Metal-Oxide-Semiconductor Field-Effect Transistors

A metal-oxide-semiconductor field-effect transistor (MOSFET) has three terminals, source, gate, and drain. In an n-MOSFET (or p-MOSFET), both the source S and drain D are N-type (or P-type) and the substrate between them is P-type (or N-type). The gate and the P-type substrate is insulated by a thin layer of silicon dioxide ($SiO_2$). Due to this insulation, there is no gate current to either the source or drain.

MOS_FET_transistors.gif MOSFET4.gif


Typically the polarities of the voltages applied to the MOS transistor are such that

$\displaystyle V_{GS}=V_G-V_S>0 \;\;\;\;\;\;\;$and$\displaystyle \;\;\;\;\;\;V_{DS}=V_D-V_S>0$ (150)

The n-MOSFET can be considered as a voltage-controlled current channel. When sufficient voltage $V_{GS}$ is applied between gate and source, the positive potential at the gate will induce enough electrons from the P-type substrate (minority carriers) to form an electronic channel called an inversion layer between source and drain, and a current $I_D$ between source and drain is formed.

The MOS transistor can be used in either analog circuits or as a switch in binary logic circuit:

$\displaystyle \left\{ \begin{array}{l}
V_{GS} \uparrow \Longrightarrow I_D \upa...
...ongrightarrow I_D \downarrow \Longrightarrow \mbox{cut off}
\end{array} \right.$ (151)

More acturately, the drain current $I_D$ and the gate voltage $V_{GS}$ can be modeled by

$\displaystyle I_D=\left\{ \begin{array}{ll}
0 & \mbox{if $V_{GS}<V_T$\ (cutoff)...
K(V_{GS}-V_T)^2 & \mbox{if $V_{GS}\ge V_T$\ (conducting)}
\end{array} \right.$ (152)


The current $I_D$ is affected by voltage $V_{DS}$ as well as $V_{GS}$. It can therefore considered as a function $I_D(V_{GS},\,V_{DS})$ of both $V_{GS}$ and $V_{DS}$ plotted below (similar to a bipolar transistor $I_C=f(I_B,\,V_{CE})$):


This function can be divided into three different regions:

In the plot of $I_D$ vs $V_{DS}$, the triode region and the saturation region are separated by the curve of $V_{GD}=V_{GS}-V_{DS}=V_T$.

Example: Assume $V_T=1V$.

Example: Assume $K=2 mA/V^2$ and $V_T=1V$, and both MOSFETs in the following circuit are in the saturation region. Find output voltage $V$.


Since both MOSFETs are in saturation region with the same $I_D$ which is determined only by $V_{GS}$ but independent of $V_{DS}$, their $V_{GS}$ must be the same. The upper MOSFET must have the same $V_{GS}$ as the lower one $V_{GS}=2V$, i.e., the output voltage has to be $3V$.

Comparison between BJT and FET

The BJT and FET can be compared with the old technology of vacuum tube. Although the specific physics of each of these devices is quite different from others, the working principles of these devices are essentially the same. In all three devices, a small AC input voltage (signal) is applied to the input terminal of the device (base, gate, or grid) to control the current that flows through the device (from collector, drain, or plate to emitter, source, or cathode, respectively), causing a much amplified voltage to appear at the output terminal (collector, drain, or plate) of the device.

tube.png tubePlot.png tubeAmplifier.png