How to make a lead-acid battery discharge protector

by:Power Kingdom     2021-06-13
Tags: How to make a lead-acid battery discharge protector When many people recycle lead-acid batteries, they charge them until they are dead. This is a very big misunderstanding, and it is extremely easy to cause over-discharge, affecting the life of the battery itself. In serious cases, the battery can be directly scrapped and there is no need to reuse it. In addition, it is generally recommended to use lead-acid batteries at a depth of about 30%, so that the battery life cycle times can reach the highest. So we can make a discharge protector so that the battery will not be over-discharged. Next, I will design the protector with a minimum voltage of 9V, but generally the termination voltage must be above 10.8V. Let us first look at the structure of the protector: as shown in the figure, the components needed in the protector are as follows: R1~R4 select 1/4W carbon film resistors or metal film resistors. Both C1 and C2 use tantalum electrolytic capacitors with a withstand voltage greater than 25V. Both VL1 and VL2 select φ5mm light-emitting diodes. VS selects the silicon steady voltage diode of 1W, 9V for use. V1 selects S8550 or C8550 silicon PNP transistor for use; V2 selects S8050 or C8050 silicon NPN transistor for use. VT1 selects 3~5A, 100V thyristor; VT2 selects 1A, 100V thyristor. We can also see the working principle of this protector according to the above figure: the under-voltage protection circuit is composed of transistor V2, Zener diode VS, thyristor VT1, VT2, capacitors C1, C2 and resistors R1~R3. The indicating circuit is composed of a light-emitting diode VL1, indicating that the battery voltage is normal, indicating light-emitting diode VL1, a battery under-voltage indicating light-emitting diode VL2, a transistor V1, a voltage stabilizing diode VS, and resistors R1, R2, and R4. When the terminal voltage of the battery GB is above 9V, VS breaks down and turns on, so that V1 is reversely biased and V2 is turned on. The high level of the emitter output of V2 provides a trigger level for the gate of VT1 through C1, so that VT1 is turned on. On, the battery supplies power to the load. At this time, VL1 lights up, but VL2 does not light up. When the terminal voltage of the battery GB drops to 9V, VS is cut off, making V2 reverse bias cut off, V1 is turned on, and the high level of the collector output of V1 provides a trigger level for the gate of VL2 through C2, so that VT2 is turned on. The charge charged on C1 is added to the cathode of VT1 via VL2, so that VL1 is cut off due to the momentary reverse voltage applied to the cathode, and the power supply path between GB and the load circuit is cut off. At the same time, VL1 goes out and VL2 lights up, indicating that the battery voltage is insufficient.
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