How to make 12 Volts to 240 Volts Inverter

How to Build a Low Inverter


Operating Principle of an Inverter:

An inverter can be taken as a crude form of UPS. Obviously the main use of an inverter is only for powering common electrical appliances like lights and fans during a power failure.
As the name suggests the basic function of an inverter is to invert an input direct voltage (12VDC) into a much larger magnitude of alternating voltage (generally 110VAC or 220VAC).
Fundamental elements of an inverter and its operating principle:

Oscillator: An oscillator converts the input DC (Direct Current) from a lead acid battery into an oscillating current or a square wave which is fed to the secondary winding of a power transformer. In the present circuit, IC 4049 has been used for the oscillator section.

Transformer: Here the applied oscillating voltage is stepped up as per the ratio of the winding's of the transformer and an AC much higher than the input DC source becomes available at the primary winding or the output of the inverter.
Charger: During power backups when the battery gets discharged to a considerable level, the charger section is used to charge the battery once the AC mains is restored.


Working Procedure:

1. As per the circuit schematic first complete the assembly of the oscillator section consisting of the smaller parts and the IC. It is best done by interconnecting the component leads itself and soldering the joints.

2. Next fit the power transistors into the appropriately drilled aluminum heat sinks. These are made by cutting an aluminum sheet into the given sizes and bending them at the edges so that it can be clamped.

3. not fit the transistors directly on to the heat sinks. Use mica isolation kit to avoid direct contact and short circuiting of the transistors with each other and the ground.

4. Clamp the heat sink assembly to the base of a well ventilated, sturdy, thick gauge metallic enclosure.

5. Also fix the power transformer beside the heat sinks using nuts and bolts.

6. Now connect the appropriate points of the assembled circuit board to the power transistors on the heat sinks.

7. Finally join the power transistor’s outputs to the secondary winding of the power transformer.

8. Finish the construction by fitting and interconnecting the external electrical “fittings” like fuses, sockets, switches, mains cord and the battery inputs.

9. An optional separate power supply circuit using a 12V/3Amp. transformer may be added inside to charge the battery whenever required ......(see diagram)








Circuit Description:


1. Gates N1 and N2 of IC 4049 are configured as an oscillator. It performs the primary function of supplying square waves to the inverter section.

2. Gates N3... N6 are used as buffers so that the circuit is not load dependant.

3. Alternating voltage from the buffer stage is applied to the base of the current amplifier transistors T1 and T2. These transistors conduct in accordance with the applied alternating voltage and amplifies it to the base of the output transistors T3 and T4.

4. These output power transistors oscillate at a full swing, delivering the entire battery voltage into the each half of the secondary winding alternately.

5. This secondary voltage is induced in the primary winding of the transformer and is stepped-up into a powerful 230 volts (AC). This voltage is used to power the output load.


Testing Procedure:

1. Begin the testing procedure by connecting a 100 watt bulb at the output socket of the inverter

2. Insert a 15 Amp./12V fuse inside the fuse holder.
3. Finally connect a 12V automobile battery to the battery inputs of the inverter.

4. If all the connections are right, the 100 Watt bulb should immediately light up brightly.

5. Keep the inverter ON for an hour and let the battery discharge through the bulb

6. Then shift the given toggle switch to the charging mode, check the meter reading,

7. The meter should indicate the charging current of the battery.

8. The meter reading should gradually die down to zero after a span of time, confirming that the battery is fully charged and ready for the next cycle.




*Disclaimer:   This project should be taken on at your own risk and is recommended for those who have experience building their own circuits. Neither the author of the article, nor allengineeringsolution4u bears no responsibility for negative outcomes to this tutorial....Good luck.

Power supply How to convert AC to DC (12v ac to 5v dc)

You need a 5v Dc but you have 12v Ac. Now what do you have to do.


The simplest circuit will be using a voltage regulator like LM 7805 These regulators are very common.
It is simple to build: connect the 12V wire at the left most terminal of the IC,


Now I'm trying to use a wall transformer to step the voltage down to 12VAC

 You need a Voltage regulator IC 
Example:

5v  regulator IC number is 7805

9v regulator IC number is 7809


It usually brings on 3 pins 


  





Pin No 1 is Voltage input

(it's maybe 12 v battery )


pin no 3  output voltage 

pin no 3 common Ground





while looking at the inscription and with the pins down. Connect the 5V output for USB port at the right most pin. Connect the ground of your 12V supply and of your USB to the middle pin or to the heatsink. You may also add two capacitors like:


now

1st step is diode bridge and ripple capacitor.
          The ripple voltage equation is:
Vripple=I2fC


Here;     

I = Load Current (1A)
 f= AC Frequency(60hz)
 C= Filter Capacitor (?Uf)


 
1000 µF at this voltage isn't terribly big. Are you limited by size or something?
To completely get rid of the ripple and produce 5 V, you need to add a voltage regulator after the capacitor.

12 VRMS = 17 VPeak, which, minus the two diode drops, is the peak DC voltage you'll see at the output of the rectifiers: 17 - 1.1 - 1.1 = 14.8 V. So there's no threat of exceeding the input limits of the regulator (35 V input).
If the ripple is 8.3 V, then the DC voltage will be varying from 6.5 V to 15 V. This is just barely high enough to feed into the regulator without dropping out of regulation, since the 7805 has about 1.5 V dropout at 1 A (depending on temperature). So yes, you should use a slightly higher capacitor (or multiple capacitors in parallel, if space is an issue).





Real life power line voltages vary from one outlet to the next, and the frequency varies by country

  • JP: 85 VAC to 110 VAC (+10%, -15%), 50 and 60 Hz
  • US: 105 VAC to 132 VAC (+10%), 60 Hz
  • EU: 215 VAC to 264 VAC (+10%), 50 Hz