Infrared emitter?

Question

Hello,

I have an infrared emitter with the following info on its datasheet:

Reverse Voltage: 5V

Continuous forward current: 150mA

Forward voltage: 1.3V typ., 1.7V max

Radiant power output: 13 - 15mW

Wavelength at peak emission: 950nm

I'm having a problem, I need to put in the maximum possible power into it so that it can work over long distances while it doesn't burn out ("For example, if I put connect it to a 5v source, it WILL burn out instantly"), so I need to know what things can I use, what resistors or transistors or voltage regulators or whatever is needed to make this infrared emitter work over long distance, because the way it's set up is just a straight connection from a 9V battery with a 470k resistor through its negative terminal hooked up right into the infrared emitter, so please help me make this thing emit infrared in longer distances.

THANK YOU VERY MUCH!

10 points assured for an answer that works!

Answer 1

First, you do know that an 'infrared emitter' is an LED that emits in the infrared band--this one specifically centered at 950nm.  So, in order to emit light, it has to be forward biased.

The first factoid on the data sheet is the reverse voltage.  Don't reverse bias it; if you do, don't do it for more than the reverse voltage rating.

Second factoid is the continuous forward current.  This is the maximum continuous forward current.  You can drive that much forward current into it...but it may not last very long if you do.  I would strongly suggest not driving it with the full 150mA.

Third factoid is the forward voltage.  It's a diode.  When it is forward biased, the voltage is going to up to the forward voltage, and then as voltage is increased the current will increase exponentially.  The quick approximation for handwaving circuit design is to declare that the current goes to infinity after the forward voltage.  If you have a constant voltage source, then you add a resistor ballast to prevent what will happen if the current does go to infinity.  The power dissipated in the resistor is wasted, so there's no reason to waste any more than necessary.  

If you have a 5V source, then 5V-1.3V (typical) = 3.7V is the voltage across the resistor ballast.  Now select the LED current--for example 100mA.  The resistor should be 37 ohms.

Finally, if you wish to have it work over a long distance, I would suggest a little more forward gain, either on the emitter, the detector, or both.  Put a lens or a parabolic reflector on it so you get a beam instead of dispersed illumination.  If the detector doesn't have a filter for ambient light, add one to improve the SNR.

Answer 2

It is obvious they don't want you to exceed 1.7 volts.  You can measure the voltage now, and add resistors or reduce the total resistance to hold the voltage to 1.7.

If the emitter produces a beam, that is the best you can do.  If it does not produce a beam, you could add a lens to focus it.

If you need to detect the beam at the long distance (other than just seeing it, the receiver wants to be sensitive to that frequency, and have sufficient amplification to do the job.

Answer 3

Ohms law to the rescue.  Hook a power supply or battery to a resistor to the emitter.  Posative to resistor to anode, negative to cathode.

To get the maximum power, the resistor you need is R=V/I, so V=your supply voltage minus 1.3v, I = 150mA.  So for a 9v battery, the resistor = (9v - 1.3v) / 150mA = 51.3 Ω (ohm.)  If you connect to 5v source, R = (5v - 1.3v)/150mA = 24.7 Ω

You can always use a larger resistor.