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You are here: Home / Battery Chargers / 4 Simple Power Bank Circuits Explained

4 Simple Power Bank Circuits Explained

Last Updated on February 23, 2020 by Swagatam 102 Comments

The article presents a 4 assorted power bank circuits using 1.5V cell and 3.7V Li-ion cell which can be built by any individual for their personal emergency cellphone charging functionality. The idea was requested by Mr. Irfan

What is a Power Bank

Power bank is a battery pack which is used to charge a cellphone outdoors during emergency situations when an AC outlet is unavailable for charging the cellphone.

Power bank modules have gained significant popularity today due to their portability and ability to charge any cell phone while traveling and during emergency requirements.

It is basically a battery bank box which is initially fully charged by the user at home, and then carried outdoors while travelling. When the user finds his cellphone or smartphone battery reaching low, he connects the power bank to his cellphone for a quick emergency topping-up of the cellphone.

How Does a Power Bank Works

I have already discussed one such emergency charger pack circuit in this blog, which used chargeable Ni-Cd cells for the intended function. Since we had 1.2V Ni-Cd cells employed in the design we could configure it to the exactly required 4.8V by incorporating 4 of these cells in series, making the design extremely compact and suitable for optimally charging all types of conventional cell phones.

However in the present request the power bank needs to be built using 3.7V Li-ion cells whose voltage parameter becomes quite unsuitable for charging a cellphone which also uses an identical battery parameter.

The problem lies in the fact that when two identical batteries or cells are connected across each other, these devices begin exchanging their power such that finally an equilibrium condition is achieved wherein both the cells or the batteries are able to attain equal amounts of charge or the power levels.

Therefore, in our case suppose if the power bank utilizing a 3.7V cell is charged fully to about 4.2V and applied to a cellphone with a drained cell level at say 3.3V, then both the counterparts would try to exchange power and reach a level equal to (3.3 + 4.2) / 2 = 3.75V.

But 3.75V cannot be considered the full charge level for the cell phone which is actually required to be charged at 4.2V for an optimal response.

Making a 3.7V Power Bank Circuit

The following image shows the basic structure of a power bank design:

Block Diagram

Power bank block diagram

As can be seen in the above design, a charger circuit charges a 3.7V cell, once the charging is completed, the 3.7V cell box is carried by the user while traveling, and whenever the user's cellphone battery goes down, he simply connects this 3.7V cell pack with his cellphone for topping it up quickly.

As discussed in the previous paragraph, in order to enable the 3.7V power bank to be able to provide the required 4.2V at a consistent rate until the cellphone is completely charged at this level, a step up circuit becomes imperative.

1) IC 555 Boost Power Bank Circuit

IC 555 base power bank smart phone charger circuit

2) Using a Joule Thief Circuit

If you think that the above IC 555 based power bank charger circuit looks cumbersome and an overkill, you could probably try a Joule thief concept for achieving quite the same results, as shown below:

Using 3.7V Li-Ion Cell

power bank circuit using 3.7V cell

Here, you can try 470 ohm, 1 watt resistor for R1, and 2N2222 transistor for T1.

1N5408 for D1, and a 1000uF/25V for C2.

Use 0.0047uF/100V for C1

The LED is not required, the LED points could be used as the output terminal for charging your smartphone

The coil is made over a T18 Torroidal ferrite core, with 20:10 turns for the primary and secondary, using multistarnd (7/36) flexible PVC insulated wire. This may be implemented if the input is from a pack of 5nos of 1.5V AAA cells in parallel.

If you select Li-Ion cell at the input source, the ratio might need to be changed to 20:10 turns, 20 being at the base side of the coil.

The transistor might need a suitable heatsink in order to dissipate optimally.

