Best way to power 9-12V 10W LED from 3S LiPO - Page 1
Forum section: electronics / beginners

author: anvoice
author_rank: Frequent Contributor
timestamp: August 30, 2017, 07:23:59 am
content: Hello, I'm trying to add some simple lighting to a small robot. The idea is to use the 3S LiPo battery (12.6V fully charged, 11.1V typical, 10.2V minimum) to power a 10W, 9-12V, 1050mA LED. I know LEDs are typically run from a constant current source, so I'd like to make one that's as light-weight as possible without being too inefficient. It's also fine if the LED loses a bit of brightness by the time the battery voltage gets close to 10V. I've found several circuits for simple current sources. So far I am looking at:- A current mirror using 2 transistors and a resistor (https://www.allaboutcircuits.com/technical-articles/the-basic-mosfet-constant-current-source/)- An op-amp current source (http://www.ecircuitcenter.com/Circuits/curr_src1/curr_src1.htm)- An LM317-based current source, with a heat sink on the LM317 (http://users.telenet.be/davshomepage/current-source.htm)- An LM350-based current source (http://www.instructables.com/id/DIY-10W-LED-flashlight/)Btw, the RC crowd seems to be into running 12V LED strips directly from their 3S or 4S LiPo batteriesMy question therefore is:Would one of these setups would be appropriate considering my criteria (1A of current, light-weight, few components, fairly efficient)? Other circuit suggestions are greatly appreciated. Thank you.

author: Zero999
author_rank: Super Contributor
timestamp: August 30, 2017, 07:51:11 am
content: When you mean a 9V to 12V LED do you mean one with a built-in driver? If so then this isn't needed. If that's the forward voltage, then you need a higher supply voltage to get a consistent brightness.The LM317 and LM350 are no goers because the drop out voltage is far too high. The current mirror is a possibility but you need to set the ratio between the reference and the output quite high, otherwise it will waste lots of power. The MOSFETs should be swapped with BJTs which make a much better current mirror.The op-amp current sink is probably the best solution here.Have you considered a switched mode LED driver?And what's so bad about LED strip? If others find it gives them satisfactory results then it seems reasonable.

author: anvoice
author_rank: Frequent Contributor
timestamp: August 30, 2017, 05:59:23 pm
content: Thanks for your reply!I mean an LED without a driver such as the one found at this link:https://www.banggood.com/10W-Warm-Pure-White-High-Brightest-Save-Power-LED-Light-Lamp-p-88169.htmlI'll look closer at the op-amp circuit.The switched-mode LED driver will probably have to be step up or step down, because the battery goes from 12.6V to 10.2V. And they're heavier/bulkier than what I had in mind.Nothing wrong with LED strips, I just assumed those don't have a driver and so it's a bit strange to power them off a non-constant voltage. If that practice is fine though, I could definitely go with that.

author: .rpv
author_rank: Contributor
timestamp: August 30, 2017, 07:21:43 pm
content: Buy o build a module with the PT4115 IC, both are on aliexpress/ebay but the module just are sold on a few versions: 300, 600 and 900 mA, but you can change the shunt resistor to deliver the needed current.Here it's a video:

author: Audioguru
author_rank: Super Contributor
timestamp: August 31, 2017, 12:07:00 am
content: The Banggood Chinese LED has no manufacturer's name nor datasheet. Of course it needs a current-limiting device (LED strips have current-limiting resistors on them) and a heatsink maybe with a fan.

author: anvoice
author_rank: Frequent Contributor
timestamp: August 31, 2017, 04:26:36 am
content: Quote from: .rpv on August 30, 2017, 07:21:43 pmThat seems to be a good way to drive this. Found a driver on aliexpress, but what will happen to its output voltage as the battery drains? Will the LED simply dim somewhat as a result?Quote from: Audioguru on August 31, 2017, 12:07:00 amDefinitely getting a fan and heatsink.

