior to power switch ON, therefore please connect the battery first before connecting it to the USB port

An LM358 has two opamps which means one opamp is wasted here and remains unused, therefore LM321 may be tried instead to avoid the presence of an idle unused opamp.

How to Set up the above USB Li-ion Charger Circuit:

That's extremely easy to implement.

  1. Apply an exact 4.2V across the supply terminals of the circuit from a variable supply source, and play with the 10k preset until the RED LED just lights up and the Green LED shuts off

  2. If you find the Green LED initially ON, then force it to shut off by moving the preset randomly, and subsequently perform and complete the above procedure.

  3. That's all, now seal the preset in this position with some suitable glue.

  4. The setting up procedure is now complete.

Constant Current CC Feature Added

As can be seen , a constant current feature has been added by integrating the BC547 stage with base of the main BJT.

Here the Rx resistor determines the current sensing resistor, and in case the maximum current limit is reached, the potential drop developed across this resistor quickly triggers the BC547, which grounds the base of the driver BJT, shutting down its conduction and charging of the battery.

Now, this action keeps oscillating at the current limit threshold, enabling the required constant current, CC controlled charging for the connected Li-ion battery.

Current Limiting not Required for USB Power

Although a current limiting facility is shown, this may not be required when the circuit is used with an USB since the USB already is quite low with current and adding a limiter may be useless.

The current limiter should be used only when the source current is substantially high, such as from a solar anel or from another battery

Please Note: 

This article was substantially changed recently and therefore the discussions in the comment may not match with the circuit diagram shown in this present updated design and explanation.

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Comments

J-boy said…
Please sir, i have three questions about the above circuit. When this circuit is charging the cellphone battery, will the LED be flashing? Or is it only when the battery is full that it starts flashing? Secondly, what is the required input voltage and current of this circuit? And finally, what is its output voltage and current?
Swagatam said…
Hello J-boy,

As mentioned in the article the LED will be lit continuously (solid) while the cell is being charged and will start flashing when the cell is fully charged.

Input should be 5V fixed, you can use a LM317 circuit at the input for allowing higher voltages.

You can remove the 22k resistor entirely for getting a good flashing effect at the full charge threshold.

output voltage will be around 4.4V, current 200mA max
J-boy said…
Thanks alot sir. I am grateful. Please sir i sent you an email at "hitman2008@live.in"
Swagatam said…
thanks j-boy,
sharing phone no wouldn't be possible, I think this is what you had requested in the mail.

We can chat here freely.
Jideofor Igwe said…
Ok. No problems. Please sir, i have a nokia charger(China made charger) that can produce 5.5v and 800ma. Can i use this charger to charge a 6v, 4.5Ah rechargeable battery? If yes, for how many hours can it charge it at a full rate?
Swagatam said…
J-boy, no you cannot charge with a lower voltage source, you will need a 7V input for charging a 6V batt.
Jideofor Igwe said…
What if i connect a Dickson charge pump voltage doubler at the output of the charger so that it can produce two times the voltage(ie 11v) and then i will use a voltage divider network to reduce the 11v to 7v, will it work normal? Or better still i will use LM317T to reduce the 11v to 7v
hello everybody

I have 12v 7.2A battery and I wanna charge my phone via usb port without using computer

I have an usb cable and want to use it in this project

how can I charge my phone directly from this battery please ??
Swagatam said…
It should be able to produce the required charging current also, which should be about 1/10th of the battery AH, then it could work
Swagatam said…
use a 7805 IC for dropping the 12V to 5V, then feed the 5v to the cellphone via the above circuit
Swagatam said…
..the 7805 will require a heatsink.
Jideofor Igwe said…
Ok. Thank you sir! Let me try and see what the current will be like
Rohith N said…
Hi sir,
the circuit is a reliable one ie charges to 100%? No exceptions in i/p current ? i/p v is 5.... can be used for any ampereage li ion batt?
Rohith N said…
As i/p v is fixed to 5v i hope i can use 7805s ckt with just ic , diode, 2 caps without current limiting Resistor?? as that of DC to DC charger & provide it to this..??
Rohith N said…
Also iam having two types of batt 3.7v & 3.6v (digi cam) li ion. can i charge ? mA may be different for both.
sir i have sent a mail regarding the substitutes for C & R in emergency lamp, havent got any reply... plz check...
Swagatam said…
Hi Rohith,
any amperage can be used provided the input also is capable of supplying that much of current and the resistor is correctly calculated.
Swagatam said…
if the li-ion is within 1ah and 2 ah a 7805 can be used and without the current limiting stage.
Swagatam said…
you can charge any battery by using the above circuit, but the current limiting should be set appropriately.

