[HTC Wildfire S] no.human.being:
Resistors don't reduce voltage. Voltage differential across a resistor only occurs as current is flowing through it and the magnitude of voltage drop is directly proportional to the current flow through the resistor. The proportionality factor between voltage drop and current strength is the "resistance".

First, the battery does not provide 3.3 V, that's the voltage that the JTAG interface requires as logic level. The battery provides something between 3.5 V and 4.0 V. Battery voltage varies with the amount of charge contained in it and the "load" the board excerts on it. That's why there will usually be voltage converters on the board that convert the varying voltage from the battery into a constant one to drive the electronics.

Second, electrical voltage is a difference in what is called electrical potential (potential is the "absolute" unit while voltage is a "relative" thing - a point in space has a certain electrical potential, the difference in potential between two distinct points in space is called a voltage).

Third, a "bipolar" transistor (one having "collector, base and emitter") is not a voltage ampllifier, but a

current

amplifier. It allows a large "collector current" (current flowing into the collector and out of the emitter) to be controlled by a smaller "base current" (current flowing into the base and out of the emitter). The voltage drop across the "base-emitter-junction" is almost constant at 0.7 V. This means that "base" will always be on potential that's 0.7 V higher than the "emitter" potential (this is an idealization that's not exactly true in reality, however the base-emitter-voltage of a transistor does not vary a lot when base current is changed so it's a good approximation). What counts is not the voltage, but the amount of current flowing into the base. The more current flowing into the base, the more current will flow through the "collector-emitter-junction" of the transistor.

Because of this behaviour if you want to control a voltage with a bipolar transistor, you usually connect the emitter to ground (since you need current flowing into the base and out via the emitter when controlling the transistor and most voltages are differential to ground), then put a resistor between your "control voltage" and the base. The resistor is the proportionality factor between the current flowing into the transistor's base and the and the voltage you apply to it. Remember that the base-emitter-region already "drops" 0.7 V, so if you want to control the transistor with, say, a 5 V signal, then you need to drop the other 4.3 V over the resistor you connect to the base. Say you want to have 1 mA of current flowing into the transistor's base (how much base current it requires can basically be derived from how much collector current you require and the characteristic current gain of the transistor), then you need a resistance of 4.3 kOhms between your 5 V signal source and the base of the transistor and through which the 1 mA will flow when you put the 5 V on.

For signals which are generated by a source with very high impedance and/or have high frequency (JTAG will probably fall into both of these categories), bipolar transistors are not suitable because they require a certain amount of current flowing into their base that high impedance sources will not be able to supply and they are too slow to switch. For these purposes you need a different type of transistor that's called field-effect transistor or simply FET (e. g. MOSFET for Metal-Oxide Field Effect Transistor - which is a bit of a misnomer since the "metal" in this case is silicon which isn't actually a metal - silicon has more physical and chemical properties in common with carbon than with any metal - carbon is also a semiconductor and recently the semiconductor industry has shown large interest in carbon since carbon-based transistors are capable of switching even faster than "traditional" silicon-based ones). The junctions in a FET are not called "collector, base and emitter", but "drain, gate and source" (FET's "drain" is equivalent to bipolar's "collector", FET's "gate" is equivalent to bipolar's "base" and FET's "source" is equivalent to bipolar's "emitter"). Contrary to a bipolar transistor, a FET is not a "current amplifier", but a "voltage-controlled current-controller". That means that in a FET, the current flowing into the drain (and out of the source) is controlled by a voltage across the "gate-source junction" (with almost no current flowing into the transistor's gate, the "gate-source-junction" has an extremely high impedance, so it's basically a "voltage only thing" with almost no current flowing into the gate). So similarly with a FET, you would usually connect the source to ground and control a current flowing into the drain via a voltage applied to the gate in reference to ground. Again, this is an inverting circuit, so you need a second FET to invert again.

No problem but electronics are more "involved" than you imagine. You definitely need at least two transistors (and additional circuitry - at least some resistors) to "cleanly" shift the logic level of a single signal.