Most efficient way to power IR LEDs? - Page 1
Forum section: electronics / beginners

author: girishji
author_rank: Contributor
timestamp: January 06, 2023, 04:17:24 pm
content: I am using 3.3v to power a bunch of IR Led's. They have forward drop of ~1.6v. If I connect 2 in series I suppose I can use smaller R and save on power (power is from battery). Is connecting Led's in series a good idea? Will using some IC with PWM save more power?

author: fourfathom
author_rank: Super Contributor
timestamp: January 06, 2023, 04:47:10 pm
content: How many LEDS and what power are you talking about? Putting similar LEDs in series is fine, but with two LEDs and the normal variation on LED voltage (and actual "3.3V" voltage) you will have an extremely hard time setting and maintaining the LED current using a fixed resistor. If that 3.3V is coming from a linear regulator you may be better off feeding the series LEDs directly from the battery (using a series resistor).But if the total LED power is significant, look into constant-current switching drivers for the LEDs. These are common and inexpensive.

author: Kleinstein
author_rank: Super Contributor
timestamp: January 06, 2023, 04:55:31 pm
content: 1.6 V would be a a very high current level, only allowed in a pulsed mode. Normally the voltage of IR LED is less than 1.6 V - more like 1.2 V for DC operation. Especially with DC operation it makes sense to use less current per LED to keep the temperature lower.2 LEDs in series makes absolute sense, if it works. 0.1 V for the control is quite low and hard. The question is also if the 3.3 V is still at that level for a fast pulse and if it stable / allways at least that high.

author: Siwastaja
author_rank: Super Contributor
timestamp: January 06, 2023, 04:56:18 pm
content: Powering two Vf=1.6V LEDs in series from 3.3V source, with 0.1V dropped on the series resistor, is of course highly efficient.But it is useless, because current regulation is colossally bad. Do the calculation yourself - calculate R for your desired current and LED Vf=1.6V. Then, change LED Vf to 1.4V, keeping the same resistor, and calculate the current!And this is exactly what happens - when LEDs heat up, their Vf drop, increasing current, causing more heating, more decrease in Vf, and thermal runaway until destruction.The solution is to regulate current; measure current and feedback the voltage using this information.Most efficient - usually a switch mode converter. If you have 3.3V available, I would put enough LEDs in series so that their Vf is always more than 3.3V (so three or four), and then use a boost converter with current feedback to power them.

author: Peabody
author_rank: Super Contributor
timestamp: January 06, 2023, 05:25:04 pm
content: Below is a circuit I came up with for a two-LED IR remote control, with current through the LEDs around 500mA. Vf per LED is about 1.7V at that current. The idea is instead of dropping 0.6V across the emitter resistor to drive the feedback NPN, one of the LEDs is also placed down there along with a small resistor so that the NPN can still be driven, but without any material voltage drop. But I haven't tried to build this circuit, and have no idea whether it works or not.Of course this is pulsed output at 38KHz, 30% duty cycle, so hopefully heat will not be a problem despite the high current. But nominally Vf does increase with current, so the LED might behave a little like a resistor and permit current damping the way a normal constant current circuit would work. But it might be very difficult to tune. I wanted it to work at battery voltage from 4.2V down to about 3.5V, and for two LEDs in series over that range you do need constant current drive of some kind.

author: Kleinstein
author_rank: Super Contributor
timestamp: January 06, 2023, 06:40:37 pm
content: That is an interesting circuit that absolutely makes sense. At the rather high current the LED voltage likely no longer has a negative TC, as much of the drop is from the ohmic component inside the LED and not the actual diode part. If really needed to work with less voltage I would consider to add another LED in parallel at the collector side.

author: girishji
author_rank: Contributor
timestamp: January 06, 2023, 08:10:52 pm
content: Didn't think about the difficulty of current regulation when putting LEDs in series and having a tiny resistor in series!I was thinking of using a battery charger (ADP5360) which has a built-in buck-boost regulator that can put out constant voltage. From what I gathered this is not the most appropriate. What I really need is a constant current boost converter. I found MAX1698, a constant current LED driver IC. It can drive a number of LEDs as long as you connect 4 of them in series. Drives from battery and Has very low power loss. Maybe this will work.

author: Peabody
author_rank: Super Contributor
timestamp: January 07, 2023, 12:34:09 am
content: Quote from: Kleinstein on January 06, 2023, 06:40:37 pmWell, parallel LEDs would certainly be easier to deal with, but that would negate the whole point of using series LEDs. Instead of having a big voltage drop across a resistor, I want to dissipate that power as additional IR, not just amorphous heat. As I like to think of it, series LEDs are almost the only free lunch you get in electronics. For the same current, you get twice as much IR - but only if you have enough headroom.The problem in my circuit is that without constant current, the resistor which just barely produces 500mA at a battery voltage of 3.5V will produce 4A at 4.2V. Not a good thing. But I think it would be very nifty if you could drive two IR LEDs in series from an 18650 battery. So I need to test whether my circuit will work.