Which emergency light are you referring to, is it from this blog?
Rohith N said…
Sir not from this blog, iam having a 6v 4ah emergency , in which the capacitor was gone firstly (2.2mF 250v) & i replaced, the result was an explosion ,by mistake i took an electrolytic one.... 8ohm 5w R was burned too. now i got a non electrolytic one(0.24mFJ 800v 1200v.dc 50khz) instead of 8ohm R i got 10 ohm 5w can i use it safely....???
Swagatam said…
yes 0.24uF/800v will work but will not charge the 6v/4ah battery. 10 ohm will do for 8 ohms

by the way i have answered your email.
Jideofor Igwe said…
Please sir, i could'nt get the uA741 8-pin op-amp. Can i replace it with LM358 or LM324 by using one of its op-amp train, since, the 358 is a dual op-amp, while 324 is quad nand op-amp?
Rohith N said…
so what can i opt for?? sir y it doesnt charge ? plz explain.....
Swagatam said…
Yes you can do it, no issues.
Swagatam said…
use a 0-6V transformer power supply with a bridge rectifier and 2200uF/25V filter capacitor, capacitive power supply will not work
Rohith N said…
sir i have charged that way, its been charging quite ok... but actually on that emergency ckt board a capacitor & resistor & some others were present , in which two things gone C & R their original values i have told u & u said 10ohm is ok but any other substitutes for C ?
Rohith N said…
Sir as a beginner i would like to know what is a capacitive power supply?
also why cant it be used for lead acid battery?
Thanks in advance...
Swagatam said…
Rohith, please refer to the first diagram in this article, you should do it in this way:

https://homemade-circuits.com/2011/12/how-to-make-efficient-led-emergency.html
Swagatam said…
capapcitive power supplies produce lower currents and higher voltages,both are not suitable for lead acid batteries.
Jideofor Igwe said…
please sir, can i use only resistor to drive a 1 watt white LED from an AC source? If yes, what should be the resistance of the resistor and if no, what should i include in it?
Swagatam said…
330 - 3.3/.3 = 1089 ohms or a 1k resistor will work

wattage will be around 100 watts.

it's better to use a cell phone charger instead., you can use a 3 ohm/2watt resistor with it

Anirudh Kundu said…
Hi sir. Thanks for helping me with the last project.
Now I was considering another project . Actually I want to make a "18650" cell charger.
Could you help me in making one?
Swagatam said…
Hi Anirudh,
please provide more details about the project.
Anirudh Kundu said…
Sir..actually I have some laptop batteries....from which I have spared the 18650 cells that are present inside the battery.
I use the 18650 for many battery operated projects....but I dun have any way to recharge them..
Don't have much money to buy one from market...so I thought of asking u to please help me in the making of a 18650, 3.7 volts cell charger.
Swagatam said…
Anirudh, you can try the circuit that's shown in the above article, it will suit your application.
Anirudh Kundu said…
Thanks a lot sir.
And I also wanted to confirm..I need 4.2 volts and 1 amp from the charging circuit. Will it deliver so? ( I have the 5v 2 amp input for the charging circuit input).
Suggest me if any change required sir.
hello Swagatam

would you check this circuit please ??

imageshack.com/a/img30/8392/pgcl.jpg

my phone doesn't recognize it although I test the out voltage is 5.07 !!!
Swagatam said…

for 1amp output replace the 3 ohm resistor with a 0.6 ohms, 1 watt resistor, rest will be as is.
Anirudh Kundu said…
Thanks a lot sir. Your replies have always been abrupt. :)
Thanks for guiding.
Harith Hamdan said…
Sir. We tried assemble your circuit but even the led is not turned on. Do you have any tips on how should we troubleshoot the circuit?
Swagatam said…
Turning the preset to and fro should cause the LED to switch ON/OFF at some specific point on the preset, if this is not happening means there's something incorrect in your circuit connections.
Kallol Bera said…
Dear Sir,
Manu thanks to you for this circuit. I have one question sir. I think you have used the OP-AMP as Schmitt Trigger. But one thing I didn't understand for the LED to blink, terminal 3 of the OP-AMP should go above 1.8V and for that to happen current through the 10k pot. should rise. So, my question is will the current through the Li-ion battery drop automatically increaing the current through the 10k pot?
Swagatam said…
Dear Kallol,

