I have a question about charging Ni-Cd battery, if question is stupid or something like that, I'm sorry but I'm only 15 and just starting my adventure with electricity.
I want to build LED flashlight, I found heatsink from old cpu and bought dc-dc boost module to change voltage and now I only need battery, I can't afford to buy 3s lipo battery pack because it also requires special charger which costs much.
So I found old cordless drill, drill is broken so I took battery pack from it and disassemble this battery, inside I found 15×1,2V some ooooold cells but every cell looks fine, I measured voltage on each cell and thats what I got: 1,2V 1,21V 0,4V 0,2V 0,2V 0,2V 1,02V 1,13V 1,11V 0,2V 0,8V 1,02V 1,15V and two 0V cells each cell has 1200mAh
Which one can I use to build working battery pack ? And how can I charge this battery pack ? I found in google how to charge Ni-Cd battery using power adapter and 2x resistor and diode but I can't work out which resistors and diode use to charge it safely.
I want to use this battery to build LED flashlight and LED I want to use is 30W 32-34V 900mAh
Any help will be very important for me because as I said I just started my education in this kind of things 🙂
Thanks a lot !!
Best Answer
This is maybe more complicated than you think.
I'm going to start way far away from charging your battery, then work around to it.
To make an LED light up you need to provide it with a voltage higher than its forward voltage. This varies by color. Red LEDs have the lowest of the visble LEDs, blue LEDs have the highest. White LEDs are either a mixture of different colored LEDs or a blue LED that lights up a phosphor mixture that the glows white. If you have a white LED, it is most likely a blue LED with phosphor.
Once you get the voltage about the forward voltage, you must limit the current. When an LED lights up, it will pretty much turn into a short circuit and try to draw all the current it can get. This will burn it out very quickly.
So, to light an LED you need to know its forward voltage (get from the datasheet or estimate based on color) and you need to know the maximum current it is allowed to consume.
From the forward voltage and the current and the voltage from your boost converter, you can calculate the value of a resistor that will limit the current properly.
Example:
Using a boost regulator that delivers 5V, a white LED with a forward voltage of 3V, and an allowed current of 20 mA.
Supply voltage minus forward voltage of the LED leaves 2V.
A resistor that would drop 2V at 20 mA can be calculated with Ohm's law E=IR. Rearranged to find R, that would be R=E/I. So, 2V divided by 20mA. Gives a resistor of 100 Ohms that you need to connect in series with LED.
You said you have a boost regulator, but don't mention the input or the output voltage. I'm going to assume 5V output from a single nicad cell (1.4V)
Be aware that the power output and the power input of such a regulator must be the same. If you need 1Watt of power output, you will have to give it 1Watt of power in.
Power is voltage times current (P=EI).
For 1Watt of power output at 5V, that would be 0.2A. On the input side you only have 1.4V from the cell. To get the same power, the regulator has to draw 0.7 A from the cell.
You have to be certain that your cell can deliver the needed current.
No regulator is perfect, so you have to expect to put more power in than you get out.
The length of time your light will operate depends on the capacity of the cell and the amount of current you draw from it.
Cells for a cordless drill are usually fairly high capacity. I'll assume 2000mAh just to have a number to use for an example.
2000mAh means that if you draw 2000mA, the battery will deliver current for 1 hour. If you draw 1000mA, it will deliver current for 2 hours. This is, of course, simplified.
Lets put everything together. An LED that draws 20mA at 3V, a boost regulator that provided 5V from a single 1.4V nicad cell, and a nicad cell with a capacity of 2000mAh.
You drive the LED from the 5V using a 100 Ohm resistor to limit the current to 20mA.
The 5V regulator provides 20mA to the LED, but draws 34mA. Since your cell can provide 2000mAh, that works out to a run time of 58 hours.
Finally, we come to your charger.
As you see, you can probably do well with one or two cells.
All you need is a simple charger.
Assuming a single cell, you need to limit the charge current. You can either use a constant current source, or a current limited voltage source.
A current limited voltage source is easiest to build. It is nothing more than a voltage source with a series resistor.
Lets take a 5V powersupply because they are common.
Lets limit it to 50mA so as to make it hard to over charge and over heat your nicad cell.
We already used Ohm's law to figure a limiting resistor for the LED. We'll use it again to figure out how to limit the charge current to 50mA.
We have 5V, and 50mA. E=IR, R=E/I. So 5V/0.05A gives a 100 Ohm resistor.
Now, lets check the power on that resistor. P=EI, 5V*0.05A gives 0.25 Watts.
Better use resistors rated for more than 0.25W. Next up is probably 0.5 or 1W
That's probably more than you wanted to know, but I think it is what you need to know.
Use only the cells that show 1.2V or more. 1.2V is the lower discharge limit for nicad cells. Any cell with a lower voltage may be damaged.
Any cell with 0V is damaged. That pretty much only happens when crystals form and grow through the internal insulation of the cell. They can be "fixed" if you absolutely have to use the cell for something, but they will be very unreliable, will not hold as much charge, and will short out again.
With your added information, we can make an estimate of the operating conditions.
Summary:
9.8V battery with 1200mAh capacity.
30 Watt LED that operates on 32V.
The LED will draw about one Ampere.
You will have to boost the voltage from the battery by a factor of three.
The booster will draw at least 3 times the LED current from the battery, so call it 3A.
1.2 Ah divided by 3A gives a run time of 0.4 hours, or about 24 minutes. With losses, less than 20 minutes.
You have to limit the current to keep the LED from burning out.
Assume 34V, reduced to 32 at 1A. That 2Watts of power you have to waste in a 2 Ohm resistor.
It would be (somewhat) more efficient to use a boost regulator with current limiting built in.
Maybe your booster does. Can't tell because you haven't posted details.