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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


The n-MOSFET can be considered as a voltage-controlled switch. 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 $I_{DS}$ between source and drain is formed. Qualitatively the conductivity between source and drain of an n-channel field effect transistor can be described as:

\begin{displaymath}\left\{ \begin{array}{l}
V_{GS} \uparrow \Longrightarrow I_{...
...\downarrow \Longrightarrow \mbox{cut off}
\end{array} \right. \end{displaymath}


More accurately, the behavior of an n-channel MOSFET can be described by the function $I_{DS}=f(V_{GS}, V_{DS})$ with a threshold voltage $V_T$, as plotted below:


This function can be divided into three different (piece-wise linear) regions:

In summary, the current $I_{DS}$ is controlled by both voltages $V_{GS}$ and $V_{DS}$, as shown in the plots above. Specifically, in the saturation region where $V_{DS}>V_{GS}-V_T$, the current $I_{DS}$ is related to $V_{GS}$ by:

\begin{displaymath}I_{DS}=\left\{ \begin{array}{ll}
K(V_{GS}-V_T)^2 & \mbox{if ...
...{GS}-V_T$} \\
0 & \mbox{if $V_{GS}<V_T$} \end{array} \right. \end{displaymath}

The triode region and the saturation region is separated by the curve $V_{DS}=V_{GS}-V_T$. In terms of the current from drain to source, this curve can also be represented by $I_{DS}=K(V_{GS}-V_T)^2=KV_{DS}^2$.

Example 1: Assume $V_T=1V$.

Example 2: 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 $I_{DS}$ 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.

next up previous
Next: MOSFET Amplifier Up: ch4 Previous: Colpitts Oscillators
Ruye Wang 2015-11-25