Wikipedia describes a Joule thief, shown below, as a circuit that can be used to provide constant load even when voltage of a battery decreases. At least that's what I think the article means.
simulate this circuit – Schematic created using CircuitLab
Is this circuit a (mostly) constant current source, or does it maintain a constant voltage? How does one calculate the current/voltage at the load given a resistance RLOAD?
You have the circuit diagram slightly wrong. The collector winding is connected to the battery terminal, not the resistor. The windings should be opposite polarity, not the same.
simulate this circuit – Schematic created using CircuitLab
It's a pity that the circuit editor only comes with a co-phased transformer, as this wiring looks untidy.
The output comes from magnetic energy stored in the core of the inductor, operation usually described as a flyback. Being an inductor, the output tries to rise to whatever voltage is required to push its current through the load. It's therefore best to call it a current source. That's why it makes a good LED driver.
During the transistor on phase, the current builds in the inductor, and at some current (supply voltage dependent) the on phase finishes. This current now flows through the output, decreasing as the voltage on the load ramps it down again. Calculating the average current is quite complicated as it depends on the on and off time, which depends on the load voltage. So the average output is not truly constant.
In addition to forcing current through the load, it also has to charge the various stray capacitances. With a very large load resistor, these capacitances may dominate the maximum voltage that could be obtained. With a high enough output voltage, the transistor will break down, so this circuit should never be run without a load.
It's basically a low-tech switching boost supply. If you correct the polarity dots on your transformer, it will self-oscillate, as an increase or decrease in collector current induces the opposite effect on base voltage. When it's on, the base voltage increases until the current stabilizes. Eventually, both sides of the transformer are pulled to saturation on the feed side, and there's no current increase to maintain the base voltage. At that point, the current through the secondary has no place to go, and the inductor lag raises the voltage to whatever is needed to get it somewhere. (Unloaded, it will probably toast your transistor.)
Note that the voltage out will be dependent on load, so it's not a regulated supply.
A joule Thief is always used to light an LED that clamps the output voltage. I think the current is higher with a higher battery voltage but it still works with a low voltage causing a low current.
Solar garden lights use a Joule Thief type of oscillating circuit including a 1.4V battery charger circuit in an IC and a small external inductor in a voltage booster circuit. The maximum current produced is less than the maximum current allowed for the LED. Then the LED operates at 3.5V and has no resistor to limit its current.
Neil_UK has already done a fine job describing everything, but I wanted to give a graphical answer to this question, to add to what Neil_UK has said.
For the following 6V Joule Thief circuit:
Here (in red) is the graph of V vs I (mA), & also efficiency (blue):
You can see that the curve is definitely not a constant current output, which would be a horizontal line for V vs I. Though the response looks somewhat linear (with a slight curve to it). This is about 250mA through 5 LED's, so very bright, with a set of 4 fresh AA cells -- heat sink mandatory for LED's, and probably the NPN N1, and Schottky diode D91 as well.
This generally matches what I see in power LED datasheets.
And here (in blue) is the graph of V vs power (mW), & also efficiency (red):
This V-vs-Power curve looks remarkably similar to the VI curve.
Note that this circuit is a much better Joule Thief than the original, because it steals even more joules per cell, going down to probably about 0.6V / 4 = about 0.15V per cell (4xAA) vs about 0.6V / 1 = 0.6V per cell for the standard version. This is because the limiting factor is the base-emitter junction of the NPN transistor. And if you build this, it probably needs at least a 220uF low ESR capacitor across V1.
So, at USB 5V, this circuit puts out 2.8W at 15.3V and 186mA with an efficiency (with real inductors) of 90.4 % (but would lose some efficiency with the necessary filtering to prevent ripple from feeding back into the USB port). The simulation capacitors are also ideal, so you'll lose a bit more there.
So, to answer the question, no, it's decidedly not even close to being a constant current source (with falling voltage), although by regulating with a microcontroller and a digital potentiometer at R1, it would be possible to implement a constant current source. Though you still need to prove stability.
Perhaps this is what the Wikipedia authors intended... At a particular voltage, the Joule Thief will produce a constant power output, which is great for driving LED's, as the forward voltage of the LED's pretty much won't matter. An additional benefit is that thermal runaway becomes impossible. Finally, if you feed the L2+NPN-base with a constant current source, that will be amplified to produce a close-to-constant-current output (for as long as the constant source maintains regulation).
