|Elliott Sound Products||AN-003|
Being a LED, they do have the rather annoying trait of being a current driven device, having a relatively low forward voltage. The current must not be allowed to exceed the design maximum, or the LED will be damaged. This requires that a current regulator must be used between the voltage source and the LED itself, so complexity is increased compared to using a normal lamp.
An alternative is to use a linear regulator, but these are very inefficient. The full current (typically around 300mA) is drawn at all supply voltages, so with 12V input, the total circuit dissipation is 3.6W. Admittedly, this is not a great deal, but where efficiency is paramount such as with battery operation, this is not a good solution. The circuit shown in Figure 1 was the result of a sudden brainwave on my part - it may have been triggered by something I saw somewhere, but if so that reference was well gone by the time I decided to simulate it to see if it would work.
Figure 1 - Ultra-Simple LED Switchmode Supply
Using only three cheap transistors, the circuit works remarkably well. It is not as efficient as some of the dedicated ICs, but is far more efficient than a linear regulator. It has the great advantage that you can actually see what it does and how it does it. From the experimenters' perspective, this is probably one of its major benefits.
One of the features of this circuit is that it will change from switchmode to linear as the input voltage falls. It still remains a current supply, and the design current (set by R1) does not change appreciably as the operation changes from linear to switchmode or vice versa.
I = 0.7 / R1 (approx.)
At higher input voltages, the circuit will over-react. Because of the delay caused by the inductor, the voltage across R1 will manage to get above the threshold voltage by a small amount. Q3 will get to turn on hard, current flows through the inductor and into C1 and the LED. By this time, the transistors will have reacted to the high voltage across R1, so Q1 turns on, turning off Q2 and Q3. The magnetic field in L1 collapses, and the reverse voltage created causes current to flow through D1 and into C2. The cap now discharges through the LED and R1, until the voltage across R1 is such that Q1 turns off again. Q2 and Q3 then turn back on.
This cycle repeats for as long as power is applied at above the threshold needed for oscillation (a bit over 5V). As shown in the table below, the circuit changes its operating frequency as its method of changing the pulse width. This is not uncommon with self-oscillating switchmode supplies.
The table above shows the operating characteristic of the prototype. I also checked the performance with an ultrafast silicon diode, and the input operating current was increased by almost 10%. The suggested Schottky diode is well worth the effort. LED current remains fairly steady at 260mA, since I used a 2.7 ohm current sensing resistor as shown in the circuit diagram.
Q1 and Q2 can be any low power NPN transistor. BC549s are shown in the circuit, but most are quite fast enough in this application. Q3 needs to be a medium power device, and the BD140 as shown works well in practice. D1 should be a high speed diode, and a Schottky device will improve efficiency over a standard high speed silicon diode. D1 needs to be rated at a minimum of 1A. L1 is a 100uH choke, and will typically be either a small 'drum' core or a powdered iron toroid. An air cored coil can be used, but will be rather large (at least as big as the rest of the circuit).
The efficiency is not as high as you would get from a dedicated IC, because the switching losses are higher due to relatively slow transitions. At best, I measured around 60%, which isn't bad for such a simple circuit. Input voltage can range from the minimum to turn on the LED up to about 16V or so. Higher voltages may be acceptable, but that has not been tried at the time of writing.
All resistors can be 0.25 or 0.5W except R1 - this needs to be rated at 0.5W. Paralleled low value resistors may be used to get the exact current you need, but always make sure that you start with a higher resistance than you think you will need. If resistance is too low, the LED may be damaged by excess current.
|Copyright Notice.This article, including but not limited to all text and diagrams, is the intellectual property of Rod Elliott, and is Copyright © 2004. Reproduction or re-publication by any means whatsoever, whether electronic, mechanical or electro-mechanical, is strictly prohibited under International Copyright laws. The author (Rod Elliott) grants the reader the right to use this information for personal use only, and further allows that one (1) copy may be made for reference while constructing the project. Commercial use is prohibited without express written authorisation from Rod Elliott.|