Using 1.5V Li-Ion Cell

power bank using 1.5V cell

The part list will be the same as mentioned in the previous paragraph except the inductor, which will now have a 20:20 turn ratio using a 27SWG wire or any other suitable size magnet wire

3) Using TIP122 Emitter Follower

The following image shows the complete design of a smartphone power bank with charger using Joule thief circuit:

Here the TIP122 along with its base zener becomes a voltage regulator stage and is used as stabilized battery charger for the attached battery. The Zx value determines the charging voltage, and its value must be selected such that it's always a shade lower than the actual full charge value of the battery.

For example if a Li-Ion battery is used, you may select Zx as 5.8V in order prevent the battery from overcharging. From this 5.8V, the LED will drop around 1.2V, and the TIP122 will drop around 0.6V, which will ultimately allow the 3.7V cell to get around 4V, which is just around sufficient for the purpose.

For 1.5V AAA (5 in parallel), the zener could be replaced with a single 1N4007 diode with its cathode towards ground.

The LED is included for roughly indicating the full charge condition of the connected cell. When the LED lights up brightly, you may assume the cell to be fully charged.

The DC input for the above charger circuit could be acquired from your normal cellphone AC/DC charger unit.
 

Although the above design is efficient and recommended for an optimal response, the idea may not be easy for a newcomer to build and optimize. Therefore for users who might be OK with a slightly low tech design but much easier DIY alternative than the boost converter concept might be interested in the following configurations:

The three simple power bank circuit designs shown below utilizes minimum number of components and can be built by any new hobbyist within seconds

Although the designs look very straightforward, it demands the use of two 3.7V cells in series for the proposed power bank operations.

4) Using Two Li-Ion Cells without Complex Circuit

Regulated power bank circuit using TIP122 emitter follower

The first circuit above makes use of a common collector transistor configuration for charging the intended cellphone device, the 1K perset is initially adjusted to enable a precise 4.3V across the emitter of the transistor.

Simple IC 7805 power bank circuit
simple power bank for charging smart phones using two 3.7V cells in series

The second design above uses a 7805 voltage regulator circuit for implementing the power bank charging function

Simple LM317 IC based power bank circuit

The last diagram here depicts a charger design using an LM317 current limiter. This idea looks much impressive than the above two since it takes care of the voltage control and the current control together ensuring a prefect charging of the cellphone.

In all the four above power bank cell phone charger circuits, the charging of the two 3.7V cells can be done with the same TIP122 network which is discussed for the first boost charger design. The 5V zener should be changed to a 9V zener diode and the charging input obtained from any standard 12V/1amp SMPS adapter.




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About Swagatam

I am an electronic engineer (dipIETE ), hobbyist, inventor, schematic/PCB designer, manufacturer. I am also the founder of the website: https://www.homemade-circuits.com/, where I love sharing my innovative circuit ideas and tutorials.
If you have any circuit related query, you may interact through comments, I'll be most happy to help!

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  1. Search Related Posts for Commenting

  2. Jayman says

    Thank you sir for this eye opening, I’m a layman but i find the 4 example very simple to do. I mean where 1N5408 and 7805IC was used.
    But I would like to integrate the TIP122 and the zener diode circuit unit in the TIP122 emitter follower circuit into it and I don’t know how to do it. Can you please give me a diagram for the integration?
    Thanks in advance.

    Reply
    • Swagatam says

      Thanks Jayman, here’s the design as er your requirement:

      power bank charger circuit

      Reply
      • Jayman says

        Thanks so much I’ve assembled the design but when tested with a multimeter, the output voltage is 4.76V it’s not up to 5v hence it’s not sufficient enough to charge my infinix Android phone.
        Thanks for your time

        Reply
        • Swagatam says

          You can remove the 7805, and use 4 or 6 1N5408 diodes in series until you have 5 to 6V at the output

          Reply
          • Jayman says

            Sorry for bothering you sir, the 7508 is now charging my phone and it increases the charge but it’s damn too slow, what can I do to correct that?

