Today I will make a simplest circuit for 100 watt inverter for generating 12v DC to
Notes.
220v AC. So input voltage 12V (battery 12V) to output volt 220V AC 50HZ, it is Easy circuit because less component to use.
- It can be used as inverters for home needs to enable light loads (electric bulb, CFL, etc) at the time of electricity failure.
- You can construct this circuit of simple inverter at a cheap rate with locally available components.
- https://advancetax.in/
1. using IC CD4047 and MOSFET IRF540
COMPONENT LIST:-
R1-330Ω
R2,R6-1K
R3,R4-220Ω
R5-390K
C1-0.01uf
C2-2200uf/25v
D1-1N4007
Q1,Q2-IRF540
IC-CD4047
TX-9-0-9(10A/5A)
12V BATTERY
LED INDICATOR
Notes.
- B1 can be a 12V/ 6Ah lead acid battery.
- Q1 and Q2 must be fitted to a proper heat sink.
- T1 can be a 9-0-9 V primary, 230V secondary, 150VA transformer .
- Do not expect much from this circuit. The is very simple one suitable for low grade applications.
- https://advancetax.in/
2. Another one is using CD4047 and 2N3055
COMPONENT LIST:-
R1 = 250K (Variable resistor/POT)
R2,R3-4.7K
R4,R5,R6,R7-0.1Ω(5W)
C1 = 0.022uF
C2 = 220uF-25V
D1 = BY127 Diode
D2 = 9.1V Zener Diode
Q1, Q4 = TIP122 Transistor
Q2, Q3, Q5, Q6 = 2N3055 Transistor
F1 = 10A Fuse
IC1 = CD4047
TX = 9-0-9v(10A/5A)
THANK YOU.......
C2 = 220uF-25V
D1 = BY127 Diode
D2 = 9.1V Zener Diode
Q1, Q4 = TIP122 Transistor
Q2, Q3, Q5, Q6 = 2N3055 Transistor
F1 = 10A Fuse
IC1 = CD4047
TX = 9-0-9v(10A/5A)
Notes.
- Use the POT R1 to set the output frequency to50Hz.
- Use a 10 A fuse in series with the battery as shown in circuit.
- Remember,this circuit is nothing when compared to advanced PWM inverters.This is a low cost circuit meant for low scale applications.
THANK YOU.......
is it working 100%
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