As the 200 watt inverter, Many people want to create a project of large power inverter about 500 watts. In order to use the multiple appliances at the same time. And can also be used on a large bus as a bus or coach, too.
Adding a 200watt into 500 watts is simple.We use the same PCB’s, and add the power output PCB. Large transformer is 500 watts or 40A, it can apply the energy output as needed.
Increased wattages ideas
We Increased electricity, by the same voltage, thus making inverter can apply the output power higher up. That adding the mosfet parallel to each other. And add transformer size as Figure 1.
How to modify to…
500W power inverter circuit using SG3526-IRFP540
As 200 watts inverter circuit. We use Q1, Q2 is the mosfet acts as a power output. That can withstand currents up to 18A. According to properties listed in the table of Figure 2. If the circuit is fully functional with maximum power of 12V x 18A = 216 watts. But in practice, the circuit should work up, it may be damaged. Therefore, it is designed to operate up to 200 watts.
Since we want the current more than 40A. But the power mosfet single, able to withstand current is 18A, therefore have to be 3 pcs. And When designing a circuit in a push-pull model thus requires a power mosfet 3 pairs.
We have the output current up to 54A which can withstand more demanding than 14A. It is good for the operation of the circuit. Because each power mosfet will not be overburdened. And the heat of the mosfet least, the lifetime of the life circuit.
In Figure 3 is the inverter circuit with a power output Increased at least 2 pairs and use the 12volts 40A transformer, so that the power output 500 watts.
Figure 3 The schematic diagram of this projects
We increase the over current checker resistors from two into tree (R22 – R23 parallel to the R8.) The operation of the circuit remains the same.
We use the same PCB as Figure 4 PCB layout and components layout in Figure 5
We set up the device the same except power mosfet (T1 and T2). The resistors R8, R13 and R15 wired jumper instead. Because the resistance both moved to the power output PCB. As Figure 5 PCB layout
Use the large wires more than about 8mm. Because it use high current load.
The parts you will need
Resistor (0.25W 1% Except where indicated)
R1______________15K_______________ = 1 pcs.
R2______________22K_______________ = 1 pcs.
R3______________2.7K_______________ = 1 pcs.
R4______________10K_______________ = 1 pcs.
R5______________12K_______________ = 1 pcs.
R6______________4.7K_______________ = 1 pcs.
R7______________47K_______________ = 1 pcs.
R8,R22,R23______0.01 ohms / 5W ______ = 3 pcs.
R9______________1K (PTC)___________ = 1 pcs.
R10_____________ 8.2 ohms___________ = 1 pcs.
R11______________16.9K______________ = 1 pcs.
R12______________33 ohms____________ = 1 pcs.
R13,R15, R18- R21_ 22 ohms___________ = 6 pcs.
R14______________ 18 ohms___________ = 1 pcs.
R16______________ 1K ________________ = 1 pcs.
R17______________ 470 ohms___________ = 1 pcs.
C1,C2____________220uF 16V____Electrolytic____= 2 pcs.
C3,C7,C9,C10_____220nF 50V____Polyester_____= 4 pcs.
C4______________1uF 50V________Electrolytic____= 1 pcs.
C5,C6___________33nF 50V_______ Polyester_____= 4 pcs.
C8___________2.2 uF 50V_______ Polyester_____= 1 pcs.
D2__________1N4148_____75V 150mA Diodes__= 1 pcs.
D2__________1N4002_____100V 1A Diodes__= 1 pcs.
D4__________18V 1W_____Zener Diodes___= 1 pcs.
D5,D6_______BY299______Didoes__________= 2 pcs.
IC1_________LM393N____op-amp IC________= 1 pcs.
IC2_________SG3526N or SG2526N IC______ = 1 pcs.
Q1-Q6______IRFP540 mosfet______________ = 6 pcs.
D1_________LED 3 mm red or as you need____ = 1 pcs.
T1________220V / 12V-0-12V / 40A transformer__ = 1 pcs.
PCBs,Heat sink, wires and others.
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