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Friday, March 1, 2019

Joule Thief With Fluorescent Ballast And Relay

Fluorescent lamp ballast or neon lamp ballast should not be thrown away because they can be used to make a joule thief. Joule thief is a simple and inexpensive circuit to increase the voltage by oscillation. It is usually used for small loads. Joule thief can be interpreted as an energy thief. This circuit is also known by other names such as blocking oscillator, joule ringer, vampire torch. The output of this circuit is high voltage reaching around 200 volts with direct current (dc). Usually it is using power transistor as switch to connect and disconnect current, and transformer with primary and secondary winding.

In this circuit there is no transformer with two windings (primary and secondary), only one winding. And it is using relay to replace power transistor. The photo below shows the arrangement of parts for this circuit. The LED lamp will be connected to those two pins of the capacitor. Because the circuit is very simple, it can be assembled without a printed circuit board (PCB) or matrix board.

Below is schematic diagram of this joule thief.

Ballasts (B) it was used for fluorescent lamp with a current capacity of 0.33 amperes and a voltage of 220 volts with a power of 15 watts.

Relay (R) with rated voltage of 5 volts, in the video the relay has 8 pins, but a 5 pins relay can do also. It functions as a circuit breaker to stop current from ballast to negative. Tested with 12 volt relay but it turns out that the power produced is lower than to the 5 volts relay. This is because the 5 volts relay draws a larger current so that the magnetic field in the coil is stronger too. Therefore the output power is also greater.

Capacitors (C) is 100 nano farads, with a voltage above 220 volts. It is collecting voltage spikes that arise when the relay cut off the current. If the current is cut off, the magnetic field in the ballast and relay coil collapses and induces high voltage in ballast coil and relay coil. This high voltage is connected in series to the supply voltage and capacitor. So that the capacitor accumulates high voltage induction plus the voltage from the supply (battery voltage).

Diode (D) 1N4007 prevents the high voltage charge in capacitor to turn back towards the ballast.

The following YouTube video shows when the circuit was tested.

If this circuit is supplied by two 9 volts batteries arranged in series, so that it has a voltage of 18 volts, it can turn on a 220 volt Philips LED lamp with 4 watts of power.

The Philips LED voltage (output) is around 56 volts with a current of 2.25 mA, so the output power or LED lamp power is 0.126 watts. Whereas the actual voltage is measured at about 17 volts with a current of 12.5 mA, so the input power is 0.213 watts, then the efficiency is around 59%. This efficiency is quite good for a very simple circuit.

Please note that not all LED bulb can be turned on by low voltage. LED bulb that cannot use dimmers, cannot work at low voltage. It is usually written as non-dimmable.

If supplied with one 9 volts battery, it is enough to turn on 40 LEDs arranged in series brightly.

For a 5 mm diameter LEDs load arranged in series, the input voltage is 9.6 volts, with an input current of 14.8 mA, so the input power is 0.142 watts. The output voltage of the LEDs series circuit is 88.4 volts with a current of 0.65 mA, so the output power is 0.057 watts. Then the efficiency is the output power divided by the input power, and the the result is 40%. This efficiency is low because the load is too small. While the power absorbed by the circuit is about the same when the circuit is given a greater load, the Philips LED bulb.

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