LED flashlight--the working principle of LED flashlight driver

After the power is turned on, VT1 is connected to the negative pole because of R1, and the voltage at both ends of c1 cannot change abruptly. VT1(b) pole potential is lower than e pole, VT1 is turned on, VT2(b) has current flowing in, VT2 is also turned on, the current flows from the positive pole of the power supply through L, VT2(c) pole to pole e, and flows back to the negative pole of the power supply. The power supply charges L, and L stores energy, and the self-induced electromotive force on L is positive on the left and negative on the right. Through the feedback of c1, the base potential of VT1 is lower than the emitter potential, VT1 enters a deep saturation state, and VT2 also enters a deep saturation state, that is, Ib>Ic/β (β is the magnification). As the power supply charges c1, the voltage across C1 gradually rises, that is, the potential of VTI(b) rises gradually, and Ib1 gradually decreases. When Ib1<When =Ic1/β, VT1 exits the saturation zone, VT2 also exits the saturation zone, and the charging current to L decreases. at this time. The self-induced electromotive force on L becomes negative on the left and positive on the right, which is fed back by c1. The base potential of VT1 further rises, VT1 cuts off quickly, and VT2 also cuts off, the energy stored on L is released, and the power supply voltage on the light-emitting tube is added to L to generate a self-induced electromotive force to achieve the purpose of boosting voltage. This voltage is enough to make the LED glow.