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1 Watt White LED Power
Supply Circuit for battery operation
This is a current
fed buck converter that is made with PNP bipolar transistors
and a N-channel MOSFET,
and operates from four AA cells (6 volts) to drive and LED
efficiently at a nearly constant power level.
Image 1. Small and cool
running, this circuit can provide
one watt or
more to a white LED.
Download
FreePC project file, gerber and png copper and
silk-screen: 1wattledbuck.zip
Introduction
I have some 1 watt warm white LEDs left over from a project and
the application for them was obvious: A better battery operated
lamp for use when the power fails, which it does frequently,
especially during the rainy season. And this one should cast
enough light with which to read without a lot of strain.
The
Circuit
This LED light uses a
buck converter to step down the voltage from four AA cells to
the lower voltage necessary to power a 1 watt warm white LED. In
this case, the LED requires 250 ma at 3.2 volts. The circuit
should close the loop around the current through the LED, since
that is what determines the amount of light emitted and is also
related to power dissipation. The astute reader will realize
that 250 ma x 3.2 volts is only 800 mW; this can be made much
higher as will be explained later, but I decided to go easy on
the battery and the LED itself.
Figure 1.
The circuit, which can be analyzed as a ring oscillator,
is a
current fed buck converter.
There are plenty of
integrated circuit switch mode power supply controllers
available in parts of the world that can handle the task quite
well. But what if you are an experimenter or student without
access to such specialized parts? A discreet component version
would be more widely accessible, and besides that, it's more
fun. As it turned out, the parts count and board space for the
discreet component power supply can be pretty small.
Long ago, I decided that a current fed buck converter is usually
the best solution for driving LEDs from higher voltages. The
control loop is very simple -running to the desired current on
each cycle of oscillation, it needed no loop filter, which made
ithe circuit inherently stable.
While looking on the web for examples to explain
the topology to a friend, I came across an
article
published by Dhananjay V. Gadre, in which he described a
simple three transistor current fed buck converter. His circuit
used two NPN bipolar transistors and one P-channel MOSFET.
Having many cheap N-channel high current MOSFETs,
costing about 16 cents each, and only a few expensive P-channel
MOSFETs, I turned the circuit "upside-down" so it uses two PNP
bipolar transistors and one N-channel MOSFET. The other
modification I made to the circuit was to add hysteresis to the
"comparator" which is made up of Q1 and Q2. The resulting
circuit is shown in Figure 1. The hysteresis helps keep
oscillation smooth until the battery voltage drops to the point
that oscillation cannot be sustained because of the fact that at
that battery voltage, less than the required current can be made
to flow through the LED, at which point the circuit goes open
circuit, placing the battery voltage across the LED, and the LED
continues to glow but grow dimmer until the battery voltage
drops below that voltage at which the LED will not light.
Start the analysis by
imagining that Q3's gate is held at battery voltage and that Q3
is conducting current through L1, the LED, and R1. Whenever the
current through R1 is sufficient to cause a high enough voltage
drop across R1 to cause significant base current into the base
of Q2, Q2 turns on, thereby turning off Q1, which in turn
results in Q3 turning off. The hysteresis provided by the
positive feedback through R3 to the base of Q2 assures that Q2
remains on until the current through R1 decreases to below the
trip point. Current, decreasing with time, continues to flow
through L1, via D1 and current, diminishing with time, continues
to be supplied by C1 to R1 as well results in the current
through R1 gradually decreasing. When it has decreased
sufficiently for Q2 to turn off (taking hysteresis into
account), Q2 does turn off, thereby turning on Q1, which in turn
turns on Q3. When Q3 turns on, current again begins to increase
through L1, the LED, and R1. The threshold at which Q2 turns on
is now higher because of the hysteresis from R3, and when that
threshold is attained, Q2 switches off and the cycle begins
anew.
