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Coming from this question, I decided to have some changes. So my case is I have an LED with this description:

  • Dimensions (diameter x height): 20 x 8 mm
  • Operating voltage: 3,4-3,6 V
  • Rated current: 700 mA
  • Power Dissipation: 3 W
  • Lighting angle: 120°
  • Luminous flux: 160 lm
  • Color temperature: 6500-7000 K
  • Mounting: Installed on an additional radiator
  • Glow Color: White

Then I have AA Li-ion 18650 3.7V, 3000 mAh rechargable battery, 21*15*10MM TO-220 aluminium radiator, TP4056 Li-ion battery charger and 2 A, 1 Ohm resistor.

schematic

simulate this circuit – Schematic created using CircuitLab

If I'll construct it this way, is everything is gonna be alright? I don't have every detail shipped yet and I'm out of time

UPD: updated my circuit by adding 1S PCB Bms protection board. Will this flashlight work properly for couple of months?

mend4x
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    Circuit diagrams are drawn with logic and current flow going from left to right. This diagram goes from right to left, so it's hard to read. Flip it around. – Pete Becker Aug 21 '18 at 12:05

3 Answers3

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Is everything OK with this circuit?

The LED will light when the switch is closed.
For a lantern or flashlight it is not okay. Mediocre at best.

Your LED "operating voltage" is too high for a Li-ion battery powered circuit. White LEDs are available with forward voltages (Vf) below 3V. An A 18650 battery discharges to 3V (see red line in discharge graph).

With your LED if the Vf is 3.5V you will have insufficient voltage when the battery is discharged to about 50%.

With a 1Ω resistor and full charge on the battery, the current will start at 700 mA (@4.2V) then quickly drop 400 mA (@3.9V) and not too long after that drop to 100 mA (@3.6V). This is why I mentioned in your previous post a battery powered lantern should a current regulator and not use a current limiting resistor.

If you use an LED with a max Vf of 3V or less it will operate down to the end of the discharge curve.

enter image description here
Source: Panasonic Li-ion NCR18650F Datasheet


If you do not have SMT capabilities then use a Cree XP-G3 (2.7V-3V) mounted to a star board.

enter image description here

Misunderstood
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You would need a boost converter in current mode. This way you will get constant LED current, whatever is the battery voltage.

  • any recommendations about which to use? – mend4x Aug 22 '18 at 07:42
  • Mmm... For 700mA output you need some serious input current when battery is low. So would go to TPS54540. If it's overkill, you can later go down to TPS54340. Current mode is tricky, but you can go to buck/boost configuration, then voltage mode would work fine. –  Aug 22 '18 at 07:49
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is everything is gonna be alright?

That depends (on what is "all right")

When the switch is closed, the LED will discharge the battery. But when the battery is nearly empty, to what voltage will it be discharged? Ideally the discharging should stop when Vbat < 3 V or so. It depends on how you want to treat the battery, do you want a long lasting battery? Then cut discharging below 3 V, if you don't care you can discharge more to 2 V.

R1 will dissipate some power especially when the battery is full because then it will be 4.2 V, not 3.7 V! Did you account for that in the dimensioning of R1?

2 A, 1 Ohm resistor

for resistors I want to see a value, 1 ohm, and a power rating like 1 W. If the resistor has a current rating maybe you have a fusable resistor? You could also just use a normal (SMD) fuse.

Note that the TP4056 needs not much more than 5 V to charge the battery. At higher voltages power the TP4056 will get too hot while charging (it will lower the current to prevent overheating though). Charging can take a long time (hours) as the TP4056 cannot charge with a very high current.

Dave Tweed
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Bimpelrekkie
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  • A mobile charger will be used for charging. I use the resistor just to downgrade 3.7V such that LED not overheats. And I can't understand about stopping discharging of battery. You mean what? – mend4x Aug 21 '18 at 09:27
  • *And I can't understand about stopping discharging of battery.* If you leave the switch on for a long time, the battery will empty. If it would discharge to 0 V then the battery could be damaged. Go to https://batteryuniversity.com/ and read the datasheet to learn how to properly use Li-Ion batteries. Misuse the battery and it will not last long. – Bimpelrekkie Aug 21 '18 at 09:31
  • *I use the resistor just to downgrade 3.7V* So then what happens when the battery is 4.2 V when it is fully charged? If you calculated with 3.7 V then the LED will overheat when the battery is full (and 4.2 V, not 3.7 V) – Bimpelrekkie Aug 21 '18 at 09:33
  • How it could be 4.2V if it's described as 3.7? – mend4x Aug 21 '18 at 09:35
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    The 3.7 V is the **average** voltage. If you work with Li-Ion batteries you should know that their voltage drops depending on their state of charge, see: https://batteryuniversity.com/index.php/learn/article/lithium_based_batteries Also the TP4056 will charge the battery to 4.2 V because that is the voltage where it is full. At **3.7 V** the battery is about **half full**. Honestly: you should already know all this before even designing something with a Li-Ion battery. Li-Ion cells must be treated correctly or they will disappoint you (stop working). – Bimpelrekkie Aug 21 '18 at 09:41
  • thanks for your help. and what's your practical advise on avoiding these situations? how can I improve the circuit such that the battery will be used properly? – mend4x Aug 21 '18 at 09:59
  • Add a battery protection module, example: https://www.ebay.com/itm/1S-2-8A-3-7V-18650-Li-ion-lithium-LiPo-Battery-Cell-Circuit-BMS-PCB-Protection/283103485274?hash=item41ea485d5a:g:HIEAAOSwPhhbIokB You connect the battery directly to this module, then use the P+ and P- connections to connect to the rest of your circuit. Your battery will now be protected from over current, over charging and over discharging. Result: a happy battery – Bimpelrekkie Aug 21 '18 at 10:07
  • @bimpelrekkie "Li-Ion cells must be treated correctly or they will disappoint you (stop working). " ... Or explode. Never forget about the exploding thing. – Jules Aug 21 '18 at 19:20