Choke circuit to run broken filament 40Watt tubelights

This simplest electronic choke can be built low cost, it can glow open tubelights too, no need of a new .

Working voltage: 170~300V for 220V tube, 80~140V for 110V tube.

You can use a tungsten bulb of 100Watt rating of 230V/110V as the series inductor, U can use Existing Choke, or use fan regulator coil, or even Soldering Iron coil(min65W). For the diodes, IN4007 has reverse voltage protection of 1000V, use this of good company(i.e, MIC, ST,BEL). Using low cost diodes can blow the capacitors. For capacitors, use Keltron or any other good company. You can use 400V capacitor if you cannot find 800V one.

PARTS LIST:
Diode: IN4007(MIC) x4
Capacitor: 2.2/800(KELTRON) x2
Neon lamp(optional)x1

Not mentioning 100Watt bulb or other inductor, as we can get it in our house.

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Mosquito Racket Repairing Diagram (Made By China)

There are these types of mosquito killer racket are easily available in market at some 150 rupees.Mosquito Killer racket .This consist an electric fence in the bat portion which has 5000-10,000 volts passing through them. This is not only mosquito repellent, it’s insect zapper. When a mosquito gets in contact the high voltage  burns it. The inner circuit consists of battery, a charging unit, and a HV inverter circuit, which I’m going to describe here.

The circuit consists of a flyback topology transformer driven by a general NPN transistor 2SD965. The feedback coil of transformer is of 6 turns, the primary is of 12 turns and the output or secondary coil is of 450 turns. When this circuit is run by 3Volts, the transformer generates about 2000-4000 volts at zero load, and the output is then coupled 3 times by using 3 IN4007 diodes and suitable capacitors, thus reaching our need of 5000-10,000 volts.

Note: All diodes used in circuit are IN4007 diode, The transistor is 2SD965, and can’t be replaced by BC548, The transformer should be ferrite core.

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Automatic LED Emergency Light

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 :

•  All are white hi bright LEDs rated for 3Volt @ 25mA

• The total current requirement is 12 X 25 = 300mA

• This current has to flow through T2 – BD140 PNP transistor

• The minimum current gain (hfe) of this transistor @ 500mA is 50

•  Hence the base current Ib requirement is Ic / hfe, 300 / 50 = 6mA

• 6. Base emitter drop of T2 at 500mA is 0.77 volt

• 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)

• Hence the bias resistance is = 6.13 / 6 = 1000ohms

• As the battery drains the final terminal voltage will be 5.4volt

• 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.

• 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 :

• The battery requires a full terminal voltage of 6.9V at this point charger should cut off.

• That is the voltage across the chain ZD1, R2 and T1 be should be 6.9 volt

• T1 be voltage of 0.7 volt plus drop across R2 and zener voltage should be 6.9V

• T1 be current = Ic / hfe

• Ic is 1.25 / 180 = 7mA

• Ibe = Ic / hfe of T1  i.e = 7 / 70 = 100uA

• Drop across R2  =1.2 X .1 mA = 0.12volt

• Hence Zener voltage = 6.9 – (0.7 + 0.12) = 6.08 the near by preferred zener voltage is 6.2 volt

• Say the battery voltage at full charge will be 7 volt with 6.2 volt zener diode

• To calculate R16 value for charging at 1 /10 th of the rated current of the battery 4.5AH / 10 = 450mAH

• Transformer 9volt AC the voltage across C1 will be 9 X 1.414 = 12.6 volt

• The drop across LM317 at 450mA current for good regulation is 3volt

• The drop across protective diode D5 is 0.7 volt.

• The  voltage available at cathode of D5 is 12.6 – (3+0.7) = 8.9volt

• The battery after fair discharge will be at 6 volt

• Hence R16 = (8.9 – 7) / 0.45 = 6 ohms

• The nearby standard value for operation is 5 ohms.

• 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.

• With this circuit over night the battery will get charged fully.

• Over charging is taken care and protected by T1

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

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Cable TV amplifier

Description :

This is a very simple cable TV amplifier using two transistors. This amplifier circuit is most suitable for cable TV systems using 75 Ohm coaxial cables and works fine up to 150MHz. Transistor T1 performs the job of amplification. Up to 20dB gain can be expected from the circuit.T2 is wired as an emitter follower to increase current gain.

TV Amplifier Active Components :

• T1, T2 = 2SC4308 or 2SC1324

Notes :

• The circuit can be assembled on a Vero board.

• Use 12V DC for powering the circuit.

• Type no of the transistors are not very critical.

• Any medium power NPN RF transistors can be used in place of T1 and T2.

• This is just an elementary circuit. Do not compare it with high quality Cable TV amplifiers available in the market.

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