Metal-Oxide-Semiconductor Field-Effect Transistors

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

Different types of MOSFET

Qualitatively the conductivity between source and drain of an n-channel field effect transistor can be described as:

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

The MOSFET can therefore 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 to form an electronic channel between source and drain, and a current $I_{DS}$ between source and drain is formed, as shown below.

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:

• Cutoff region: When $V_{GS}<V_T$, no current flows through S and D, due to the two pn-junctions, i.e. $I_{DS}=0$, independent of $V_{DS}$. This is illustrated in part (b) of the figure above.

• Triode region: When $V_{GS}>V_T$ and $V_{DS}<V_{GS}-V_T$ (i.e., $V_{GS}-V_{DS}>V_T$), some electrons in the p-type substrate (minority carriers) are pulled toward the gate to form a inversion layer close to the gate to form an n-type channel with certain resistance between S and D.

  The current $I_{DS}$ increases proportionally to $V_{DS}$, with a linear coefficient $r_{on}=\Delta V_{DS}/\Delta I_{DS}$ (Ohm's law), and also nonlinearly as $V_{GS}$ increases (to pull more electrons toward the gate to enhance the conductivity of the n-channel).

• Saturation region: When $V_{GS}>V_T$ but $V_{DS}$ increases further, the voltage between gate and e-channel close to the drain $V_{GD}=V_{GS}-V_{DS}$ becomes small and the e-channel close to the drain narrows. In particular, when $V_{DS}>V_{GS}-V_T$, the voltage between gate and drain is:
  
  $$V_{GD}=V_{GS}-V_{DS}<V_{GS}-(V_{GS}-V_T)=V_T$$
  
  
  In this case, the e-channel at the D end is nearly closed (pinch-off), $I_{DS}$ is saturated and maintains at a constant value independent of $V_{DS}$ (as higher $V_{DS}$ tends to draw more electrons toward the drain on the one hand but also enhance the pinch-off effect on the other). Now $I_{DS}$ is only affected by $V_{GS}$, i.e., the transistor behaves like a voltage controlled current source, as illustrated in part (c) of the figure above.

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

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

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 $V_T=1V$.

• when $V_{GS}<V_T=1V$, the MOSFET is in cutoff region with $I_{DS}=0$ independent of $V_{DS}$.
• when $V_{GS}=2V > V_T=1V$ and $V_{DS}<V_{GS}-V_T=1V$, the MOSFET is in linear or triode region with $I_{DS}$ affected by both $V_{GS}$ and $V_{DS}$.
• when $V_{DS}>V_{GS}-V_T=1V$, the MOSFET is in saturation region with $I_{DS}$ determined only by $V_{GS}$.

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

• BJT has a low input resistance $r_{in}$. But as MOSFET's gate is insulated from the channel ( $r_{in}>10^{11} \Omega$), it draws virtually no input current and therefore its input resistance is infinity in theory.
• BJT is current ($I_b$ or $I_e$) controlled, but MOSFET is voltage ($V_{GS}$) controlled. Consequently, the power consumption of MOSFETs is lower than BJTs.
• MOSFETs are easy to fabricate in large scale and have higher element density than BJTs.
• MOSFETs have thin insulation layer which is more prone to statics and requires special protection.
• BJTs have higher cutoff frequency and higher maximum current than MOSFETs.
• MOSFETs are much more widely used (especially in computers and digital systems) than BJTs.