The fact that the loop is closed around LED current is
significant. The light output of an LED is nearly linearly
proportional to current and is specified in data sheets, while
light output as a function of voltage is fairly nonlinear and is
not a controlled parameter. It should be clear that the LED
current is equal to the base-emitter voltage of Q2 divided by
R1. In the case of a 2.2 ohm resistor, this came out to 0.6V/2.2
ohms = 272 ma, which is close to what was observed. See Figure
2, below. The base-emitter voltage of Q2 reduces
approximately 1.8 millivolts for each degree of temperature
rise. This "thermal drift" is wholly acceptable as a trade off
to obtain the simplicity and economy of the circuit. Yes,
the circuit could be made to have very low drift at the cost of
increased complexity, but a drift of 16 millliamps over
temperature will not be noticed visually.
If you decide to build this circuit, you can choose the current
through the LED by selecting an appropriate value for R1. The
value of the resistor is found by dividing the peak base-emitter
forward voltage of Q2 by the desired current. For a
2N2907, the peak base-emitter forward voltage will be
approximately 0.65 volts in this circuit.
In general, this circuit is very forgiving
and will work with a wide range of component values. This lends
itself to being a "junkbox project", in other words, a project
built mostly from parts on hand.
Then 2N2907 should be recognized as being an easily-substituted
part. A 2N3906, BC556, BC557, or similar should work, though I
did not test these specific transistors in the circuit.
The MOSFET needs to have a fairly low Vth
(gate threshold voltage) to assure that with as little as 4
volts of gate drive, the FET remains saturated. It should also
have a low of an on resistance to minimize losses, especially at
lower input voltages. The maximum drain voltage is equal to the
maximum battery voltage plus the forward drop of D1.
D1 is a one amp Schottky diode to minimize forward voltage drop,
and thus the losses in D1. A three amp Schottky would probably
improve efficiency a little bit by lowering the forward drop a
little more. If you don't have a Schottky diode, then try a fast
recorvery diode of the type used in switching power supplies,
just make sure the current rating is high enough.
The inductor is discussed a little bit below, but for now, it
should be sufficieint to say that the exact inductance is not
critical.
Performance
Figure 3. LED power (brown line) and
input power
(blue line) as a function of battery voltage.
As can be seen in Figure 2, the performance
was as good as I had hoped for, staying above 80% over the range
of 4.5 volts to 6 volts. My intention was to operate this from
four AA cells, and 80% efficiency is pretty good at such a low
voltage without using any exotic components, unless you think of
Schottky diodes as being exotic. At this time, I need to point
out that the components in power section of the circuit are all
critical to efficiency. At first, I used a small bobbin type
inductor that measured 200 micro henries and 0.7 ohms as L1, but
after measuring very disappointing efficiencies, I replaced it
with a 150 micro henry 0.2 ohm inductor and efficiency improved
significantly.
An undesirable but
acceptable side effect of creating hysteresis with a voltage
that is proportional to battery voltage is that it reduces the
regulation. I think this is the reason for the increase in
current through the LED as the input voltage is raised.
Figure 3. LED
current (brown line) and input current
(blue line) as a
function of battery voltage.
Figure 3 is more or
less proof that this is a buck converter. Notice that as the
increasing battery voltage passes the voltage at which
regulation occurs, the battery current begins to decrease. It
cannot be seen from this chart, but the peak current through the
inductor continues to increase with increasing battery voltage.
By the way, though performance was measured at
over 10 volts battery voltage, I do not advocate operating C1, a
10 volt tantalum capacitor, above 7.5 volts. The circuit will,
as shown, operate beyond 7.5 volts, but a capacitor with a
higher voltage rating should be used. I operated this circuit
continuously from an AC line powered power supply for one week
after assembling it with no problems at all. That represents
many times the total hours I would expect to use this during
power failures, and it represents a small fortune in replaceable
AA cells, so I take that test as enough to show reliability
sufficient for my needs.
Below is an efficiency calculation at 4.32 volts
input.
In = 4.32V x 222 ma = 959 mW
out = 3.117V x 241 ma = 749 mW
Efficiency = 100 x 749/959 = 78% (pretty good!)
The oscillation frequency is mainly a function of
battery voltage as well as the inductance, the hysteresis
imposed by R3, the LED voltage drop, and to some extent, the
value of C1.