The voltage at pin3 will drop and will be below the reference level at pin2 while the discharged battery is being charged.
This will create a low at pin6 which will keep the LEd and the PNP latched ON for supplying the charging voltage to the batt.
As the batt becomes fully charged, pin3 voltage will rise and at some of time cross pin2 potential, reverting the pin6 output to high, switching off the PNP and the charging voltage.
Kallol Bera said…
My heartiest thanks to you sir.
Swagatam said…
you are welcome!
Rushi Palkar said…
Sir, what is the function of 10k variable resistor in this circuit?
Swagatam said…
It's for setting the full charge cut-off level of the battery.
Rushi Palkar said…
sir, usb negative and ground wire in this circuit are separate??
Rushi Palkar said…
Since I am making a power bank, will you upload a circuit schematic of dc-dc booster? which will convert 3.7v li-on battery to 5v or greater (I will use 7805 to make it 5v) and output current of 2a
Swagatam said…
apply an external 4.2V DC from the battery side in the last diagram, adjust the preset such that the LED just lights up...that's all the circuit set now.

a DC/DC booster circuit can be seen in the following link

2.bp.blogspot.com/-rQ2R8YaFF8c/U372ZScaZdI/AAAAAAAAG8o/nL51vjRHAkk/s1600/speed+breaker+boost+circuit.png
Swagatam said…
USB negative is connected with the circuit and the battery negative.
Swagatam said…
....through the 3 ohm resistor, do exactly as shown in the diagram..
Rushi Palkar said…
how many amps is needed to charge a 3.7v li-ion battery?
and tell me value of resistor (3ohm in circuit) so i can get 2a output current
Swagatam said…
could be at the rate of 1C meaning at the same rate that's printed on the battery,

the resistor value could be found with the following formula

0.6/AH,

or simply, the BC547 and the 3ohm can be completely removed....current limiting is not so crucial
Rushi Palkar said…
See this circuit diagram I have edited your circuit schematic as per my need. Tell me is this circuit correct or not.
https://docs.google.com/file/d/0B3aDGghiCQzzamJVbWRkdDlQbjg/edit
Rushi Palkar said…
Sir, will you give me the formula to find the output voltage in this circuit so ican replace 10k pot and put a keys for charging 1.2v, 2.4v,3.7,.......batteries
Swagatam said…
Rushi, you can go to this site and find the values:

www.raltron.com/cust/tools/voltage_divider.asp

input V will be your 12v, 2.4v or 3.7v etc.

output V will go to the pin3

R1 and R2 will be across positive and ground in the circuit with the junction going to the pin3.

Your link above is not opening, change the status to "sharing"
Swagatam said…
Use R1 any arbitrarily selected value....keep R2 blank and select output just above the pin2 reference voltage. that is just above 1.9V

Rushi Palkar said…
https://drive.google.com/file/d/0B3aDGghiCQzzamJVbWRkdDlQbjg/edit?usp=sharing
Rushi Palkar said…
sir i knw voltage divider but I want to replace 10k pot and to replace i want formula to find value of resistances between pin3 and gnd and pin3 and vcc
to get a specific voltage for different batteries.
Rushi Palkar said…
and we can charge any battery with this circuit-- Li-Polymer ni-Cad ,etc????
Rushi Palkar said…
okay okay now I Got It sorry
Swagatam said…
a preset is always a better choice, fixed resistors will need to be calculated and fixed extremely accurately.

yes lipo, Nicd batts can also be charged with the above set up
Swagatam said…
.........your diagram is correct!
Rushi Palkar said…
Thank you for your help.
Regards.
Rushi Palkar said…
do you take classes???
I want to learn from you.
Swagatam said…
Sorry, taking special classes will not be possible for me, you can learn by posting comments and getting replies from me here.
Dinko Protrka said…
Hi,
I will be using 5V 1.5A adapter to power this,how can i increase output current to 1A instead of 200mA
Swagatam said…
Hi, use a different resistor in place of the shown 3 ohm....for 1 amp, the calculations will be as follows:

