With high power white LED driver portable lighting equipment

Many people believe that white LED is the future trend of lighting. In series with the multiple leds can replace incandescent bulbs and compact fluorescent lamps. High power white LED to maximum brightness ( About 40 lumens) It takes three. About 6 V dc voltage and current of 350 ma.

in portable applications, white LED is often by sealed lead-acid (has a 12 v standard output SLA) Batteries. Shown in the figure circuit using the input voltage of 12 v battery to power a group a bunch of white LED. It is characterized by low cost, high efficiency, brightness is not affected by the battery voltage, have dimmer function, and to protect the battery.

the drive circuit in the configuration of booster and adopted the SG1524 pulse width modulation ( 脉宽调制) Switching regulator ( U1) 。 This configuration make the U1 produces the maximum output voltage of about 40 V, can drive up to 11 1 w white LED series group. Due to their high power consumption, the LED must be equipped with corresponding radiator.

for a given operating frequency, the design of the drive includes inductor, the input capacitor, output capacitor, switch transistor, the output diode choice. Working frequency is:

in this case, the choice of frequency is 100 KHZ. The larger the frequency, inductance is smaller, but the greater the switching loss.

the battery voltage to 13 when fully charged. 2 v, about 10 when completely discharge. 8 V。 Through the LED voltage should be high enough to be under different input voltage LED forward biased. To ensure this, the duty cycle is:

VO = output voltage by LED series; VD = diode pressure drop; VIN = minimum SLA voltage; VDS = MOSFET pressure drop.

for an 8 - LED series, schottky diode VO = 28. 8 v, v VIN = 11, VD = 0。 4V。 Ignore the VDS and the duty ratio is 62. 3%. U1 has two independent switch transistor, each can provide current 100 ma, the runtime's biggest duty ratio is 45%. In order to achieve the required duty cycle to the two transistors in parallel. Due to the current more than 100 ma needed for the LED, so want to plus a MOSFET.

the L1 may be calculated the average of inductor current value:

if dIL inductor current ripple is a certain percentage of the average current, then the maximum value of the inductor current is:

if the ripples is 40% of the average current, ILpk = 1. 12 A。 Inductance value, therefore, is:

in this case, the formula 5 shows the VIN for 11 v, the inductance of the minimum value is 184. 3? 吗? H。 Output capacitance value depends on the output voltage ripple, and the input voltage value depends on the current the most value.

in order to guarantee the stability of the lighting, LED by the current monitoring must be and remain the same. To do this, the current can be made of the R8, R11 and U2b conversion into voltage, and through U1 error amplifier inverting side feedback. This kind of negative feedback can adjust duty ratios steady current through the LED. LED brightness varies with the R11. Operational amplifier U2a and R9, R13, R14 and R15 to monitor the battery voltage, as long as the battery voltage is lower than 11 v closed when the LED, thereby preventing the depth of the battery discharge.

the author: Santosh Bhandarkar, scientists and engineers, Indian Space Research Organization, Email address: santybh @ yahoo. com