Description.

The circuit diagram of a blinking LED that can be operated from the mains supply is shown here. The working of this circuit is as follows. When mains power is switched ON the capacitor C2 charges through resistor R1 and diode D1. When the voltage across the capacitor becomes greater than 32V ( break over voltage of diac DB3) the diac conducts making  the LED blinks and  capacitor C1 discharges through the LED.This cycle repeats again and again .

Circuit diagram.

Notes.

• 230V is potentially dangerous , be very careful while handling this circuit.
• The timing  of this circuit depends on resistor R1 and capacitor C1.

This is a modified version of the circuit Super bright LED Night Light published here. This circuit is submitted by Mr. Seetharaman and its full credit goes to him.

This is the circuit of a well tried and reliable 230 Volt AC mains operated 24 LEDs (super bright 50mA LEDs). While practically compare the brightness between this circuit and 11watts tube, the LED light is much better. The layout is made in such a way, you get uniform illumination. A photograph of the cicuit is also given in this post.

Hi All Readers of CircuitsToday,

As there is serious discussions going on about our previous Automatic LED Emergency Light Circuit, Mr.Seetharaman has come up with a modified version of the same which answers many of the doubts raised in our Comments section.

Here follows Seetharaman’s Description about the Modified Automatic LED Emergency Light.

Dear John
Since there were lots of doubts from our readers on the LED emergency light, I have written a detailed letter to two of our readers. Of course few modifications are required, which I have indicated on the drawings and on calculations how I arrived at it.  I thought of enclosing it to you so that it can be kept in some library of our site for any ones reference. For each circuit we can have a similar detailed theory, may not be with all calculation, just general operation theory of each part of the circuit. You may think of it. As a first feed back I am sending this.  I have a feeling that most of the people would have made the mistake in base emitter of BD140, the lead out are against normal convention.

To understand the above circuit in a better way, it can be divided into two parts.

1. LED lamp circuit
2. The Battery charger circuit

LED Lamp circuit

1. All are white hi bright LEDs rated for 3Volt @ 25mA
2. The total current requirement is 12 X 25 = 300mA
3. This current has to flow through T2 – BD140 PNP transistor
4. The minimum current gain (hfe) of this transistor @ 500mA is 50
5. Hence the base current Ib requirement is Ic / hfe, 300 / 50 = 6mA
6. Base emitter drop of T2 at 500mA is 0.77 volt
7. With the fully charged battery at 6.9volt terminal voltage (for cycle operation use) the voltage available across the new bias resistance is (6.9 – 0.77)
8. Hence the bias resistance is = 6.13 / 6 = 1000ohms
9. As the battery drains the final terminal voltage will be 5.4volt
10. The bias resistance will be (5.4 – 0.77) /  6 = 770 ohms Hence a 680 ohms was preferred for bias resistance with drained battery also it will give enough brightness.
11. The very important information about BD140 is, as you view the pins, metal portion of the transistor facing down left is emitter centre collector and right is base. Most of the constructors make this mistake, relying on the convention that left base and right emitter. If you have made this mistake please correct it.

Once this portion is checked for reliable operation we will proceed to charger portion.

The Battery charger circuit

1. The battery requires a full terminal voltage of 6.9V at this point charger should cut off.
2. That is the voltage across the chain ZD1, R2 and T1 be should be 6.9 volt
3. T1 be voltage of 0.7 volt plus drop across R2 and zener voltage should be 6.9V
4. T1 be current = Ic / hfe
5. Ic is 1.25 / 180 = 7mA
6. Ibe = Ic / hfe of T1  i.e = 7 / 70 = 100uA
7. Drop across R2  =1.2 X .1 mA = 0.12volt
8. Hence Zener voltage = 6.9 – (0.7 + 0.12) = 6.08 the near by preferred zener voltage is 6.2 volt
9. Say the battery voltage at full charge will be 7 volt with 6.2 volt zener diode
10. To calculate R16 value for charging at 1 /10 th of the rated current of the battery 4.5AH / 10 = 450mAH
11. Transformer 9volt AC the voltage across C1 will be 9 X 1.414 = 12.6 volt
12. The drop across LM317 at 450mA current for good regulation is 3volt
13. The drop across protective diode D5 is 0.7 volt.
14. The  voltage available at cathode of D5 is 12.6 – (3+0.7) = 8.9volt
15. The battery after fair discharge will be at 6 volt
16. Hence R16 = (8.9 – 7) / 0.45 = 6 ohms
17. The nearby standard value for operation is 5 ohms.
18. At the end point of battery 5.4 volt the maximum charging current can be of (8.9 – 5.4) / 5 = 0.7 amps well within the higher charging limit of the battery.
19. With this circuit over night the battery will get charged fully.
20. Over charging is taken care and protected by T1

Hope with the above guide line you can make your light work successfully.

More Modification!!!

Dear Readers,

Mr. Seetharaman has further modified this LED Emergency Light with an under voltage cut-off protection to protect battery from deep discharge. Once the battery terminal voltage falls below 5.7 volts the LEDs will be switched off. Take a look at the modified circuit shown below.

Description.

The circuit shown here works from 24V DC and can flash two 24V bulbs alternatively. The circuit is nothing but an astable multivibrator operating at 1Hz and 50% duty cycle. The lamp to be flashed is connected in place of the collector resistor (Rc) of a typical design. 2N3055 transistors which are well known for their ruggedness are used here. They can handle collector currents up to 10A. So a wide range of lamps (in terms of wattage) can be connected to this circuit.

Circuit Diagram.

Notes.

• Assemble the circuit on a good quality PCB.
• Use 24V DC for powering this circuit.
• Lamp L1 and L2 can be 24V/25W type.
• It is advised to fit the transistors with heat sinks.
• Flashing rate can be modified by adjusting the values of R1, R2, C1 and C2.

Description.

This is the circuit diagram a powerful LED flasher. High power LEDs are very common in the market now and a 1W high power LED is used here. Transistors Q1 and Q2 are wired as an oscillator which produces positive pulses of width 20ms @ ½ Hz. Transistor Q3 and MOSFET Q4 inverts this pulse and MOSFET Q5 drives the LED D1.

Circuit diagram .

Notes.

• LED D1 is a 1W high power LED.
• Capacitors C1 and C2 must be rated 10V.
• Circuit can be powered from two AA cells connected in series.
• The peak LED current is set to around 250mA.