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Charger for lithium-ion battery



Today, many users have accumulated several working and unused lithium batteries that appear when replacing mobile phones with smartphones.


When using batteries in phones with their own charger, due to the use of specialized microcircuits for charge control, there are practically no problems with charging. But when using lithium batteries in various homemade the question arises, how and how to charge such batteries. Some people believe that lithium batteries already contain built-in charge controllers, but in fact, they have built-in protection schemes, such batteries are called protected. The protection schemes in them are mainly intended to protect against deep discharge and overvoltage when charging above 4.25V, i.e. This is an emergency protection, not a charge controller.

Some "friends" on the site here will also write that for a little money you can order a special board from China, with which you can charge lithium batteries. But this is only for fans of "shopping." It makes no sense to buy something that is easy to assemble in a few minutes from cheap and common parts. Do not forget that the ordered fee will have to wait about a month. Yes, and the purchased device does not bring such satisfaction as made do it yourself.

The proposed charger is able to repeat almost everyone. This scheme is very primitive, but completely copes with its task. All that is required for high-quality charging of Li-Ion batteries is to stabilize the output voltage of the charger and limit the charge current.

The charger is characterized by reliability, compactness and high stability of the output voltage, and, as is known, for lithium-ion batteries this is a very important characteristic when charging.

Charger diagram for li-ion battery

The charger circuit is made on an adjustable voltage regulator TL431 and a medium power bipolar NPN transistor. The circuit allows you to limit the charging current of the battery and stabilizes the output voltage.


The role of the regulatory element is the transistor T1. Resistor R2 limits the charge current, the value of which depends only on the parameters of the battery. A 1 watt resistor is recommended. Other resistors may have a power of 125 or 250 mW.

The choice of transistor is determined by the required charging current set to charge the battery. For the case under consideration, charging batteries from mobile phones, one can use domestic or imported NPN transistors of medium power (for example, KT815, KT817, KT819). With a high input voltage or when using a low power transistor, it is necessary to install the transistor on a radiator.

LED1 (highlighted in red in the diagram) is used to visually signal the battery charge. When you turn on a discharged battery, the indicator glows brightly and dims as it charges. The indicator light is proportional to the battery charge current. But it should be noted that with the LED completely dying out, the battery will still be charged with a current of less than 50mA, which requires periodic monitoring of the device to prevent overcharging.

To improve the accuracy of monitoring the end of charge, an additional option for indicating the battery charge (highlighted in green) on the LED2, low-power PNP transistor KT361 and current sensor R5 has been added to the charger circuit. The device can use any variant of the indicator, depending on the required accuracy of the battery charge control.

The presented circuit is designed to charge only one Li-ion battery. But this charger can also be used to charge other types of batteries. It is only necessary to set the required output voltage and charging current.

Charger manufacture

1. We purchase or select from available components for assembly in accordance with the scheme.

2. Assembly of the circuit.
To check the operability of the circuit and its settings, we assemble the charger on the circuit board.



The diode in the battery power circuit (negative bus - blue wire) is designed to prevent the discharge of a lithium-ion battery in the absence of voltage at the input of the charger.

3. Setting the output voltage of the circuit.
We connect the circuit to a power source with a voltage of 5 ... 9 volts. With the trimming resistance R3, we set the output voltage of the charger in the range of 4.18 - 4.20 volts (if necessary, measure its resistance at the end of the setting and put the resistor with the desired resistance).

4. Setting the charging current of the circuit.
Having connected the discharged battery to the circuit (as the LED turns on), we set the charging current value (100 ... 300 mA) using the tester R2. With an R2 resistance of less than 3 ohms, the LED may not light.

5. We prepare a board for mounting and soldering parts.
We cut out the required size from the universal board, carefully process the edges of the board with a file, clean and tidy the contact tracks.


6. Installation of a debugged circuit on a working board
We transfer the parts from the circuit board to the working one, solder the parts, perform the missing wiring of the connections with a thin mounting wire. At the end of the assembly, we thoroughly check the installation.


