D.C Ballast

Circuit diagram

INTRODUCTION:

In the world of electronics, there is a great importance of dc ballast .We need this kind of device for getting our requisite ac current from dc input .The availability of ac power supply is not as good as dc supply .The places where we don’t get ac current for our practical use, we can get ac supply by using dc ballast. In this case we use dc current as input .Again, in many electronic devices such as gas-discharge tube, we need high voltage to get light. So DC ballast is very useful .

OBJECTIVES:

The main objective of dc ballast is to create such a high voltage enough to light a gas discharge tube .Gas discharge lamps generate light by sending an electrical discharge through ionized gas .In the operation of gas discharge tube, the gas is ionized and free electrons are accelerated  by the electrical field in the tube . For this we need very high voltage as input. The required high voltage is provided by dc ballast using low voltage dc supply .A dc ballast provides the features of low cost and higher reliability .Here we attempted to convert low dc voltage to high ac voltage to light a gas discharge bulb using dc ballast.

 

 

 

APPARATUS REQUIRED:

1. E core

2.Transistor CTC2233(2 pieces)

3. Inductor

4. Ceramic capacitor (2 pieces - 1 micro farad, 0.01 micro farad)

5. Resistor (1 piece-300ohm)

  OPERATION:

         When we apply dc voltage as input, then we get the current divided in two branch; minimum current pass through the resistance R1 and maximum current pass through the inductor L5, as we know that dc is short there .Then the maximum current will flow through the coil L1 and L2 within a short difference of time .It is an astable multivibrator .So, it has no stable state .It switches back and forward  from one state to other .As, the capacitor C1 started to charge and we get a induced voltage in L3 because of L1 .

                                                   When the current from L1 passes through the transistor U2, it becomes on and then the U1 is in off situation .In other side, the current also flow in L2 .As a result C1 started to charge from opposite side become discharged from the first side .Here, we also get a negative  induced voltage in L3 .As a result U1 becomes on .It is a continious process .Then, this found ac voltage will be steped up in L4 coil .A capacitor C2 is connected, so that ac current can pass easily in that output circuit. Thus we get our required output ac voltage.

EXPERIMENTAL RESULTS:

Input  voltage, Vi=12V;

Output voltage, Vo=1587.4V;

Input  current, Ii=1.85A;

Output current, Io=12.60mA;

Input power, Pi=22.2W;

Output power, Po=20.1 W;

Vc= 11V;

Ve= 0.075V;

Vb = 3.8V;

Vbe = 3.8V;

Vce = 11V;

Vcb =  15V;