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I have a 3V 700mA LED (Datasheet). I am operating it with LR1130 cells.

There are 3 cells connected in series to give 4.5V. Now as I am driving the LED at higher voltage, it becomes hotter and it also burns if I keep it turned on for a while.

Actually I want to get as much as brightness from it without burning it. That is why I am not using any resistor between the cells and the LED because it reduces the brightness.

Is there any SMD LED Driver IC or small circuit that can drive this 3V LED at 4.5V but power it in pulses at high rate so that the LED does not burn. or is there any other sophisticated circuit available for this purpose?

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  • \$\begingroup\$ No. Burn is a function of average power and that produces heat and accelerates failures. You must enhance cooling. and not exceed abs max current rating. But you can regulate with a MOSFET ADJ. LDO to desired specs \$\endgroup\$
    – Tony Stewart EE75
    Commented Sep 7, 2021 at 12:13
  • \$\begingroup\$ The coin cells already have resistance in the ESR of the chemical interface due to small size \$\endgroup\$
    – Tony Stewart EE75
    Commented Sep 7, 2021 at 12:14
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    \$\begingroup\$ "Is there any SMD LED Driver IC..." That question is off-topic (we don't recommend specific products) but we can help with a circuit design. \$\endgroup\$
    – Null ♦
    Commented Sep 7, 2021 at 12:15
  • \$\begingroup\$ That datasheet is pretty crap. What's the luminous intensity specified in candela at a given current? That's all that matters. If you use some older technology LED then it will not shine as brightly. \$\endgroup\$
    – Lundin
    Commented Sep 7, 2021 at 12:42
  • \$\begingroup\$ Why do you need the pulses? If you don't drive it all the time, it will get dimmer. Have you considered adding a heatsink? \$\endgroup\$
    – Stack Exchange Supports Israel
    Commented Sep 7, 2021 at 16:19

3 Answers 3

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Your LED has a maximum current rating: 700 mA. 3V just happens to (on average at room temperature) be the voltage at which this current flows.

So, the moment you use a 1.5× higher voltage, you forced an exponential amount of additional current through your LED. That leads to a quadratic (of that exponentially increased current!!) increase in power converted to heat. You're literally burning the LED yourself. (It probably only took some time because LR1130 can't source much current themselves. You're using a totally inadequate type of battery.)

Can't do that. Your maximum current is rated because that's the current the LED can withstand continuously.

Shorter pulses of course mean that there's less current flowing on average: you get a duty cycle \$\eta\$ < 100%.

Sadly, that not only reduces the amount of heat, but also the amount of light emitted.

So, no, there's no "clever circuit": You need to drive your LED as close as possible to these 700 mA. That is what the resistors are used for! You really need to read up on why series resistors are necessary; the "better" alternative to a series resistor is a constant-current supply (as opposed to the constant-voltage supply a battery is).

Also, your choice of cell: that's a coin cell. It might be able to deliver 50 mA without damage to it, with luck. This is a comically wrong battery type fore someone who wants to source 700 mA!

So, basically, everything you chose was wrong :( :

  1. You need batteries that can actually source 700 mA. Yours can't.
  2. You need a way of limiting the current through the LED. A series resistor is an easy and often sufficient way of doing that.
  3. You can't get more light out of the LED than it's rated for without risking damage. That's why it's rated.

Probably, due to your inadequate batteries, you never actually saw the LEDs at full brightness, and its your batteries, and not your LEDs, that were broken.

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  • \$\begingroup\$ Although pulses of higher current do result in the same average current and therefore the same average light output, there's research which shows that our eyes are fooled by the pulses of brighter light. So this could be somewhat effective at creating an apparently brighter light source - but the drawbacks of the pulsed light often outweigh the slight benefit of increased perceived brightness. \$\endgroup\$
    – brhans
    Commented Sep 7, 2021 at 12:53
  • \$\begingroup\$ A key word here is "risking". You can absolutely drive some things past their ratings... of course they may go bang or go fizz or just stop working. And if they do work their lifetime may be shortened. For serious production don't do it. For non-critical DIY projects done once in a while, though... I seem to recall that scientists push datasheet limits all the time when working on new discoveries. Of course the risk of failure is acceptable to them. \$\endgroup\$
    – Stack Exchange Supports Israel
    Commented Sep 7, 2021 at 16:23
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LEDs are driven with current, NOT with voltage. If its (LED's) threshold is 3V, then anything over 3V connected to it will act as short circuit with infinite (in practice, finite, but well over any spec) current.

You always* therefore need a current limiting resistor. If your supply is 4.5V and LED drops 3V, then you will have 1.5V across the resistor. Picking different values for the resistor you can set different current through the resistor = current through LED.

For example, 1.5k resistor will give 1.5V across resistor, so it's 1mA of current through the resistor = 1mA through LED. Take half the resistance (for example two 1.5k in parallel will give 750Ohm) and you have 2mA through LEDs and so on.

How to get maximum brightness: If you want 700mA through the resistor with 1.5V voltage drop, it's 1.5V/0.7 ~ 2.15 Ohm. Get something slightly larger than that, you don't want to exceed max current.Also, your resistor should be able to take such current, which is not little! 1.5V*0.7A = 1.05W! That's a lot! You should probably connect a few resistors in parallel, will be easier.

And no, you will never achieve brightness like what you had when it burned, because it was already out of the spec. There is a limit to how bright it can be without burning.

*you can have a special LED constant current driver without "visible" resistor, but it's a whole another circuit and it's worth a separate discussion; may actually be feasible in your case.

EDIT If you're using a coin cell, you can get 5mA, MAYBE 10mA out of it, it just can't source more. If you have AA/AAA, maybe you can pull a few dozen mA, don't count on more either.

enter image description here

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  • \$\begingroup\$ may actually be feasible in your case: you can't get 700 mA out of a coin cell! \$\endgroup\$
    – Marcus Müller
    Commented Sep 7, 2021 at 12:16
  • \$\begingroup\$ I haven't noticed coin cell. Apologies. I thought it was an unspecified source. Then, well, 10mA is probably the maximum, if not 5mA \$\endgroup\$
    – Ilya
    Commented Sep 7, 2021 at 12:16
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Best bet is an adjustable super low LDO circuit with trimmer and caps and a much better battery like a Li Ion cell.

Li Ion secondary 18650 3.7 to 3.0 V, 2200+ mAh = 7.7 Wh
Alkaline Primary LR1130(3x) 4.5 to 2.7V, 50 to 80 mAh= 0.2 Wh

These are SMD, so it can be made on a small board with copper pour for heat sink. LED will also need > 4 sq in (ca. 26 cm²) of metal heatsink surface area > 2W with suitable design.

enter image description here

https://www.n-redc.co.jp/en/pdf/datasheet/r1173-ea.pdf

You probably discovered that if you tested these batteries long enough while burning your fingers, they would only last for 10 minutes until getting much dimmer then last longer very dim. This is very inefficient.

A slightly more efficient method is a Buck Regulator with very low dropout that can regulate voltage with temperature compensation of LED or current regulation. You can buy these too.

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