PhysicsProf: I am going to make a few comments about your suggestions in advance of anyone trying them out. The purpose of commenting ahead is to hopefully shed some more light on the way the Joule Thief works. 2. Replace the 1 Kohm resistor to the base with 500ohms; take measurements. This can be expected to have a marginal effect on the operation of the circuit. Lowering the base input resistor will mean that the transistor switches on slightly faster. Because the JT does a "snap" when the transistor switches on and off, you are talking about changing the timing by a few microseconds at most. Also lowering the base resistor will increase the power dissipation slightly. I suspect that lowering the base input resistor will increase the operating frequency of the JT by a very small amount. I should also state that I am assuming that the 1K resistor is low enough in value to fully switch on the transistor so changing the resistor to 500 ohms will not make much of a difference. In the unlikely case that the 1K resistor does not not in fact fully switch on the transistor, then things change. When you make a Joule Thief you want the transistor to act as an ON-OFF switch, you don't want the transistor to work in partial conduction mode. 1. Replace the 10ohm + 100 ohm resistors with a 1-ohm resistor, and take measurements. (Makes V*I more straightforward also.) In the past I made reference to the JT as a transformer and made reference to changing the load on the secondary affecting the impedance match and the power transfer. Those statements were incorrect because it's not really a transformer. The real way of looking at the JT is that it is a circuit that operates in two cycles, first cycle charges the core up with magnetic energy and then the second cycle discharges that energy stored in the core, and then the process starts over again. The following discussion assumes that the collector LED has been removed and the only output load is on L3. The key point here is that once the core has been charged up with energy, then you discharge that finite amount of energy. So if you change the load on the L3 secondary coil, you will be able to affect the rate that the energy discharges, but not the amount of energy itself that's available to discharge into the load. So with those points in mind, it looks like reducing the load to an LED in series with a one-ohm resistor will speed up the discharge cycle considerably. This shortened discharge cycle will therefore increase the operating frequency of the JT and as a result increase the average power consumption of the JT. However, the overall average-power-in to average-power-out efficiency is not likely to change considerably. The most important thing to learn here is that the amount of energy per individual discharge cycle should not change substantially when you go from a (LED + 110 ohm) load to a (LED + 1 ohm) load. It all depends on the amount of current flowing through L2 the instant before the transistor switches off. If the amount of current is the same then the amount of energy stored in the core will be the same. Thus you are dealing with a fixed amount of energy that is available in the charged-up toroidal core of the Joule Thief. MileHigh
« Last Edit: 2011-02-08, 20:14:47 by MileHigh »
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