Assume that the circuit is ready to work, that is, capacitors C1 and C2 are fully charged. At time t=0, the power tube V1 is turned on, and the current I1 starts from the positive pole of the capacitor C1, as shown by the hollow arrow, flows through the power tube V1, the BA of the primary winding N1 of the transformer Tr, and returns to the negative pole of C1 until t= t1, forming a positive half wave, as shown in Figure 4(b). At t=t1, V1 is cut off due to the disappearance of the positive trigger signal. #6000w inverter 48v At this time, the positive trigger signal is added to the control pole of V2 to turn it on. The current I2 starts from the positive electrode of the capacitor C2 and flows through AB of the primary winding N1 of the transformer Tr. As shown by the solid arrow in the figure, it can be seen that the current direction is opposite at this time. The current I2 flows through the transformer and flows through the collector-emitter of the power tube V2 back to the negative electrode of the capacitor C2 until t=t2. The trigger signal disappears and ends, and this process forms a negative half wave, as shown in Figure 4(b). After repeating the above process, a series of continuous sine waves are formed.

Please kindly take a view and better to understand us:

48v off grid inverter