Wednesday, April 10, 2013

Buck Converter 1 Watt White LED Driver

This is an example of efficiently driving a 1 watt white LED from a 12 volt
battery using a buck converter. The LED could simply be connected with a
series resistor to get the desired current, but the efficiency would be
only 25% since the resistor would drop 9 volts while the LED only requires
3. The buck converter provides about 90% efficiency.

The idea is to establish a circulating current through the inductor, diode
and load, while the switch replenishes the lost load energy on each cycle.
The duty cycle of the switch will be the output voltage divided by the input
voltage, or about 3/12 (25%) in this case. Its actually a little greater
since there is a small (2.2 ohm) resistor in series with the LED that drops
about 0.5 volt, so the total load is about 3.7 volts and the duty cycle is
around 31%. The circuit could also be used to charge AA batteries from a
12 volt source with adjustment to the duty cycle.

The driver section uses a CMOS hex inverter (CD4069) where two of the
inverters form an oscillator with 31% duty cycle at about 11.5 Khz,
or 66us off time, and 21uS on time for the MOSFET switch. The remaining 4
inverters are used in parallel to provide additional drive current to
the gate of the MOSFET. The duty cycle can be adjusted with either the
15K or 20K resistors.

The minimum inductor value was worked out from E = L * di/dt and a LED
current of 250mA. The minimum value is where the current falls to 0 during
the switch off time, or 66uS. The peak inductor current would then be twice
the average or 500mA and the inductor will charge from 0 to 500mA in 21uS.
So, di/dt is 0.5 /.000021 = 23810 amps per second. The inductor voltage
(E) will be 12 minus the load voltage 3.7 or 8.3 volts and the minimum
inductor value L will be 8.3 / 23810 = 0.35 mH. The actual value used should
be somewhat higher to avoid the current falling to zero and to avoid large
peak currents and possible saturation. The example here uses a approximate
2 mH inductor so the change in current is about 100mA and the peak current
is lower at about 300mA. The current waveform is shown in the LTspice
picture below. Notice the current ramps from about 50mA below the average
current to about 50mA above the average or about 100mA total change. The
15 ohm resistor in the LTspice picture represents the LED plus a 2.2 ohm
resistor. The MOSFET is represented by the SW (switch) component, and the
drive circuit by the V3 symbol.

The inductor (pictured below) should be rated for saturation current of more than the peak current, or maybe 300mA in this case. The toroid inductor used is fairly large for the task measuring about 1.5 inches diameter with 20 turns of #18 wire. The core is conductive so it probably should be taped in case the wire insulation fails. The picture shows the naked core for illustration. A smaller core with an air gap could be used to avoid saturation, but would require more wire which would add to the losses due to the wire resistance. Another approach is to use a higher frequency so smaller inductors can be used. But this will add to losses since there would be more switching transitions per unit of time, which adds to the loss. The diode is a VSK330 schottky 3 amp variety for low loss, but most any 1 amp rectifier could be used with somewhat less efficiency. The IRFZ44 MOSFET is also an overkill rated at 50 amps max but very low on-resistance of only 28 milliohms. A much smaller device could be used, but I dont have the numbers. Note the circuit has no regulation, so the 12 volt input should be stable. If the battery voltage varies, the duty cycle and LED current should be set using the highest expected supply voltage. 


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