            Reply
            • Swagatam says

              Try two 7805s in parallel and check the response, but make sure to put both over the same heatsink

            • Jayman says

              Thank you sir, after using two 7805 in parallel the charging speed showed a positive response.
              At that time the battery pack voltage was at 7V and the output voltage was 4.9V.

              But when the phone was charged to about 54 percent, the battery voltage as decreased to 6.1v and the output voltage was 4.1v respectively.
              At this point the phone only shows charging but didn’t increase again.
              Then I noticed that both the 1N5408 and the 7805 that was generating heat before is not heating up again but was cold.

            • Swagatam says

              Jayman, it is the internal circuitry and connections of the phone which is causing the battery to switch at different charging rates. If it would be an open battery you wouldn’t have had such issues.
              If your 7805s heats up then it may not be supplying proper current for the phone battery.
              In that case you will have to replace it with a 5 amp regulator such an LM338 regulator circuit with heatsink or the one that is shown below:
              https://www.homemade-circuits.com/lm317-variable-switch-mode-power-supply/

      • Novice says

        Sir, I’m confused a bit at the zener diode point. I want to connect a red LED to it as shown in the original design before this integration was done. But which terminal of the diode will I connect to the red LED and which terminal will go out to the negative point for charging the cell from the zener diode?

        Reply
        • Swagatam says

          Novice, you can connect the LED in series with an appropriately selected zener diode, just make sure that together they produce an output voltage of 8.4V for the battery pack. The longer leg (anode) of the LEd will to transistor base

          Reply
          • Jayman says

            I couldn’t get LM 338 only LM 317 is available at my reach here and the maximum amp that it could give is 1.5 can I connect it in parallel just like I did for 7805?

            Reply
            • Swagatam says

              That will give you with great difficult near 3 amps if the heatsink is really large. make sure to connect both of them over a common heatsink.
              Before that, test each of the ICs separately through their standard variable voltage regulator designs, as explained below, if both the ICs work correctly, then you can put both of them in parallel with the resistor/pot network.

              How to Use LM317 for Making a Variable Power Supply Circuit

            • Jayman says

              I’ve tried the LM 317 the output current of 1 is 2 so two of it gave 4A.
              But the problem is not solved yet in the sense that all these transistors works by stepping down the input current to around 5/6V depending on the input but let’s say generally by reduction of 2v from the incoming one.
              So once the input voltage reach 6 or 5v the output voltage will also drop to 4v or lesser making it insufficient enough to continue charging the phone.
              So I think a voltage booster is actually what’s needed to make it work continuously so that the output voltage can always be between 5 to 6v.

              So I’ve some scrab power bank panels with me since it has a ready made boost converter coil, I’m thinking of a way to use the coil in my own circuit so that my output voltage can always be 6v maximum to charge the phone

            • Swagatam says

              You will need a buck converter version for that. You can try the following circuit in hat case:

              LM317 Variable Switch Mode Power Supply (SMPS)

  3. Hameed Jamiu says

    In the second to the last diagram, can we reduce the input voltage to 5v/2A so that normal phone charger can be used to recharge the cell.
    And can the batteries be more than two let’s say for example arranging 5 in parallels in 2 segments and then connect the two back in series so as to increase the mAh of the battery to up to 10,000 mAh and more while the voltage will remain at 7.4V. is that possible and does it need any additional components to achieve that?

    Reply
    • Swagatam says

      5V cannot be used to charge two 3.7V cells in series, that is why 8.4V is shown in the second last diagram.
      Yes you can connect as many cells you like, in series/parallel connections to increase the mAh capacity of the power bank as per the required specifications.

      Reply
      • Jayman says

        But at the end of the page you recommend 12v charger won’t it cause a damage to the circuit?

        Then there’s no indicator led in the circuit how do I know when the cell is fully charged/discharged?

