Battery charger working.

Anybody know working of this circuit???

nikhildascl said:

Anybody know working of this circuit???View attachment 77864

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This is a circuit diagram of an indicator + 2 Amp charger to sense/charge battery of 12 volts. Working principle of this circuit is the battery voltage compared with the reference voltage. By using the LM324 IC, which is a Low Power Quad Operational Amplifier. The output of OP-amp is used (better use transistor switch) to turn on and turn off the LEDs. Voltage level indicated by four lights led. When the voltage is below 11.5 volts then the red led light and buzzer sound indicate that the battery needs charging.

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The Lead acid battery charger circuit consists of a standard step-down 12V AC (2-amp) transformer and a bridge rectifier comprising diodes D1 through D4. Capacitor C1 smoothes the AC ripples to provide a clean DC for charging the battery. The battery voltage analyser circuit is built around the popular quad op-amp LM324 that has four separate op-amps (A through D) with differential inputs. Op-amps have been used here as comparators. Switch S2 is a push switch, which is pressed momentarily to check the battery voltage level before charging the battery.

The non-inverting terminals of op-amps A through D are connected to the positive supply rail via a potential divider chain comprising resistors R1 through R5. Thus, the voltage applied to any non-inverting input is the ratio of the resistance between that non-inverting terminal and ground to the total resistance (R1+R2+R3+R4+R5). The resistor chain provides a positive voltage of above 5V to the non-inverting inputs of all op-amps when battery voltage is 12.5V or more. A reference voltage of 5V is applied to the inverting inputs of op-amps via the 5V zener diode ZD1.

When the Lead acid battery charger circuit is connected to the battery and push switch S2 is pressed (with S1 open), the battery voltage is sampled by the analyser circuit. If the supply voltage sample applied to the non-inverting input of an op-amp exceeds the reference voltage applied to the inverting inputs, the output of the op-amp goes high and the LED connected at its output lights up.

The different levels of battery voltages are indicated by LED1 through LED4. All the LEDs remain lit when the battery is fully charged (above 12.5V). The buzzer connected to the output of IC1 also sounds (when S2 is pressed with S1 kept open) as long as the voltage of battery is above 9.8V. If the voltage level goes below 9.8V, the buzzer goes off, which indicates that it’s time to replace the battery.

See More http://powersupplycircuit.net/lead-acid-battery-charger-circuit.html

https://tinyurl.com/cjcrpms

The resistor and capacitor on the right are the "battery". Turn it on, let it charge up to 10V, turn it off and press the voltage level indicator switch to check. First LED lights at 10V. That circuit simulator is a good resource in general.

There is a description of your circuit at **broken link removed** that explains it.

Note that the voltage level indicator will only display meaningful results if the charger is OFF.

As for parts:
X1, S1, S2 are as described.
D1-D4 is a bridge rectifier and C1 is a filter to flatten out the rectified voltage.
R1-R5 and the op amps are for battery voltage sense; the op amps are used as comparators.
R6 and ZD1 provide a 5V (max) reference voltage for comparison. Voltage along R1-R5 determined by battery voltage, compared to 5V reference and used to light indicators.
LEDs and buzzer provide information to a human (probably).
Fuse protects circuit (e.g. from shorts or bad batteries).

See the simulation I linked to for an animation of this all.