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I have the below LED and a rechargeable battery with the specifications below. To make an emergency LED light myself, I guess I don't need a resistor connected in series since battery voltage (3.7V) is almost equal to operating voltage of LED (3.2 V to 3.6 V). Please confirm.

Also, I believe the LED will automatically draw the current it needs to lit since the battery capacity is 2600 mAh and the LED operates with a minimum current of 350 mA. Or do I need to add any component to limit the current drawn from the battery?

LED
Power: 1 W - 3 W
Model Name: CREE XPE-R3
Emitted Color: Cool White (6500k)
Brightness: 1 W: 122LM   3 W: 320LM
DC Forward Voltage (VF): 3.2 V-3.6 VDC
DC Forward Current (IF): 350 mA ~ 1000 mA
Maximum Pulse Voltage: 3.8 V
Maximum Pulse Current: 1200 mA


Ultrafire 18650 Battery
- Capacity: 3600 mAh (actual 2600 mAh)
- Voltage: 3.7 V
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    \$\begingroup\$ Since LEDs have very low resistance (since they're diodes) it'll draw as much current as possible, which may exceed the maximum value it can handle. Anything to limit current flow with a resistor is a good idea; current limitin resistors are typically used anywhere from 220 to 1k ohms. \$\endgroup\$ – ezra_vdj Oct 22 '15 at 9:18
  • \$\begingroup\$ If you are already in a emergency situation and only have the battery and the LED, yes, it will work without a resistor. But you will be pushing the LED slightly beyond its limits and so, will reduce its life time. It will also produce a lot of heat, so if you can lit it a little and then let it cold for a moment, it is better than leaving it on all the time. If you are preparing yourself for a future possible situation, then it is a better a idea to limit the LED current somehow (as explained in the various answers). \$\endgroup\$ – ricardomenzer Oct 22 '15 at 12:12
  • \$\begingroup\$ Take into account that fully charged Li-Ion batteries usually have voltages in the ~4V range. Unless its internal resistance is relatively high (2Ω or so) you’ll kill the LED. \$\endgroup\$ – Michael Oct 22 '15 at 14:42
  • \$\begingroup\$ Since those LED's seem to only cost a dollar or two, why not just try it? Make sure the LED is on a fire proof surface (and properly heatsinked), hook it up and see what happens. Use a ammeter to see how much current it's using to see how badly you're overdriving it. Or, buy the cheapest flashlight you can find that uses one of those LED's and see what they do for current limiting -- they really cut costs to the minimum on those cheap flashlights, so if a component is not needed (even if it only costs a penny), they won't use it. \$\endgroup\$ – Johnny Oct 22 '15 at 15:54
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    \$\begingroup\$ That's because the battery used is a coin cell with 30 ohm esr. Like as if you have a resistor in between. \$\endgroup\$ – Passerby Oct 22 '15 at 23:32
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Of course you need a resistor - the LED rated forward voltage could be as low as 3.2 volts and the battery is 3.7 volts - what happens under these circumstances - either the battery collapses to 3.2 volts or the LED burns with over current. Do the correct thing and calculate a resistor value.

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  • \$\begingroup\$ Thanks for the reply. By applying ohms law, (3.7 - 3.2)/.350 = 1.428 ohms. Should i get 1.428 ohms resistor? Do we have resistors around this value? \$\endgroup\$ – Bala Oct 22 '15 at 9:38
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    \$\begingroup\$ A 1.5 ohm resistor will give you 333mA. A 1.5 ohm in parallel with 30 ohms gives you 1.42857 ohms. I've seen people using constant current drivers for this type of application. \$\endgroup\$ – Andy aka Oct 22 '15 at 10:17
  • \$\begingroup\$ I built something similar with three 0.5R resistors. Running the thing at slightly less than its rated current is a good idea to prolong its life. \$\endgroup\$ – pjc50 Oct 22 '15 at 15:38
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You need at least one resistor for both the reasons that have been named by Andy aka and ezra_vdj:

  1. Without a resistor the current would flush through the diode and burn it. There is no regulation in the diode; it is a passive element. Think of it as a mill driven by a river. If you don't limit the current of the water, the wheel will spin to fast, destroying your mill's mechanical parts.
  2. To keep the analogy, without limiting the voltage, your waterwheel will not be able to withstand the pressure of the water and its shovels will break off.