author: .rpv
author_rank: Contributor
timestamp: August 31, 2017, 05:19:21 am
content: Quote from: anvoice on August 31, 2017, 04:26:36 amThat will depend on the vf of your led's, this IC it's a buck converter so the output voltage will be less as the input, so I got 10w led's advertised as 9-12v but the vf was under 10v (something about 9.2-9.7v), if you have this led's too won't be a problem (and for this case you'll need to have some kind of battery protection otherwise you might damage your battery if the voltage goes too low), also there is 10w led's with vf>12v, with this type you'll be able to use like 10% of your battery at most, when the battery voltage be less of the vf of your led's the light it'll be very dim if any.BTW, with this IC you can put a potentiometer from the dim pin to ground and the IC will give you dimmer capabilities.

author: anvoice
author_rank: Frequent Contributor
timestamp: August 31, 2017, 07:44:01 pm
content: Thank you. I don't know for sure what the Vf is going to be yet, I'll have to see when the LED arrives. As far as battery protection goes, do you mean simple over-draining damage or will the circuit circuit somehow damage the battery once it gets below a certain voltage? For the first scenario I plan to have various indicators, including a buzzer when voltage gets too low.I think I'd rather use PWM if I do want dimming since I prefer remote control of the lighting.

author: .rpv
author_rank: Contributor
timestamp: September 01, 2017, 12:54:30 am
content: Quote from: anvoice on August 31, 2017, 07:44:01 pmYou also can use PWM on the dim pin of this IC, it's very versatileBut if your led's have a vf over 12v it'll be better to use a boost led driver, something like the XL6001 IC, here's a video (on spanish):

author: anvoice
author_rank: Frequent Contributor
timestamp: September 01, 2017, 08:16:04 am
content: Nice. I'll have to see I guess what the Vf is on that LED once it arrives. As far as the XL6001, I'll have to build that circuit myself since I can't seem to find one on ebay or ali. Would it work with the fully charged battery though (12.6V)?

author: Zero999
author_rank: Super Contributor
timestamp: September 01, 2017, 11:10:39 am
content: The forward voltage will depend on the current.The XL6001 is a boost converter IC. It can only increase the voltage, not decrease it. If the battery voltage rises above Vf plus a Schottky diode drop, at the set current, the LED current will rise above the regulator's set point. Whether or not this damages the LED depends on if the temperature of the die and the power dissipated.I say test it. Measure the LED current, using a Hall effect sensor, when the battery is fully charged, to see if it exceeds the specification. Replacing the Schottky diode with an ordinary silicon diode (higher forward voltage) and/or adding a resistor in series with the LED could help, at the expense of efficiency. Another option is a transistor in series with the low side of the LED, configured as a constant current sink, with the current set to a higher value than the boost regulator. It will then behave as a linear regulator, if the battery voltage rises too high. Note that these tweaks may prevent or reduce the efficacy of the PWM dimming feature, when the battery voltage is high.

author: IanMacdonald
author_rank: Frequent Contributor
timestamp: September 01, 2017, 01:40:27 pm
content: The volt drop of a white LED is slightly less than the 3.7v nominal output of a lithium cell, so the two work quite well together with just a small series resistance.A somewhat better arrangement though, is a transistor with the LED in its collector, with a resistor from base to supply.Because the transistor's collector current is determined by the base current, and this changes far less than that in a series resistor, the brightness will be more stable over the battery voltage range.If we assume a gain of 100 and require 1000mA to the LED, then we need 10mA to the base, which is 1k2 at 12v. You would need to select the resistor to give the right current though, as transistor gains do vary quite a bit. Start with higher and work down. A small power transistor is ideal, and it and the LED will need a heatsink at this current level. You will probably find that 500mA is more than adequate drive.The advantage of this arrangement, crude as it sounds, is that it has a very low series voltage loss.

author: anvoice
author_rank: Frequent Contributor
timestamp: September 01, 2017, 08:03:47 pm
content: Quote from: Hero999 on September 01, 2017, 11:10:39 amWill do. I'll run some tests when the LED gets here.Quote from: IanMacdonald on September 01, 2017, 01:40:27 pmThanks for the info. I'll explore these options as well. If I'm lucky I might be able to build it out of components I have on hand, I have some power transistors lying around.

