LED flashlight driving circuit principle

A cheap LED flashlight has appeared on the market. The front end of this flashlight has 5 to 8 high-brightness light-emitting tubes and uses 1 to 2 batteries. Due to the use of ultra-high-brightness light-emitting tubes, the luminous efficiency is very high and the working current is relatively small. In actual measurement, the current of a flashlight with 5 AA batteries is only about 100 mA. It is very energy-saving. If a large-capacity rechargeable battery is used, it can be used continuously for more than ten hours. The author bought one. After disassembling the front end, the circuit diagram was drawn based on the actual object, as shown in Figure 1.

Working principle:

After the power is turned on, VT1 is connected to the negative pole because R1 is connected to the negative pole, and the voltage across c1 cannot change suddenly. The potential of VT1 (b) pole is lower than that of e pole, VT1 is turned on, and current flows into VT2 (b) pole, VT2 is also turned on, and the current flows from the positive pole of the power supply through L and VT2 (c) pole to e pole, and then flows back to the negative pole of the power supply. The power supply charges L, 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, and VT1 enters deep saturation state, and VT2 also enters a deep saturation state, that is, Ib>Ic/β (β is the magnification factor). As the power supply charges c1, the voltage across C1 gradually increases, that is, the VTI(b) pole potential gradually increases, and Ib1 gradually decreases. When Ib1<=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, and is fed back by c1. The VT1 base potential rises further, VT1 is quickly cut off, and VT2 is also cut 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, achieving the purpose of boosting. This voltage is enough to make the LED light up.