The battery will discharge as needed to supply the current demand of the load impedance that you are using. The trick in the joule thief circuit is that it uses a technique that allows a battery to discharge to a much lower state than usual. While a 1.5 volt battery would usually be considered unusable at 1 volt the joule thief circuit continues to use the battery down to a level of perhaps 0.3 volts. Thus the name implies that you are able to steal a bit more energy from the battery.
Generally most joule thief circuits are simply a bad designed boost convertor (Thereby giving the illusion of powering devices that require much higher than the voltage of the battery), if you choose the right components this can operate to very low source voltages in the order of hundreds if mV (As long as the transistors are able to turn on and off still)
The frequency is harder to control as it is based on vary basic components, variences in the inductance, capacitors, temperature etc will all have an impact in a simple circuit like this and if your driving a few LEDs or something then who cares.
If you need Ghz frequency then you should stop looking at joule thief circuits and focus your search more on RF LC tank oscillators and a RF transistor to go with it, your inductors will be very different from the joule thief
This question is really above my technical ability to accurately answer. I have a feeling that it would have to be drastically changed and redesigned to operate at that frequency. The question would have to include what your application is and what are the other parameters of what you are trying to accomplish. The joule thief as designed seems to be a power circuit to use the energy in a battery to drive a small load like an LED. None of these parameters would require a 1 gHz oscillator. Keep in mind also that most modern high energy density batteries do not tolerate a very deep discharge and would be damaged if left on an joule thief circuit to the final fail voltage.
an intresting question 'do they drain batteries faster than usual'
theres quite a few 'traps' with things like this, like making parallel battery packs of 3's and 4's..you get less of the capacity for the third cell and even less out of the 4'th cell than you do on paper, even working it out in watts is ermmm..misleading greatly.(conclusions on this were a definite make two packs of 2 in parallel and use them individually for a much longer run time).
e.g. you use the joule theif to get say 30 minutes more out of the battery, but it takes 50 minutes off the original run time adding up to you abused a cell at the risk of the bang to a much much lower voltage but allas..it ran for much less time over all.
so, does a joulle theif take more than it would to run than to not use it? to broaden, booste power supplies take more current in amps to 'step up', the higher the booste the higher amp draw at the imput..any one?
