Most silicon transistors have an emitter to base reverse voltage rated at 5 volts maximum, with some transistors rated as high as 6 volts. Old germanium transistors were higher than this, but when the manufacturers began to make silicon transistors, they apparently gave up having a higher emitter to base reverse breakdown voltage so they could gain performance in some other area. As a result, almost all modern silicon transistors are rated for 5 or 6 volts.
What happens if this voltage is exceeded, and the emitter to base junction breaks down and conducts current? I thought that the junction would act as a zener diode and as long as the power didn’t cause overheating, it would not do any harm. Then one day someone in the Usenet Newsgroups said that when the emitter to base junction breaks down and conducts current, it permanently damages the transistor by reducing its current gain. I didn’t believe this, so I put a transistor on a power supply and ran just a few milliamps through the emitter to base junction in the reverse direction for a few minutes. I measured the transistor’s current gain before and after I did this, and I found that the current gain afterwards was much lower than it was before, something less than half of the original current gain.
I was really surprised. I had no idea that the transistor’s current gain would be affected by such a low power and small current. But one has to remember that most transistors are used as amplifiers and their emitter to base junction is never reverse biased, so there is no chance of this damage happening. When a transistor is used for switching, the E-B junction can be sped up by putting reverse voltage on it. Or if a transistor is used in a circuit with inductance, the inductor can put reverse voltage on the E-B junction. This is the case with the Joule Thief.
The conventional Joule Thief runs on only 1.5V, and the highest voltage anywhere in the circuit is across the LED, which is less than 5V with a single LED under normal conditions. The feedback winding is typically the same number of turns as the primary, so whatever voltage is on the primary is reflected in the feedback winding, even though the polarity is reversed. This means that when the primary’s peak voltage is 5V, the feedback’s peak voltage is a negative 5V.
As long as the LED is in the circuit and puts a limit on the maximum coil voltage, the transistor’s emitter to base junction will not have to handle more than 5 volts. If the LED is removed or there are two or more LEDs in series, the voltage across the emitter to base junction will increase, because the feedback winding reflects the voltage in the primary.
The feedback winding is connected to the battery through a resistor, so it will have 1.5 volts added to its voltage. In other words, if the peak voltage across the feedback winding is negative 5V, then the voltage relative to negative is a negative 3.5V, after adding the battery voltage. The peak voltage across the feedback winding would have to exceed -6.5V for the negative voltage on the base to exceed -5V. When we design the JT circuit we want to make sure that the negative voltage on the base never exceeds -5V.
We can allow the peak voltage across the primary winding to go higher than 6.5V if we change the feedback winding so it has less windings than the primary. If the feedback winding has half as many turns, then the voltage across the feedback winding will be half of the primary voltage. But since most transistors used in a conventional JT can handle only 20 to 50 volts, we can’t allow the primary to feedback turns ratio to be too high because the primary voltage would then be above the transistor’s collector breakdown voltage.
If the output voltage must be high to drive multiple LEDs, then it is easy to wind a third high voltage winding on the coil, which will have many turns of fine wire. This voltage will be ten times higher than the primary if it has ten times the number of turns. Of curse the current will be proportionately lower, too.
Back to experimenting…