Dear sir, Swagatam
I have a 2N6292 . My friend suggest me to make the simple fixed voltage high current DC power supply to charge the 12 v battery. He had given the attached rough diagram. I don't know anything about the above transistor. Is it so ? My input is 18 volt 5 Amp transformer. He told me to add 2200 uF 50 Volt capacitor after rectification. Is it works ? If so , is there any heat sink necessary for transistor or/and IC 7815 ? Is it stops automatically after battery reaches 14.5 volt ?
Or any other alteration needed ? Please guide me sir
Yes it will work and will stop charging the battery when around 14 V is reached across the battery terminals.
but I am not sure about the 1 ohm base resistor value...it needs to be calculated correctly.
The transistor and the IC both may be mounted on a common heatsink using mica separator kit. This will exploit the thermal protection feature of the IC and will help safeguard both the devices from overheating.
The shown single transistor high current battery charger circuit is a smart way of charging a battery and also achieving an auto shut off when the battery attains a full charge level.
The circuit is actually a simple common collector transistor stage using the shown 2N6292 power device.
The configuration is also referred as an emitter follower and as the name suggests the emitter follows the base voltage and allows the transistor to conduct only as long as the emitter potential is 0.7V lower that the applied base potential.
In the shown single transistor high current battery charger circuit, the base of the transistor is fed with a regulated 15 V from the IC 7815, which ensures a potential difference of about 15 - 0.7 = 14.3 V across the emitter/ground of the transistor.
The diode is not required and must be removed from the base of the transistor in order to prevent an unnecessary drop of an extra 0.7 V.
The above voltage also becomes the charging voltage for the connected battery across these terminals.
While the battery charges and its terminal voltage continues to be below the 14.3 V mark, the transistor base voltage keeps conducting and supplying the required charging voltage to the battery.
However as soon as the battery begins attaining the full and above 14.3 V charge, the base is inhibited from a 0.7 V drop across its emitter which forces the transistor to stop conducting and the charging voltage is cut off to the battery for the time being, as soon as the battery level begins going below the 14.3 V mark, the transistor is switched ON again...the cycle keeps repeating ensuring a safe charging fr the connected battery.
Base resistor = Hfe x battery internal resistance