1 watt LED power supply circuit

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 2N3904, 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.

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.

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.

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