How to calculate the power dissipation in a transistor

You don't need to forward bias the B-E junction. Whether you do depends on what you want the transistor to do. To keep it off, you want to not forward bias the B-E junction. As for the C-B junction, keeping that reverse biased is fundamental to how BJTs operate.

A BJT is basically a reverse biased junction that can be made selectively leaky. You apply a voltage accross C-E and with the base open nothing happens. The reverse biased junction doesn't allow any (except for small leakage we will ignore) current to flow. However, the special property of a BJT is that a little current thru the base messes up the insulating capability of the reverse biased junction. The gain, and hence the useful properties, of a BJT come from the fact that is only takes a little current to muck up the reverse biased junction such that it allows a lot more C-E current to flow. This ratio of C-E current to B-E current is the basic gain of the transistor. It can be as low as 5-10 in big mongo power transistors and 100s in high gain signal transistors.

The B-E junction also looks like a diode to the external circuit. It will have a forward voltage drop when conducting just like a regular diode. In silicon, this is 500-750 mV for most non-extreme applications.

If you want to use the transistor as a switch (either as full off or full on as you can make it) then you have to make sure there is no base current in the off case, and plenty enough to support the desired collector current in the on case. Driving the base to the emitter voltage is a good way to make sure the transistor is off. To turn it fully on, you need to provide at least 1/gain of the desired collector current.

In other cases, a BJT might be used in "linear" (it's often rather non-linear, but this is the term used to mean in-between mode or not-switch mode) mode, like a audio amplifier. In that case you want to always keep it somewhat on and have the input signal change its operating point. If done right, this can amplify the signal. Different configurations give you voltage gain, or current gain, or some combination. In these cases, biasing the transistor refers to keeping it somewhere in the middle of the operating range so that a little input signal can change the output both ways. Biasing is basically setting up the DC operating point.

If you wish to avoid holy wars you'll need to avoid making simplistic and incomplete statements :-).

Bipolar transistors are current driven.

MOSFETs are voltage driven.

In both cases the spread of parameters during manufacturing is such that a circuit will almost always rely on feedback to produce a given voltage or current gain.

MOSFETs tend to be slightly more costly at the very bottom end for "jelly bean" applications. But, for switching more than a few ~100mA, MOSFETs are usually as cheap or cheaper than functionally equivalent transistors, are easier to drive from a uC (microcontroller) as a digital switch than bipolar transistors and tend to have very significantly superior on characteristics.

An "on" bipolar transistor exhibits a saturation voltage. This can be several tenths of a volt and to get it much under 0.1V usually requires a high base to collector current ratio that is undesirably high. At 1 A a 0.1 \$V_{sat}\$ (saturation voltage) dissipates 0.1 W and is the equivalent of a R = V/I = 0.1/1 = 100 \$m\Omega\$ transistor. But at 10A the figures are 1 Watt dissipation and 10 \$m\Omega\$. The 0.1V is very difficult to achieve at higher current levels.

The \$R_{DSon}\$ (Drain-Source on resistance) of MOSFETs is typically under 0.1 \$\Omega\$ and you can get devices with 10 \$m\Omega\$ or even sub 1 \$m\Omega\$.

As switching speeds rise MOSFETs need a gate driver to charge and discharge the gate capacitance. These can be relatively cheap.

More soon ....