In this article I have explained a 5 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
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) Simple Power Bank Circuit using 18650 Li-Ion Batteries
The following figure shows the simplest and the best power bank circuit design that you can build and implement quickly.
Parts List
- Resistors are 5 watt CFR
- R1, R2 = As per the calculations
- C1, C2 = 100uF/25V
- D1 = 9V zener diode 1 watt
- D2 = 6V zener diode 1 watt
- T1, T2 = 2N3055
- P1, P2 = 1k preset
- SPST ON/OFF Switch = 1
The circuit incorporates two identical emitter-follower transistor regulator stages, whose output voltages can be adjusted and fixed using presets.
One regulator stage forms the charger section of the power bank, while the second regulator stage constitutes the charging output for charging a mobile phone.
Input Stage for Charging the Internal Power Bank Battery
One regulator circuit can be seen configured around T1, R1, D1, P1, C1 which forms the input side of the power bank.
This input side is supposed to be supplied with an external 12V DC for charging or topping up the internal Li-ion Batteries of the power bank, whenever they are exhausted.
R1 supplies the base voltage and current for T1. This voltage is stabilized to around 9V by the zener diode D1.
Since T1 is configured as an emitter-follower, its emitter terminal follows its base voltage producing around 9 - 0.6 = 8.4V at the emitter end. The 0.6V is the internal drop of the transistor.
This 8.4V from the T1 emitter supplies the charging voltage for the two 18650 batteries connected in series.
The above 8.4V specifies the full charge voltage for the two batteries (4.2V + 4.2V).
However, since there's no over charge voltage cut-off in this T1 configuration, it is recommended to keep the battery full charge level a shade lower than 8.4V.
Ideally this can be reduced to around 8.2V for an optimal charging of the power bank batteries without the danger of overcharging them.
The 8.2V can be set by appropriately adjusting the preset P1.
Output Stage for Charging External Mobile Phone
The circuit configured around T2, R2, D2, P2, C2 forms the output stage of the power bank.
This stage supplies the output voltage for charging an external mobile phone.
The working of this T2 emitter follower regulator stage is exactly similar to the T1 stage explained above.
However, for this regulator the output voltage is optimized with a 5V output which becomes perfectly suitable for charging any mobile phone or smart phone.
The setting up of the 5V output is precisely done using the P2 preset and the D2 zener diode.
Power bank Resistor Calculations
The resistors R1 and R2 decide how much current can be achieved across the emitter side of the transistors.
These output currents determine how fast the internal battery of the power bank and the external mobile phone can be charged.
R1 and R2 can be calculated using the following formula.
R1 = [Input Voltage - (0.6 + Battery Voltage)] x hFE / Max Charging Current
Assuming the battery voltage to be 7.4 V (3.7 + 3.7), Max charging current to be 1 amp and hFE for 2N3055 = 70, we can solve the R1 value as shown below:
R1 = [12 - (0.6 + 7.4)] x 70 / 1 = 280 Ohms (Nearest Standard Value = 270 Ohms)
Power = 12 - (0.6 + 7.4) x 1 = 4 watts or higher.
R2 = [Input Voltage - (0.6 + Battery Voltage)] x hFE / Max Charging Current
Assuming the battery voltage for charging an external cell phone is 5V, hFE = 70 and Max charging current = 1 amp
R2 = [8.2 - (0.6 + 5)] x 70 / 1 = 182 Ohms (Nearest Standard Value = 180 Ohms)
Power = [8.2 - (0.6 + 5)] x 1 = 2.6 watt or higher.
2) IC 555 Boost Power Bank Circuit
3) 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
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
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
4) 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.
5) Using Two Li-Ion Cells without Complex Circuit
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.
The second design above uses a 7805 voltage regulator circuit for implementing the power bank charging function
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.
UNOM Terhile says
There is this power Boaster been made in Nigeria and the boaster is used on generators or inverters with 1000watts output. Please would you share with me the possible design of this Boaster circuit and how it works
Swagatam says
Power booster is possible only through a transformer for all regular loads, but if you want to power only resistive loads then you can use capacitors.
KENJI says
Sir in the ic 555 boosted circuit, can i put 6x 3.7v 18650?
Swagatam says
Hi Kenji, yes you can, however first it is important to setup and optimize the circuit using a single cell…
Kenji says
Sir the 5v from the phone charger is for charging the battery in the ic boosted circuit?
Swagatam says
Yes, that’s correct.
KENJI says
Another question sir can increase the voltage of the battery to 8.4v, it will not affect the circuit?
Swagatam says
You can use it, but If you have 8.4V source then the boost converter might not be needed, you can use the 8.4V directly to charge a cellphone via a regulator??
KENJI says
Ok thanks
KENJI says
What is the output current, and how do i increase it, im planning to use this circuit in my project thnaks
Swagatam says
Power is upto 3 amps, you caan increase it upto 6 amps by using TIP35 transistor.
KENJI says
Ne555 will work here? Srry im new with electronics and i dontwhich 555 should i buy, Thanks
Swagatam says
Any IC 555 can be used here, all will work.
KENJI says
Its should be cmos version?
Swagatam says
Yes, CMOS will be better since it will consume less current…
KENJI says
Can it charge 2 cellphone at same time?
Swagatam says
Please try with a single phone first, if it works then we can easily upgrade it for 2 phones.
KENJI says
Sorry but I can’t find a ba159, what can i use?
Swagatam says
Try FR107 or any fast recovery diode if your inverter frequency is higher than 500Hz, if it’s less then use 1N5402
Kenji says
i cant find 50uh inductor, 100uh will work?
Swagatam says
The coil will actually need to be experimented and set by trial and error. Start with 100 turns of enameled copper wire over a ferrite rod or ferrite ring, and check the boosted voltage level and also make sure the driver transistor is not heating up. If the transistor heats up, increase the number of turns until heating stops. But if the voltage is not getting boosted that would indicate a faulty 555 configuration.
Kenji says
i made the circuit but the voltage continues to rise and when i put a load the voltage drop to around 3.3 volts
Swagatam says
The 555 boost converter will need to be optimized correctly. Which load did you use, and which battery did you use (mAH rating?)