0.6/1 = 0.6 ohms...so use a 0.6 ohm resistor in place of 3 ohm for allowing 1 amps....wattage of the resistor will be 0.6 x 1 = 0.6 watts. a 1 watt can be used.
Rushi Palkar said…
Sir, I want a 32-0-32 transformer but I cant get it so can you send me a circuit schematic of power supply to power this circuit
https://drive.google.com/file/d/0B3aDGghiCQzzWWpZQUJJcHRZX00/edit?usp=sharing
Swagatam said…
Rushi, use a 24-0-24V transformer, after full bridge rectification and filtration the output will become well over 32-0-32V
Rushi Palkar said…
Sir, I cant get more than 12-0-12 transformer
Swagatam said…
12-0-12v will give 18V RMS, you can use it with a lower amp output....boosting this further won't be possible...
Rushi Palkar said…
I want a circuit that will convert or reduce 230vac to 45vdc
Rushi Palkar said…
Sir i saw all your schematics related to li-ion battery charging and one doubt came in my mind which circuit will be best for charging li-ion battery through usb?
Swagatam said…
you can try this circuit:

https://homemade-circuits.com/2012/03/how-to-make-simple-12-v-1-amp-switch.html

adjust R6 and R7 for getting the required 45V, 7805 stage can be removed
Swagatam said…
any circuit that cuts off the supply at the specified full charge threshold of the cell can be used.
Rushi Palkar said…
Sir please send me a schematic of a circuit which will boost battery voltage to 5v
Swagatam said…
Rushi, you can try the following circuit

https://homemade-circuits.com/2012/10/1-watt-led-driver-using-joule-thief.html
Rushi Palkar said…
sir i have a resistive touch pad it has 4 pins. how can i use it like mouse for pc. please give me a schematic.
Swagatam said…
Rushi, it looks complex to me, presently I do not have any idea regarding it...
Rushi Palkar said…
okay, thanks for your reply.
Rushi Palkar said…
sir, www.circuitdiagram.org/images/adjustable-power-supply-circuit-7805.gif
this circuit can really give 12v output if that input is 9v? I am confused in that circuit 7805 is used and how can be the output can be varied upto 12v giving 9v input?
Swagatam said…
Rushi, that's a rubbish circuit, 7805 is a fixed regulator IC, it cannot be used like a variable regulator.
Rushi Palkar said…
sir i am making a digital clock using 8051 micro controller no other ic is used in that schematic any other ic is needed for giving pulse? shopkeeper said me that it might require 1 more ic for pulse
Swagatam said…
Rushi, I think external clock is not required, it should be coded in the chip itself...
Swagatam said…
...OK sorry, it seems you do require another clock timer IC for it....a DS12C887 Real Time Clock may be
Rushi Palkar said…
i think it needs to be programmed
Swagatam said…
yes it could be so!
Nurv Ellasar said…
good day sir,

i have a 3.8v cellphone battery and will your updated design still be compatible with my battery to charge?

thankz, more power
Nurv Ellasar said…
Good day,
I have a 3.8v 1500mAh cellphone battery. Which will be the best to use the LM358 or the 741? Any other tips?
Thankyou more power.
Swagatam said…
good day Nurve, yes you can try the last circuit, calculate the 3 ohm resistor as per your battery AH rating with this formula

R = 0.7/batt AH
Swagatam said…
Hello Nurv, LM358 will be more appropriate since it will operate even with voltages as low as 3V, IC 741 is not specified for working below 4.5V
Nurv Ellasar said…
good day sir,

im kinda confused your text mentioned BC557 but your diagram shows BC547. which of the two will i use? and lastly how much of the resistor i use according to your formula 0.7/1.5 = 0.466?

thank you more power
Swagatam said…
BC557 is used in the first diagram, BC547 is used in the last diagram, so it is BC547 as per the last diagram

yes 0.466 or 0.5 ohms is correct.
Amit Kumar said…
Hello Sir , can i use bc557 instaed of 2n2907pnp for above 2nd circuit, above 2nd circuit is suitable to charge nokia 3.7v900mah battery
Thanks in Advance
Swagatam said…
Hello Amit,