author: Doctorandus_P
author_rank: Super Contributor
timestamp: January 07, 2023, 03:01:29 am
content: How critical is it to save more battery energy?I like the current source / sink around the MOSfet, and with two diodes in series you have twice the light output for the same current.Efficiency for IR leds probably also goes down with higher current just as with regular LED's, so you may be able to generate more light with 4 LED's instead of two by reducing the current though each branch.Other options are* Finding LED's with higher efficiency.* Lowering the LED current as far as works for your application* Using lenses or mirrors to focus the light.* Reducing the amount of time the LED's are on.But how much energy goes though the LED's compared with the rest of the circuit?Usually the LED's are on for only a very short time, an other parts consume more energy.You also have to measure currents in the different parts. You can't rely on datasheets.* Is your uC really sleeping as much as it should and does it get down to it's advertised sleep current?* Lowering the voltage for the uC may help a bit in getting it's current down further.* Things like a leaky electrolytic capacitor can be the biggest power consumer for really low power circuits.For really lowpower circuits you have to look at every part of your circuit.

author: JustMeHere
author_rank: Frequent Contributor
timestamp: January 07, 2023, 07:34:47 am
content: Check out this datasheet:https://www.onsemi.com/download/data-sheet/pdf/nsi45015w-d.pdf

author: Kleinstein
author_rank: Super Contributor
timestamp: January 07, 2023, 10:23:07 am
content: The idea with a 2nd parallel LED on the collector (actually drain as it is a MOSFET) side was in addition to the LED on the source side. So it would be 3 LED total and 2 LEDs running with half the current to get lower voltage drop.For IR leds in pulsed mode and thus not much heating it is not that clear that the efficiency goes down with current. There are even IR LEDs that get more efficient at high current, kind of approaching laser mode operation with additional stimulated emissions, but without a resonantor. One still pays the price of higher voltage.In things like an IR remode it is not just about the energy consumption, but also about the available current. 500 mA even pulsed is already quite a low for a small battery.Optics can do a lot, but a very focussed beam can also be inconvenient.

author: girishji
author_rank: Contributor
timestamp: January 07, 2023, 07:28:16 pm
content: These LED driver ICs are slow to turn on. NSI4501x variants take milliseconds. MAX1698 takes 15us. Haven't found a nanosec turn on IC. I need to keep the IR LEDs turned on for a few microsec max. Whatever power these ICs save is offset by slow turn on time.I have a Buck converter available directly from battery charger, but max is 3V. Anyway I can work with this directly, add some feedback, and build a constant current circuit for LEDs? I would like to connect some LEDs in series, at least 2 or 3. Any pointers in this direction appreciated.

author: fourfathom
author_rank: Super Contributor
timestamp: January 08, 2023, 01:28:05 am
content: Quote from: girishji on January 07, 2023, 07:28:16 pmAn easy solution is to use a boost regulator with good transient-current response (or with big output capacitors), put your LEDS in series with a current-limiting resistors, and switch the string with a fast FET switch (N-channel between the bottom of the string and ground). If you pick the boost voltage to give you a reasonable current-limiting resistance the overall efficiency can be pretty good.

author: Siwastaja
author_rank: Super Contributor
timestamp: January 08, 2023, 11:21:20 am
content: Oh, you are doing modulation at high frequency, maybe communication over light, or time-of-flight distance measurement?Then the two options are:1) constant current source, LEDs shorted by a MOSFET2) constant voltage source + series resistor, LED string disconnected by a MOSFETI did use the latter, simple approach in a 3D time-of-flight camera design, as per the image sensor chip manufacturer example circuits. Modulated with 20MHz square wave, finding the fastest possible small N-MOSFET (without going to "RF" transistors), and coming up with fast enough gate drive for it was pretty important.But with series resistor approach, you need some 20-30% of voltage headroom. The MOSFET I used had significant Rds_on, though, and it has positive thermal coefficient, and was thermally coupled to the LED heatsink. I calculated that the increase of Rds_on by temperature compensated almost half of the LED Vf drop due to temperature, much better than nothing.

author: fourfathom
author_rank: Super Contributor
timestamp: January 08, 2023, 04:59:18 pm
content: Quote from: Siwastaja on January 08, 2023, 11:21:20 amI was going to recommend option #1, but didn't the OP want to minimize power consumption?Another option is a switched constant-current source -- a simple one with a MOSFET and bipolar transistor base-emitter sensing a small shunt resistor can work well into the MHz region. A 5V logic-level driver feeds the gate resistor and the bipolar collector pulls down the gate voltage as needed to regulate the current. This is a linear current-regulator with all the power-dissipation issues and VBE temperature sensitivity you would expect.And as long as the current-level doesn't need to be precise, your option #2; the switching boost regulator and a current-limiting resistor for the MOSFET-switched LED string ought to work adequately.

