Introduction to the principle of a 3.7V Lithium Polymer Battery LED lamp driver circuit

　　After the 220V AC power is reduced and current-limited by capacitors C1 and R1, the AC voltage at points A and B is about 15V. It is rectified by VD1~VD4. A DC voltage of about 14V is obtained on C2 as high-brightness light-emitting diodes VD5~VD8. The working voltage of the light-emitting diode is about 14mA. Since capacitor C1 does not consume active power and the power consumed by the bleeder resistor is negligible, the power consumption of the entire circuit is approximately 15×0.014≈0-2 (W).

　　This corridor light made of high-brightness light-emitting diodes turns off during the day and lights up automatically after dark. The electricity consumption per year does not exceed 2 degrees, and the service life can reach 10 years.

　　How the circuit works

　　The circuit of LED corridor light is shown in the figure below. The circuit consists of a capacitor step-down circuit, a rectifier circuit, an LED light-emitting circuit and a photoelectric control circuit.

　　After the 220V AC power is reduced and current-limited by capacitors C1 and R1, the AC voltage at points A and B is about 15V. It is rectified by VD1~VD4. A DC voltage of about 14V is obtained on C2 as high-brightness light-emitting diodes VD5~VD8. The working voltage of the light-emitting diode is about 14mA. Since capacitor C1 does not consume active power and the power consumed by the bleeder resistor is negligible, the power consumption of the entire circuit is approximately 15×0.014≈0-2 (W).

　　In order to further save electric energy and extend the service life of high-brightness light-emitting diodes, a photoelectric control circuit composed of photoresistor R2, resistor R3 and transistor VT1 is added to the circuit. At night, the resistance of photoresistor R2 can reach more than 100K. After the voltage at both ends of C2 is divided by R2 and R3, the DC bias voltage provided to the base of VT1 is very small, and VT1 is cut off, which has no impact on the work of the light-emitting diode; during the day, due to the photoelectric effect, the resistance of R2 can Reduced to less than 1OK, then VT1 is turned on and close to saturation. Since the maximum current through C1 can only reach 15mA, due to the shunt of VTl, the voltage on C2 can drop below 4V.

　　Because the working voltage of each light-emitting diode must reach more than 3V before it can start working, and four series connections must reach more than 12V, so the light-emitting diodes cannot emit light normally at this time.

　　So, after the VT1 shunt is short-circuited, why does the power consumption of the circuit become smaller? This is because the capacitor voltage reduction circuit can basically be regarded as a constant current source when the output voltage is not too high, and its power consumption is proportional to the load resistance. After VT1 is shunted to reduce the voltage on C2, it is equivalent to the load The resistance decreases, so the power consumption decreases. Take the AC voltage across A and B falling to 5V as an example. Assume that the AC current passing through C2 reaches 15mA. At this time, the power consumption of the circuit is about 5×0.015≈0.08 (W). Therefore, taking an average of 12 hours of lighting every night as an example, the electricity consumption in a year is: (0.2×12+0.08×12)×365≈1.2 (kWh), that is, 1.2 kilowatt hours of electricity.

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