I received a message asking several questions about the Joule Thief. The asker asked to be explained in layman’s terms, so I’ll try to keep it understandable. First off, I do not use a Spice program to simulate the circuit. I used LTSpice a few years ago, but not any more. Why should I when it takes a few minutes to connect five parts and I can then get the Real Thing? Also, always keep in mind that a Spice program uses a component that is not like a Real World component. A real world transistor might have a current gain of from 100 to 300, or even a wider spread, 100 to 500. I don’t think LTSpice gives you that kind of real world simulation.
Q1. How did I measure the efficiency of a Joule Thief?
A1. I used the supply voltage times the supply current to get input power (the regulated power supply is set for 1.5V and its meter tells me the current). For example 1.5V times .075 A (or 75 milliamps) equals .1125 watts. I measured the voltage across a 1 ohm current sense resistor in series with the LED. For example .017V times 1 ohm gives .017A. I multiplied this current times 3.3V for the LED forward voltage to get the output power to the LED. For example .017 times 3.3 equals .0561 watts. Then I divided the output power by the input power to get the efficiency. For example .0561 divided by .1125 equals .4987. Multiply this by 100 percent gives 49.87 percent efficiency.
Q2. Do you have any recommendations on measuring efficiency, with a DMM, without a regulated power supply?
A2. You can measure the battery voltage with a DMM, at the point where the battery connects to the circuit. Another 1 ohm resistor can be used in series with the battery to allow the DMM to measure the voltage across it and then get the current, same as above example. We have to remember that there is about 100 mA supply current, and that means about 100 millivolts or 1/10 volt drop across this 1 ohm resistor. That’s about 7 percent of the 1.5V supply, and it’s too much V drop. We can lower this by using a lower value resistor. A 1/2 ohm resistor will give about 50 millivolts drop, which is better. I used 1/10 ohm resistors, which give only 10 millivolts drop. That’s less than 0.7 percent of the supply voltage.
Q3. I know that the 2N3904, 2N2222 and 2N4401 are all different types of NPN transistors but what particular characteristic makes one more efficient over the other (in this application)?
A3. The transistor should have high current gain at high currents and very low collector voltages. A maximum collector current of 500 milliamps is a good starting point, the more the better. It must also have a low Vce(sat) at high currents. I look at the graph and see if the transistor’s collector saturation voltage is 1/4 volt or less at currents of at least 250 mA, but hopefully at closer to a half amp. The BC337 does not have any graphs for Vce(sat) but every BC337 that I have used can handle current better than a PN2222A or 2N4401, so I recommend it for JT use.
Q4. You also mentioned the JT eval website on scribd, so you may or may not be able to answer these questions but: http://www.scribd.com/doc/35050054/Joule-Thief-Eval 1. What exactly is a figure of merit? Is this just another name for efficiency?
A. I cover the stuff in this document in my recent blog.
Q5. How does the capacitor (base capacitor from the JT eval) allow the transistor to switch on and off faster? What are the consequences of this and why?
A5. The author explained this in the document at the link you gave. But I have my opinion about that, and it is covered in my recent blog (see the link above).
Thank you for the thoughtful and thought provoking questions. I hope my answers are correct, or if not, reflect what results others may get. As with the other author, his mileage varied because of his situation. Your mileage may differ if you do the same experiment.