This is the Simple 100W inverter 12V to 220VAC Circuit Diagram 2. To lower or raise any specific voltage is better adapted from a transformer, but this component does not operate in direct current, which is available in battery or vehicle. Therefore we must put an oscillator that generates an alternative current.
The Integrated circuit (CD4047) is an oscillator that has the output inversely to one another. This means that while if one is high, other states are low and vice versa. The signals are too weak to drive the transformer so that the driver applied consists from three transistors in the chain. Diode in parallel with the end of each transistor prevent reverse current produced by removing the coil current that can burn the transistor. 5A diode placed in parallel with the power line creates a short circuit when the polarity is accidentally reversed, causing the fuse to skip. The 50K preset adjusts the oscillator frequency, which is directly proportional to the frequency of the AC produced in the transformer. In order for to work stable the oscillator is provided a 220 Ω current-limiting resistor and Zener 9.1V with a filter capacitor.
Common transformer is used to make a lower voltage, but in this circuit will be used inversely (step up). The resulting transformer calculation here is 220v and v 9.3+9.3V, but finding a suitable transformer which sold in stores is difficult, so here which commonly used voltage 9V+ 9V. With this voltage the capacity should be at 100VA.
This 100W inverter circuit need power supply : V max: simple 12V DC, I max: 10A
100W Inverter Calibration: Simply power the system and put a frequency counter or oscilloscope at the output of the transformer. Turn the preset of 50 kΩ located at 4047 until the measured frequency is 50Hz. After this the calibration is complete.
IMPORTANT:
This 100W inverter equipment generates alternating current whose waveform is square. This is because the transistors are arranged in cut / saturation. There is no problem for this inverter to supply resistive equipment such as welders, light or source. But for TV set or video recorder which is used as the reference frequency and wavelength of the network may not be working correctly.
The Integrated circuit (CD4047) is an oscillator that has the output inversely to one another. This means that while if one is high, other states are low and vice versa. The signals are too weak to drive the transformer so that the driver applied consists from three transistors in the chain. Diode in parallel with the end of each transistor prevent reverse current produced by removing the coil current that can burn the transistor. 5A diode placed in parallel with the power line creates a short circuit when the polarity is accidentally reversed, causing the fuse to skip. The 50K preset adjusts the oscillator frequency, which is directly proportional to the frequency of the AC produced in the transformer. In order for to work stable the oscillator is provided a 220 Ω current-limiting resistor and Zener 9.1V with a filter capacitor.
Simple 100W inverter 12V to 220VAC Circuit Diagram 2
Common transformer is used to make a lower voltage, but in this circuit will be used inversely (step up). The resulting transformer calculation here is 220v and v 9.3+9.3V, but finding a suitable transformer which sold in stores is difficult, so here which commonly used voltage 9V+ 9V. With this voltage the capacity should be at 100VA.
This 100W inverter circuit need power supply : V max: simple 12V DC, I max: 10A
100W Inverter Calibration: Simply power the system and put a frequency counter or oscilloscope at the output of the transformer. Turn the preset of 50 kΩ located at 4047 until the measured frequency is 50Hz. After this the calibration is complete.
IMPORTANT:
This 100W inverter equipment generates alternating current whose waveform is square. This is because the transistors are arranged in cut / saturation. There is no problem for this inverter to supply resistive equipment such as welders, light or source. But for TV set or video recorder which is used as the reference frequency and wavelength of the network may not be working correctly.
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