Two IR LEDs in series, powered by 18650 - Page 1
Forum section: electronics / projects

author: Peabody
author_rank: Super Contributor
timestamp: June 12, 2020, 02:51:34 am
content: A while back I built an IR remote for my Roku box. It used a single LED and a Darlington transistor, with everything powered directly (no regulator) by an 18650. The Darlington was probably not a good idea, and I've thought about modifying the circuit to replace it with a mosfet. But I also wondered about using a second LED in series with the first one, so as much battery power as possible would be converted into signal power instead of just wasting it as heat in current-limiting resistors. The attached circuit is where I am at the moment. The LEDs would probably be Vishay TSAL6100, and the mosfet would be one of those SOT-23 jobs with vanishingly small RDSon.But I wonder if this is a practical design. It seems the current is going to vary a lot depending on the battery's charge state. Has anyone here done something like this? Does it seem to work ok?Also, I'm assuming the Arduino GPIO could drive the mosfet gate at 38K without issues. Does that seem right? I guess it would depend on gate capacitance?Anyway, before I start ordering parts, any comments or suggestions would be appreciated.

author: Buriedcode
author_rank: Super Contributor
timestamp: June 12, 2020, 03:15:44 am
content: I don't see why using two IR emitters in series would be a problem, their Vf is 1.6V max, so would still be fine when your 18650 is <3.2V. Of course as your battery voltage decreases, their output will, but this isn't that easy to work out by how much as their forward voltage will decrease with decreasing current. Also, as its a remote, almost all protocols have relatively low duty cycles, both to save the batteries in the remote, and to allow for higher peak power, increasing the remotes range.For arguments sake, lets say your 18650 is fully charged to 4.2V. And the Vf worst case for the emitters is 1.4V. 4.2 - (2*1.4) / 4.7 = ~300mA through the emitters. That isn't massive wit regard to a lot of remotes. Many common remotes have a single emitter, running off two alkaline cells, with a 10-22R resistor.One thing I would check, preferably actually test, is the MOSFET's turn on time. I have used several of the modern wonders that are SOT23 MOSFETS with single, or just double digit miliohm on resistance. They *tend* to have quite long turn on/off times, as well has higher gate capacitance (I sayinghigherbecause plenty have tiny gate capacitance so work nicely up to hundreds of kHz). The carrier is 38kHz, but for reasons above might not be 50% duty. Say its 25%, for a carrier period of 26.3us, the pulse width is ~6.6us. As the GPIO will be limited in its driving capability that 6.6us has to take into account not just the turn-on delay, but the gate charging.You could probably simulate it in LTspice, as there are plenty of MOSFETs models in there. Ultimately, though, its not like you're stunting amps here, or constantly transmitting for minutes at a time - I can't see anything popping, it just might produce a carrier with a low duty cycle. But things to think about! If in doubt, knock up a dirty prototype and test/measure. Discretes are so cheap you probably have to order 10 at a time anyway.

author: Doctorandus_P
author_rank: Super Contributor
timestamp: June 12, 2020, 05:40:48 am
content: Voltage drop of LED's varies a lot with the color of the LED. IR LED's can be at around 1V (and temperature depentent etc).If you put shunt resistor R1 on the Bottom side of the IR LED's you can put a bipolar transistor over it that pulls the gate to GND as a current limit, but make sure first that Ugs of the FET is low enough to tolerate the 600mV lower voltage drop.Also:For Bipolar transistors Ube is relatively constant at around 600mV, but in saturation the Collector can drop to as low as 50mV to 200mV above the Base. Sometimes bipolar transistors circuits are switching faster by placing a scottky diode over the Base and Collector, which pulls away the base current when the collector is nearing saturation.

author: Peabody
author_rank: Super Contributor
timestamp: June 12, 2020, 02:46:39 pm
content: Quote from: Doctorandus_P on June 12, 2020, 05:40:48 amThanks very much, but I should have said at the outset that I am not formally trained in this stuff. Could you possibly post a hand drawing illustrating these suggestions?

