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IC 555 Li-Ion Battery Charger Circuit

Last Updated on June 14, 2025 by Admin 17 Comments

This IC 555 Li-ion battery charger circuit was build to charge a couple of series Lithium cells (3.6 volts each, 1 Amp Hour capacity).

The charger operates by supplying a short current pulse through a series resistor and then monitoring the battery voltage to determine if another pulse is required. The current can be adjusted by changing the series resistor or adjusting the input voltage. When the battery is low, the current pulses are spaced close together so that a somewhat constant current is present. As the batteries reach full charge, the pulses are spaced farther apart and the full charge condition is indicated by the LED blinking at a slower rate.

IC 555 Li-Ion Battery Charger Circuit

 

A zener voltage reference (2.5 volts) is used on pin 2 of the comparator so that the comparator output will switch low, triggering the 555 timer when the voltage at pin 7 is less than 2.5 volts. The 555 output turns on the 2 transistors and the batteries charge for about 30 milliseconds. When the charge pulse ends, the battery voltage is measured and divided down by the combination 22K, 8.2K and 620 ohm resistors so when the battery voltage reaches 8.2 volts, the input at pin 7 of the comparator will rise slightly above 2.5 volts and the circuit will stop charging.

The proposed IC 555 Li-ion battery charger circuit could be used to charge other types of batteries such as Ni-Cad, NiMh or lead acid, but the shut-off voltage will need to be adjusted by changing the 8.2K and 620 ohm resistors so that the input to the comparator remains at 2.5 volts when the terminal battery voltage is reached.

For example, to charge a 6 volt lead acid battery to a limit of 7 volts, the current through the 20K resistor will be (7-2.5)/ 20K = 225 microamps. This means the combination of the other 2 resistors (8.2K and 620) must be R=E/I = 2.5/ 225 uA = 11,111 ohms. But this is not a standard value, so you could use a 10K in series with a 1.1K, or some other values that total 11.11K

Be careful not to overcharge the batteries. I would recommend using a large capacitor in place of the battery to test the circuit and verify it shuts off at the correct voltage.

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Tagged With: 555, Battery, Charger, ion

About Admin

Hey friends, Thanks a bunch for stopping by this site! I am an engineer with a Bachelor of Engineering in Electronics and Telecommunication. One of my passions is gathering information from all sorts of electronics books and tutorials. I then take that information and compile it into a language that is super easy to understand. My goal is to make those complex electronics circuit concepts and technical terms much more accessible for all the new and budding electronics engineers out there. I can also design customized circuit diagrams as required by the users.
If you have any questions related to this field, please do not hesitate to drop a comment! I am always here and ready to help you out with any queries you might have. I cannot wait to hear from you!

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r k
6 years ago

which pin of IC-555 is grounded?

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Author
Admin
5 years ago
Reply to  r k

pin#1

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Marjan Markovski
8 years ago

tnx. I have tested it but the diode does not turn off even when battery is full. I have even put two full batteries in a serie and the same again

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Author
Admin
8 years ago
Reply to  Marjan Markovski

replace the 22k resistor with a preset and adjust it until the LED just turns ON at full charge….

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Marjan
8 years ago

dear I finally made the circuit but when put the battery after a few seconds the transistor 2907 started smoke a little and was very hot. should I put some cooling or change with some other

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Author
Admin
8 years ago
Reply to  Marjan

yes you can add a heatsink or replace the 2N2907 with BD140, I am glad you could make it successfully

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Marjan
8 years ago

Dear, what number is the pin of ic555 that is connected with the pin 4 of ic741?

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Author
Admin
8 years ago
Reply to  Marjan

it’s pin#1, and also connect pin#4 of ic555 with its pin#8

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Marjan Markovski
8 years ago

can you help me because I don’t know how, Im not too good in electronics. Thanks

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Author
Admin
8 years ago
Reply to  Marjan Markovski

please use Ohms law to calculate the value of the required resistor in place of 18 ohm

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Marjan Markovski
8 years ago

*higher capacity

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Marjan Markovski
8 years ago

Sorry but allso need to ask you If I should change something else too because my battery is bighed capacity (samsung 3.8v 2600mAh) Thanks

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Author
Admin
8 years ago
Reply to  Marjan Markovski

you may have to calculate and replace the 18 ohm resistor appropriately

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Marjan
8 years ago

just one more question please, what value should be the two resistors to make charger for one mobile battery?

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Author
Admin
8 years ago
Reply to  Marjan

please calculate it as per the indicated example and formula….

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Marjan
8 years ago

what is the value of the other capacitor?

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Author
Admin
8 years ago
Reply to  Marjan

0.047uF

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