# Transform 8~13V to 12V

I'm trying to build a diving flashlight and now I'm stuck with the electrical part.

I have a $12$V $10$W high brightness LED, and am now trying to decide which is the best option for the batteries. For now, I guess the best choice is a pack of $18650\; \left(3.7\; V\right)$.
Found a package with $10$ batteries with nominal capacity* of $4200$ mAh for $$15$. So, the plan is to put $3$ groups of $3$ batteries in series. This way, it should provide $\left(3\times4.2\right)=12.6\;$V at full charge* and $\left(3*2.7\right)=8.1\;$V at minimum charge*, and a nominal $\left(3\times 4200\right)=12600\;$mAh of capacity*. To maintain the brightness of the LED constant, is there an easy way to build a regulator with input $8 \sim13$V and output $12\;$V without losing much power? Or another better idea? UPDATE I don't have so much skill about electronics, and don't know exactly how to find the correct datasheet, but the LED is this: *: Edited and replaced with correct terms • The number in mAh is not current. – immibis Aug 21 '15 at 9:32 • Posting a link to the datasheet for your LED would avoid people wild guessing what you really need. Is it a bare LED? a LED with integrated resistor? A LED with integrated regulator? Too few information for meaningful analysis. – Lorenzo Donati Aug 21 '15 at 10:50 • If you're saying you found 10 18650 cells specified with a capacity of 4200mAh each, do not use those. They are almost certainly incorrectly specified, they are a fire hazard, and even in the best case will not perform to expectations. The current highest "standard" (tops at 4.2V) 18650 Li-ion cell is the Panasonic NCR18650B, which is rated at 3400mAh. Anything higher is almost certainly making incorrect claims, especially at$15 for 10 - at retail you can expect more like \$15 for 2 at best. – Bob Aug 21 '15 at 17:33
• What value would you put on your life? How about your ability to breathe? (Li-ion fires can release HF gas, which does a nice job of permanently damaging lungs) Do you need full function of your hands? (battery explosions can mangle hands rather badly) -- "don't want to spend too much" is not a good excuse for skimping on safety. An underwater explosion is even more potentially hazardous. Waterproofing won't help - this chemistry is perfectly capable of exploding on its own without any external interaction. – Bob Aug 21 '15 at 17:49
• By your own admission you are a newbie in electronics, so avoid Li-ion batteries like the plague. Stick with NiMH. Their energy density (Wh per kg) is not as big as Li-ion, but they are rugged and tolerate a fair bit of abuse. Their nominal voltage is ~1.2V, so you can build a battery pack of 10 in series and recharge them as a single battery.... – Lorenzo Donati Aug 21 '15 at 20:52

Update: After reading some materials, especially this "Internal Resistance of an LED as a function of Temperature" paper, it is clear that LED resistance will be lower in higher temperature. The paper shown the LED in test (5mm green LED) has resistance approximately 2000 Ohm in -38 degree Celsius, then reducing to approximately 500 Ohm in 78 degree Celsius.

As resistance decreases, it is better to use Current Regulator than Voltage Regulator..

## How to build the constant current LED driver circuit?

1. Determine the current at operating voltage stated by manufacturer. For example If we have 12 Volt/10 Watt LED array, we expect the operating current = 0.833 A.
2. Use current regulator circuit, make sure the output gives constant 0.833 A at various input voltage source. Examples of the ICs/circuits:
3. If you would not like to build circuit, you may purchase the LED driver, this is the example of the product.
• Nice! For all this options, could I use two (or two groups) 18650 batteries in series as the power source, or the draining rate could cause problems? – elias Aug 27 '15 at 13:42
• For the current, using more than 2000mAh li-ion battery for 833mA load is okay. For this application, always use 1A fuse in series with the battery, because it can explode with short circuit. For rechargeable Li-ion, never over discharge the battery because it can be damaged permanently. You could (better) charge a partially discharged li-ion battery – Oka Aug 27 '15 at 13:58
• Thanks for answer. My doubt still is: Why they say I should not regulate voltage? Because if I put a very low (tested with 7.5V), the LED brightness is very low. I think (as the current) the voltage also need to be regulated to ~11V. – elias Aug 28 '15 at 4:30
• By using current regulator, the voltage will increase/decrease automatically to maintenance constant current. Example: if you use 700mA LED driver for your LED, the voltage will reach approximately 10v at normal temperature, as long as the input voltage fits the driver specification – Oka Aug 28 '15 at 5:31

# DO NOT USE A VOLTAGE REGULATOR

You should never voltage regulate a LED. As an LED is a constant voltage device, voltage regulation is meaningless. The Power to Voltage graph (PV graph) is non-linear (or even polynomial) and very sensitive. A tiny over-voltage would produce a huge increase in power (destroying your LED).

