An Inverter is a small circuit that converts direct current (DC) to alternating current (AC). The power of a battery is converted into main voltages or AC voltage. We can use AC power for electronic appliances like television, mobile phones, computers, etc. The basic function of an inverter is to convert DC to AC. A step-up transformer is used to create main voltages. In this article, we are going to design a 100 watts inverter circuit.
Block Diagram of 100 watts inverter circuit
The blog diagram of the inverter is given below
The block diagram explains that the battery supply is given to the MOSFET driver where it will oscillate the DC current and the resulting oscillate current is given to the step-up transformer then the step-up transformer gives AC Power.
Components Required for 100 watts inverter circuit
- IC – CD4047 – 1
- MOSFET IRF540 -2
- Resistor – 330 ohm – 1
- Resistor – 220 ohm – 2
- Resistor – 1k ohm – 2
- Resistor – 390k ohm – 1
- Diode – 1N4007 – 1
- Switch – 1
- Capacitor – 0.01uF – 1
- Capacitor – 0.1uF
- Transformer – 150VA – 1
- 12v Battery-1
Circuit Diagram of the inverter circuit
Working Detail of 100 watts Inverter circuit
CD4047 is a multi-vibrator with very low power consumption. It can operate in a mono-stable multi-vibrator and also a stable multi-vibrator. In a stable multi-vibrator mode, it operates in free-running or gate-able modes and also provides good stable frequency stability. It generates a 50% duty cycle that creates a pulse that is applied to the inverter circuit. This is the main use of IC CD4047.
IRF540 is an N-channel MOSFET that is mainly used in switching regulators, switching converters relay drivers, etc. The reason behind using MOSFET in the INVERTER circuit is a high switching transistor that works in very low gate drive power and has high input impedance.
Explanation of 100 watts inverter circuit
- In this 100 watts inverter circuit diagram, we observe that the 12V battery is connected to the diode LED and also connected to the pin8, PIN7, PIN9, and PIN12 of the IC. The positive power supply is also to pins 4,5,6 and 14 which are stable and complement the stability of the IC CD 4047. The Basic function of the Diode in the circuit is to prevent reverse current.
- CD4047 IC works in the stable multi-vibrator mode To work it in a stable multi-vibrator mode we need a capacitor that is connected between the pin1 and pin3 of the IC. Pin2 is connected through the resistor and a variable resistor to change the output frequency of the inverter. Pins 10 and 11 of the IC are connected to the gate of the IRF540. The output frequencies of pins 10 and 11 generate frequency with a 50% duty cycle.
- The output frequency is connected to the MOSFET through 220-ohm resistors which will help to prevent the loading of the MOSFET. The main AC current is generated by both MOSFET (IRF540). Here MOSFET acts as two electronic switches.
- The output AC is given to the step-up transformer of the secondary coil from this coil to increase AC voltage. These voltages depend on the power of the transformer.
- In this way, DC is converted into AC Current.
- The battery can be a 12V/ 6Ah lead-acid battery.
- MOSFET must be fitted with one proper heat sink to protect.
- Transformer can be a 9-0-9 V primary, 230V/110v secondary, or 150VA transformer.
- There is a very simple one suitable for low-grade applications not for high watts output.