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Dom> Blog> LED light circuit for "squeezing" dry battery power

LED light circuit for "squeezing" dry battery power

April 25, 2023
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This LED lamp uses only three transistors, and only needs 0.71V voltage to start LED illumination. A 5th alkaline battery can be used to continuously illuminate the lamp for 170 hours, which greatly improves the battery utilization.
The power supply voltage of the LED lamp is set according to the voltage range of the alkaline battery of No. 5 from 0.9V to 1.7V. Figure 1 shows the measured battery power supply. This circuit can operate to a 0.71V supply voltage.

The circuit is shown in Figure 2. The forward current of the LED follows the supply voltage change. The transistors Tr1, Tr2 form a current mirror circuit to stabilize the bias (DC operating point). During the turn-off of Tr3, the bases of Tr1 and Tr2 are slightly reversed, and no current flows through R1. As a result, the circuit can oscillate stably over a wide temperature range and battery voltage, as well as the error range of the components. Fig. 3 is a collector voltage waveform of Tr1.

The DC resistance of the inductor L1 in Figure 2 is 4Ω. Its large resistance limits the current when Tr3 or LED is shorted.

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