Many people consider white LEDs
to be the future of lighting. Strung together,
several LEDs can replace an incandescent
lamp or a compact fluorescent
lamp. High-power white LEDs need a dc
voltage of about 3.6 V at a current of
about 350 mA to achieve full brightness,
about 40 lumens.
In portable applications, white LEDs are
often powered by sealed lead-acid (SLA)
batteries having a typical output of 12 V.
The circuit shown in the figure takes this
12-V input and uses it to power a string of
white LEDs. It features low cost, high efficiency,
constant intensity independent of
variations in battery voltage, dimming
capability, and battery protection.
The driver circuit uses an SG1524
pulse-width modulation (PWM) switching
regulator (U1) operating in boost configuration.
This configuration enables U1 to
produce a maximum output of about 40 V,
which can drive a string of up to 11 serially
connected 1-W white LEDs. Because of
the high power dissipation, the LEDs must
be mounted with a suitable heatsink. The
design of the driver involves the selection
of an inductor, input capacitor, output
capacitor, switching transistor, and output
diode for a given operating frequency.
Operating frequency is:
A frequency of about 100 kHz is chosen
for this example. Higher frequencies
allow smaller inductances, but they also
increase switching losses.
The battery's voltage is 13.2 V when
fully charged and about 10.8 V under full
discharge conditions. The voltage across
the LEDs should be high enough to forward-
bias the LEDs under varying input
voltages. To ensure this, the required
duty cycle is:
where VO = output voltage across the
LED string; VD = diode voltage drop; VIN =
minimum SLA voltage; and VDS = MOSFET
voltage drop.
For an eight-LED string, VO = 28.8 V,
VIN = 11 V, and VD = 0.4 V for the Schottky
diode. Ignoring VDS, the required duty
cycle is 62.3%. U1 has two independent
switching transistors, each capable of
supplying about 100 mA and
operating with a maximum duty
cycle of 45%. To achieve the
required duty cycle, the two transistors
are connected in parallel.
Since the LEDs need current
higher than 100 mA, an external
MOSFET is required.
To compute the value of L1,
start with the average inductor
current:
If the ripple in the inductor current
dIL is a certain percentage of
the average current, the peak
inductor current is:
Assuming 40% ripple over the
average current, ILpk = 1.12 A.
Therefore, the inductance is:
In this example, Equation 5
gives the minimum value of inductance as 184.3 μH, where VIN is 11
V. The output capacitor's value depends
on the ripple allowed on the output voltage,
while the input capacitor's value
depends on the current peak.
To ensure constant illumination, the
current through the LED must be monitored
and maintained constant. To do
this, the current is converted to voltage
by R8, R11, R12, and U2b and is fed
back to the inverting terminal of U1's
error amplifier. This negative feedback
adjusts the duty cycle to maintain the
current through the LEDs. Varying R11
provides dimming of the LEDs.
Op-amp U2a and R9, R13, R14, and
R15 monitor the battery voltage and
switch off the LEDs whenever the battery
voltage falls below 11 V, thereby preventing
deep discharge of the battery.