Powering
LEDs from batteries presents some special problems. The thing
about LEDs is that they have a high voltage drop for a semiconductor.
For blue, purple, UV, and white LEDs this is around 4 volts
and
for others it's around 3. This voltage drop must be overcome
before the LED can produce light output. What is worse is
that as
the batteries are drained, the load voltage decreases making the
problem even worse. With LED flashlights becoming more and
more
common, these are problems designers are dealing with in some
interesting ways. Here I'll show you a few circuits that are
in
some commercial products to give you an idea of how the voltage can be
boosted for the purpose of driving the LEDs.
Walkway Marker
Our
first example comes form a solar powered walkway marker.
These
are the little lights with the stake on the bottom that you can push
into the ground along your driveway or sidewalk and have the solar
panel on top. The solar cell charges a AA NiCd battery during
the day and at night the battery powers the LED. The circuit
board in this particular model was
originally designed to hold a pair of 5mm amber LEDs, but the
manufacturer apparently found a source of higher power 10mm amber LEDs
and the final product only needs one of these. Due to the
limited space, many of the components are surface mount.
The transistors are both 2N3904 equivalent surface mounts.
Unfortunately, the capacitor is also surface mount and is
unmarked.
The charging
circuit is fairly simple and has a photovoltaic solar cell to charge
the battery and a diode to prevent the battery from powering the cell
when it's dark. Now moving along, there is a cadmium sulphide
(CdS) photo resistor, a 10k resistor and a 1k resistor that forms a
voltage
divider at the base of Q1. When light
hits the photo resistor, it has a low resistance which is amplified by
the transistor. The collector is tied to the base of the left
hand transistor, so when it's on it clamps its base to ground and
prevents it from oscillating. When it's dark and the CdS Cell
has a high resistance, the right transistor is off which
allows the rest of the circuit to begin oscillating.
Now,
you may think that during all this time the LED would be on, but
remember what I said earlier about the voltage drop, which in this case
is 3v. To the 1.5v battery this looks like an open circuit.
This
is where our oscillator comes in. The oscillator is formed by
the two inductors, Q2, the 12k resistor, and the capacitor.
When the oscillator first starts, the top of the capacitor is
sitting at battery voltage, having already charged through the
inductors. This is carried to the base of Q2 through the 12k
resistor and causes it to turn on. Current from the battery
now begins flowing through the top inductor and current from the
capacitor through the bottom inductor. Once the capacitor is
discharged, Q2 is turned off, since it's basically clamping on its own
base. Now the funny thing about inductors is that
the stored magnetic flux collapses or flies back once power is
removed. This causes electricity to flow back in the opposite
direction and at a higher voltage.
So at the anode
of the LED and collector of the transistor, the voltage was pulled down
to 0 while the transistor was on. Once the transistor is off,
the inductors fly back and the voltage at this point rises quickly
until it's actually higher than when it started and high enough to turn
on the LED. This is because not only is the flyback voltage
in the coils higher than the original battery voltage, but it is also
in series with it.
You will also notice that the
emitter-collector of Q2 shorts directly across the LED. This
causes the LED to turn on and off, but it does it so fast, ultrasonic
in fact, that it looks as though the LED is constantly on. This is also
why that if you wanted to do something else with the circuit, such as
charge a capacitor, you would still need a diode to prevent it from
discharging back through the transistor.
This will
continue until the
battery is drained to less than .6 volts, which is the voltage drop of
the junction of a transistor.
LED Flashlight
Next we
have the circuit from an actual LED flashlight. This circuit
is so small that it fits inside the "bulb base" behind the reflector in
a 2C cell flashlight. All of the parts are surface mount
except the LEDs, the 2N3904 and the coil. The LED in the
schematic is actually 8 5mm ultra bright white LEDs.
Unfortunately, once again we have an unmarked surface mount
capacitor.
Instead of a CdS cell, a flashlight of
course has a switch. In our walkway marker circuit, one of
the transistors took the place of the switch, so the oscillator section
was really 2 coils and a transistor. Now instead, this
oscillator has one coil and two transistors. The coil is also
different. It is a larger wire wound I core instead of the
type that resembles a resistor. A larger coil and transistor
means more power since it has to drive 8 LEDs instead of just one.
Plus the white LEDs have a 4v voltage drop instead the 3v
drop
of the amber one from the walkway marker.
In this
circuit, the 2N3904 is the power transistor and the 2N3906 is the
driver. When switched on, the capacitor begins charging
through the coil and the 6.8k resistor. Once it
reaches .6v, current also flows through the base of the 06, turning it
on. Once on, the 04 is also turned on through the 330 ohm
resistor. During this time, the LED has been off,
even though we have 3v, since this is still not enough for the 4v
voltage drop of the LEDs. If they were say red LEDs, they
would be glowing since they only have a 3v voltage drop. At
any rate, it wouldn't matter since now that the 04 is on, this creates
another current path and would start oscillation in the coil anyway.
At the same time, the decrease in voltage at the collector of
the 04 is passed by the capacitor to the base of the 06, causing them
both to turn on harder. Also, more current is drawn from the
coil, causing an increase in flux. Since the capacitor cannot
pass DC, this state cannot continue and eventually the transistors are
forced off. Another way to look at it is that the capacitor
becomes fully charged and the side towards the base of the 06 is
charged positive; and positive on the base of a PNP doesn't do anything
but turn it off.
We now end up in the same place as
we did in the first circuit - excess flux in the coil. Once
again, the collapsing flux creates a flyback pulse which is in series
with the battery. Once this flyback ends, the voltage again
reverses and this is carried back through the capacitor, turning the
transistors on once again.