BC557 will burn due to high charging current requirement of the cell phone, you can try TIP127 or BD140 etc instead
Kalarav said…
Hi , Sir
I have two 3.7 v battery and i connected it in series so it will be 7.4v so how can i use this circuit

thanx in advance...
Swagatam said…
Hi Kalarav,

you can use the last circuit from above for your application without any change, just the input voltage will need to be increased from 5V to 10V
Amit Kumar said…
Hello Sir , what is the use of 1N4148 diode, can I replace it with 1N4007 diode
Swagatam said…
Hi Amit, yes you can do it
Amit Kumar said…
Thanks Sir Ji
विक्रम संवत 2072 भारतीय नववर्ष और नवरात्री के पावन पर्व पर हार्दिक मंगल शुभकामनाएं |
Swagatam said…
Thanks Amit, wish you the same.
Shane Addinall said…
I have a 3.7v 20700mah battery pack (Lithium ion cells) I require a charging circuit that will deliver 2.5amps / 3amps from a 5v 3amp charger. Your assistance would be appreciated as using the current charging circuit 4.4v 1amp takes too long to charge. Your help you'd be appreciated 
Swagatam said…
connect the 5V source to the battery pack through a 6A4 diode, this will immediately make the 5V 3amp charger compatible with your battery since the diode will drop the excess 0.7V and feed the correc4.3V for the batt pack.


If you intend to have an automatic cut off when the battery gets fully charged you can opt for the last circuit shown in the above article.
Shane Addinall said…
Thank you very much, Swagatam, is it possible to build a circuit that will step up 3.7v to 5v @ 3Amp/h?
Swagatam said…
You are welcome Shane, yes it can done using a 555 boost converter circuit but the output wattage will never exceed the input wattage

wattage refers to V x I specs of the battery

so if the input 3.7 x 3 = 11.1 watts

the output would become 5V @ 11/5 = 2.2amps
KUNTAL PATEL said…
hello everyone,
i try to simulate this circuit in NI multisim but i found that the LED doesnt glow. please give me sollution.

thank you...
Rahul Singh said…
Tried the circuit with LM358, the led glows very little but never flashes, can you please provide the circuit with LM358.

Thank You.
Swagatam said…
the LED is not supposed to flash, it will stay solid when the over charge is reached.

reduce the LED series resistor to 470 ohms or 220 ohms and check the response.
Surya said…
Hi bro,
In the second diagram above the usb input and battery output places are opposite according to first diagram, does it works? please clear my doubt. Nice blog though.
Swagatam said…
yes bro, the positions are correctly shown in the second diagram and will work as explained....but remember to connect the battery first and then switch ON the USB supply in order to enable the circuit respond correctly
sir, what should i do to increase the current to 1 - 1.5 Amps so as to increase the charging speed ??
Swagatam said…
reduce the value of the 3 ohm resistor proportionately, this will allow more current to the output
Ezequiel said…
Hi, I've got a nokia 3V7/1Ah battery. What is important if I want to choose an alternative transistor of 2N2907? The IC Max and the HFe, right?
Ezequiel said…
The 3 ohm resistor on the below circuit is 1/2W too or it doesn't matters?
Swagatam said…
yes the current (Ic) rating is important, make sure it's rated at above 1amp...hFe is not so important
Swagatam said…
multiply 0.6 with the intended charging current and it will tell you regarding the wattage of the resistor
Hi Sir. Thank you for linking me here. Is it possible to output a higher current to charge the battery? And also is there another circuit to modify the output current from the battery to the load. For example we want to charge the battery by 1a and output 500ma. In our study, we arebusing the battery as a powerbank. Could that be possible for a 1860R 3.75v 24000mah battery? Thanks in advance!
Swagatam said…
Hi Pinky, you can replace the 2N2907 with a TIP127 and increase the current handling capacity of the circuit upto 3amps.
If the voltage of the powerbank is compatible with the load, then the current factor can be ignored...you can use any magnitude of current it won't affect the load or the powerbank...just make sure the voltage specs are matching

if the voltage specs do not match in that case you employ the following concept for restricting the current to a recommended level:

https://homemade-circuits.com/2013/06/universal-high-watt-led-current-limiter.html
Thank you so much Sir Swagatam. I tried the circuit but I only got 3.5v output where I read that it has to be 4.4v. I measured it in the negative side of the diode and to the positive.
Swagatam said…
Pinky, which circuit did you try?....please make the last circuit in the above article, the first one might not work correctly....
I did exactly in the 2nd circuit. I used a 1.5v batterry and used a booster to have 5v and used it as a source.
Swagatam said…
the circuits are intended for genuine 5V supply...a modified input will not do....what's the current level of your 1.5 input source??

