Introduction
Suppose you’re making a project that operates on DC but suddenly you got the realization that you need an AC component for your circuit. May be an AC light or an AC motor. What would you do? Now, you need an AC supply for that but your device has a DC supply. So, you require an AC inverter circuit. With some modification, the following circuit can also be adopted in UPS and other DC to AC converter applications.
To understand the whole circuitry we are here with this exciting “Simple Inverter Circuit”, which requires fewer components that are easily available at cheap prices. We are using a 15W LED bulb as our load but you can use any comparative AC load.
Hardware Components
The following components are required to make the Inverter Circuit
| S.no | Component | Value | Qty |
|---|---|---|---|
| 1. | Transistor | MJE13007 | 2 |
| 2. | Center tapped Transformer | 12-0-12V | 1 |
| 3. | LED Bulb | 15w | 1 |
| 4. | Resistor | 330 Ohm | 2 |
| 5. | Battery | 12V | 1 |
MJE13007 Pinout
For a detailed description of pinout, dimension features, and specifications download the datasheet of MJE13007
Inverter Circuit
Working Explanation
When we supply 12V to this Simple Inverter Circuit One of the transistors goes into the conductive stage. Therefore, one-half of the coil of the transformer conducts. The transistor T1 remains conductive until the breakdown occurs. Now transistor T2 conducts. This whole process of Push-Pull Configuration repeats again and again. From the coil of the transformer, we get the 220V output.
Application and Uses
- It can be used to operate AC devices.
- You can make this simple project to operate the AC lights or bulb.
- With some modification, it can be used in power supplies, UPS circuits, etc.
