I often read where people add the rectifier and filter capacitor between the coil/transistor and the LED. They claim that it increases the efficiency. I just want to clarify what I see is happening as the Joule Thief is operating. My observations are based on what I see on an oscilloscope connected to the JT, and other measurements.
In a JT, the coil stores the energy that has to go somewhere when the transistor turns off, and this gets transferred to the LED. During this time the transistor is open, it as if it was disconnected, and it does not dump the energy to ground. If you add the Schottky diode and filter capacitor, the schottky does not increase the amount of energy in each pulse that comes from the coil.
When your JT is running, each pulse of energy from the coil starts at a low voltage and high current and gets transferred as a higher voltage at a lower current. To my thinking it’s a pulse of energy, and whether you transfer it directly into the LED or through the diode and filter capacitor, which turns the pulses into filtered direct current, the energy from the coil is the same. But the diode has a voltage drop and when you find the power by multiplying the current times this voltage drop, this is a loss of power that ends up heating the diode – it is wasted power.
There may be other factors, such as how the LED responds to pulsed DC and filtered DC. The filtered DC could be brighter because of something that is caused by the LED’s response. But the way I see it is the LED is already a diode, and the circuit does not have to have a rectifier and filter capacitor to make it more efficient.
In my recent blog, I compared the conventional JT with my Supercharged JT, and gave the results of the actual light output measured with the luxmeter. The SJT is much more efficient, and is a simple circuit using inexpensive parts.
One might ask why I don’t compare the light output of the SJT with a conventional JT with Schottky and filter capacitor. I can’t see why I would need to because I believe the SJT has to be more efficient because of the lower current and therefore lower loss in the (Schottky or regular) diode.
Update Jan 6 2012 Comparison W & W/O Schottky & Filter Cap
I used a slide switch to switch the LED between the conventional (without) and the 1N5817 Schottky diode and filter capacitor. First, the components I used. The coil was a 1/2 inch high mu core with 13 turns solid twisted pair from a cat5 cable. The LED was blue, with a 1 ohm current sensing resistor in series. The transistor was a BC337-25. The filter capacitor was a 100 uF, and the supply had a 100uF bypass cap. The supply voltage was 1.5V for both tests.
The LED current was 19.7 mA and the supply current was about 95 mA without Schottky. When I switched to the Schottky and cap, the LED current went up to 24.7 mA, and the supply current went up to about 98 mA. This is a substantial gain in brightness with only a small increase in supply current. This agrees with the claims by others that the Schottky diode and filter cap increase the brightness of the LED.
However I must remind everyone that my Supercharged Joule Thief does even better with easier to get parts and less expense.
Update Jan 10 2012 – QS left a comment; Thank you. I’m happy to see he is somewhere, because Bill and I were seeing his emails bounce because his email inbox was full.
A quick comment re the rectified / filtered LED discussion.
My personal take is that directly driving the LED means the max LED current has to be double the ‘average’ current (or thereabouts, depending on the duty cycle etc). The LED’s light-output efficacy drops as driving current goes up, getting only around 85% increase for each doubling of current.
Also, a higher voltage is required to ‘push’ the current through the device.
Taken together, this will more than offset the 0.35-volt loss through the schottky diode.