The charger can be assembled in any convenient way, including wall mounting. When installed without errors and serviceable parts, it starts working immediately after switching on.


When connected to a charger, a discharged battery begins to consume maximum current (limited by R2). When the battery voltage approaches the set one, the charge current will fall and when the voltage on the battery reaches 4.2 volts, the charging current will be practically zero.

However, it is not recommended to leave the battery connected to the charger for a long time, because he does not like recharging even with a small current and can explode or catch fire.

If the device does not work, then it is necessary to check the control terminal (1) of TL431 for voltage.Its value must be at least 2.5 V. This is the smallest allowable value of the reference voltage for this chip. The TL431 chip is quite common, especially in computer PSUs.
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36 comments
I'll go for the sake of a transformer
Okay, I see, the author about the trimmer is not going to answer. Therefore, for the rest I will tell you why I asked this: such trimmers usually have a small allowable current through the engine, some 30 mA, some 100 mA. Therefore, you can not put them in the current circuit in such an inclusion as the author.
Quote: lihvin
Remind your simple and clear question, is this this:
Quote: Ivan_Pokhmelev
1. The circuit on T2 is stuck thoughtlessly.

No, to quote, so completely, without trimming:
1. The circuit on T2 is stuck thoughtlessly. Why? Think for yourself.
You were given the opportunity to think, you did not use it. ((
Quote: lihvin
Quote: Ivan_Pokhmelev
Quote: lihvin
You submitted incorrect information, nothing more.

Once again I am convinced that you have not learned to read! And you see only what you want to see and what suits your "conclusions".
I see only what you write, and you write false information.
I will explain for the most dull, not knowing the basics of mathematics: your text
check control terminal (1) of TL431 for voltage. Its value must be at least 2.5 V. This is the smallest allowable value of the reference voltage for this chip.

means that a voltage of 3 V is an excellent value, and 2.45 V is worthless. In fact, everything is exactly the opposite!
Quote: lihvin
Why are you, such a versatile super competent specialist and clairvoyant, asking?

Once again, you have shown your technical illiteracy: perhaps this will be news for you, but in this case there are quite a lot of multi-turn trimmers. I am interested in the specific type used by you. Or hesitate to call?
Author
Quote: Ivan_Pokhmelev
Another question: what specific type of resistor is used as R2?

Why are you, such a versatile super competent specialist and clairvoyant, asking? You all know everything and everyone.
You already examined this tuning multiturn resistor under a microscope (even took a photo), studied all the documentation, consulted in Oxford and prepared some next comments with a refutation to my answer. Go ahead.
Moreover, I already answered this question.
Author
Quote: Ivan_Pokhmelev
Quote: lihvin
You could not answer your own simple and clear question.

Remind your simple and clear question, is this this:
Quote: Ivan_Pokhmelev
1. The circuit on T2 is stuck thoughtlessly.

Quote: lihvin
Quote: Ivan_Pokhmelev
2. For the most stubborn, not understanding what they are talking about and not wanting to understand what is wrong with the scheme on T2.

And then similarly for 4 days ...
I will answer the rest of the question as follows. Reread my answer, from which the quote was taken, in its entirety, without choosing or shifting the meaning of the written word.
Author
Quote: Ivan_Pokhmelev
Quote: lihvin
You submitted incorrect information, nothing more.

Once again I am convinced that you have not learned to read! And you see only what you want to see and what suits your "conclusions".
Another question: what specific type of resistor is used as R2?
Quote: lihvin
But as I see it, the recommended “4th paragraph after the memory scheme”, you have not read. And it says: "... an additional option for indicating the charge has been added to the charger circuit .... Any version of the indicator can be used in the device ...."
Quite right, and it says: "in the memory circuit added". This indicator is included in your scheme, without any indication that you need to eat from a completely different source. And the fact that you are not at odds with Ohm’s law follows from the fact that you could not independently answer the simple and clear question .
Quote: lihvin
Although they couldn’t, traditionally, do without numerous insults,

Give me an example where I insult someone. I follow this, and if you take sarcasm about illiteracy as an insult, then this is your personal problem.
Quote: lihvin
I agree with you, and what's new here?
Just what your guess is:
Its value must be at least 2.5 V. This is the smallest allowable value of the reference voltage.
false. You submitted incorrect information, nothing more.
Author
Quote: Ivan_Pokhmelev
For the most stubborn, who do not understand what is at stake and do not want to understand what is wrong with the circuit on T2.