author: Zero999
author_rank: Super Contributor
timestamp: September 01, 2017, 08:49:25 pm
content: Quote from: IanMacdonald on September 01, 2017, 01:40:27 pmOther than what you've just said, another problem is the Hfe also has a positive temperature coefficient. It shouldn't result in a thermal runaway situation here, but care must be taking that the collector current doesn't exceed the LED's maximum rating, when the operating temperature is high. I'd prefer to add an emitter resistor and suitable op-amp and, if required, buffer transistor to make a proper current sink.Here's a design which uses both of the op-amps on the LM358 in parallel, to give as much base drive, as the IC can muster. Q1 needs to have a high enough gain to give the required current, otherwise an emitter follower can be added. Vref can be a potential divider, connected to: the battery (in which case the current is proportional to the supply voltage), a zener diode (better, use 5.6V for the highest stability) or a reference IC such as rhe TL431. D1a to D1c represents the LED, which will actually be many LED dies connected in series and parallel.Quote from: anvoice on September 01, 2017, 08:03:47 pmActually forget what I said about adding resistors, a silicon diode or linear regulator. Build the SEPIC converter circuit on page 8 of the data sheet. This IC should still be fine for 1A, at the voltage difference you're using, even though the application circuits only suggest 300mA. Failing that, the SEPIC circuit should work with other boost converter LED driver ICs. Either way, I'd certainly run such a cheap LED at a lower current, than its rating.http://www.xlsemi.com/datasheet/XL6001%20datasheet.pdf

author: anvoice
author_rank: Frequent Contributor
timestamp: September 02, 2017, 06:31:06 am
content: Quote from: Hero999 on September 01, 2017, 08:49:25 pmFrom what I understand, the SEPIC circuit's voltage will be constant regardless of input within a certain range, making it a sort of automatic select step down and step up converter? That sounds like the perfect way to drive this particular LED, assuming a Vf of 12V. With the PT4115 it would only work if the Vf is around 9V, otherwise the LED will gradually dim as battery voltage goes down.

author: Zero999
author_rank: Super Contributor
timestamp: September 02, 2017, 09:25:29 am
content: Quote from: anvoice on September 02, 2017, 06:31:06 amYes, that's the idea: a SEPIC regulator will work a higher input, as well as lower input voltages.

author: Audioguru
author_rank: Super Contributor
timestamp: September 02, 2017, 02:18:28 pm
content: The 3-cells lithium battery is 12.6V only when it is fully charged. Its voltage drops as it discharges and when it reaches about 9.6V then it is almost dead. Your Chinese high power LED actually has 9 LEDs, 3 in series and 3 rows in parallel and might use 9V or 12V.How will you charge the battery? You must use a balanced Lithium battery charger circuit to avoid an explosion or very hot fire. The balancing detects the voltage of each cell and insures that the weakest cell is not dangerously over-charged.If you use an active current regulator then its current will over-discharge and ruin the battery. Then you will need a circuit to disconnect the active current regulator when the battery voltage drops to about 9.6V.

author: anvoice
author_rank: Frequent Contributor
timestamp: September 03, 2017, 08:13:23 pm
content: Quote from: Audioguru on September 02, 2017, 02:18:28 pmI'll charge the battery using a dedicated balancing battery charger (already have it). As far as disconnecting the circuit and protecting the battery, I'll have various alarm methods to tell me when the voltage gets below approximately 10V so I can manually disconnect it. By that time a transistor driving the active current regulator will have disconnected the LED itself. Driving a LiPo to 3.2V per cell is in general not recommended, I'm thinking 3.4V at the lowest.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: September 04, 2017, 02:37:25 am
content: Why 3 cells when 1 will do?https://www.eevblog.com/forum/projects/building-a-simple-switching-circuit/msg1252706/#msg1252706(Real answer: you want a SEPIC converter for ~1:1 conversion ratios.)Tim

author: anvoice
author_rank: Frequent Contributor
timestamp: September 04, 2017, 06:34:18 am
content: If possible, I'd like to power everything from the on board battery, which is 3S (motors want that).Yes, Hero999 already suggested the SEPIC circuit to me. It's what I'll build if the Vf of the LED turns out to be 12V.