### 17Led emergency light using cellphone battery and charger

This emergency lighting system uses LED and powered by a 3.6V cellphone battery or any battery of the same voltage. The battery is charged by a cellphone charger with a current of 350mA. During the charging process, transistor Q1 is off so are the LEDs. When the power is cutoff (brownout), transistor is energized an delivers  25mA current to the four LED. The emergency light consumes about 0.5W and a fully charged battery can last up to 5hours.
 schematic diagram
Part List:
D2, D3, D4, D5 - preferably white LEDs
Q1 - 9012 or any general purpose pnp transistor
cellphone battery or similar
Battery charger
1000uF /10V capacitor
4.7k and 220 ohms 0.25W resistor
5 ohms 1W resistor
 CS9012 pins
 LED pins
The emergency lighting is a low power type and good only for  short duration of brownouts. You can modify the circuit to increase its power capacity by replacing its parts. You can also  use the 6V emergency lighting system that has higher power and can last about 14 hrs ( maximum).

#### 17 comments:

1. Hi this is Kumar and can i use any one npn or pnp transistor and how much the circuit costs with out battery and its charger

2. Kumar, regarding about the cost

1.) transistor,diode,cap
,resistors - \$0.35
2.) 4 white LED -\$1.5
3.) + pcb and labor + time

but if you want much cheaper, I suggest replace white LED with other color that are much cheaper.

the above price list is based on price of electronic parts in our area.

3. I want to know that, why 220 ohm and 4.7k resistances are connected to the transistor base.What are their functions?

4. Atish good day!
thanks for asking...
the purpose 220 and 4.7k resistor combination divides the voltage from the charger. Since the value of 4.7k resistor is very much larger than 220 ohms, it follows that the voltage across 4.7k is much large than 220. In this condition, transistor Q1 is off.
The 4.7k resistor also limits the total current that will flow across each LED.
actually you can replace the 220 ohms to any value provided that the chosen value is very much smaller than 4.7k ohms. 200,330 and 470 ohms will do.

22o ohms resistor ensure that LED will not light up when there is supply in the charger. In other words, it serves as a switch that automatically turn on LED when brownout occur.

5. Thanks DH-gm!!
I have made the circuit.
I have some more questions:
1)What is the flowing path of discharging current? it confusing me.
2) Please tell me more how 22o works as switch?

6. Atish good day!
This is how the circuit work.
When power from charger is available, the voltage across 4.7k ohms with respect to ground is approximately equal to the voltage from the charger which is about 5V.Since the voltage from the base of transistor Q1 is much higher compared to its emitter which is only about 3.6V, meaning the transistor is off and LED also is off.
When the charger is off(brownout), the voltage across 4.7k will have lower value thus turning on the transistor and as well as the LED.

Answer for Question 1:
Actually i did not get what you are trying to ask..But i will try to answer.
Charger is ON, current will flow from charger ,
1.small amount of current will flow directly to 220 ohms and 4.7k resistor
2.majority of the current will flow to 1N4001 to the battery.This time it charges the battery and NO current will flow to the transistor.
Charger is OFF:
1.No current will flow from charger via 220 ohms resistor.
2.Current will flow from battery to emitter of transistor and small amount of current will flow to 4.7K resistor and majority of current will flow to the LED.

Answer for Question 2:
During the charger is on,the current will flow from the charger to 220, then to the 4.7K ohm resistor and then to ground.In this combination, voltage across 4.7K is approximately equal to charger voltage, thus LED is off.
Removing the 220 ohms makes the voltage across 4.7k lower and making the LED light.
What i am trying to say is the 220 ohms when removed makes the LED light up ALWAYS.
Actually the word "SWITCH" is wrong since what i am trying to imply is that the 220 ohms is LED light CONTROL.

If my answer is not clear to you, dont hesitate to ask for clarification. I will try to answer and expalin it well.tnx

7. What is the function of the 1000uF capacitor?

8. Amrita good day!
with or without a 1000uF capacitor, the LED emergency circuit still works.
Upon designing this simple circuit, during my testing when the supply is cutoff(simulating brownout) i noticed voltage spikes. with this observation, i decided to add a capacitor in parallel to the battery to filter the spike to protect both the battery and transistor. But if you are so sure that the charger you are using is regulated, you can remove the capacitor..
the capacitor thus not play a very significant role in the circuit, thus keeping or removing it is up to you.
tnx.

9. uhm. About the charger, how can i connect it to the circuit? I mean what i have here is an ordinary Nokia charger.

10. You told that we have to use npn transistor but in ckt diagram you mentioned the pnp transistor.correct it. Thank you

11. Error and correction:

You can only use PNPtransistor such as 9012 or any similar general purpose transistor. Just follow the circuit above.
tnx.

12. Can I use more number of LED's using the same circuit. What changes should I make to increase the number of LEDs.

13. could you suggest away to connect a 6v 4.5 ah battery to drive a 3 watt led through a driver circuit,,,a schematic would be wonderful..

1. http://www.simple-electronics.com/2011/10/6v-battery-charger-circuit-for-lead.html
it will help you to charge 6 V battery and you can combine with LED driver as you can find at http://www.simple-electronics.com/search/label/LED/driver

14. PLEASE HELP ME

THE LED IS ALWAYS ON WITH OR WITHOUT THE SUPPLY

WHAT IS THE PURPOSE OF THE ARROW NEAR THE LED?

15. ^
it's ground symbol CMIIW

16. You said on March 22, 2011 at 4:11 AM (see above), "When the charger is off(brownout), the voltage across 4.7k will have lower value thus turning on the transistor and as well as the LED.
"
But how the battery voltage will reach that 4.7K crossing that reverse connected (wrt battery) diode?