a penlight cell will not work, the 1.5v AH needs to be thrice that of the cell phone battery AH level....
Our source is from a heat harvester where we will boost its voltage to 5v. It depends to the amount of heat wether what current will be produced. Do we need to have current regulator in order to use this circuit? And also will this circuit work for our purpose? I'm really having a hard time to look for a good circuit for protection to the battery.
Swagatam said…
current should be 2 to 3 times more than the charging current only then the voltage will step-up and charge the battery....otherwise the voltage will keep dropping
devi said…
Hello, sir,

If you don't mind, can you guide me to add a piezo buzzer in the second circuit? I need a warning sound when the battery fully charged.

Thank you in advance.
Swagatam said…
Hello Devi,

you can add it across pin6 of the IC and ground line of the circuit
Usman Saleem said…
I need some help regarding this :

What would be actual current at emitter of 2907 as in thread you have mentioned it should be 4.30v. I am using simple cell phone charger with 5v output.


To set preset I have followed your instruction and connected the charger (5v) - 1n7001 - 1k resistance to make it 4.30v am I doing it in wrong way?


if 5v input is set to 4.30v what will be prest values should be at pin 2 of 741 and pin3 ?

I have designed and tested last corrected circuit diagram connected battery of 3.20v (approximately 40% discharged) turned on input and checked with voltmeter nor the LED is turned on not the battery is being charged!

Is that possible my IC is damaged? as during scolding heat issue or once i have accidentally touched both pin 3 and 2!

Please help me I am extremely committed myself to do this as a challenge !

I have all parts in extra quantity with me ! I can re-design if some chip is required to update.
Swagatam said…
If you have used a 1K resistor in series with the 5V input then the battery will never charge because the resistor will drop the entire current going to the battery.

use a single 1N4007 diode with the 5V supply and connect this supply from the battery side first.

Next adjust the preset such that the red LED just shuts off.

now seal the preset.

remove the 5V supply and the diode, and replace it with any battery which may require charging.

connect the 5V supply with the diode from the other side that is from the USB side and check the procedures....
Swagatam said…
...please do everything as explained above, if you do anything against this then the circuit will not respond.
Linas Neas said…
Dear sir, what is the purpose of transistor BC547, and resistor of 3 ohms? Whether is't enough op-ams pin of 4?

Another question: is this circuit for charging super capacitor for 2.7 V voltage?
Swagatam said…
Dear Linas, BC547 and 3 ohms is for limiting current to the battery (constant current)

2.7V super capacitor can be directly charged from a 3.7V li-ion battery (from cell phone) through a 1N4007 diode.
Irfan Ahmad said…
Dear sir i want to make a power bank using 3.7v rechargable cells. Can u help me to make a circuit for power bank. I want the output 5v 2amp... thanks in advance
Swagatam said…
Dear Irfan, I'll to include the article soon in my blog...
Irfan Ahmad said…
Thanks sir... i m waiting for that...!!!
And sir give me favor to inform me on this email id...
Irfan45@msn.com
Swagatam said…
Irfan, I have published it here:

https://homemade-circuits.com/2015/11/make-this-power-bank-circuit-using-37v.html
romeo gasilan said…
This is nice circuit but is the charging current is fixed at 200ma? How can i increase that to 500MA or 1A?? Im planning to put this as a charger of powerbank rated 3.7V
Swagatam said…
thanks, you can upgrade the circuit to any desired current level by changing the 3 ohm resistor through the following formula

Resistor = 0.7/max charging current
Arun Kumar said…
Who many input voltage and amps charging battery 3.7v 1000ma to 3000ma? What Mobile Phone charger used this circuit? How adjustable preset voltage? please answer this question
Swagatam said…
you can use 1amp to 3 amp charging current for the mentioned batts respectively....use the mobile charger which are used normally for smart phones, the circuit explanation is already given in the article above.
Debanjan Mitra said…
Sir, i want to made a powerbank for my smartphon where i will used 6V 5.5Ah lead acid battery as a source and output should be 500mA please help me, Is any modification needed in this circuit?
Swagatam said…
Debanjan, the current is not important, it's the voltage that should be compatible with the phone's charging voltage spec. You can use a 7805 IC circuit with your battery for implementing it as explained here:

https://homemade-circuits.com/2012/03/how-to-make-simple-dc-to-dc-cell-phone.html
Debanjan Mitra said…
i already done this by 7805 and 7806 voltage regulator but there was some problem charge become very slow and the below 500mAh batteries are not charged, phon just on and off automatically. one more qus, Is pulse is needed to good recharge the Li-ion battery?
Swagatam said…
try reducing the series limiting resistor value to 2 ohm or to 1ohm and see the response, and make sure the 7805 is mounted on a large heatsink.

