Most common low-power supplies are flyback circuits, and some LED power supplies are single-ended or push-pull, all in single-ended form, where the magnetic core can only utilize the first quadrant. However, push-pull and bridge circuits use the first and third quadrants of the magnetic core.

Therefore, with the same core and frequency, the power of push-pull and bridge circuits will be greater compared to the others.

Previously, the push-pull self-excited circuit used a transistor as the main driving device, but there were some drawbacks, such as small amplification, difficult driving, large forward voltage drop, and susceptibility to high-frequency self-excitation and damage.

However, with the emergence of MOSFETs, they have a greater advantage in high-power scenarios. So we use MOSFETs for push-pull self-excitation. How does it work?

Let's explore around the circuit diagram below:

When switch S1 is closed, the relay coil is energized, and the power supply +V is supplied to the center tap of N1 N2 (transformer) through inductor L1. In addition, the current also flows through capacitor C1 and resistor R2 to the center tap of N3, N4.

At this point, the current in N3, N4 flows through resistors R3, R4 to the gates of Q1 and Q2.

If we were to think in the usual way, we might think that Q1 and Q2 would conduct and burn out, but that's not the case.

The reason is that N1 and N4 are of the same polarity, as are N2 and N3. However, in practice, the inductance, internal resistance, and other parameters of the two sides will not be the same, including the on-resistance of the MOSFETs.

Let's assume that Q1 conducts first. Then, there will definitely be a current on N1, which will immediately induce current/voltage on N3, N4. Since N4 is of the same polarity, it will superimpose current/voltage on the gate-source of Q1 and rapidly amplify it, causing Q1 to conduct completely.

N3, on the other hand, is not of the same polarity, so it will also generate an opposite electromotive force (from weak to strong), causing the gate-source of Q2 to produce negative pressure and decisively turn off.

After Q1 conducts, the winding of transformer N1 reaches magnetic saturation, and the current increases. At this point, the magnetic flux no longer increases, and N4 cannot induce current/voltage to N3, N4. N1 will rapidly generate a strong reverse electromotive force, and the electromotive force of N4 will also reverse, causing Q1 to cut off.

At this point, the current on N1 reverses, becoming the same polarity as N3. N3 will then provide current/voltage to Q2 through resistor R3, and Q2 will immediately conduct.

The next steps are the same as before.

When the winding of transformer N2 reaches magnetic saturation again, N2 also generates a reverse electromotive force, causing N3 to also generate a reverse electromotive force, and then Q2 is turned off, while N4 induces current causing Q1 to conduct.

This cycle repeats itself, creating oscillation.

In fact, the role of inductor L1 is to decouple and limit the peak current. This prevents the circuit's no-load current from being too large, which could otherwise cause the switching MOSFETs to overheat and be damaged.

It should be noted that boost rectification is used here. In order to reduce the turns ratio of the transformer and the low AC voltage, it is not easy to breakdown the enameled wire, reducing the danger.

When the push-pull power supply is operating, only one of the two symmetrical power switch tubes generally conducts, so its conduction loss is small. Coupled with its simple structure and high utilization of transformer magnetic cores, push-pull power supplies are widely used in low-voltage, high-current applications.

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