        Reply
        • Swagatam says

          I have recommended 12V with a TIP122 emitter follower circuit, as shown in the 2) example from the following article:

          https://www.homemade-circuits.com/how-to-make-simple-dc-to-dc-cell-phone/

          for indication you can add the following circuit across the battery terminals:

          Battery Full Charge Indicator Circuit using Two Transistors

          Reply
  4. Bright says

    Hi
    I have a problem with my power bank since the inductor stoped connecting then I used wire to connect it and since then the power bank stopped charging instead it keeps reducing instead of charge
    Please sir tell me what to do

    Reply
    • Swagatam says

      The inductor is one of the the main components in the power bank, if you short it, it will never work

      Reply
  5. Danny mac-donald says

    hello,i have batteries ,wire ,and a battery booster can i use the booster to create a gate way for charging the battery and charging of phone….?!

    Reply
  6. ID.et says

    I have a battery of 7.4v and 1000mAH, is it possible to build power bank with 5v output,, If possible, can I get a circuit diagram please!

    Reply
    • Swagatam says

      You can use a 7805 IC to get 5 V output

      Reply
  7. Desmond says

    Nice design sir…. the output is 5v how many Ampere??

    Reply
    • Swagatam says

      Thanks Desmond, it will depend on the input current and the coil dimensions

      Reply
  8. abba says

    charging current at inputs seems the same as that of 6v/200mAH, is this the case ? if yes then charging 3.7v lithium battery with 200mah will take a very long time to charge. please explain.

    Reply
    • Swag says

      yes it will take long, but will charge it safely…you can increase the current if you wish

      Reply
  9. abba says

    Hi swagatam
    my question is what is the principle behind switching off of the 6v/200mAH bulb as the battery get charged?
    thanks a lot

    Reply
    • Swag says

      Hi Abba, as the battery reaches full charge level, the TIP122 emitter also reaches a point where it is almost equal to the base voltage which stops the transistor from conducting any further and also the battery stops accepting any further current, together this stops any current from flowing into the battery and thus shuts down the bulb

      Reply
  10. abba says

    hi Swagatam
    I appreciate your effort for explaining the circuit and the concept behind them. my question is can you please elaborate more on how diode 1N4007 can be use to keep the transistor BC557 off during the charging phase of the battery(first circuit)?
    thanks in advance

    Reply
    • Swag says

      Thanks Abba, the 1N4007 from the TIP122 emitter applies a direct positive voltage on the base of BC557 which blocks the negative supply from the 10K resistor and thus shuts down the base bias for the BC557

      Reply
    • abba says

      thanks a lot, i get the concept now. my other question is how can i calculate the value of the individual resistors in a circuit especially if the circuit is a complex one.
      thanks in advance.

      Reply
      • Swag says

        It can a very long discussion not possible to do it through this comment box, you can perhaps join the forum and learn from the scratch…in short basically the base resistor for a transistor is selected as explained in the following article

        https://homemade-circuits.com/2012/01/how-to-make-relay-driver-stage-in.html

        for an IC it is as per the datasheet of the IC.

        Reply
  11. Noah says

    Dear sir
    I want to make a boost converter using 6v battery. Can you help me to make a boost converter circuit with an output voltage and current of 12v and 1.5A to power my 12v LED Bulb.
    Thanks in advance

    Reply
    • Swag says

      Noah, you can study the following article and accordingly build the design as per your requirement

      https://homemade-circuits.com/2015/10/calculating-inductor-value-in-smps.html

      Reply
    • Noah says

      Thanks Swag

      Reply
      • Swag says

        you are welcome!

        Reply
  12. wakkas says

    Hi…In your first circuit ….TIP 122 beside the diode BA159 heated very much after using heat sink……how can i reduce the heat of TIP 122

    Reply
    • Swagatam says

      Hi, the frequency will need to be optimized correctly for getting the desired results and to keep everything cool….connect a voltmeter across C, the keep tweaking the pots until you find maximum voltage across C.

      this will ensure the most efficient working of the TIP122

      I hope you have the current limiting transistor also installed as shown in the design.

