You can make an **LED Emergency lamp **easily using the LED Panel available now. The lamp turns on when the power fails and turns off when the power resumes. When the mains power is available, the 12 V 7.5 Ah battery charges. Since LED panel is used, the emergency light can have a back up time of many hours. You can make 9 Watt or 18 Watt Emergency lamp using the circuit presented here.Its Backup time will be 10 Hours.Its cost will be around Rs. 950. Of these, the battery alone costs Rs.800. If properly maintained, the battery will last more than 4 years.

*Designed and developed by D.Mohankumar*

**You need the following components**

1.15 Ohms 2 Watt resistor – 1

2. 470 Ohms 1 Watt resistor -1

3. IN 4007 Diodes – 3

4. BD 140 PNP Transistor – 1

5. 1000 uF 25V Electrolytic capacitor – 1

6. 12-0-12, 1 Ampere Step down Transformer – 1

7. LED Panel – 0.5 Watt 18 Nos. – 1

8. Push switch with lock – 1

9. 12V 7.5 Ah Lead Acid battery (UPS type) – 1

**LED Panel**

Different types of LED Panels are now available and you can purchase the cut pieces as per your requirement. Generally, 1 Foot LED strip costs around Rs.100. LED panel has an array of White LED chips connected in sets having three LEDs each with a 200 Ohms series resistor. It requires 12 volt 1 Amps power supply to light all the LEDs with full brightness. Each set of three LEDs may be 0.5 watt or 1 watt and are serially connected with the 200 ohms resistor. In this circuit, a Strip of 18 LEDs is used each having 0.5 watt power. So it gives 9 Watts power.

**How the circuit works?**

Its working is too simple. The PNP transistor BD 140 is working like a switch to turn on / off the LED panel. Power for charging the battery is derived from a 12-0-12 V, 1 Ampere Step down transformer. Diodes D1 and D2 are rectifiers to convert low volt AC to DC. Capacitor C1 removes AC ripples from DC to make it clean. Resistor R1 is the current limiter for charging. A 15 Ohms 2 Watts resistor is used here. As a rule, the charging current, should be 1C for a 10 hours charging. So here 7.5 Ah battery requires 750 Milli Ampere current for charging. The formula for determining the charging current is

Ah of the battery x 0.1

So 7.5 Ah x 0.1 = 0.75 A or 750 mA.

Now how much current the 15 Ohms resistor gives.

Input voltage / Resistor value in Ohms x 1000

12V / 15 X 1000 = 800 mA or 0.8 Amps.

You can increase the value of R1 to 22 Ohms, 27 Ohms or even 47 Ohms for low charging current.

The LED panel has 18 half watt LEDs in 6 sets. Each set has three LEDs with a 200 Ohms common series resistor. So it takes 60 mA .

12 V / 200 = 0.06 A or 60 mA. So all the 6 sets consumes 360 mA (6 x 60).

When the mains power is available, diode D1 forward biases to charge the battery. At the same time, the base of PNP transistor will be held high by the DC from mains power. So T1 will be off. LED panel thus remains off.

When the mains power fails. D1 reverse biases and the base of T1 turns low through R2. T1 then turns on and the battery current flows through its emitter- collector direction to light the LED panel.

**Backup Time **

The Backup time and Battery capacity can be calculated as

Battery capacity = Watts / Battery Voltage x Hours

Here the load is 9 Watts and battery is 12 volts. Suppose we use it for 10 hours. Then

9W / 12V x 10Hr = 7.5Ah. So the Emergency light turns on for 10 Hours even though the brightness falls in the final time.

You can also use the same circuit for 6V 4.5 Ah battery. But in this case, you have to select a 6 volt LED panel or use ordinary 5 mm White LEDs (2 in series with a 100 Ohms resistor). The 6V 4.5 Ah battery needs 400 mA current using the 15 Ohms resistor. So use a 6-0-6 , 500 mA transformer.

You can add a **Battery Monitor** circuit to the battery. When the battery is full, the Bicolour LED lights as Green. When the battery voltage drops below 9 volts, the bicolour LED turns Red indicating the need for charging.

Thanks, sir. Very lucid and the calculations are a great help.