Diagram of a simple battery charger

 Violation of the battery operating mode (due to improper operation of the vehicle relay regulator, or long-term storage) almost always leads to sulfation of the plates. As a result, the internal resistance of the battery increases, and even when charged, it cannot produce the necessary starting current. The simplest way to resuscitate such a battery is charging in training mode, when during one period of mains voltage the battery is charged with a current 5-10 times lower than the battery capacity, within one half cycle, and discharge with a current 50-100 times lower than the battery capacity.

Usually, after ten hours of this mode, most sulfated batteries return to normal.

The figure shows the diagram of the simplest device that implements such a mode. During the positive half-cycle, an opening voltage appears on the basis of the compound transistor, which is set by resistors R1 and R2. The transistor is opened and a charging current is supplied to the battery through it. The magnitude of this current depends on the degree of opening VT1, and therefore on the position of the R2 engine. The charging current flowing through the battery is measured by the P1 ammeter.

With the transition of the mains voltage through zero, the VT1 transistor closes, and during the negative half-wave of the mains voltage, the battery is discharged through the powerful resistor R3.

The P2 voltmeter is used to monitor the voltage on the battery. It should not be allowed to be more than 14V.

If the battery is strongly sulfated, its internal resistance will be high, and even with a small charging current, an increased voltage (16-17V) will drop on it, this should not be allowed, and at the first stage of resuscitation, you need to set the current with the R2 resistor so that the voltage on the battery will be no more than 14-14.5V, and then, after 15-30 minutes, gradually increase the current, observing that the voltage does not exceed 14 V.

The maximum current that this device produces is up to 15 A, if you need to accelerate the charging of the battery, you can set the current to 10-12 A. But at the same time, you need to make sure that the electrolyte does not boil (remove one of the caps, and if active bubbling is visible, reduce the current to such a level that it is not there).

D242 diodes can be replaced with any other diodes with a current of at least 10 A, for example, KD213, D243, KD202. The KT827 transistor can be replaced with the KT825, but at the same time the polarity of the connection of the diodes, P1, P2 and the battery can be changed.

The P1 ammeter has a current of up to 3-5 A, but its scale needs to be recalibrated, because its readings will be 2.5 times lower, that is, if the ammeter shows 3 A, then in fact it is 7.5 A. Voltmeter of any direct current. The voltmeter readings do not need to be corrected, but they will be real only when the battery is connected.

As a basis for the transformer, a TS200 power transformer (TC 180 is also used) from old tube TVs is used. It is necessary to remove all its secondary windings, then wind new ones - two windings of 40 turns each (on different coils of the transformer). And then connect them in the same way as the mains windings are connected.

The transistor and diodes should be on the heat sinks, especially the transistor. As a heatsink for the transistor, you can use the metal case of the device, but not connect it to other circuits, or isolate the transistor with dielectric gaskets (mica).

For diodes, a metal bracket with an area of at least 50 cm2 can be used as a heat sink, which is fixed inside the case on insulating racks so that it does not come into contact with the device housing.

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