Over the past few weeks, I’ve been asked several times if I knew where you could buy “a 12-volt LED.” That’s a little like asking me where you can buy 10 yards of rivets. Light-emitting diodes are not so much characterized by voltage as they are by current. Can you make an LED (or even better, a series string of them) run on 12 volts? Sure, just so long as you have a way of controlling the current.
In the early 1900s, Henry Round, at the Marconi Labs in England, noticed that an electric current passed through certain materials (silicon carbide) he was experimenting with in radios would glow faintly blue. Decades later, Oleg Vladimirovich Losev, a Russian experimenting with these new “semiconductors,” was able to improve greatly on Round’s experiments, but was unfortunately killed in the WW-II siege of Leningrad before he could widely distribute his knowledge.
In 1961, Bob Baird and Gary Pittman of Texas Instruments (re)discovered that passing current through a diode made of gallium arsenide gave off infrared light, and the following year Nick Holonyak at General Electric was able to vary the proportions of gallium and arsenic to make a deep ruby red visible light emitting diode (LED). Ten years later, George Craford (one of Holonyak’s graduate students) added some other elements and made the first yellow, orange and bright red LEDs.
In 1990, Shuji Nakamura of Nichia, in Japan, went back to the Round formula of silicon carbide but soon discarded it in favor of indium gallium nitride and perfected the blue LED. Not long after some bright engineer thought of coating that blue LED with white phosphor, and so was born the white LED. Somebody got to playing around with the stewpot and found a mix of aluminum, gallium and nitrogen that would glow white without the use of phosphors.
Just for grins, and before we get into how to use them, it might be informative to see the list of LED colors and what sort of stuff is used to make them. We can mix and match any or all of these elements: aluminum, arsenic, carbon, gallium, indium, nitrogen, phosphorus, silicon, selenium and zinc. For the colors these elements make when combined, see the table.
So, we have colors of the rainbow including red, orange, yellow, green and blue. Seems to me that’s a pretty fair assortment, plus white, of course. Each of these is semiconductor material, which means we can “dope” them with a small amount of impurity to make either a “p” type semiconductor or an “n” type semiconductor. According to what we already know, a microscopic sandwich of p and n type material makes a pn junction, or diode. Thus, by passing current through the diode in the forward direction, we can get it to glow in its characteristic color. But how much current? Aye, there’s the rub.
LED Limits
LEDs are not only current-driven, they are current-limited. That is, the little microscopic grain of semiconductor salt can only handle so much current before it goes to that great lighthouse in the sky, leaving behind that puff of holy smoke as it departs. Do a quick perusal of the web sites of several catalogs, and you’ll see that most of the small LEDs are specified at a current of 20 milliamperes (20 thousandths of an ampere, or mA). At this current, red lamps have a voltage drop of around 2 volts, orange 2.1, yellow 2.2, green 2.3, blue 3.8, and white 3.3.
At first blush, this would suggest that one could string six of the red lamps in series and have a “12 volt” bulb. Um, no. In the first place, the data sheets say that the nominal 2 volt rating at 20 mA can vary from 1.8 to 2.2 volts within any batch of lamps. In the second place, when you go from a resting battery voltage of 12.6 to a full-charge battery voltage of 14.4, those LEDs are going to be severely over-powered and will, in all probability, burn out. Incandescent lamps are somewhat forgiving of varying voltages; solid-state lamps are not.
So the problem resolves itself with something that approximates a constant current to the lamp(s) over a varying voltage. Let’s see now, Ohm gave us a way of changing voltage to current, called a resistor. Let’s suppose (Schematic A) that our battery can go from 12.6 volts at rest to 14.4 volts at full charge and that we have a single red (20 mA) LED with a nominal forward voltage of 2 volts. To keep the current below 20 mA, I’ll choose a resistor so that the maximum battery voltage won’t over-current the lamp. Ohm says that the resistor is going to have to drop (14.4 battery volts minus 2 volts of LED voltage) 12.4 volts at 20 mA. Ohm’s Law says that the resistor has to be of value R = E/I, which is 12.4/0.02 = 620. Simple, put a 680 ohm (nearest standard +/-10% value) in series with the LED and problem solved. How much current will the LED draw when the battery returns to 12.6 volts? Ohm to the rescue says that the lamp will draw (12.6 – 2)/680 about 16 mA at minimum battery volts. Experience says that the human eye will have great difficulty distinguishing the difference between an LED at 20 mA and one at 16 mA. They’re the same brightness for all practical purposes.
LEDs in Series
Now let’s get fancy. Suppose that we want to put a string of LEDs in series and use a resistor to limit the current. How about three blue LEDs and the same 12-volt battery setup? Three blue lamps in series will have a nominal voltage drop of (3.8 x 3) = 11.4 volts. If we choose the same maximum current of 20 mA and the same maximum battery voltage of 14.4 volts, plugging these numbers into the above equation says that the resistor should be 150 ohms. However, returning the battery to 12.6 volt rest now says that the current is going to drop to 8 mA. I will tell you that there is a noticeable drop in brightness between 20 mA and 8 mA.
What to do? Individual resistors for each LED will work. Reducing the number of LEDs to two will work. And, a “current source” that can work on a (12.6 – 11.4) 1.2 volt difference will also work. What might this current source look like? An “upside down” PNP transistor makes a very fine current source and will work to around 0.7 or 0.8 volts difference between input and output.
Schematic B shows the details of our current source. R2 is chosen to flow about a milliampere of current through D5 and D6, setting the voltage at the base of Q1 at about 1.2 volts below the battery voltage. This 1.2-volt drop will remain relatively constant with battery voltages from 12.6 to 14.4 volts. If the base of Q1 is a constant 1.2 volts below the battery voltage, then the emitter of Q1 will be a constant 0.6 volts below the battery voltage. Given that we want 20 mA to flow to our diodes, Ohm (that pesky rascal again) tells us that R3 has to be 0.6/0.02 or 33 ohms to force 20 mA of current through the three blue LEDs shown.
The nice thing about Schematic B? It didn’t take into account what the nominal voltage drop across the LEDs was. You can put red, green, yellow, blue or white LEDs in this circuit, and it will be quite happy to provide 20 mA to any color or combination of colors.
So Todd, Bill, Jerry and Carl, what I told you in the newsgroups in February really does work as advertised. And now the whole world knows the secrets that I laid on you. LEDs really are magical little devices that replace a whole bunch of those burnout prone incandescent bulbs, and (if run within ratings) they have a theoretically infinite lifetime.
So you want to play around with some of those “super-power LEDs” that are blindingly bright and can be used for wingtip and panel flood lighting? Stay tuned. We’ll investigate some of those (surprisingly cheap) high-power LEDs in a future article.