author: Peabody
author_rank: Super Contributor
timestamp: June 12, 2020, 03:04:37 pm
content: Quote from: Buriedcode on June 12, 2020, 03:15:44 amWhat I have now is a single emitter, a Darlington, and a 10R resistor. But I had forgotten that a Darlington doesn't fully saturate, and there is always a one-diode drop across it. So while it "wastes" less base current, it makes up for that by dissipating that diode drop on the full load current. So then I thought maybe just a regular NPN, but they still have some drop at saturation, and wasted base current (wasted in the sense that the base current doesn't flow through the LED). So I thought a mosfet would essentially eliminate any drop in the LED circuit, and would only waste whatever is needed to deal with gate capacitance (which, however, as you point out, could be significant). The thing is - adding the second emitter is as close as I'll ever get to a "free lunch" in electronics. You get twice the IR power on the same current. Well, not free in terms of cost and parts count, but you know what I mean.What do most remotes use - a mosfet or bipolar? I've only had one that I remember having two emitters, and it was a universal remote.QuoteWell, I'm an amateur, and have never done LTspice. I guess I should look at that at some point.Thanks very much for your comments.

author: Kleinstein
author_rank: Super Contributor
timestamp: June 12, 2020, 03:34:23 pm
content: There is no need to use a super low R_on MOSFET if there are some 10 Ohms in series anyway. With a larger R_on just reduce the resistance. So one could probably accept some 5-10 Ohms R_on.For the driver I would consider a BJT with limited base current. So more like just a BJT with a base resistor of some 500 Ohms to the IO pin, if the µC can deliver some 5-10 mA (if not one could consider an extra transistor). When not in saturation a BJT can switch reasonable fast. The current will vary a little with temperature and individual gain, but not too much and with higher temperature more current may be wanted. An individually adjusted resistor may not be so bad. The good thing would be that the supply voltage would have less effect.For the drive signal a 25% on/off ratio is good - going to 50% does only give some 40% more amplitude, but twice the average current. So the shorter on time is not just to compensate some possibly longer turn off time, but also because of the amplitude of the Fourier component that goes up like the sine of Pi times the PWM ratio.

author: ledtester
author_rank: Super Contributor
timestamp: June 12, 2020, 03:40:12 pm
content: I wouldn't worry too much about wasting power when blinking the LEDs.Your battery life will be determined mainly by how much current is used when in standby -- i.e. waiting for key presses.

author: ledtester
author_rank: Super Contributor
timestamp: June 12, 2020, 03:49:26 pm
content: Came across this idea for waking up an Arduino from deep sleep when a key from a keypad matrix is pressed:https://forum.arduino.cc/index.php?topic=431029.msg2971395#msg2971395

author: Peabody
author_rank: Super Contributor
timestamp: June 12, 2020, 10:16:01 pm
content: I think I already have the sleep situation optimized. I've removed the regulator and power LED, and since a keypress is what wakes up the processor, there are no oscillators running during sleep. There are limits to what I can accurately measure in terms of low current, but I measure about half a microamp during sleep, which I think is pretty good if the processor is alive at all.I know the current draw during IR transmission isn't really critical. I can just recharge the 18650 a little more often if needed. It's more the principle of the thing - to get it as good as it can be.

author: bin_liu
author_rank: Contributor
timestamp: June 13, 2020, 02:58:02 am
content: Power consumption is inevitable. The problem is how to work reasonably. My suggestion is to use a constant current circuit. I used STM8S003f3 a long time ago to make a 10m*10m electronic fence. For example:

author: Peabody
author_rank: Super Contributor
timestamp: June 13, 2020, 02:38:56 pm
content: Quote from: bin_liu on June 13, 2020, 02:58:02 amI will look again at a constant current solution. I had concluded I wouldn't be able to use two LEDs in series and have it work down to a battery voltage of 3.5V or so. But I will try out your circuit.

author: bin_liu
author_rank: Contributor
timestamp: June 15, 2020, 12:55:54 am
content: I don't know what type of infrared emission tube you are using, such as this IR333-A, if the current is not large, two series work is also feasible.Really, if the emission current is large, it is recommended to use two parallel work.

author: Buriedcode
author_rank: Super Contributor
timestamp: June 15, 2020, 03:17:32 pm
content: A constant current circuit seems a bit over kill for a remote. It's not visible illuminate that has to be consistent, and a constant current circuit will just drop voltage across a resistor and transistor. As your cell discharges, the output will go down, but unless you're trying to make the remote work at extended range, I don't see how it'll matter. Often the range of remotes is more determined by the receiver (those TSOPxxxx receivers are awesome) and ambient lighting, and given the low duty its why most remotes just have one emitter running at rather high power.Try it, maybe with a constant current circuit, and test the range. You might find you can get away with a lower current that you think, which would keep the Vf < 1.4 allowing for two LED's. But if the range is fine with one anyway, then it doesn't matter. And if the range isn't good enough, you can play with the resistor to determine emitter current.