Triode Region MCQ Quiz - Objective Question with Answer for Triode Region - Download Free PDF

We have in general

$\varphi \left(\mathrm{x}\right)=\frac{1}{\mathrm{q}}\left[{\mathrm{E}}_{\mathrm{i}}\left(\mathrm{b}\mathrm{u}\mathrm{l}\mathrm{k}\right)-{\mathrm{E}}_{\mathrm{i}}\left(\mathrm{x}\right)\right]$

For eg ${\varphi }_{\mathrm{S}}=\frac{1}{\mathrm{q}}\left[{\mathrm{E}}_{\mathrm{i}}\left(\mathrm{b}\mathrm{u}\mathrm{l}\mathrm{k}\right)-{\mathrm{E}}_{\mathrm{i}}\left(\mathrm{s}\mathrm{u}\mathrm{r}\mathrm{f}\mathrm{a}\mathrm{c}\mathrm{e}\right)\right]$

and ${\varphi }_{\mathrm{F}}=\frac{1}{\mathrm{q}}\left[{\mathrm{E}}_{\mathrm{i}}\left(\mathrm{b}\mathrm{u}\mathrm{l}\mathrm{k}\right)-{\mathrm{E}}_{\mathrm{F}}\right]$

→ coming back to our gate voltage relationships, we begin by noting that ${\mathrm{V}}_{\mathrm{G}}$ in the ideal structure is dropped partly across the oxide and partly across the semiconductors.

${\mathrm{V}}_{\mathrm{G}}=\mathrm{\Delta }{\varphi }_{\mathrm{o}\mathrm{x}}+\mathrm{\Delta }{\varphi }_{\mathrm{s}\mathrm{e}\mathrm{m}\mathrm{i}}$

We take $\varphi =0$ references to be in the bulk. Thus the voltage drop across the semiconductors is simply:

$\begin{array}{l}\mathrm{\Delta }{\varphi }_{semi}=\varphi \left(x=0\right)-\varphi \left(bulk\right)\\ =\varphi \left(\mathrm{x}=0\right)\\ =\mathrm{\Delta }{\varphi }_{\mathrm{s}\mathrm{e}\mathrm{m}\mathrm{i}}={\varphi }_{\mathrm{S}}\end{array}$

Now, we turn to finding ${\varphi }_{\mathrm{o}\mathrm{x}}$ in terms of ${\varphi }_{\mathrm{S}}$_. An ideal oxide with no carrier or charge centers in the oxide has constant E–field in it. Thus

${\mathrm{E}}_{\mathrm{o}\mathrm{x}}=\frac{-\mathrm{d}{\varphi }_{\mathrm{o}\mathrm{x}}}{\mathrm{d}\mathrm{t}}=\mathrm{c}\mathrm{o}\mathrm{n}\mathrm{s}\mathrm{t}\mathrm{a}\mathrm{n}\mathrm{t}$

And $\frac{\mathrm{d}{\mathrm{E}}_{\mathrm{o}\mathrm{x}}}{\mathrm{d}\mathrm{x}}=0$

Therefore,

$\mathrm{\Delta }{\varphi }_{\mathrm{o}\mathrm{x}}=\underset{-{\mathrm{t}}_{\mathrm{o}\mathrm{x}}}{\overset{0}{\int }}{\mathrm{E}}_{\mathrm{o}\mathrm{x}}\mathrm{d}\mathrm{x}={\mathrm{t}}_{\mathrm{o}\mathrm{x}}.{\mathrm{E}}_{\mathrm{o}\mathrm{x}}$

where ${\mathrm{t}}_{\mathrm{o}\mathrm{x}}$ is oxide thickness

Now we can relate the electric field density across the oxide semiconductor interface using the well known boundary condition on fields normal to an interface between two dissimilar materials