you can also try a PWM in conjunction with a 7805 and see how it works. You can use the following circuit for feeding the PWMs to the 7805 input:

https://homemade-circuits.com/2012/01/how-to-build-simple-pwm-controlled-dc.html
Arun Kumar said…
Who many battery 3.7v battery 1000ma to ? This circuit supportable upto battery amps
Debanjan Mitra said…
Thank you sir i will try both circuits, Good night
Hasan Basri said…
how to set max of current up to 2A ?
Hasan Basri said…
how to set up to 3A, i need to charger 4 Pararel 3.7 Battery..?
thanks.
nguyenthe toan said…
Hello sir,
I am an electric and electronics final year student. My final year thesis is portable smart solar battery charger for Li-ion battery(3.7V/1200mAh) from solar panel 5V/1000mAh which have LEDs indicator, overvoltage and overcurrent protection, etc. I was hoping sir can help me on how to make a solar charger smart. I mean both theory and PCB circuit.
That is what I have thought about developing so far sir. But i am not sure about the complexity of it therefore i am open to any new suggestion to improve this design.
Thank you for your quick feedback and I truly appreciate your assistance sir. Have a great day sir.
Swagatam said…
nguyenthe toan, you can try the following design and see how it performs:

https://homemade-circuits.com/2015/10/smart-emergency-lamp-circuit-with.html
Faruqi Mohani said…
Good day sir, I want to make a circuit with automatic cutoff when battery reached fully charge and need the output with 5v and 1A. I use samsung li-ion battery with 3.7v and 1000mAh. For the input i use samsung adapter with 5v and 2A.
Is it ok for me to use this circuit?
Is need to me to add regulator IC7805/LM368 in this circuit? Or no need because of the input 5v?

And lasty, i am not understand how to setting preset by connecting into the battery.

I hope you reply this comment. I really need your help sir.
Thank you sir.

Swagatam said…
Faruqi, you can use the mentioned adapter for charging your battery.

7805 will not be required if the adapter is rated at 5V
use a 1 ohm 1 watt resistor for the current limiting resistor.

the setting up procedure is already explained in the article, please read it carefully. you will need a power supply adjusted at 4.2V for the setting up and this must supply must be fed from the battery side.



Shubham Wani said…
Hello Sir,
I am trying to make the circuit for mobile to mobile charging, How can I modify this circuit to charge another mobile or which circuit I should use sir. Mobile A as a charger and Mobile B as a client. Both are having micro-usb connector at their ends. I tried using OTG (On to Go) connector but it is not working in all the device and current rate is also very low.

Hope for your positive and quick reply sir..
Thank you..
Swagatam said…
Hello Shubham, the above circuit won't be required for mobile to mobile charging application....a boost converter circuit could be tried for the purpose, and to implement it successfully the charger mobile battery must be rated higher in terms of its AH rating.

the supply from the charger mobile will go to the boost converter and the output from the boost converter to the client mobile....that's all would be needed.
Feathers said…
Hello, sir. We are having difficulties in making this circuit work. We do not know if we are having trouble with the variable resistor. Also, can we use the digital voltmeter to the "To Cell Phone" instead of a discharged battery to know if a "battery" would charge since the voltage reading, if we are right, will increase in the DVM.
Swagatam said…
Hello feathers, the circuit above is very simple and does not require any complex setting up procedures...please mention the problem specifically i'll try to help.

sorry I could not correctly understand your second question regarding usunf DVM with cell phone, how can that be possible?? please clarify
putera irsyad said…
Hello sir. Can u help me. Can I set a current to get 1ampere using the circuit u give and what component should I change to get 5volt 1ampere.

