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Simple Power Bank - Full Circuit Diagram and Construction Details

The article presents a handful of 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 modules have gained significant popularity today due to their portability and ability to charge any cell phone while traveling and during emergency requirements.

A power bank is basically a battery bank box which is initially fully charged by the user at home, and then carried by the user during the course of a journey. And when the user finds his cellphone battery reaching low, he connects the power bank to his cellphone for a quick emergency top-up charging of the attached 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:

Basic 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.

Using IC 555

A simple boost charger circuit could be used for this purpose as explained in the following paragraph:

Power bank circuitusing IC555 boost circuit


The inductor L is made by using 5 turns of 22SWG super enameled copper wire over any suitable toroidal ferrite core.

We can see two variable resistors (presets) included in the design, these are required to be optimally set for acquiring the most effective and efficient performance outcome from the boost charger circuit.

The voltage across "C" is the output which is used for charging the external device, and in our case the voltage here must be fixed at around 5V.

The entire setting procedure for the above boost power bank circuit can be learned from the following article:  Calculating Inductor Value in SMPS

The SPDT switch is used for toggling the high power 3.7V Li-ion battery across the charger input supply and the cellphone battery. While at home, the SPDT could be positioned towards the upper charging input connected with a fixed 4.1V charging supply. Once fully charged the unit could be used as a power bank for charging any cellphone simply by toggling the SPDT towards the lower IC IC 7(555) boost circuit.

The charging input is preferably obtained from any standard SMPS cell phone charger unit, and through a simple LM317 based power supply. Make sure the supply si fxied at 4.1V so that the cell can never get overcharged.

So the above takes care of a power bank using a 3.7V cell, the Ah level depends on the load specs, or the cellphone required to be charged, for smart phones the Ah level should be preferably  above 5000mAH.

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

Simple Power Bank Using a single 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 charger circuit using 1.5V Li-Ion 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

How to Charge the Joule Thief based Power Bank

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

How to Charge the Joule Thief based Power Bank


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.

Using Two Cells without Involving any Complex Circuit



Simple Power Bank using two 3.7V Li-Ion Cells

The first circuit above makes use of a emitter follower 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 Power Bank Charger using 7805 IC Regulator


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


Simple Power Bank with LM317 current control

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 three above power bank 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.

Need Help? Please leave a comment, I'll get back soon with a reply!




Comments

  1. 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-)

    ReplyDelete
  2. 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...

    ReplyDelete
  3. 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

    ReplyDelete
  4. 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

    ReplyDelete
  5. 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....

    ReplyDelete
  6. 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

    ReplyDelete
  7. the formula for calculating the resistor is

    R = 0.6 / max current output (amps)

    ReplyDelete
  8. 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.

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

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

    ReplyDelete
  11. 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.

    ReplyDelete
  12. LED will require an external circuit for the indications... which can make the circuit more lengthy...

    ReplyDelete
  13. 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

    ReplyDelete
  14. 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.

    ReplyDelete
  15. 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

    ReplyDelete
  16. ...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

    ReplyDelete
  17. 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.

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

    ReplyDelete
  19. 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....?

    ReplyDelete
  20. 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

    ReplyDelete
  21. 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?

    ReplyDelete
  22. 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

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

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

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

    ReplyDelete
  26. it's same as described above...4.2V

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

    Thank you.

    ReplyDelete
  28. 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-

    ReplyDelete
  29. 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

    ReplyDelete
  30. sir this circuit can give cAPAcity of 10000 mah ?

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

    ReplyDelete
  32. 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.

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

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

    ReplyDelete
  35. 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?

    ReplyDelete
  36. 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

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

    ReplyDelete
  38. 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

    ReplyDelete
  39. 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

    ReplyDelete
  40. give all the specifications correctly

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

    ReplyDelete
  42. you can try a buck converter circuit instead

    ReplyDelete
  43. 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

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

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

    ReplyDelete
  46. 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 )

    ReplyDelete
  47. Chinwike, I could not understand your question...which transistor are you referring to??

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  48. 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

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  49. 7(555) indicates a CMOS version, please refer to the diagram and see the number on the IC.

    ReplyDelete
  50. 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. .

    ReplyDelete
  51. 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.

    ReplyDelete
  52. 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

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  53. 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.

    ReplyDelete
  54. 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......

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  55. 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.

    ReplyDelete
  56. 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 ......

    ReplyDelete
  57. 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

    ReplyDelete
  58. 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

    ReplyDelete
  59. 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

    ReplyDelete
  60. 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

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  61. 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

    ReplyDelete
  62. 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.

    ReplyDelete
  63. 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.

    ReplyDelete
  64. 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

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  65. 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.

    ReplyDelete
  66. 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

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

    ReplyDelete

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