Here a simple 1 watt LED driver circuit based on a modified joule thief design, runs on a 1.5V single AA or AAA cell.
I got the concept of this circuit few years back from a LED torch, which was powered by a single AA battery. This circuit is pretty efficient and it can power a 1 watt white LED quite brightly.
This LED driver was constructed on a little PCB, unfortunately I don't have the the torch any more to show you some pictures.
1 watt LED driver circuit diagram and part list
Part list >
- One BC557 transistor
- One BC337 transistor
- One 47kOhm resistor
- One 33pF capacitor
- Inductor coil, 40 turn on a 5x2mm ferrite toroid.
- One 1N4148 Diode
- One 47uF electrolytic capacitor
- One 1 watt white LED
- A battery of cource, AA or AAA NiMH/Alkaline or Zinc Carbon
You can make the Inductor coil easily by winding 30-40 turns of enameled copper wire on a small toroid salvaged from a old CFL, or even you could wrap the wire on a 5mm Iron screw.
The circuit diagram >
Working of this 1 watt LED driver circuit > Two transistors, BC557 and BC337 along with the 33pF capacitor and 47k Ohm resistor forms a simple oscilator circuit, supplying the inductor coil a pulsed DC current.
The coil produces high voltage pulses, the high voltage pulses got filtered by the 1N4148 diode, smoothed by the 47uF capacitor and eventually lights up the 1 watt LED.
You can construct this circuit easily on a strip board, a 1.5cmx2cm piece of strip board is enough.
Hi,
This circuit can also charge Ni-MH cells, using 2 * 1.2V 1.3Ah cells in series;
the charge parameters could be as under:
2 * 1.2V 1.3Ah in series equal 2.4V 1.3Ah 3.12Wh.
Both cells would need 130mA current at 2.7volts.
To accomplish this---
1. The diode should be changed to 1n4007,
2. A 2.7V zener should be connected across C2 in reverse bias and
3. A single alkaline AA battery/ 'D' size standard 1.5v cell can be used.
The whole thing can be built in a small box and could be suitably carried outdoors for use.
Yes, it could be used to charge NiMH or Li-Ion batteries with little modification. Thanks for the feedback, I always appreciate that.
Using 3V (two standard AA cells in series) as input source for the circuit, we can make a power bank circuit which can output 5V and current anywhere between 500mA to 600mA.(provided the source is rated for a couple of amps)
This way a small phone worth Rs.2000 can be charged from this circuit with an internal battery rated at 1.5Ah max.
The li-ion battery calls for 150mA charge current and the internal circuit(not the li-ion protection circuit) being the voltage processing and filtering circuit would call for another 200-300mA current.
So, the cell phone calls for a total of 350-500mA of current at 5V at its input charging socket (depending on remaining charge left and charge to be filled in, the current draw may vary).
To accomplish this---
1. The input should be rated at least 3A. So 3V@3A equals 9Watts of input power. A single alkaline cell can provide that much of current.
2. Next follows the coil. A 24/25awg super enameled copper wire should be used for winding the toroid. The number of turns is simply a matter of experimentation by trial and error.
3.The oscillator transistors should be rated at least 2 amp max. The base resistor is also a matter of experimentation(higher the value, lower the output current and vice-versa) this defines how hard the transistor will switch on and off.
The transistors will need heatsinking if they are switched too hard.
4. The rectifier diode---general purpose 1n4007 will work best. One could also use fast recovery diodes rated at 1-2 amps.
5. Not forgetting the Zener Diode, if we miss connecting this component the output voltage will not stay constant. The output will be full of spikes and unclean DC waveform.
The zener should be rated at 5.1V@1W.
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Now comes the calculation:
1. The output of the power bank as said should have 5V at max. 600mA / 0.6A.
So. 5V at 600mA equals 3W.
To get the 600mA we have to rate the oscillator components at a maximum of 1-2amps to maintain a smooth oscillation.
Again, 3W being the maximum power of the powerbank, this dictates the input current draw at 3V.
So, 3W at 3V means a current draw of 1amp maximum, along with switching losses the current may go upto 50-100mA more (varying from design to design).
Therefore the input current should be rated for 3amps at 3Volts.
Now, having a maximum of 1.1A current draw at 3V means 3.3watts of power draw.......only when the output is loaded fully at 600mA at 5V........this is the situation when the cell phone is discharged.
2. Calculating the Run-time, we have---
The battery current 3 amps divided with the maximum current draw with losses coming up to 1.1 amps,
3 / 1.1 = 2.73 Hours or 164 minutes.
In general considering factors such as temperature, load conditions, battery aging,
etc.
A run-time of 2 Hours can be expected from the circuit with fresh batteries connected.
This circuit should be mounted in a closed box to avoid interference and can be used outdoors.
Thanks
Hi,
I've built this power bank a week ago that's why i could describe it.
One more thing, most circuits on the internet show people making power bank with 1.5V as input to the circuit.
Here the circuit's working doesn't last as expected, though they never tell the runtime, their circuits work, provide the output of 5volts but with very negligble amount of current say 100-200ma.
Their cell phone just reads charging but i'm dead sure their charging wont last more than 5 minutes or so.
What matters more is the output current of the powerbank, this tells how slow / fast your cell phone will charge and also the backup it can provide.
The Differential voltage also matters. If this is kept small the eaiser it gers to convert to the required voltage with less hassle.
1. Web circuits show 1.5V as input.
So 5v - 1.5v equals 3.5v. This is input to output differential voltage.
2. My circuit shows 3V as input.
So 5v - 3v equals 2v.
The circuit is possible to build with 3V.
In reality,
the much more efficient power banks are the ones that use lithium ion batteries. Since a single Li-ion cell is rated at 3.7v nominal, if we see the differential voltage, we have:
5v - 3.7v = 1.3v.
This voltage is the least of all, meaning mine and the web circuits as described previously.
Therefore, manufacturers have much less hassle while building such power banks.
Anyway, i have got succeeded in making my own power bank with 3V as input supply!!!and it worked!!!