Finally! Although they could not, traditionally, manage without numerous insults, they matured (the truth is more expensive) and already at the end of the prepared answer (thanks on behalf of the readers), the question finally came up. And ask right away: "I do not agree with the supply voltage of the memory due to the additional circuit on T2" where then take 7 extra comments.
With Ohm's law, in the calculation of the voltage drop, and you are in full order. But as I see it, the recommended “4th paragraph after the memory scheme”, you have not read. And it says: "... added to the memory scheme additional charge indication option .... In the device, you can use any version of the indicator .... "In this regard, I wonder how selective your attention is. One diode installed on the board and not in the circuit, you recalled in the comments three times. And the fact that the parts block is on T2, stipulated as replaceable and missing on the board You have not noticed many times ....
Author
Quote: Ivan_Pokhmelev
For the most stubborn and not understanding what they are talking about, writing:
check control terminal (1) of TL431 for voltage. Its value must be at least 2.5 V. This is the smallest allowable value of the reference voltage for this chip.
In fact, the typical value of the reference voltage is approximately 2.5 V and can deviate in the operating temperature and current range by ± 0.1 V, that is, take values ​​from 2.4 V to 2.6 V.

I agree with you, and what's new here?
Author
Quote: Ivan_Pokhmelev
Quote: lihvin
As R2, a multi-turn resistance of 10 ohms is installed.

Lying is bad.

You are right about this. Apparently I did not find the 10th and put more. But you must admit that this does not change anything, because Only part of the tuning resistance set during tuning is involved in the operation. For reference, there is a program on the Internet that determines reg. car number photographed from the side.
For the most stubborn, who do not understand what is at stake and do not want to understand what is wrong with the circuit on T2.
To obtain a voltage of 4.2 V on the battery, you must have at least 4.8 V.
The resistance of the resistor R5 is 22 Ohms, therefore, at a current of 0.1 A, 2.2 V will fall on it, and at a current of 0.3 A - 6.6 V. And this means that the minimum the supply voltage at a current of 0.1 A should be at least 7 V, and at a current of 0.3 A - at least 11.4 V. And where is the (5 ... 9) V declared by the author?
I highly recommend that the author carefully study Ohm's law.
For the most stubborn and not understanding what they are talking about, writing:
check control terminal (1) of TL431 for voltage.Its value must be at least 2.5 V. This is the smallest allowable value of the reference voltage for this chip.
In fact, the typical value of the reference voltage is approximately 2.5 V and can deviate in the operating temperature and current range by ± 0.1 V, that is, take values ​​from 2.4 V to 2.6 V.
Quote: lihvin
As R2, a multi-turn resistance of 10 ohms is installed.

Lying is bad.
Author
Quote: Ivan_Pokhmelev
Why did they choose such a dubious circuit without stabilizing the charge current? Another penny transistor regretted?

For you it is doubtful, but for me it is simple, proven and reliable. A stable voltage is important for charging lithium batteries (it is provided), and possible current fluctuations as they are charged are compensated by a large reserve of charging current (up to 0.7C). Why complications of a workable circuit. But as soon as you collect it, you will definitely improve and tell how.
Author
Quote: Ivan_Pokhmelev
Let's continue the questions: in the diagram you have picky resistor R2 (3 ... 8) Ohm (we will not find fault with the fact that in the series E24 there is no such rating), but at the installation - tuning as much as 47 ohms. What is it for?