author: Siwastaja
author_rank: Super Contributor
timestamp: January 08, 2023, 05:08:25 pm
content: Quote from: fourfathom on January 08, 2023, 04:59:18 pmIt probably is the minimum power consumption solution, when one needs to modulate quickly, at least in theory; power to the short circuit is Rds_on * I, which can be very small. OTOH, the series resistor solution would waste at least 20-30% on the R, depending on how important the current stability is.But the devil is in the details; for fast switching and minimum power, one needs to compare full designs and not just generic ideas.

author: Kleinstein
author_rank: Super Contributor
timestamp: January 08, 2023, 05:18:51 pm
content: For the very high current range the LED can be voltage controlled. There is enough internal resistance to compensate the negative TC part of the diode. So the current control with the LED in place of a resistor works OK. The question is just if one also wants to run the 2nd LED also with that much current, or split the current over 2 LEDs to get away with less voltage. Even 250 mA is likely high enough to alow direct parallel connection and still get OK current sharing. So the circuit shown in the earlier post cited below has a good chance to work, though it is not super fast on turn on. The turn off part may anyway be slow from the LEDs. At least some of the IR LEDs are slow and would need the short to turn them off faster.Quote from: Peabody on January 06, 2023, 05:25:04 pm

author: fourfathom
author_rank: Super Contributor
timestamp: January 08, 2023, 06:28:38 pm
content: Quote from: Siwastaja on January 08, 2023, 05:08:25 pmYeah, I realized that after I posted. But as you mention, the details can be problematic.

author: Peabody
author_rank: Super Contributor
timestamp: January 08, 2023, 06:30:49 pm
content: For high speeds, would it make sense to replace the mosfet with a bipolar transistor? There would be more voltage drop, and heat, but a lot less capacitance to deal with. So just a standard 2-NPN constant current setup. I'm hoping these issues don't present themselves at 38K. :-)

author: fourfathom
author_rank: Super Contributor
timestamp: January 08, 2023, 07:10:54 pm
content: Quote from: Peabody on January 08, 2023, 06:30:49 pmYes, the 2-NPN should be faster than the NMOS/NPN CC source. Here's a pspice sim of that if the OP is interested. I show a 100ns on-pulse, driven from a 3.3V logic-level, three random LEDS, and two 2N3904s in a CC configuration, about 65mA. I added a shunt resistor across the LEDS thinking that it might speed-up the LED turn-off, but I'm just guessing about that (I've never seriously played with those LED characteristics before). The Q1 transistor may be undersized -- that depends on the duty-cycle and the Vdrop across it. That 10V supply could be lowered, depending on the actual LED characteristics.

author: girishji
author_rank: Contributor
timestamp: January 08, 2023, 11:37:14 pm
content: I do not need to switch this at high frequency. The LEDs (say 40 count) stay on for 1 usec before switched off for several millisec. I will supply about 1mA only to each LED. I have 3V constant voltage available from a buck converter running off of a LiPo battery. Design should be optimized for saving power.If I understand correctly I should get a buck-boost converter IC? Can I get as high a voltage I can get (that agrees with input current requirements)? I then need a low-side N MOSFET to turn off power.I don't quite follow the constant current solution diagram. Not sure what the 2 npn's doing current regulation. I need to think this through.

author: fourfathom
author_rank: Super Contributor
timestamp: January 09, 2023, 03:33:50 am
content: Quote from: girishji on January 08, 2023, 11:37:14 pmThat two-transistor constant-current design will certainly work at lower switching speeds, I just wanted to show that it requlates very quickly.Here is just the constant-current design all by itself. This is perhaps the simplest constant-current design you will ever find. It's not a switching current regulator, it's fully linear and will dissipate power determined by the current and whatever voltage is across the regulator (Vsupply minus the LED Voltage in the previous example). But it's a good circuit to study: The bottom transistor stays off until the voltage across that 10 Ohm resistor rises to about 0.6V (the Vbe of the transistor). At that point the lower transistor begins to conduct, shunting current from the base of the upper transistor. This provides negative feedback and so limits the current. The main drawback (besides the power dissipation) is that Vbe is temperature-sensitive (about -2 mV per deg C), and so the current varies with temperature. Often this amount of variation doesn't matter.And I'm not suggesting that you need the constant-current regulator. It's quite possible that a fixed-voltage switching supply and a diode string with series resistor and MOSFET switch to ground will meet your needs.

author: girishji
author_rank: Contributor
timestamp: January 09, 2023, 11:09:07 am
content: Thank you for the explanation. Now I can reason how the transistors are working in tandem.(Also, I like your Simpson quote. When I studied in US many years ago, there were RadioShack.)