The past couple of months have seen us do a rather neat navigation light design using some new high-power, high-efficiency light-emitting diodes (LEDs). After much playing around over the last couple of months, I have come up with what I consider an optimum design to get from the 12-volt aircraft battery bus to the LED, wasting as little power as possible in the process.
The classic way of forcing an LED to light up from a battery is by putting a series resistor between the battery bus and the LED to limit the current to the LED to the rated value. Lets see what that might look like using the red LED as an example. Two of these LEDs in series at their rated current (700 mA) will drop 4.8 volts. If we don’t want to over-current them, we can use the high-charge battery bus limit of 14.4 volts. This means that our resistor will have to drop (14.4 – 4.8) 9.6 volts. Once again, brother Ohm tells us that the resistor will have to be (9.6/700 mA) 13.7 ohms. I cant buy 13.7-ohm resistors, so I have a choice of running them a little over current with a 12-ohm resistor or a little under current with a 15-ohm resistor. That will give us either 800 mA (12 ohms) or 640 mA (15 ohms). Lets err on the side of safety and use the 15-ohm resistor.
So we are delivering (4.8 x 0.7) 3.4 watts of power to the diodes but wasting (9.6 x 0.7) 6.7 watts of power in the resistor. Thats not exactly a paragon of efficiency. Moreover, what happens when the battery bus drops to 12.5 volts when not being charged? This drops the current to about 500 mA and noticeably dims the lights.
An Elegant Solution
Resistors are a cheap but inelegant way of solving the problem. A much better way of doing things is with a switching power supply. A switching power supply takes advantage of the fact that a chopped (square wave) battery output run through an inductor and capacitor to average out the voltage takes a high voltage at low current and transforms it to low voltage at high current. Efficiencies of 85% to 90% are quite common with switchers. Instead of burning about 7 watts in a resistor, our switcher burns about half a watt as heat, and the rest as power directly to the LEDs.
We have our choice of devices to do the switching, but for my money it is a lot less expensive to buy a low-power switcher and boop (thats a technical term, you’ll get used to it) the current up with an external low-cost, high-power transistor. Low-power switchers are tolerant of wide variations in component tolerance. In particular, the 78S40 has been around since dirt, is widely available, cheap in low quantities, and is foolproof in design.
The high-power pass transistor is half a buck, same for the inductor, and the rest of the circuit is nickels and dimes of resistors and capacitors. Figure that a $5 bill will give you silver change back.
Now Im not going to teach you how to design switching power supplies. As a matter of fact, Motorola wrote a 36-page applications note (AN920, widely available to download) back in the late 80s that is the bible on how to use the little 78S40. Couple that app note with an ironclad rule that we have here in the RST labs that if we have to do a second calculation of the same design, it gets converted to a spreadsheet. It takes 12 relatively complex calculations to do a switcher design, and each step takes about 5 minutes on a hand calculator. Thats an hour by my watch, and each step has the chance of error in data entry or calculation. A spreadsheet takes about the same time to construct and can do the calculations perfectly every time.
Because the green LEDs have a relatively steep voltage-current curve, that makes it easy to over-current the lights and burn them out. To avoid this, Im going to waste a little power in a ballast resistor to smooth that curve out. Each power supply will have a different output voltage to match the design voltage of the three LED colors. We can take care of this in a simple single-resistor value change.
Without going too deeply into the design, here are a few things to note:
1. I needed a fractional-ohm resistor from the +12 bus in to the emitter of Q1. Rather than buy a large and expensive fractional ohm resistor, I chose to parallel half-penny quarter-watt, half-ohm resistors to get the value I wanted. These resistors set the output short-circuit limit current to keep oopsies from blowing out the LEDs (if you catch it in time to turn it off).
2. R10 needs to be 12K ohms for the red diodes and 18K ohms for both green and white diodes.
3. All resistors are quarter-watt unless otherwise noted. C1 is a disk ceramic (noncritical). C2 is an electrolytic, 16 volt rating. I used two equal value cheap 1-watt carbon film resistors at R6 and R7 as ballast resistors in series to dissipate 2 watts.
Now for the serendipity of the design. On a long trip a few weeks ago, I noticed that a lot of truckers have gone to a rapid flasher on their LED turn signals. My wife, Gail, and I remarked to each other that the flashing was annoying as the devil, but it got our attention; it flashed into my mind that I would like to have annoying navigation lights. This particular device has an internal op-amp that was going to be unused, so why not make an annoyance circuit and flash these LEDs? The right lower corner of the schematic shows the circuit-it flashes the LEDs on and off at about three times a second.
This was so much fun that Im seriously thinking about doing an LED replacement for the landing light. Im hiney deep to a small hippo in trying to understand the FAR requirements for landing light intensity. Look for at least a mention of LED landing lights in a future column. Until then, shiny side up, pointy end forward, and have fun with it.