On the working board there is a tuning resistance of 472 (4700ohm or 4.7k), in the last photo this is visible. In the diagram, this is R3. As R2, a multi-turn resistance of 10 ohms is installed.
Author
Quote: Ivan_Pokhmelev
Quote: lihvin
If you install 4.2 V on the terminals for connecting the battery

Not so in your text:
set the output voltage of the charger in the range of 4.18 - 4.20 volts
According to your scheme, "output +" is the upper output of R3, "output -" is the lower output of R4.

The output voltage of the battery charger is removed from the terminals for connecting the battery. And it doesn’t matter what is inside the memory. If there is no diode in the circuit, it will be as you wrote, if it is, it will be different, but the terminals will be the same.
Author
Quote: Ivan_Pokhmelev
Quote: lihvin
A diode in the battery circuit appeared in connection with the use of a charger (charger) in practice, as part of another circuit. For this scheme, it is optional,

However, you included it in the description:
The diode in the battery power circuit (negative bus - blue wire) is designed to prevent the discharge of a lithium-ion battery in the absence of voltage at the input of the charger.


To this question, I also answered yesterday: "I noticed it in the photo of the memory when editing the job description. Therefore, it is not on the diagram." To explain the inconsistency, after detection and under the same photo, indicated the purpose of this diode. That you have no riddles left. And the use of this diode is at the discretion of the manufacturer.
Author
Quote: Ivan_Pokhmelev
Quote: lihvin

6. "Its value shall not be less than 2.5 V. ...."

We will not answer the question?
Yesterday I answered this question. I repeat:
If you read the last paragraph of the description, it begins "If the device does not work, then ..." and then commented.
And if it’s not clear, I’ll try to supplement it.
"If the device does not work, then it is necessary to check the control terminal (1) of TL431 for voltage. Its value must be at least 2.5 V. This is the smallest acceptable value of the reference voltage for this microcircuit. "At a lower voltage, it is necessary to select a resistor R1, the value of which will provide the TL431 cathode current of more than 1 mA, or check the validity of components.

"And to the question on the circuit on T2 also we don’t want to answer? "

1. Why "too", I already answered 7 of your questions (half of which was disclosed in the description of the memory).
2. You still have not been able to formulate your question, what do you dislike there. Maybe if you read the 4th paragraph after the memory scheme, then the question will disappear.
Why did they choose such a dubious circuit without stabilizing the charge current? Another penny transistor regretted?
Let's continue the questions: in the diagram you have picky resistor R2 (3 ... 8) Ohm (we will not find fault with the fact that in the series E24 there is no such rating), but at the installation - tuning as much as 47 ohms. What is it for?
Quote: lihvin
If you install 4.2 V on the terminals for connecting the battery

Not so in your text:
set the output voltage of the charger in the range of 4.18 - 4.20 volts
According to your scheme, "output +" is the upper output of R3, "output -" is the lower output of R4.
Quote: lihvin
A diode in the battery circuit appeared in connection with the use of a charger (charger) in practice, as part of another circuit. For this scheme, it is optional,

However, you included it in the description:
The diode in the battery power circuit (negative bus - blue wire) is designed to prevent the discharge of a lithium-ion battery in the absence of voltage at the input of the charger.
Quote: lihvin

6. "Its value should be at least 2.5 V. This is the smallest acceptable value of the reference voltage for this chip." "And what is the largest then? Datasheet read or - Another masterpiece of teacher creativity.

We will not answer the question?

And we also do not want to answer the question on the scheme on T2?
Author
Quote: To Delusam
[/ i] -
I ask the author to answer for the accusation against Ivan.
Quote: lihvin
Your comments are thoughtlessly written. Why? Think for yourself.

The question was posed to them correctly ..

Quote: Ivan_Pokhmelev
What voltage will be on the battery when the voltage reaches the voltage of 4.2 V indicated by you on the battery and the diode (not reflected in the diagram, but present in the assembled product) connected in series?