To counter the problems you have some options:

  1. Use a voltage divider with a resistor to define the current.
    Or use a DC-to-DC converter, which might save you some energy, probably prolonging battery lifetime (take its losses into account).
  2. Use a resistor in-line with the LED to limit the current.

The values for the resistors can easily be calculated by Ohm's law, and the solution should be easy to see when you draw a simple circuit diagram.

Good luck.

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  • \$\begingroup\$ Thanks kamuro. If i choose option(2), a 1.5 ohms resistor should work right? \$\endgroup\$ – Bala Oct 22 '15 at 10:02
  • \$\begingroup\$ Looks good to my eyes. \$\endgroup\$ – kamuro Oct 22 '15 at 10:22
  • \$\begingroup\$ @Satheesh if you got the circuit nearby and on something like a breadboard, just experiment with different values to give sufficient brightness. Ideally, due to the 350-1000ma forward current range, and 3.7V supply, you're looking for resistance anywhere from 5-10ohms. \$\endgroup\$ – ezra_vdj Oct 22 '15 at 10:31
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    \$\begingroup\$ (@kamuro should be able to comment anywhere now my dude) \$\endgroup\$ – ezra_vdj Oct 22 '15 at 10:36
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Your battery may be nominally 3.7 V, but its voltage is likely to vary over a range more like 3.2 V to 4.2 V depending on its state of charge. There is also likely to be variability in the LEDs.

You might just about get away with a single resistor solution, but the light output will vary massively with the state of battery charge. If you want consistent brightness you will need a switched mode LED driver (see for example Switch mode LED torch).


OK, since you asked about the resistor solution.

In order to protect the LED we need to ensure that under the worst-case scenario the current is no more than the maximum rated. The worst-case scenario is if the battery is fully charged and the LED's forward voltage is at the bottom of its specified range. From some googling it seems a fully charged "3.7 V" lithium polymerbattery is about 4.2 V.

Thus, an LED forward voltage of 3.2 V, battery voltage of 4.2 V, resistor voltage 1 V, current 1 A, and a resistance of 1 Ω. The power dissipated in the resistor is 1 W.

So make sure you use a resistor of at least 1 Ω and a suitable power rating (at least 1 W for a 1 Ω resistor, at least 0.5 W for a 2 Ω resistor, etc.), and you should not damage the LED. Beyond that it's largely a matter of experimenting to determine what value gives you the desired balance of brightness and battery life.

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  • \$\begingroup\$ Thanks Peter, this is my first try on electronics. So, want to go with resistors to have some basic understanding. Hope the only disadvantage with resistor is the light output may vary depending on the battery charge and i assume it is safe and wont fire. Please correct me if i am wrong. \$\endgroup\$ – Bala Oct 22 '15 at 11:59
  • \$\begingroup\$ Thanks Peter. The answer above from Tom says we can use 3W resistor(actual in place of 0.175W). Could you please explain how did you arrive at 1w for 1Ω resistor, at least 0.5W for a 2Ω resistor? Just want to make sure i understand what you saying..Sorry if my question is silly. Thanks in advance. \$\endgroup\$ – Bala Oct 22 '15 at 14:09
  • \$\begingroup\$ a 3W resistor is overkill. \$\endgroup\$ – Peter Green Oct 22 '15 at 14:10
  • \$\begingroup\$ So, would you pls explain how did you arrive at 1w for 1Ω resistor or 0.5W for a 2Ω resistor? \$\endgroup\$ – Bala Oct 22 '15 at 14:12
  • \$\begingroup\$ To calculate the power in the resistor we can use the equation p=v²/r . In our worst case scenario the voltage across the resistor is 1V. So p=1/r . That works out to 1W for a 1Ω resistor and 0.5W for a 2Ω resistor. \$\endgroup\$ – Peter Green Oct 22 '15 at 14:16
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Here's a visual representation of the calculations and values that others have discussed:

Current Limiting Resistor Calculations

First, take the minimum forward voltage of the LED (3.2 V). Which leaves 0.5 V across the current-limiting resistor.

You can then calculate the resistor value using Ohm's law and your desired current (for example 350 mA).

