anything except the burnt-out incandescent light bulb. I didn’t hesitate at
all to crack this burnt-out bulb and
solder a VF = 3V, IF = 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.
The circuit uses ATtiny11 (U1) as a
charge pump and operates as follows.
February 2008 33