Nuts and Volts - February 2008

anything except the burnt-out incandescent light bulb. I didn’t hesitate at all to crack this burnt-out bulb and solder a V_F= 3V, I_F= 10mA white LED in its place to the remaining two whisker electrodes of the bulb (Figure 2).

The labor was absolutely minimum since the only thing I had to take care of was soldering the LED’s anode pin to the bulb’s anode whisker.

The brightness of this single LED isn’t really comparable to the original bulb, but it’s still useful at night and only consumes 3V* 10 mA = 30 m W (as compared to the original bulb at 3V at 500 mA = 1,500 mW — an energy savings of 50 times!).

Before making this particular LED flashlight, I searched on the Internet to see if there were 3V 20 mA LEDs. Even though the difference is only 10 mA, this specific requirement is important because most household flashlights use 3V battery cells, and the LED’s brightness is roughly proportional to its current. So, a higher current rating would double the brightness.

I found only one manufacturer who offered such an LED. The datasheet listed it as typically 3.0V at 20 mA. So I purchased 15 of them from a distributor. Unfortunately, after testing them, I found that none of them reached 20 mA at 3V. All 15 pieces only had the same 10 mA at 3V. When I questioned this on the company’s website, their engineer admitted it was a “datasheet error.” (Very frustrating!)

Make it Brighter — Parallel Three LEDs

The straightforward way to increase brightness is to connect LEDs in parallel. The brightness more than doubled ( 90 mW). I did have to use a small perfboard to solder the three LEDs in parallel; the subsequent supporting and connecting mechanism between the battery and the board also had to be addressed. This second LED flashlight was better than the first one, but it was still not brighter than the original.

■ FIGURE 2. This is the easiest way to make an LED flashlight.

In order to outperform the traditional flashlight in brightness, the battery voltage would have to be raised to more than 3V. That’s why most of the commercial LED flashlights use 4.5V batteries (three AA or AAA cells). I happened to have a fancy 4.5V bulb flashlight that contained a battery compartment for three AA cells. The bulb consumes 300 mA at 4.5V, or 1,350 mW. Such huge current is not good for battery longevity, so I wanted to change it. This seemed to be the best place for using 3.5V, 20 mA LEDs.

One problem that must be solved is the voltage gap between the 3.5V LED and 4.5V battery. I have seen some commercial LED flashlights where the 4.5V battery voltage is directly applied to the parallel 3.5V LEDs. That’s not a good engineering practice. It will overtax the LEDs and shorten their life.

The simplest solution is to add a current limiting resistor. As shown in Figure 3, this resistor is in series with three parallel LEDs. While each LED consumes 3.5V at 20 mA, the resistor Rs consumes the remaining 1V at 60 mA, so Rs = 1V/60 mA. Choose a close nominal value of 15Ω — that’s the main design of my third LED flashlight.

This flashlight (Figure 4) looks much brighter than the original one. Its total power consumption is 4.5V* 60 mA = 270 m W — only 1/5 of the original — so with this smaller current, the battery life will be several times longer, and higher lighting efficiency is achieved.

Nevertheless, we should note that resistors do waste some energy to protect the LEDs. In this example, there is always 1/( 4. 5) or 22% energy waste to resistance heating. The utility efficiency is 78%. We’ll see how to deal with this problem later.

■ FIGURE 3. Parallel three LED flashlight with a current limiting resistor.

BY G.Y. XU

Digital Flashlight Design Experiment

I use the phrase “digital flashlight” to categorize any flashlight that contains a microcontroller (MCU). Of course, digital flashlight, must use LEDs, but an LED flashlight is not necessarily a digital one.

Figure 5 is the circuit for a digital flashlight design experiment. It contains only a few components in addition to the LEDs. Let’s draw some guidelines from this experiment. Note the supply voltage is 4.5V, and the interrelated component values are already chosen for the best fit. The LEDs are connected in serial instead of parallel.

Basically, the circuit is a voltage booster or step-up switching voltage converter, which converts the 4.5V supply to higher voltage for the serial LEDs. The number of LEDs should be two or more.