It's good practice to calculate the power through the resistor:

P(W) = I x V.... or P(W) = I^2 x R
0.350 A x 0.5 V = 0.175 W

You'll be fine with a standard 3 W resistor.

Do you have a heat sink for your Cree LED? They have a tendency to run quite hot. If you have the capability, I would recommend either constant current or PWM dimming as it's more efficient.

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  • \$\begingroup\$ Thanks so much for the detailed explanation. Yes, the led comes with a heat sink. \$\endgroup\$ – Bala Oct 22 '15 at 11:55
  • \$\begingroup\$ No worries, there are quite a few design options when it comes to high-power LED lighting. I would highly recommend looking into PWM dimming using a MCU and MOSFET. Or alternatively, you could use a high power constant-current driver IC. Such as this from TI: ti.com/product/lm3404hv/description . I was contracted to design and build some bespoke bar-lighting last year, and found the PWM option to be the most energy efficient, cheap and customisable. Though it comes with more of a learning curve. \$\endgroup\$ – Tom Wilson Oct 22 '15 at 12:05
  • \$\begingroup\$ the calculation says 0.175W and you mentioned standard 3W resistor is fine. Do you mean we can use 3w resistor in the place of 0.175w? \$\endgroup\$ – Bala Oct 22 '15 at 12:06
  • \$\begingroup\$ Yeah, you'll be fine with a 3W. If your power calculation starts going above 2W, I'd recommend a resistor with a higher power rating. \$\endgroup\$ – Tom Wilson Oct 22 '15 at 12:12
  • \$\begingroup\$ You may find this guide useful: instructables.com/id/Circuits-for-using-High-Power-LED-s \$\endgroup\$ – Tom Wilson Oct 22 '15 at 12:13
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If educational understanding is your main goal then the biggest fattest white LED which your 3 to 4 V battery might power is a very bad choice due to the heat sinking requirement, fat wires needed, rapid depletion of battery voltage, and comparable practicalities which distract you from learning about electrical circuit design. I'd suggest buying a ten-pack of bog standard "ultrabright" blue LED rated for >20mA continuous operation (at about 3.2V) and a bog standard resistor of about 33 Ohms 1/4 Watts rated will do (of the 50 ish Ohms needed, the mid sized blue LED provides "a few" Ohms). From there you can see it work on day one and then safely try out such things as a transistor regulated contant current source.

The same circuit which works at 20mA with a small blue will work at 20 to 30mA with your fat white. More than that will be too bright to look at directly.

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Try using a 1N914/1N2002 in series in a forward conducting situation. It will drop .7 volts. This negates the loss due to a resistor. I have 'stacked' 4 LEDs, they work fine at 12 volts, with no resistors, and have for years. Watch out for over voltage situations.

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  • \$\begingroup\$ What voltage source? Perhaps you were saved by the internal resistance of a battery? \$\endgroup\$ – Peter Mortensen Oct 22 '15 at 21:59
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What if the voltage of the battery is lower than the diode's voltage drop?

The simple equation for an LED resistor does not work anymore.

It happens that you need a resistor in almost all cases. Even when you have the same voltage. Here's why:

Let's assume that we have a diode with the following I-V characteristics>

Enter image description here

For currents under 10 mA, the characteristics become linear, while the drop is the inverse of the internal resistance. At 20 mA, the drop is 2 V, and at 10 mA the drop is 1.95 V.

The internal resistance is:

Rint = (V1-V2)/(I1-I2) = (2 V - 1.95 V)/(20 mA - 10 mA) = 5 Ω

The internal voltage is:

Vint = V1 - (I1*Rint) = 2 V - (20 mA * 5 Ω) = 1.9 V

If we connect 0 Ω resistance we get a current of 20 mA @ 2 V. If the voltage source increases to 2.05 V, a small increase of 50 mV, the LED's current is

ILED = (2.05 V - 1.9 V) / 5 Ω = 30 mA

This says that a small increase in voltage will result in big change in current. Because of this, it is better to put a resistor with a higher value, for a better current control. A voltage drop of 10 mV on a resistor of 100 Ω gives 100 uA which is little current.

Therefore the conclusion is that we need to put a high voltage drop acroos the resistor to enable constant current through the LED.

Rresistor >> Rint (the resistance of the resistor is fairly high in comparison with the internal LED resistance).

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