Here is what I measured:
Vin
F
4.12V
9.6 kHz
5.9V
15.28 kHz
Light output is important -after all, making light is this
device's sole purpose. The LED is rated at 75 Lumens typically
at 350 ma. According to the manufacturer's chart, at 250 ma,
light output is 75% of that at 350 ma, with no optical elements
between the LED and the rest of the world to attenuate the
output, we can take 75 Lumens x .250/.350 = 56 Lumens output.
Assembly
The LED is a 1 Watt Dominant Semiconductor NPF-RSD-MCPCB warm
white LED. It is important to keep the LED from getting too hot.
As with all semiconductors, operating life decreases
exponentially with temperature and at some temperature,
catastrophic failure tends occur. A minor consideration in this
case is that the amount of light for a given current also falls
off slightly with increased temperature.
This individual LED drops 3.12 volts at 241 ma, which equals 752
milliwatts. An LED that has a higher forward voltage drop will
dissipate more power because the current is nearly constant.
The manufacturer recommends that junction temperature be
maintained at less than 100°C. This is not to prevent
catastrophic failure as much as to assure that the LED does not
"wear out", or grow dim too quickly.
The thermal resistance of the junction to the case is 10° C per
watt, so with 752 milliwatts dissipation, the LED junction will
be 0.752 watts x 10 ° C/watt = 7.5 °C hotter than the heatsink.
To keep the junction at 100° C or lower the heatsink surface
must be less than 100 C - 7.5° C = 92.5° C.
If the maximum air temperature is 40° C, the maximum rise of the
heatsink over ambient is 92.5° C - 40° C = 52.5 C.
I asume a very low thermal resistance between the LED case
and the heatsink because I soldered the LED directly to the
heatsink, forming a eutectic bond. Such a bond of the full
contact areas is low enough to ignore compared to the
thermal resistance of the heatsink.
To get a temperate rise of 52.5° C with a dissipation of 752
milliwatts requires the heatsink to have a thermal resistance
from the LED mounting surface to ambient of less than 52.5° C/
0.752 watts = 69.8° C/watt. A pretty easy thing to accomplish.
Image 2.
The copper was removed in one long strip to isolate the
positive and negative terminals
of the board and the LED soldered directly to the copper.
The bare
connecting wires shown in this photo were later replaced
with the
red and black insulated wires visible in another photo.
I took a 3 cm x 3 cm piece of 0.5 mm thick
FR-4 printed circuit board material, both sides of which appear
to to be clad with 1/2 oz copper, and used a hand-held routing
tool to remove a thin strip of copper cladding from one side of
the board, as shown in Image 2. I then soldered the LED directly
to the copper.
Image 3. The
insides of the light before closing the cover. As a favor to
whomever might
want to service this in the future, I glued a copy of the
schematic and the
parts placement drawing on the inside of the back cover.
As can be seen in
Image 3, the piece of copper clad board holding the LED and the
power supply board were glued to the inside of the front of a
plastic box. A holder for four AA cells is held in place with
two self tapping screws.
The lamp was operated with the back cover removed, using a bench
power supply in place of the AA cells. After 30 minutes the
temperature of an electrode of the LED was measured to be 58° C
and the ambient air temperature was measured at 29° C, which
means that the simple heatsink achieved a thermal resistance of
29° C/0.752 W = 38.5° C/W, which is better than the required
69.8° C/W. Even though the thermal resistance probably increased
when the cover was put on the box, the junction temperature is
sure to be comfortably less than 100 C.
Image 4. The
finished light, ready for emergency use.
The light, shown in Image 4, is compact, being about twice the
size of the battery holder for the 4 AA cells, and can stand on
edge to illuminate a large area fairly evenly in the case of a
power outage.
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Contents ©2011, 2012, 2013 Richard Cappels All
Rights Reserved. Find updates at www.projects.cappels.org
First posted in May,
2011, Revised February, 2012. Corrected
replacement transistor remarks February & September, 2013
(Thank you twice, Mr. DeMoya).
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email to me at projects(at)cappels.org. Replace "(at)" with "@"
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