100w inverter circuit

Depiction 

Here is a 100 Watt inverter circuit utilizing least number of components.I think it is very hard to make a not too bad one like this with advance less components.Here we utilize CD 4047 IC from Texas Instruments for creating the 100 Hz heartbeats and four 2N3055 transistors for driving the heap. 

The IC1 Cd4047 wired as an astable multivibrator produces two 180 degree out of stage 100 Hz beat trains.These beat trains are preamplifes by the two TIP122 transistors.The out puts of the TIP 122 transistors are opened up by four 2N 3055 transistors (two transistors for every half cycle) to drive the inverter transformer.The 220V AC will be accessible at the optional of the transformer.Nothing complex simply the basic inverter standard and the circuit works awesome for little loads like a couple of knobs or fans.If you require only a minimal effort inverter in the area of 100 W,then this is the best. 

Circuit Diagram with Parts List. 

100-w-inverter-circuit http://www.projectsworld.wordpress.com 

Notes. 

A 12 V auto battery can be utilized as the 12V source. 

Utilize the POT R1 to set the yield recurrence to50Hz. 

For the transformer get a 9-0-9 V , 10A stage down transformer.But here the 9-0-9 V winding will be the essential and 220V winding will be the optional. 

On the off chance that you couldn't get a 10A evaluated transformer , don't stress a 5A one will be quite recently enough. Be that as it may, the permitted out put power will be lessened to 60W. 

Utilize a 10 A wire in arrangement with the battery as appeared in circuit. 

Mount the IC on an IC holder. 

Remember,this circuit is nothing when contrasted with cutting edge PWM inverters.This is an ease circuit implied for low scale applications. 

Configuration Tips. 

The most extreme permitted yield energy of an inverter relies upon two factors.The greatest current rating of the transformer essential and the present rating of the driving transistors. 

For instance ,to get a 100 Watt yield utilizing 12 V auto battery the essential current will be ~8A ,(100/12) on the grounds that P=VxI.So the essential of transformer must be evaluated over 8A. 

Likewise here ,every last driver transistors must be appraised over 4A. Here two will lead parallel in every half cycle, so I=8/2 = 4A . 

These are just harsh estimations and enough for this circuit.