Editing equivalent oxide thickness

{{title|Equivalent Oxide Thickness (EOT)}}{{confuse|oxide thickness|l1=Oxide Thickness (t<sub>OX</sub>)}}
'''Equivalent Oxide Thickness''' ('''EOT'''), represented by <code>t<sub>eq</sub></code> or <code>t<sub>OXE</sub></code>, is the [[gate oxide thickness]] of the SiO<sub>2</sub> layer of a [[transistor]] that would be required to achieve similar capacitance density as the [[high-κ]] material used.

A [[gate dielectric]] with a [[dielectric constant]] that is substantially higher than that of SiO<sub>2</sub> will initially have a much smaller equivalent electrical thickness. This key feature allowed for the industry to continue on with [[Moore's Law]]. As the semiconductor industry began to experiment with transitioning from a SiO<sub>2</sub> gate oxide to a [[high-κ]] material, EOT can be used to quickly compare those materials using existing SiO<sub>2</sub>-based models.

== Equation ==
One can treat [[MOSFET]] behavior like two parallel plate capacitors,

:: <math>C_{ox} = \frac{A\epsilon_r\epsilon_0}{t_{ox}}</math>

Where <math>A\epsilon_r</math> is the relative [[dielectric constant]] of SiO<sub>2</sub> in our case. Therefore one calculate the equivalent oxide thickness as,

:: <math>\frac{A\epsilon_{SiO2}\epsilon_0}{t_{ox}} = \frac{A\epsilon_{high-\kappa}\epsilon_0}{t_{ox-high-\kappa}}</math>

:: <math>t_{ox-high-\kappa} = \frac{\cancel{A}\epsilon_{high-\kappa}\cancel{\epsilon_0}}{\cancel{A}\epsilon_{SiO2}\cancel{\epsilon_0}}t_{ox} = \frac{\epsilon_{high-\kappa}}{\epsilon_{SiO2}}t_{ox}</math>

Note that the dielectric constant SiO<sub>2</sub> is 3.9

:: <math>t_{ox-high-\kappa} = \frac{\epsilon_{high-\kappa}}{3.9}t_{ox}</math>

Where <code>t<sub>oxe</sub></code> is the equivalent oxide thickness, <code>ε<sub>high-κ</sub></code> is the [[dielectric constant]] of the [[high-κ]] material used, and <code>t<sub>ox</sub></code> is the physical oxide layer thickness.

== Example ==
For example, consider [[Hafnium Dioxide]] (HfO<sub>2</sub>) which has an <math>\epsilon_r = ~24</math> (subject to variations in temperature). A layer of just 6.15 nm  in thickness [[Hafnium Dioxide]] (HfO<sub>2</sub>) would result in an equivalent SiO<sub>2</sub> oxide thickness of around <math>t_{ox} = \frac{24}{3.9}6.15\text{ nm} = 1\text{ nm}</math>. This is indeed the material used by [[Intel]] following their transition to [[high-κ]] at the [[45 nm process]] node.