Lead acid batteries are normally used for heavy duty operations involving many 100s of amps. To charge these batteries we specially need high current battery chargers rated to handle high ampere charging levels for long periods of time. Here we discuss one such automatic lead acid battery charger with high Ah charging capacity.
This high power lead acid battery charger circuit can be used for charging large lead acid batteries in the order of 200 to 500 Ah, the design is perfectly automatic and switches of the power to the battery and also itself, once the battery gets fully charged.
Simulation and Working
Looking at the shown high current schematic for charging large batteries, we can understand the circuit idea through the following simple points:There are basically three stages in the shown configuration viz: the power supply stage consisting of a transformer and a bridge rectifier network.
A filter capacitor after the bridge network has been ignored for the sake of simplicity, however for better DC output to the battery one can add a 1000uF/25V capacitor across the bridge positive and negative.
The output from the power supply is directly applied to the battery which requires to be charged.
The next stage consists of an opamp 741 IC voltage comparator, which is configured to sense the battery voltage while it is being charged and switch its output at pin #6 with the relevant response.
Pin #3of the IC is rigged with the battery or the supply positive of the circuit via a 10K preset.
The preset is adjusted such that the IC reverts its output at pin #6 when the battery becomes fully charged and reaches about 14 volts which happens to be the transformer voltage at normal conditions.
Pin #2 of the IC is clamped with a fixed reference via a voltage divider network consisting of a 10K resistor and a 6 volt zener diode.
The output from the IC is fed to a relay driver stage where the transistor BC557 forms the main controlling component.
Initially, power to the circuit is initiated by pressing the "start" switch. On doing this, the switch bypasses the contacts of the relay and powers the circuit momentarily.
The IC senses the battery voltage and since it will be low during that stage, the output of the IC responds with a logic low output.
This switches ON the transistor and the relay, the relay instantly latches the power via its relevant contacts such that now even if the "start" switch is released, the circuit remains switched ON and begins charging the connected battery.
Now as the battery charge reaches about 14 volts, the IC senses this and instantly reverts its output to a high logic level.
The transistor BC557 responds to this high pulse and switches OFF the relay which in turn switches of the power to the circuit, breaking the latch.
The circuit gets completely switched OFF until the start button is pressed once again and the connected battery has a charge that's under the set 14 volt mark.
How to set up.
It's very easy.
Do not connect any battery to the circuit.
Switch ON power by pressing the start button and keep it depressed manually, simultaneously adjust the preset such that the relay just trips or switches OFF at the given rated transformer voltage which should be around 14 volts.
The setting is complete, now connect a semi discharged battery to the shown points in the circuit and press the "start" switch.
Due to the discharged battery, now the voltage to the circuit will drop under 14 volts and the circuit will instantly latch, initiating the procedure as explained in the above section.
Circuit Diagram for the proposed battery charger with high ampere capacity is shown below
Using two opamps:
An alternative way of achieving battery charging for a lead acid battery with high amperage can be observed in the following diagram, using a couple of opamps:
At low battery level, the lower opamp output remains high disabling the upper transistor relay driver, which allows the battery to initiate its charging process through the N/C contact of the relay.
As the charging voltage rises, the potential at (-) input pin of the lower opamp exceeds its (+) input level, which switches OFF the lower BC547, however the upper opamp output being still at logic zero the relay sustains its N/C relay position and carries on charging the battery.
At the full charge level when the potential at (+) input pin of the upper opamps tends to exceed its (-) input pin, the output of the upper opamp goes high, switching ON the relay and toggling its contacts to N/O.
This action switches OFF the charging current for the battery.
In this condition, a positive feed back from the 1K/10K resistive divider is applied to the (+) input pin of the upper opamp, which latches the opamp hard, and this in turn latches the relay in the N/O position.
Now, if any intended load is operated through this battery, and as it goes through a discharging process....after a period of time the battery voltage tends to drop to a point wherein the potential at (-) input of the lower opamp gets lower than its (+) input pin....this instantly causes its output to go high, switching ON the lower BC547.
The BC547 grounds and breaks the feedback latching potential and also the base triggering voltage of the relay driver transistor.
The action switches OFF the upper relay driver transistor allowing the relay contacts to revert to its N/C position, yet again initiating an automatic charging of the attached battery.
The set up of the above circuit can be visualized in the following video which shows the cut off responses of the circuit to the upper and the lower voltage thresholds, as fixed by the relevant presets of the opamps
Using a BJT only for high Ah Charging
The following explanation details how a battery may be charged effectively without using any IC or relay, rather simply by using BJTs, let's learn the procedures:
The idea was suggested by Mr. Raja Gilse.
Charging a Battery with a Voltage Regulator IC
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
Charging with an Emitter Follower Configuration
Yes it will work and will stop charging the battery when around 14 V is reached across the battery terminals.
However 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