EDIT:

As Temlib says. Another reason that you should never drive an LED with a constant voltage source is that LEDs are made of semiconductors. Semiconductors when heated will reduce in resistance.

Given that

As the Resistance decreases, the power (Ohmic heating) output increases, this leads to further reduction in resistance, which leads to more power ....

This process is called Thermal Runaway, and is a very common cause of LED failure.

# Use a Constant Current Source

The correct way to drive an LED is to use a CURRENT regulator, as the the Power to Current relationship is near linear, and you can much more accurately control the power being delivered to your LED (and prevent it from releasing the magic smoke).

• why the only correct answer is the last one? – Vladimir Cravero Aug 21 '15 at 9:33
• which specifically would be the most appropriate for this case? Can I build one with a few components? – elias Aug 21 '15 at 14:49
• Whilst it is very easy to build a linear constant current source, I would advise against it. It is much better (efficient) to buy a LED driver IC to integrate into your circuit. Make sure to match your with your LED. br.rsdelivers.com/product/recom/rcd-24-035-vref/… – Aron Aug 21 '15 at 16:46
• Some LEDs have some negative thermal resistance coefficient which means that, driven with a constant voltage, the current augment as it self-heats, provoking a thermal runaway. – TEMLIB Aug 21 '15 at 17:54
• @TEMLIB thanks. I've edited your comment into the main answer. – Aron Aug 22 '15 at 5:43

OK, there are two separate issues here:

(a) You have a LED. Therefore you need to drive it with a circuit to regulate current, be it a plain resistor, a linear current regulator or a switching current regulator (AKA LED driver).

Because of the high LED power a resistor is not a good choice. Linear regulators need the battery voltage to be always higher than the LED forward voltage and dissipate the extra power as heat. Switching regulators exist in various types (eg. boost, SEPIC and buck) and are more efficient.

(b) You want to decide on battery voltage so that circuit (a) is as simple, as cheap or as efficient as possible (add your own criteria here).

There are three possible design choices for the battery voltage $V_\text{bat}$:

1. $V_\text{bat}$ is always below the LED forward voltage ($V_\text{f} = 12$V in your case). You need a boost converter.
2. $V_\text{bat}$ is sometimes below, sometimes above $V_\text{f}$. You need a SEPIC converter or buck-boost converter.
3. $V_\text{bat}$ is always above $V_\text{f}$. You can use a buck converter, or a linear regulator if $V_\text{bat}$ is close to $V_\text{f}$, but not too close (they need 1 to 2 V of extra voltage).

Now as for the current limit circuit, there are again three choices:

• you can use a switching voltage regulator (boost, sepic or buck) to set $V_\text{out} = V_\text{f} +$ 1 to 2 V followed by a linear current regulator. But the losses of the two regulators will add up to 20-40%.

• you can use a switching converter (boost, sepic or buck) that has a built-in output current limit. Because of this feature it's probably sold under the name of LED driver. Losses will be lower, 10-20%.

• for the buck case only: calculate the power dissipated as heat in a linear regulator: $P_\text{reg} = I_\text{LED} \times (V_\text{bat}^\text{max} - V_\text{f})$ (using volts / amperes, not mA). If $P_\text{reg}$ is less than 1 to 2 W, you might be better off using a linear regulator only. That's because switching regulators have an efficiency of 80-90% and will waste 1 to 2 W anyways. If it's more than that, you should use a switching converter as described above.

You will find that buck LED drivers are easy to find. Boost and SEPIC LED drivers are harder to find and must sometimes be built from scratch. Buck and boost voltage regulators and linear current regulators are also easy to find.

