Don’t throw away old Laptop battery; it can be reused for other purposes. I hacked one Laptop battery and salvaged six high efficient Samsung Lithium Ion batteries each rated 3.6V, 4300 mAh. It is worth Rs. 1000. Few days before, I found that, my Laptop is not getting enough back up and shutdown immediately when the charger is removed. So I replaced the battery at the cost of Rs.1200. Due to curiosity, I just opened the case of the old battery pack and found that, the batteries inside can be reused. After testing, I found that only one battery out of the six is damaged. All the other five are in good condition. That damaged battery created the problem in the battery pack and prevented the remaining five batteries from their work. So actually this is happening in our Laptop. If one battery fails, we have to replace the battery pack. If the good batteries are salvaged from the old pack, we can reuse them for many different applications like LED lighting, Mobile charging, making power supply etc. So I spend Rs.1200 for the new battery pack but I salvaged 5 good batteries worth Rs. 1000 from the old battery pack. So my actual expense is only Rs.200. So do not discard the old Laptop battery. Tear it, salvage the individual batteries and reuse. You can save money and can reduce the e-waste.
What I found inside the Laptop battery pack is that , six 3.6V Lithium Ion batteries are arranged and interconnected to make a Battery pack of 10.8V which has a power of 47 Wh. The connection is interesting. The 6 batteries are arranged in three sets each having two batteries connected in parallel. Then the three parallel sets are connected serially. So this arrangement gives 10.8V (3.6V x 3 = 10.8V) to the battery pack. This increases the efficiency. Along with the battery pack, there is a circuit board to control the charge / discharge cycles of the batteries.
Power of the battery pack is 47 Wh. Each battery is 3.6V at 4300 mAh. The six batteries are arranged in three sets having two batteries connected in parallel. So there are three sets each having 3.6V. So total voltage of the pack is
3.6V x 3 = 10.8V
Its current rating is 4300 mAh which is 4.3 Amps. So the power is
P = V x I = 10.8V x 4.3 = 46.44 Watts.It is shown a 47Wh
First I separated the individual battery from the pack by tearing the pads soldered. So all the six batteries are free.
Checked the battery specifications. It is 3.6V, 4300 mAh original Samsung battery. But the Polarity (+ and -) marking not given. So I tested with an LED to identify the + and -. Marked the polarity in all batteries.
Then tested individual batteries. Two batteries have 3V, two have 2.5V, one has 2.8V and one has 1.9V. So the 1.9V battery is a damaged one and cannot be reused. So discarded. The remaining 5 batteries have sufficient voltage to charge fully. So decided to reuse them to make an 18V Regulated power supply with 1 ampere current output.
Made a simple arrangement to make the Regulated power supply. See diagram below
The five 3.6 V batteries are connected serially. So in the fully charged state, the pack can give 18V at 4300 mAh. The first socket in the unit is for charging the battery. It needs 18V power supply for charging. Then there is the switch . When it is closed, the 18 V from the battery packs goes to the Voltage regulator ICs. IC 7812, IC 7809 and IC 7805 gives corresponding 12V, 9V, and 5V oputputs which can be tapped from the corresponding sockets. Just after the switch, 17 volt is available in the socket, directly from the battery pack. Actually 18 V comes from the battery pack, but some voltage drop occurs through the LED .
A simple Power on Monitor cum Voltage indicator circuit is given in the power supply. So when the switch is closed and if the battery pack voltage is higher than 12V, LED lights. So this gives two indications. 1. Power on status and all sockets have voltage 2. Voltage in the pack is higher than 12V and no need for charging. If the battery pack voltage drops below 12V. LED will not light and indicate the need for charging.
The working of the circuit is simple. The Cathode of 12V Zener diode is connected to 18V positive rail. The Anode of Zener is connected to the Positive of LED. The negative of LED is connected to 470 Ohms resistor that goes to the negative rail. So when the output from the battery pack is abouve 12V, Zener conducts and LED lights. When the voltage from the battery pack drops below 12V, Zener turns off and LED remains dark.