So my understanding is that FET's are voltage driven, which means the more volts you shove up their gate, the more volts flows from the source to the drain.
The more voltage you apply on the gate (with respect to the source), the more
current can flow through the MOSFET in the source-drain circuit.
This assumes that the connections to the source and drain are such that current can flow through into the MOSFET's drain and out of its source (i.e. in the "output circuit").
In nearly all practical circuits, external components will limit the current flow in the output circuit. This is also true of transistor circuits (where the output circuit is the collector-emitter path).
Often the MOSFET or transistor will be
saturated, which means that it is conducting fully because it is heavily biased. The current flow in the output circuit is set by external components, usually by the resistance between the drain (or collector) and the positive supply rail.
This description applies to an N-channel MOSFET or an NPN transistor. For a MOSFET, the gate is driven positive relative to the source, and "output" current flows in the drain and out the source. For an NPN bipolar transistor, "input" current flows in the base and out the emitter, causing "output" current to flow in the collector and out the emitter.
This current is "conventional" current - current that flows from positive to negative. It's the opposite of the actual flow of electrons and is used for convenience because the arrows in the component symbols match conventional current.
BJT's are current driven so the more amps you feed the base the more amps are conducted from the collector to the emitter.
Yes. Again the collector current is limited by external components but below this limit, that statement is true.
1. If this is true (is it?), why is the bias of a BJT specified in V (i.e. 0.7V)?
The base-emitter junction of a BJT is a diode junction. Look at the voltage vs. current graph for a diode junction. Go to
https://en.wikipedia.org/wiki/Diode#Current.E2.80.93voltage_characteristic and scroll down to the graph. The point marked Vd is around 0.7V (anode positive relative to cathode).
At this voltage, the diode will conduct
some current. If it's a base-emitter junction of a transistor, it may conduct, say, 1 mA. If the transistor has a current gain of 100, this will cause 100 mA to flow in the collector circuit (assuming the collector current is not limited to some lower value by components in the collector circuit).
If you now apply a small signal to the base, superimposed on the steady 1 mA current, the base current will rise and fall along with the signal. On positive peaks of the signal, the base current might rise to 1.5 mA, and on negative peaks of the signal, it might fall to 0.5 mA. The collector current will also vary between 150 mA and 50 mA in sympathy with the base current. You have gain, or amplification, of an AC signal.
This circuit is called a Class A amplifier. A Class A amplifier keeps the active device conducting at all times. In this example the collector current ranges from 50 mA to 150 mA but never drops to (or near) zero. The transistor is biased part-way into its "linear region", and the input signal causes it to move up and down through its linear region, but never outside it.
It's actually more accurate to say that the BJT is operated with a base bias
current of 1 mA. This yields a base-emitter voltage of about 0.7V because of the diode characteristic, but it's the current that is important. Other types of junction transistors (Schottky transistors and germanium transistors) will have base-emitter voltages less than 0.7V when biased into their linear regions. But bipolar transistors are current-controlled devices and it's the base-emitter
current, not the base-emitter voltage, that is important.
This description covers BJTs when biased into their linear region, where they can amplify an AC signal reasonably linearly. In other words, the collector current is a reasonably accurate reproduction of the base-emitter current (amplified by a factor determined by the transistor's current gain). This is used when a signal must be amplified accurately, with minimal distortion; for example, amplifying an audio signal from a microphone to increase its voltage.
But transistors can also be used as switches. In this case, the transistor is switched between two states: OFF, where the base-emitter voltage is usually zero, and no current flows into the base and no current flows in the output circuit, and ON, where base current (and collector current) are present.
Usually, in the ON state, the transistor is
saturated, which means that so much current is forced into the base that the transistor conducts as hard as it can, and the collector current is limited by other components in the circuit. This is called a "saturated switch". For example, the base current might be 3 mA, but the collector current might only be 40 mA because it is limited by resistance between the collector and the positive supply. The transistor might have a current gain of 100 but it cannot cause 300 mA to flow in its collector circuit. It is therefore "saturated". This arrangement is used, for example, when a transistor drives a relay coil.
2. Do FET's require biasing as well? Is it specified in Amps?
MOSFETs and JFETs are both voltage-controlled and their bias is specified in volts at the gate relative to the source. If a MOSFET or JFET is operated in its linear region, a steady bias voltage must be applied to the gate, to cause drain current to flow, and variations in the gate-source voltage will cause corresponding variations in the drain current. Just like the BJT example, this is a Class A linear amplifier.
MOSFETs require a positive voltage on the gate relative to the source (this is called "enhancement mode"). JFETs require a slight negative voltage on the gate relative to the source (this is called "depletion mode" and is comparable to the way electron tubes aka valves operate). JFETs are often used as small-signal Class A amplifiers and preamplifiers in audio and other applications.
MOSFETs are almost universally used as switches. In this case the gate voltage is driven between 0V (or close to it) (which turns the MOSFET OFF) and a positive voltage of between 3V and 10V depending on the MOSFET's gate voltage requirements (which turns it ON). With enough gate-source voltage, the MOSFET will saturate and conduct heavily in its drain-source path.
Some MOSFETs have extremely low Rds(on) values, i.e. when they are saturated, the resistance between drain and source is very low, and they can conduct dozens or hundreds of amps without dissipating much power, because the low resistance means there is only a small voltage drop between drain and source, so power, which is voltage x current, is kept low.
When used in this way, MOSFETs can switch extremely heavy currents ON and OFF, as long as the time spent in the linear region is minimised. In other words, the gate voltage needs to jump very quickly between 0V (approx) and, say, +10V. Confusingly, this actually requires a high gate current. Not because the MOSFET is current-driven - it isn't; it's voltage-driven - but because of the capacitance between the gate and source of the MOSFET, which can be quite high, especially for heavy duty MOSFETs. To make the MOSFET switch between OFF and saturated in a very short time, the gate-source voltage needs to change very quickly, which means that the gate-source capacitance must be charged and discharged very quickly, which takes a lot of current. MOSFET gate driver ICs are available with output currents of 1A and more.
This description relates to N-channel MOSFETs and JFETs. Polarities are reversed for P-channel devices.
That was probably a lot more than you wanted to know. I like to be thorough