I apologize for missing a few questions, as I don’t have the opportunity to be on the site all day. Then to the answers.
Vladimir Nikolayevich!
Please explain why the line you quoted is an accusation against Ivan P., and his line in the comments is "1. The circuit on T2 is stuck thoughtlessly. Why? Think for yourself." is a pedagogical masterpiece?
If the "local authority" cannot clearly formulate the question and therefore makes puzzles, he asks everyone where his hands are growing (already suspiciously), and is only interested in the rating in the commentators of the month, then why spend time answering such questions. Therefore, what is the question - such is the answer.
I will answer specific questions about the device.
1. A diode in the battery circuit appeared in connection with the use of a charger (charger) in practice, as part of another circuit. For this circuit, he not requiredbecause The battery can be removed after charging until the charger is turned off. Blame it, but noticed it in the photo of the memory when editing the job description. Therefore, it is not in the diagram.
2. If you install on the terminals for connecting the battery 4.2 V before the charge, then after the charge it will be so.
3. The diode is different, with similar characteristics. I can’t see these days, the memory works out of town. Yes, in the circuit it is not needed.
4. "2. The charge current depends on the voltage of the power source, this should not be forgotten." Do not forget. To charge the battery, the maximum allowable current (limited by R2) is important, and then it will only decrease. It can be set at the highest voltage IP. (see 3 paragraph from the end of the description).
5. "3. The lack of shutdown when the charging current drops below (0.07 ... 0.1) C requires attention, you can’t leave the battery for charging at night, you need to look after it and turn it off in time." (see paragraph 2 from the end of the description, apparently did not have the strength to read it).
6. "Its value should be at least 2.5 V. This is the smallest acceptable value of the reference voltage for this chip." "And what is the largest then? Datasheet read or - Another masterpiece of teacher creativity.
If you read the last paragraph of the description, it begins "If the device does not work, then ..." and then commented.
I essentially didn’t find any questions, and let the commentator continue to count the rivets of the tank ...
No answers? Let's continue the questions:
Its value must be at least 2.5 V. This is the smallest allowable value of the reference voltage for this chip.
And what is the largest then? Did you read the datasheet or is it your personal observation of the environment?
Quote: To Delusam
Judging by the appearance, this is 1N4007.
This is one of the most popular diodes in this case, but maybe some other is sealed.
Judging by the appearance, this is 1N4007.
"-
It is strange that it is not in the scheme. Not everyone is able to track this incident at the assembly room.
I ask the author to answer for the accusation against Ivan.
Quote: lihvin
Your comments are thoughtlessly written. Why? Think for yourself.

The question was posed to them correctly ..

Quote: Ivan_Pokhmelev
What voltage will be on the battery when the voltage reaches the voltage of 4.2 V indicated by you on the battery and the diode (not reflected in the diagram, but present in the assembled product) connected in series?
Once you have nothing to answer, let's continue to ask. )))
What voltage will be on the battery when the voltage reaches the voltage of 4.2 V indicated by you on the battery and the diode (not reflected in the diagram, but present in the assembled product) connected in series?
What type of this diode?
That is, in fact, there is nothing to answer?
Author
Quote: Ivan_Pokhmelev
1. The circuit on T2 is stuck thoughtlessly. Why? Think for yourself.
2. The charge current depends on the voltage of the power source, this should not be forgotten.
3. The lack of shutdown when the charging current drops below (0.07 ... 0.1) C requires attention, you can’t leave the battery for charging at night, you need to look after it and turn it off in time.


Your comments are thoughtlessly written. Why? Think for yourself.
1. The circuit on T2 is stuck thoughtlessly. Why? Think for yourself.
2. The charge current depends on the voltage of the power source, this should not be forgotten.
3. The lack of shutdown when the charging current drops below (0.07 ... 0.1) C requires attention, you can’t leave the battery for charging at night, you need to look after it and turn it off in time.

We advise you to read:

Hand it for the smartphone ...