I have a torch in which 3 AAA cells are put in (1.6 (new AAA) * 3 =4.8v). The problem is these are not long lasting and i have to change them after few days. What i am looking for is a rechargeable option. I cannot use 1.2v Ni-Cd(1.2 * 3 = 3.6v) regargeble as the voltage is too low and led is dim. Recently i extracted 18650 (LGABD11865) batteries from my Old Dell Laptop.

I used powerbank module to test it out with my torch light. The output is slighltly over 5v (around 5.11) and the light consumes around 1.3Amps. When the voltage is reduced to 4.8v (checked through power supply) the amps are reduced to 1 Amp which is reasonable as it wont consume more battery and the torch wouldnt heat up.

Can someone please help me in this regard i have very basic knowledge of Electronics i need 4.8v CV with atmost 1 Amps. Furthermore the torch can only take 1 18650 Battery along with 1 tp4056 charging module or small powerbank module (usb type A port removed). enter image description hereenter image description here

Update: I used 3 new AAA batteries in the casing which came with the torch. The voltage when measured showed 5.01v but the current at full brightness was 0.5amps. I need a resister MOSFET anything which can lower the current input to the torch. At present when connected with 5v supply the torch takes around 1.23 amps.

  • \$\begingroup\$ If the 5v regulator can remain in the circuit, you may wish to consider a flashlight driver based on the ubiquitous AMC7135 low drop out linear current regulator. Your best options would be 2 suitably protected 18650s in series with a buck type current regulator or 1 suitably protected 18650 with a boost type current regulator. Originally the flashlights protection from thermal runaway may have been the internal resistance and diminishing output of the battery, so make sure you don't neglect current control if you're switching to a low impedance battery type. \$\endgroup\$
    – K H
    Jan 22, 2021 at 1:06
  • \$\begingroup\$ Battery protection circuits are sometimes in cells, and sometimes in the device that contains the cells. Since you are taking cells from an unknown device, you should assume they are unprotected until you confirm otherwise, and never charge or discharge a lithium ion battery without suitable protection circuits. \$\endgroup\$
    – K H
    Jan 22, 2021 at 1:07
  • \$\begingroup\$ Also just be aware that there are tons of available compact LED drivers like the luxeon Buckpuck and Boostpuck and their knockoffs, and there exists a subset of flashlight parts that can be bought individually, including casings, lenses, LEDs, drivers, controllers and switches. Parts and machining cost money, so before you buy stuff to modify an existing flashlight, you may wish to check out Convoy to see if you can get a better whole flashlight for cheaper than the cost of the mod. An S12 is 30 maple dollars and compares well against much more expensive lights. \$\endgroup\$
    – K H
    Jan 22, 2021 at 1:25
  • \$\begingroup\$ On the other hand, if getting into electronics is your goal, messing with flashlights is good experience, and the biggest challenge I find with considering a mod to an existing light, is getting enough room in the casing for whatever parts you need to add. If your original light didn't come with a current regulator board, it usually doesn't have room or threading allotted to mount that extra board, so if you want to consider changing the battery, you also have to find enough room for the extra circuits in the springs or by shrinking the battery. \$\endgroup\$
    – K H
    Jan 22, 2021 at 1:30
  • \$\begingroup\$ Have you taken apart the lens end of your flashlight to see what drives the LED? \$\endgroup\$
    – K H
    Jan 22, 2021 at 4:39

1 Answer 1


You say that your knowledge of electronics is very basic, so I shall not recommend a bespoke design. It’s normal to use a constant-current driver for LEDs, not least because the current is very sensitive to the voltage applied. If you can find a constant current driver that fits your requirements then you’re all good, although I think you may struggle because of the relatively low supply voltage. As an alternative you might try using a resistor to limit the current: you want to supply 1A to the LED and the resistor should drop (5.1 - 4.8) = 0.3V. Using Ohm’s law, R=V/I so you’ll need a 0.3 ohm resistor with a power rating of at least 0.3W. The nearest thing you’ll easily find is a 0.33 ohm part with 0.5W power rating. If you place this in series with the LED you should be ok. I note that 1A is a lot to ask from AA cells, so you may well be overdriving the LED which will shorten its life. If you use a larger value resistor you’ll get slightly lower output but longer battery life, perhaps try 0.47 ohm. As an aside, there are some good reasons why resistors have these oddball values - look up ‘preferred value resistor’ if you’d like to know more.

  • \$\begingroup\$ Using a resistor with such small resistance may not successfully achieve current regulation of the LED. The resistor must waste enough power that it heats up enough in comparison to the LED that it's positive temperature coefficient cancels out the negative temperature coefficient of the LED, or thermal runaway will occur. For very small voltage drops, a low drop out linear current regulator is best. This can be aided by keeping the resistor thermally close to the LED's heat sink to ensure the resistor heats up from the LED's dissipated power as well. \$\endgroup\$
    – K H
    Jan 22, 2021 at 1:12
  • \$\begingroup\$ Quite so, but this may be beyond the capabilities of the OP, with all due respect, and the OP has indicated that the LED is ok in the short term at least, I suspect because the 5V boost converter is current-limiting, do any additional resistance will improve stability further. Also, protection against thermal runaway doesn’t rely on the tempco of the resistor. \$\endgroup\$
    – Frog
    Jan 22, 2021 at 4:28
  • \$\begingroup\$ Thermal runaway is caused by the decreasing resistance as the LED heats up increasing current and causing a compounding effect. If the temperature coefficient of the entire circuit comes out negative, you get thermal runaway. This only matters with tight voltage drops like the one here, but also matters more for power LEDS. \$\endgroup\$
    – K H
    Jan 22, 2021 at 4:33
  • \$\begingroup\$ To see this effect in action, you can take a power LED mounted on an aluminum star and apply a constant voltage to get it's rated current. Then leave the voltage applied and monitor current while the LED heats up and burns out. Then try adding a very small resistor in series, suitable to cause a .2 or so volt drop, and repeat the procedure with larger resistors until the LED no longer burns out. This is the destructive testing way to find an adequate temperature coefficient compensation. \$\endgroup\$
    – K H
    Jan 22, 2021 at 4:37
  • \$\begingroup\$ Thankyou All for your informative responses. I would like to add a suggestion. I just found out a powerbank module on Aliexpress which has output 5v/1A and can be connected with 18650 battery. Will this work if i connect it with my LED? Will itlimit current input? And Frog you are probably right about shortening life of my LED because it does heat up if we use it at full power with 1amps. Instead of limiting current i would want to decrease voltage which would automatically decrease its current input. Is there a way to make it 4.6v - 4.8v? I wouldnt go below 4.6v as it dims the light. Thankyou \$\endgroup\$
    – Zulqurnain
    Jan 22, 2021 at 20:17

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