Have you wondered how you would charge your cell phone from normal everyday battery alkaline batteries, you know the non-rechargeable kind that everyone keeps in the refrigerator? I was asked recently about how to do this and it got me
thinking. I’m not an alkaline battery expert, so I had to get on-line and do a little research to figure out which solution works best. In addition, it turns out to be a cool electronic circuit project.
As it turns out, charging your cell phone is quite easy. You need just a few batteries, a few electronic parts and a cell phone of course.
Parts List (super simple charger):
Description : (Reference Designator | Manaufacturer Parts List | www.Digikey.com Part number)
- 0.5amp Linear Regulator : (U1 | L78M05CV | 497-2957-5-ND )
- bypass capacitor : (C1 | K104Z15Y5VE5TL2 | BC1148CT-ND )
- Bypass capacitor : (C2 |K104Z15Y5VE5TL2 | BC1148CT-ND )
For those that want a little more of a challenge, here is A DIY Regulator part list and schematic
Parts list (DIY Regulator):
- 4.7 ohm R1 regulator :(R1 | RC12JB4R70 | RC12JB4R70-ND )
- 2watt 6V Zener :(Z1 | 1N5340BRLG | 1N5340BRLGOSCT-ND )
- 3A power transistor :(Q1 | TIP31A | TIP31A-ND)
- Current limit transistor :(Q2 | 1N5340BRLGOSCT-ND | 1N5340BRLGOSCT-ND )
- 0.68ohm 2w current limit resistor :(R3 : :RSF200JB-73-0R68 | 0.68W-2-ND)
- 10uF Bias capacitor :(C1 | UMP1A100MDD | 493-15546-ND)
- 100k Bias Resistor :(R2 | CFR-25JB-52-100K | 100KQBK-ND)
To maximize efficiency and keep things simple, both circuits work best if the input battery voltage is approximately 6v. I will explain this further below.
In addition, since we are transferring energy from the Household batteries to the Cell Phone battery, the first step is to figure how much energy (in Joules) is needed to charge the Cell phone. Then we simply determine how many batteries are needed to be used to meet this energy.
How much energy is needed to charge a cell phone?
Energy is defined as the integral of power over time. The equation we will use is shown here:
Energy is measured in the Joules Units = Amps * Voltage * seconds
My model 6 iPhone has an internal lithium ion battery totaling approximately 1400 milli-amp hours (1400mAH = 1.4 Amps). This battery has an average voltage of 3.2v. The total energy to charge the model 6 iPhone using the equation above gives the following:
16128 (joules) = 1.4 Amps * 3.2v * 1 Hour * 60 Seconds/Minutes*60 minutes/hour
So, to be able to get a full charge, we need our battery capacities to be greater than this energy amount. We also know that the iPhone will try to pull up to 1amp of current during charging to minimize the charge time. We need to take this into account when determining the capacity of the batteries.
How much energy does a battery Hold?
There are lots of flavors of house hold batteries, which is the best to charge your cell phone? In my search on the web, I found the following capacity information about house hold alkaline batteries.
Battery Model – Capacity in mAh – Capacity in Joules – Average Cell Voltage
- AAA – 1150mAh – 5071Joules – 1.225v
- AA – 2122mAh – 9360J – 1.225v
- C – 7800mAh – 34398J – 1.225v
- D – 17000mAh – 74390J – 1.225v
- 9V – 400mAh – 12960J – 1.225v
The list above is when the batteries are being drained by 50mA or less. When currents up to 1A are used for charging an iPhone, the battery energy will be Approximately 30% lower.
So we have a few options to charge our phone. One C or D battery would work. Three AA batteries would work. Four AAA probably won’t work to fully charge an iPhone. Two 9v batteries would work. As you can see there are several ways to do this, but for the simple circuit above to work, we also need for the total battery voltage to be close to 6v. The way this can be done is to stack 4 AA,C,or D batteries in series (4 x 1.225v ~= 6v). The 9 battery can also work but in this case, they should be placed in parallel. However, the circuit I show above does not work well for 9v batteries. The power fet will get too hot and you will waste a lot of battery energy in heat. Basically, the circuit works best by stacking four of AA, C, or D type cell batteries types in series.
You can configure different types of batteries in series to do the job, but it is important not to allow them to get too hot during charging. If the batteries get too hot to touch during discharge, you should shut the circuit down.
How does the circuit work?
The super simple linear regulator maintains a 5 v output voltage up to 0.5 amps of load current. However, because the drop out voltage of this linear regulator is approximately 1.5 volts the output may be closer to 4.5v when charging.
The DIY regulator works by regulating the output from Vin minus the voltage across resistor R1 and will clamp near 6v if the input is above 6V. The benefit of this circuit is that the charge time is less because I adjusted the current limit to be about 1 amp. The output voltage is the input voltage minus the voltage across R1 minus the 0.7v base to emitter voltage of the transistor. (Vout = Vin – Vr1 – Vbe). All transistors have a 0.7v drop across the Vbe junction. The 5 ohm resistor was chosen to ensure we would regulate around 5v even when the input voltage higher is higher than 6V. z1 is used to clamp the output voltage near 5.3 volts when the input is above 6v . However, Above 6.5volts, the zener Z1 will get extremely hot. The base current with this transistor required for 1A if Ib = Iout/beta. Beta for this transistor is about 10. Assuming the input is 6v, and a 1 amp load current, the current through R1 will be 100mA. The voltage drop across R1 is 0.5v. Any excess current will go through the zener. I used a transistor with as high of a beta as I could find so that I would not be wasting too much current in the zener diode. The zener diode effectively provides the regulation of the output voltage
Q2 and R3 provide 1Amp current limiting to the circuit. The Iphone is a USB compatible device and will try to pull as much current as possible from the batteries (up to 2.1amps). Therefore, some amount of current limiting is needed. Q2 and R3 provide this limiting to the system. When the current reaches approximately 1amp through R3, the Q2 transistor will turn on and pull the Base of Q1 low reducing its current. This will limit the current to 1Amp.
Have fun . Please let me know if you have questions or comments.