I came across this guy’s EMP blog describing his experiments with various Joule Thiefs, and the results he got. I couldn’t find an email address or how to contact him, so right now I just have to say it in my blog. He has been methodical and taken measurements while he has experimented and he shows the graphs of his results.
He has done some research, and viewed some Joule Thief info online, one of them was my blog. I’m proud to say that he thought highly of my Supercharged Joule Thief. But I was concerned that he might damage something when he said that the transistor was overheating when he used three cells: the battery voltage was higher than the LED voltage.
When experimenting with Joule Thiefs, there are a few things that are important to remember. One is that the LED is connected almost directly to the battery. The resistance of the coil’s primary winding is very low, less than 1 ohm typically. One end is connected to the battery, the other end to the LED, so it is almost the same as a direct connection to the battery. The blue or white LED’s forward voltage is about 3.2 volts. If the battery voltage is increased to 3V, the LED will start conducting, even with the transistor disconnected. This means the battery voltage is too high.
Two fresh alkaline cells in series will give 3.1 to 3.2 volts, which will light the LED very brightly. Two used alkaline cells at 1.4V each will give about 2.8V, which should light the LED dimly, and may work okay. Or two NiMH or NiCd rechargeable cells at 1.4V maximum each will give 2.8V or less, and should work okay. If the LED still lights when the base of the transistor is disconnected, then the voltage is probably too high.
Another item that is important is the maximum negative voltage on the base, which is 5V or 6V for most silicon transistors. The emitter to base junction of most silicon transistors will start to break down at about 7 to 10V, and acts like a Zener diode. But this permanently damages the transistor, the current gain will be permanently lowered.
Also, as the higher voltage causes current to flow in the LED, the LED’s resistance goes down, which loads the rest of the circuit, and prevents the transistor from switching on and off. Instead, the current through the1k resistor just turns the transistor on and it gets hot from excessive current.
Emitter To Base Breakdown
The voltage across the feedback (base) winding is the same as the voltage across the primary (collector) winding, but it is negative, and the battery voltage, typically 1.5V, is subtracted from it. So if there are two LEDs in series, the total peak voltage across them is typically 4 to 4.5 volts each, or a total of 8 to 9V. Subtract 1.5V, and it’s 6.5 to 7.5 volts. This is more than the 5 to 6V maximum rating, and may damage the transistor.
This indicates that it’s a bad idea to put two or more LED in series; instead the LEDs should be put in parallel. If the LEDs must be put in series, then the number of turns on the feedback winding should be reduced. Half as many turns means half the voltage.