Then, I have a generator which can generate 20-50volt 400mA.if I apply the supply as a input for the circuit.can it work? Please help me sir.This for my assignment. Can u email the answer at irsyadkanashii@gmail.com.
Swagatam said…
Hello Putera, for converting 20 to 50V 400mA into 5V 1amp you may have to employ a buck converter, and then the output from this buck converter could be applied to the above mentioned circuit.
Sherwin Baculi said…
Hi there. I want to make my 18650battery into 11.1v . Do you have a balance circuit on it?
Swagatam said…
what is your exact requirement? a 3.3V to 11V converter or 11v to 3.3V charger?....please clarify
Firkan Hartanto said…
Sir, can i charge 18650 with this circuit ? Tks
Firkan Hartanto said…
Should i change ic 741 with LM358 or not sir ? Tks a lot
NoName said…
is this only limited to 5V output?
sudhanshu said…
i have 3.7 volt battery.. can i input 3.7 volt in this circuit.
Swagatam said…
yes you can!
morteza said…
I made this circuit but both LEDs are illuminate
Swag said…
connect a zener diode in series with the pin#6 of the IC, between the LED junction and pin#6, see example here

https://homemade-circuits.com/make-6v-4ah-automatic-battery-charger/
morteza said…
If I do this it will work correctly? because I measured output volyage andit was 2 volts.
Swag said…
2V where? are you getting it after the zener connection?

after connecting the zener diode only one LED will be ON.
Morteza said…
No sir I didnt connect zener diod but without it I recieve 2 volt output and cellphone dont charges,please help me ,I really need this circuit.I want to build a circuit that connects between mobile charger and mmobile and cuts off charging when it fully charges.
Swag said…
connect the zener diode as I advised in my previous comment and check again....at any instant only one LEd should be ON, two cannot be ON, because the 741 output will be either "high" or "low" at any instant, never both..
jindro said…
Sir good afternoon! can I charge 2 li ion batts. in parallel? is it better to use lm 321?
Swag said…
Hi Jindro, you can use two Li-ion in parallel with this charger. LM321 is good, but 741 will also work equally well...
Solomon said…
Hello sir, am requesting a circuit from you: a 12 to 5v 2A or 2.5A usb Charger circuit with for charging phone from 12v source
Swag said…
Hi Solomon, you can try the following concept and slightly modify the coil and the R8/R9 values as per your required specifications

https://homemade-circuits.com/5v-pwm-solar-battery-charger-circuit/
Michael said…
I've got a question that Google and YouTube hasn't been able to answer. Is there a way of wiring 3, 18650 cells to get two simultaneous voltages. I'm trying to get 3.7 volts and 11.1 volts from the same three cell bank. Something like a series and parallel, probably with some diodes, type of setup maybe? Or would it be better to just go with series and run a step down converter for the 3.7 volt side of my project?
Thanks in advance,
Maintenance Mike
Swag said…
nothing complex about it..you will have to connect them in series, the end terminals will allow you to get 11.1V, while the 3.7V can be extracted from the bottom most battery which has its negative acting as the ground...I hope you go it.
Michael said…
Awesome, why didn't I think of that. And also, thanks for the quick response. I see a portable stereo headphone amp / usb power bank in my near future. Woo whooo!
Swag said…
Glad you liked my suggestion, wish you all the best...
nitin chaudhari said…
Sir, I have li-ion mobile battery 3.7v . Can I directly connect it to 5v mobile chargers' terminal to charge it or it requires to connect a diode??
Swag said…
Nitin, If you are not using an automatic cut off circuit then you may have to include two 1N4007 diodes in series, so that the 5V drops to a safe 3.9V level.
A single diode will also work, but in that case you will have to monitor the charging voltage and switch it off as soon as 4.1V is reached.
nitin chaudhari said…
Sir, can I use above circuit to charge a spare li-ion mobile battery using mobile charger with 5v and 2A output (as written on charger, another charger noted as 5v and 1 A)???
Also I don't have device to set the value of preset ?
Can u suggest some fixed value of resistor??
Swag said…
Nitin, preset is the easiest device to set, calculating fixed resistor practically can be very tedious, so you will have to use a preset here for the adjustments.

yes any 5V input source will work. as long as the current is not higher than the Ah level of the battery
miguel said…
The transistor 2N2222 is a NPN ...the correct transistor is a PNP like BC557...
With a NPN, as showed, the current max is 10mA with a PNP up to 100mA ...
The schematic is wrong
Swag said…
You seriously need to learn electronics...I can help you understand if you want.

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