      Reply
    • wakkas says

      yes………but till now i am facing another problem……output voltage is shown 4.44 volt and phone is showing charging….but charging level will not increased ……..what will be the problem……

      Reply
    • Swagatam says

      please use an ammeter in series with the positive of the charging line, and check the current level……it should be at least 50% of the battery AH rating….if not then rewind the boost inductor using two or three parallel wires.

      Reply
    • wakkas says

      bro till now i am facing the same problem after increasing the inductor turn……and the current remain 0.07A after increasing…..now how can I increase the current ……

      Reply
    • Swagatam says

      it can be difficult for me to troubleshoot your circuit without checking, if you are not able to optimize the boost circuit you can simply use a LMN317 power supply, set it at 4.1V and charge a Li-Ion cell which has an AH eating much higher than your cellphone battery, and use the set up as the power bank

      Reply
  13. biju thomas says

    Hi. My power bank circuit is complaint. Its battery is 3.7v but the output is required 5v1A. .plz help me..plz given a simple booster circuit diagram.am not a professional. .

    Reply
    • Swagatam says

      you can use any IC 555 based boost converter circuit for this….I have already posted this in my website…you can search it using the search box at top right.

      Reply
  14. Admin says

    sir… is this circuit using cmos version of 555. from the sheet normal ne555 need minimal voltage 4.5 volt… or did you ever succeed with normal ne555

    Reply
    • Swagatam says

      7(555) indicates a CMOS version, please refer to the diagram and see the number on the IC.

      Reply
  15. Chinwike Nwosu says

    Sir, what's the rating for the transistor that's suitable to stop the power bank from turning of while a phone is charging (the voltage of the 5 power bank battery is 3.7v each )

    Reply
    • Swagatam says

      Chinwike, I could not understand your question…which transistor are you referring to??

      Reply
  16. meysam786 h says

    thank you

    Reply
  17. Heidar Yaghoobi says

    THANK YOU SO MUCH SIR
    FOR YOUR USEFUL CIRCUITS AND EXPLAIN .

    Reply
    • Swagatam says

      You are welcome

      Reply
  18. Unknown says

    Hello Sir I'm now in 5th from ECE branch i wish to be design a Power bank. Sir… Of 10000mAh capacity Sir… Help me Sir

    Reply
    • Swagatam says

      Hello, please provide all the technical details of the power bank, as per your specifications.

      Reply
  19. Ansar Ahmed says

    Sir there is a large power loss in case of second circuit using ic 7805. Is there any way to minimize the loss?

    Reply
    • Swagatam says

      you can try a buck converter circuit instead

      Reply
      • Novice says

        Sir, I’ve used the 1N5408 and 7805 circuit. It indicates charging but the charging is not increasing but rather decreasing and the 7805 IC generate much heat. What should I do to correct the errors?

        Reply
        • Swagatam says

          Novice, if your mobile phone is showing “charging” then it must keep charging, it shouldn’t decrease. You can put one more 1N5408 diode at the OUT pin of the 7805 and check the response. The 7805 will heat up since it is a linear IC, you will have to use a heatsink with it to control the heating. If you want minimum heat dissipation then you may have to replace the 7805 circuit with the following one:

          LM317 Variable Switch Mode Power Supply (SMPS)

          Reply
  20. Prasad P. says

    Sir IAM USING LGABB41865 BATTERYS
    OF 5 BATTERYS SINGLE BATTERY HAVE
    2600mah TOTAL 5 BATTERYS mah of 13000 mah because please give circuit for that power bank charging and out of 1amp and 5volts please help me sir

    Reply
    • Swagatam says

      give all the specifications correctly

      Reply
  21. ochin chukwuma says

    Thanks a lot sir, nice design. I can't find the charging unit and battery connector on the 3rd simply diagram.

