Current fed push-pull inverter circuit diagram
Source: InternetPublisher:elleny Keywords: Circuit diagram inverter circuit WIDTH Updated: 2020/06/12
The current -fed push-pull inverter circuit is shown in Figure 1. In the figure, the DC voltage is sent to the center tap of the transformer Tr through the inductor L1. L1 and the capacitor C2 connected across the two ends of the primary winding of Tr form a formal resonant circuit. R1 and R2 , C1 form a starting circuit, the principle is the same as Figure 2. Due to the positive feedback effect of Np and Nb, VT1 and VT2 are driven to conduct alternately.
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In this circuit, the highest voltage that the switching transistor collector can withstand is approximately π times the DC voltage VDC. For the United States, Japan and Taiwan where the mains voltage is 110V/120V/127V, this circuit is still suitable. The output of the transistor in this circuit is a sinusoidal voltage, the switching loss is small, and the output at both ends of the secondary NS of the transformer is also a sinusoidal voltage. Even if the load is open-circuited and short-circuited, and the load changes greatly, the inverter can still work continuously. In Figures 1 and 2, even if one lamp fails, the circuit can still work normally.
The electronic ballast with two lamps produced by Motorola in 1996 adopts this circuit mode. The original lamp tubes are two 32W cold cathode T8 tubes, so there is no need for an auxiliary winding to heat the filament. The specific circuit is shown in Figure 2 Show.
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In the figure, C1, R1 and VD1 form the starting circuit. The high-frequency inverter circuit is composed of VT1, VT2, transformer Tr, C2, etc. The transformer provides positive feedback, causing VT1 and VT2 to turn on and off alternately. This circuit is relatively simple and does not use many components. In the above 2×32V circuit, Tr can use EE35 cores and L1 can use EE19 cores. The volume and weight are not large, and the material cost is not high. The specific circuit is shown in Figure 2.
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