• Do you have any schematic for a boost converter appropriate for this case? – elias Aug 21 '15 at 17:51
• Here is one of many ICs that could be used: ti.com/product/lm3421 . It's not a simple circuit to make though, especially because it must be done on a custom PCB with adequate layout re: noise & stability. – Damien Aug 21 '15 at 18:11
• Fortunately you might just be able to use this ready-to-use driver if you accept to reduce the LED power to 7W (600 mA @ 12V): eu.mouser.com/Search/Refine.aspx?Keyword=709-LDB-600L – Damien Aug 21 '15 at 18:11
• (It will also be vastly cheaper for just one unit, than making your own) – Damien Aug 21 '15 at 18:13
• Another option: eu.mouser.com/Search/Refine.aspx?Keyword=LDH-45A-700 – Damien Aug 22 '15 at 7:32

Very simple solution is to use higher voltage and then chopping down the voltage with DC-DC BUCK converter or using the lower voltage and Boosting it with BOOST converter. But if you have variable voltage source that is way too fluctuating, just as 8V-13V in your case, You can use Buck-Boost converter IC. Such IC have embedded Buck and Boost converter in them, that way buck-boost converter checks the input voltage; if its lower than needed voltage then Buck-Boost converter operates in boost conversion mode, and if input voltage is greater than needed output voltage then it operates in buck conversion mode leaving you a constant output voltage. The biggest advantage of Switching regulator ( Moreover Buck-Boost converter) is they are highly efficient.

Some switching regulators have programmable Input voltage. That means if you have a li-po battery of 4.2v (full -charge stae) and you are powering your flashlight, it will loose its energy and ultimately voltage. Lipo battery shouldnot be discharged below 3.3~3.0 V. So, if you set your controller to 3.0V undervoltage Lock out, then your Switching regulator stops the conversion mode at 3.0v leaving your battery in safe condition.

• Cool. Is easy to build one of these or is better to buy? By the way, do you have any circuit diagram? – elias Aug 21 '15 at 6:14
• -1 Voltage regulators are unsuitable for driving a LED. This answer is just plain wrong. – Aron Aug 21 '15 at 8:13
• If the array has a built-in current limiter, why add a voltage regulator upstream? You'd only decrease the overall electrical efficiency. – Aron Aug 21 '15 at 9:48
• @Aron the answer does not explicitly suggest using a voltage regulator. Buck or boost converters can perfectly well work as current sources. I modified a badly designed torch which had LEDs in parallel, putting them in series and adding a boost converter to supply 30mA at roughly 42V. It is pretty straightforward, and efficient. – Oleksandr R. Aug 21 '15 at 10:34
• @arjun my point is that it is difficult to do this with voltage a voltage regulator without some kind of Ohmic (parasitic) load/losses. – Aron Aug 21 '15 at 11:36

A LED is non-linear as stated before and the current can change dramatically around the forward voltage. Instead, also stated before, a constant current source should be used. Depending on the forward current of your LED I would suggest implementing the constant current source using a JFET with a resistor in it's feedback(google it) in this case, because of the relatively low voltage range. This will cost you only 2 components.

• The only problem with this answer is that it would use a linear regulator, which of course has a linear losses. For a battery powered system, if improperly sized, can lead to efficiencies of the ~50% range easily. It would also require the batteries be placed in series to bring the voltage high enough for linear regulation, which can be dangerous for an unbalanced Lithium based battery (as opposed to cell). – Aron Aug 26 '15 at 15:16
• Good point, the power dissipation could exceed that of the LED itself. Of course, like always in electronic design, a trade-off between effort and quality is in order. If we are talking problems, there is never an ideal solution because there will always be an extremely more complex but energy saving/accurate solution. – Armannas Aug 26 '15 at 16:17

I think try to combine your batteries in a way that the resulting voltage is always less or higher than your lights voltage. Then you can use dc-dc converter that either boost or decrease voltage.
It must not be in the middle of your battery voltage change range. I would create 2 groups of 4 (instead of 3 of 3) which is 4.2 * 2 = 8.8V

• This answer does not take into account how LEDs work. – Aron Aug 21 '15 at 9:49

There are two DC/DC options: SEPIC or Buck/Boost. Both don't require input voltage to be only above or only below your LED voltage.

If you just want to buy one, take buck and operate it as buck/boost by connecting it's output to battery minus and LED's plus and buck's GND to your LED's minus. Make sure the PS holds 3A current and 30V voltage.

If you are building the PS by yourself, build a SEPIC, it's much more fun.

• By the way, bot will suck the batteries to dry with high efficiency. – Gregory Kornblum Aug 21 '15 at 6:00