    Reply
    • Swagatam says

      thanks Ochin, the charging of the 3.37V cells can be done through any standard adapter externally.

      in the last paragraph I have mentioned the word "LM317 charger" with reference to the cellphone which is supposed to be charged using the shown emergency power packs.

      the battery of the power pack needs to be charged from an external charger which could be your cell phone charger unit

      Reply
  22. Arhyel Thlama says

    Well done sir, I tried to charge my power bank with combination of three cells from laptop battery, I don't know their capacity. So my power bank leds that used to be off, now is always on blinking as if it's charging whether is not connected to a charger. Please do you know what might have happen to my charger and how-to rectify the problem?

    Reply
    • Swagatam says

      Arhyel, what is the input specification of your power bank?…this specification will help us to know how much input is required to be fed safely to the unit…if it was more than then your PB could be damaged

      Reply
  23. Angelo Dionisio says

    Can i replace 3.7v battery to 6v lead acid? It is compatible??

    Reply
    • Swagatam says

      yes you can do that

      Reply
  24. Bhavesh Nandaniya says

    How much baat including in perelal and improve the nex by nex Mobil charj

    Reply
  25. Swagatam says

    you can calculate it through the link given in the article….or do it with some trial and error….please read the full article for knowing the details.

    Reply
  26. yash sharma says

    sir how i can check a 7805 ic

    Reply
    • Swagatam says

      you can do it by actually supplying an input voltage across the specified pins and checking exactly 5V across its output pins

      Reply
  27. Auto Zone says

    respected swagatam

    my powerbank charging only 10 scecond then auto cut off why?
    if i can make another one with its 2 battery (3.7v*2ps)–if that batteries are sutable for ur circuite used LM317 ic –is this ic will heat-

    Reply
    • Swagatam says

      Hi AZ, check the voltage with a voltmeter while the cell is charging, it could be reaching the full charge value quickly…this may be because your charging current is high,,,,check this first

      Reply
  28. Michael Hiro says

    I would like to know what are the theories used in creating a power bank?

    Thank you.

    Reply
  29. prince kumar says

    And also please tell me how much volt requierd for charging a 18650 cell

    Reply
    • Swagatam says

      it's same as described above…4.2V

      Reply
  30. prince kumar says

    Sir curcuit is good but a bit complicated
    And also please tell me which type of bulb used in curcuit
    A tourch bulb ?

    Reply
    • Swagatam says

      Prince, you can use a 3V torch bulb rated at 200mA or above.

      Reply
  31. romeo gasilan says

    Sir this powerbank is 3.7V to 5V/2A ? Right? So how about if the battery is at 4.2V What will the output .. ?

    and also i want to rem0ve the bulb …and replace it with led and resistor c0nnected to supply batt. Or output ? How can i done this?

    Reply
    • Swagatam says

      Romeo, you can adjust the PWM pot to get a precise 4.2V

      the bulb is a safe and an easy option to verify the charging condition…LEd indicator will not work in that position and might require a complex circuit using an opamp

      Reply
  32. Anil Kumar. K says

    Hi Swagatham
    For Ni-Cd cells, the charging current is calculated by deviding 'Battery capacity Ah/10'.

    How can I calculate the safe charging current and charging timeof Li-Ion and Li-Po batteries….?

    Reply
    • Swagatam says

      Hi Anil,

      as a rule of thumb for Li-ion and Lipo batts you can use a charging current as high as their AH level…so for example if the AH of the battery is 2AH then the charging current could be 2amps, but the temperature of the batt could be crucial and will need to be monitored manually or by some automatic method

      Reply
  33. Pawan Ahire says

    Hi sir,
    Can you please help me to make a 3.7v mobile battery level indicator, I want to do this by using ATtiny85, for this I have use concept of voltage divider, so I have taken Vin as 4v, R1 as 130 ohm and R2 as 910 ohm for 3.5v Vout.
    Thanks.