$\left({\mathrm{D}}_{\mathrm{s}\mathrm{e}\mathrm{m}\mathrm{i}}-{\mathrm{D}}_{\mathrm{o}\mathrm{x}}\right){|}_{\mathrm{o}-\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{t}\mathrm{e}\mathrm{r}\mathrm{f}\mathrm{a}\mathrm{c}\mathrm{e}}={\mathrm{Q}}_{\mathrm{o}-\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{t}\mathrm{e}\mathrm{r}\mathrm{f}\mathrm{a}\mathrm{c}\mathrm{e}}$

where $\mathrm{D}=ϵ\mathrm{E}$ and ${\mathrm{Q}}_{\mathrm{o}-\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{t}\mathrm{e}\mathrm{r}\mathrm{f}\mathrm{a}\mathrm{c}\mathrm{e}}$ interface is charge/area located at the interface ${\mathrm{Q}}_{\mathrm{o}-\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{t}\mathrm{e}\mathrm{r}\mathrm{f}\mathrm{a}\mathrm{c}\mathrm{e}}=0$ for ideal case.

$\begin{array}{l}{D}_{ox}=D_{semi}{|}_{x=0}\\ \Rightarrow {\mathrm{E}}_{\mathrm{o}\mathrm{x}}=\frac{{ϵ}_{\mathrm{s}\mathrm{i}}}{{ϵ}_{\mathrm{o}\mathrm{x}}}{\mathrm{E}}_{\mathrm{s}\mathrm{e}\mathrm{m}\mathrm{i}}\end{array}$

Thus, from (1)

$\mathrm{\Delta }{\varphi }_{\mathrm{o}\mathrm{x}}=\frac{{ϵ}_{\mathrm{s}\mathrm{i}}}{{ϵ}_{\mathrm{o}\mathrm{x}}}{\mathrm{t}}_{\mathrm{o}\mathrm{x}}.{\mathrm{E}}_{\mathrm{s}\mathrm{e}\mathrm{m}\mathrm{i}}$

Now gate voltage ${\mathrm{V}}_{\mathrm{G}}$, is

${\mathrm{V}}_{\mathrm{G}}={\varphi }_{\mathrm{s}}+\frac{{ϵ}_{\mathrm{s}\mathrm{i}}}{{ϵ}_{\mathrm{o}\mathrm{x}}}{\mathrm{t}}_{\mathrm{o}\mathrm{x}}{\mathrm{E}}_{\mathrm{s}\mathrm{e}\mathrm{m}\mathrm{i}}$

${\mathrm{E}}_{\mathrm{s}\mathrm{e}\mathrm{m}\mathrm{i}}$ can be shown to be equal to $\sqrt{\frac{2\mathrm{q}{\mathrm{N}}_{\mathrm{A}}}{{ϵ}_{\mathrm{s}\mathrm{i}}{ϵ}_{\mathrm{o}}}{\varphi }_{\mathrm{s}}}$

So ${\mathrm{V}}_{\mathrm{G}}$ can also be rewritten replacing ${\mathrm{E}}_{\mathrm{s}\mathrm{e}\mathrm{m}\mathrm{i}}$

Now from (2)

${\mathrm{V}}_{\mathrm{G}}={\varphi }_{\mathrm{s}}+\frac{{ϵ}_{\mathrm{s}\mathrm{i}}}{{ϵ}_{\mathrm{o}\mathrm{x}}}.{\mathrm{t}}_{\mathrm{o}\mathrm{x}}.{\mathrm{E}}_{\mathrm{s}\mathrm{e}\mathrm{m}\mathrm{i}}$

Substituting value, using ${\varphi }_{\mathrm{S}}={\varphi }_{\mathrm{F}}$   we have

$\begin{array}{l}{V}_G=0.298+\frac{11.7}{3.9}\times \left(0.1\times {10}^{-4}\right)\times \left(9.56\times {10}^3\right)\\ =0.298+0.2868\\ ={\mathrm{V}}_{\mathrm{G}}=0.5848\mathrm{V}\end{array}$