LED flashlight drive circuit schematic diagram

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 luminous tubes, the luminous efficiency is very high and the working current is relatively small. The actual measurement uses one AA battery and 5 flashlights, and the current is only about 100mA. Very power saving. If you use a large-capacity rechargeable battery, it can be used continuously for more than ten hours, so I bought one. After disassembling the front end, a circuit diagram was drawn based on the actual object, as shown in the figure.

working principle:

After the power is turned on, VT1 is connected to the negative pole because R1 is connected, and the voltage at both ends of c1 cannot change suddenly. The potential of VT1(b) pole is lower than that of e pole, VT1 is turned on, 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 the e pole, and then flows back to the negative pole of the power supply. The power source charges L, L stores energy, and the self-induced electromotive force on L is positive on the left and negative on the right. After the feedback effect 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 amplification factor). As the power supply charges c1, the voltage across C1 gradually increases, that is, the potential of VTI(b) gradually increases, and Ib1 gradually decreases. When Ib1<=Ic1/β, VT1 exits the saturation zone, and VT2 also exits the saturation zone. , 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 base potential of VT1 further rises, VT1 quickly cuts off, 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, achieving the purpose of boosting the voltage. This voltage is enough to make the LED glow.