I'm new to electronics, my background is mostly on the software development side... so go easy on me, as this might be a really stupid question. I just want to be sure what is safe to assume and what not.

My project consists on emulating the IR signal sent by the RC controller of a toy helicopter so I can then command it with my hand using a LEAP motion device.

The IR signal works perfectly, but I was not being able to match the range the RC controller was capable of getting. Anything like a computer monitor in between was enough for the helicopter to fall to the ground.

I purchased the brightest IR leds I could get (TSAL6100, 1.35V@100mA) and I was definitely getting some better range than my previous IR led but still nowhere near as much as the RC controller.

Since the IR is always blinking (and not constantly on), I assumed "it wouldn't hurt much" if I started reducing the resistors and try to get more brightness (that I've been checking with a cellphone camera). I was still not being able to match that of the RC controller.... so then I started increasing voltage with a separate power source, since arduino max output is 5V and RC controller using 6 AA batteries must be somewhere around 7V

I know the dangers of doing this, but I went ahead anyway and had a 12V power source without any resistor, making those 3 IR leds blink and now I get the same and even farther range than the RC controller..... so far none has burned, however, how crazy am I for doing this ?

What would be the correct way of achieving the exact same performance as the RC controller...

Will the IR leds blinking at 38kHz ever burn out under these conditions ?

Only under these conditions I am able to see through my camera that the led brightness matches that of the RC controller..... so how are they doing it "properly" ?

Thanks in advance.

  • \$\begingroup\$ What does the driver stage look like? Can you measure current waveform on a scope? The important property is current, not voltage. It is important to understand the various current properties that are listed in the datasheet and check them with your measurements \$\endgroup\$ – jippie Sep 1 '14 at 18:13
  • \$\begingroup\$ given my lack of knowledge and experience with electronics, I'm not exactly sure on where to start to perform such measurements to be sure I'm under acceptable conditions. also my measuring tools at the moment are limited to a computer, arduino controller and probably making use of a IR receiver I have but that's about it. \$\endgroup\$ – Eric Castro Sep 3 '14 at 8:57
  • \$\begingroup\$ You should get at least a(n autoranging) multimeter. Doesn't have to be an expensive one, as long as you leave the really cheap ones in the shop. A multimeter is a must have if you want to troubleshoot electronics. \$\endgroup\$ – jippie Sep 3 '14 at 15:44
  • \$\begingroup\$ Op most likely has the led controlled by a transistor controlled by the Arduino IR remote library. No special driver stage. A picture/schematic would help. \$\endgroup\$ – Passerby Oct 4 '14 at 14:57

These LEDs are specialized on short pulses of high current, so pretty much what you're doing. The datasheet suggests that 100µs pulses can be driven by currents as high as 1A (pulse peak, see "basic characteristics" -> "Radiant Intensity").

With 38kHz your pulses are on the order of 15µs only, so you might be getting away with even higher currents. But note that the datasheet doesn't give any info above 1A (see figure 3, which might be of interest to you).

Your way of reaching those currents is dangerous for the life of the LED though. Don't drive them with 12V without a current limiting resistor, even if the resistor limits the current to something high like 1A. That's just how LEDs work, their differential resistance isn't linear. What's probably saving you right now is the batteries' inability to deliver such high currents in so short pulses. You're effectively not driving them with 12V, because your batteries' voltage drops temporarily. That's not so good for the batteries either, depending on their chemistry, or anything connected to the batteries (might brown-out). You could add a fat 47µF ceramic capacitor to buffer the energy in front of your output stage. That might help burn those LEDs :).

The reason you are getting worse results than the controller might also have to do with the angular distribution of the emitted light. The LEDs you're using are pretty focussed (see fig. 9), their half-intensity angle is just plus/minus 10°. Maybe you'd be better off with an LED that has more spread?

| improve this answer | |
  • \$\begingroup\$ Great clarifications, I understand that my mistake isn't really on the amount of stress I'm putting into the LEDS but they way I'm doing such thing - I should be focusing on getting stable current even if it's really high, and not just playing around with voltage. However I wasn't using any batteries but a 12V adapter directly connected to the power outlet. Does that change anything? I have added two more LEDS in parallel, although results have not changed much. Since I didn't know the math behind, I guess I needed someone to help me understand the µs values in the datasheet, so you the man! \$\endgroup\$ – Eric Castro Sep 3 '14 at 8:54
  • \$\begingroup\$ @EricCastro: No, the response of the 12V AC/DC adapter will also be too slow (probably even slower than a battery). For short-time energy buffering like this, you'll need capacitors. \$\endgroup\$ – DerManu Sep 3 '14 at 20:27

The "usual" way to drive IR LEDs in this way is with a transistor and current limiting resistor. The LEDs are often in series rather than parallel.

As you are driving the LEDs with short pulses, you can put more current (eg 1A) than when you are driving them continuously. Size the resistor to the current you want (use ohms law) and work out the voltage drop across it. Then add up the voltage drop across the LEDS. Note that as you increase the current the voltage drop increases across the LEDs. There will be a graph in the data sheet that shows it. Don't be surprised if it's between 2 & 3V @ 1A.

Once you have the total voltage make sure your power supply is slightly higher. Stick a big capacitor next to the LED driver to act like a local reservoir for the current pulses (if you don't you may drop some current on the way from the power supply and that'll mean you won't be as bright as possible).

I'm not sure how you are driving the LEDs but when you setup for high current pulses make sure the LEDs can't be powered continuously (ie your driver output goes high while you are debugging) else you will blow the LEDs. A simple way is to prevent this is to use a capacitor (ie 100nF) in series with the base of the transistor / your driver output which will block a steady state signal, but will be transparent to your signal. You'll also probably need a resistor there to ground too to keep everything playing nicely.

| improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.