    Reply
    • Swagatam says

      Hi Pawan, sorry, I do not know how to do it with a ATtiny85….

      Reply
  34. Seelamsetti Manikanta says

    good post sir, what is the max current of the last current control circuit, can we build boost converter using attiny micro controller and a 7 segment display interface , i want to increase efficiency because i used 6*1.2v 2700mah which doesent charge my 2000mah android a single time the 7805 getting heated up and a lot of power is wasted there, i bridged 2 7805 ics together but there is no improvement the charging current is very low 150ma i connected directly 4 cells without any regulator and the phone got charged with with 1500ma current but due to voltage fluctuations the charge controller in the mobile disconnecting the power source. any help is appreciated thank you sir.

    Reply
    • Swagatam says

      Thanks Seelamsetti,

      just use 4 of the 1.2V cells in series and connect the output directly to your cellphone, no need of using a 7805 IC, as shown below:

      https://homemade-circuits.com/2012/11/homemade-cell-phone-emergency-charger.html#

      yes MCU can be used for making a power controller circuit

      Reply
    • Swagatam says

      …if your cellphone is disconnecting the power due to fluctuations then you can add another pack of 4 cells in parallel or try bigger 1.2V cells

      Reply
    • Seelamsetti Manikanta says

      thank you sir

      Reply
  35. adesina temiloluwa says

    i have been waiting for so long for this kind of circuit
    i will definately build it sooner n give updates thanks for this BOSS SWAGATAM

    Reply
    • Swagatam says

      you are most welcome Adesina!!

      Reply
  36. Angelous Chavez says

    Sir the circuit above is so interesting.

    Is there any idea that can modefie or improve the above circuit like for example adding of a small solar panel to charge up the battery 3.7v.

    Reply
    • Swagatam says

      Thanks Angelous, yes definitely you can use a suitably rated solar panel and use it to charge the 3.7V cells.

      Reply
    • Angelous Chavez says

      sir why did you used bulb? it is possible to change that into the red LED?

      Reply
    • Swagatam says

      Bulb will pass sufficient current for charging the battery and also indicate the situation, LED will not pass sufficient current and will not serve purpose, it will remain lit forever, not allowing the battery to get charged.

      Reply
    • Swagatam says

      LED will require an external circuit for the indications… which can make the circuit more lengthy…

      Reply
  37. Mayank Padm Bhushan says

    Thanks sir for this circuit
    What is the output amp in 1st circuit
    I want 1 amp output from 1st circuit

    Thank you very much

    Reply
    • Swagatam says

      Mayank, you can calculate and change the 0.6 ohm resistor for fixing the desired current capacity of the circuit…accordingly the cell mAH will also need to be upgraded to the intended level

      Reply
      • Unknown says

        sir this circuit can give cAPAcity of 10000 mah ?

        Reply
    • Swagatam says

      the formula for calculating the resistor is

      R = 0.6 / max current output (amps)

      Reply
  38. Irfan Ahmad says

    Also tell where charging pin or female usb connector is attached…?
    and which point is used to connect the cells…
    And how much cell we connect in this circuit…

    Reply
    • Swagatam says

      Irfan, Everything's shown in the diagram, I am sorry if you are not able to understand the diagram then it would mean that you are very new in the electronics field, and this circuit can be extremely difficult for a new comer…so I think you should rather try the other 3 circuits instead of the first….

      Reply
  39. Irfan Ahmad says

    Thanks sir.. there is some confusion for me.
    1- Is it control the discharge levels of Li-ion cells..
    2- is it control the over charging of Li-ion cells…
    3- is it delivers the 2amp for smart phone sharge
    Once again thanks for the circuit… B-)

    Reply
    • Swagatam says

      Irfan, the cell will not discharge unless connected to a cell phone

      the cell will not overcharge but when the indicator lamp shuts off you should remove the input supply…

      2amp may be obtained if the cell is rated at 3000mAH

      Reply


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