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GATE-2012 ECE Q26 (electronic devices)

Posted By Krishna Sankar On October 27, 2012 @ 2:38 pm In GATE | No Comments

Question 26 on Electronic Devices from GATE (Graduate Aptitude Test in Engineering) 2012 Electronics and Communication Engineering paper.

## Solution

To answer this question, had to dig up the copy of Solid State Electronic Devices, Ben G Streetman, Sanjay Kumar Banarjee (buy from Amazon.com [1], buy from Flipkart.com [2]) and am referring to the content from Section 3.3 Carrier Concentrations.

Electrons in solids obey Fermi-Dirac statistics.  The function $f(E)$ Fermi-Dirac distribution function, gives the probability that an available energy state at $E$ will be occupied by an electron at absolute temperature $T$,

$f(E) = \frac{1}{1+e^{(E-E_F)/kT}}$.

The quantity $E_f$ is the Fermi level and $\begin{array}k&=&\mbox{8.63x10^{-5} eV/K} & = & \mbox{1.38x10^{-23} J/K}\end{array}$ is the Boltzmann’s constant.

The concentration of electrons in the conduction band is,

$n_0 = \int_{E_c}^{\infty}f(E)N(E)dE$, where $N(E)dE$ is the density of states $cm^{-3}$ in the energy range $dE$.

The integral is equivalently stated as,

$n_0 = N_cf(E_c)$,  where $N_c$ is the effective density of states at conduction band edge $E_c$.

The Fermi function at $E_c$ is approximately,

$\begin{array}f(E_c)&=&\frac{1}{1+e^{(E_c-E_F)/kT}}&\simeq e^{-(E_c-E_F)/kT}&\end{array}$.

So in this condition the concentration of electrons in the conduction band is,

$n_0 = N_ce^{-(E_c-E_F)/kT}$.

Similarly, the concentration of holes in the valence band is,

$p_0 = N_v$1-f(E_v)$$, where $N_v$ is the effective density of states at valence band edge $E_v$.

The term,

$\begin{array}1-f(E_v)&=&\frac{1}{1-1+e^{(E_v-E_F)/kT}}&\simeq e^{-(E_F-E_v)/kT}&\end{array}$.

So in this condition the concentration of holes in the valence band is,

$p_0 = N_ve^{-(E_F-E_v)/kT}$.

The product of $n_0$ and $p_0$ at equilibrium is a constant for a particular material and temperature even if doping is varied.

$\begin{array}{lll}n_0p_0&=&N_ce^{-(E_c-E_F)/kT}N_ve^{-(E_F-E_v)/kT}\\&=&N_cN_ve^{-E_g/kT}\end{array}$where

$E_g = E_c-E_v$ is the gap between the conduction band and valence band.

Similarly, for an intrinsic material, the product of  $n_i$ and $p_i$  is,

$\begin{array}{lll}n_ip_i&=&N_ce^{-(E_c-E_i)/kT}N_ve^{-(E_i-E_v)/kT}\\&=&N_cN_ve^{-E_g/kT}\end{array}$

For an intrinsic material, the electron and hole concentration are equal i.e. $n_i=p_i$.

The constant product of electron and hole concentration can be written as

$\Huge{n_0p_0 = n_i^2}$

With the above understanding, let us try to find the solution to the problem. In our question,

$\begin{array}n_i&=&10^{10}&\mbox{per cm^3\end{array}$ and $\begin{array}n_0&=&10^{19}&\mbox{per cm^3\end{array}$.

The hole concentration is,
$\begin{array}{lll}p_0&=&\frac{n_i^2}{n_0}\\&=&\frac{10^{20}}{10^{10}}\\&=&10\mbox{ per cm^3}\end{array}$.

The volume of the source region is

$\begin{array}{lll}\mbox{volume}&=&10^{-18}\mbox{ m^3}&=&10^{-12}&\mbox{cm^3}\end{array}$.

The number of holes is,

$\begin{array}{lll}\mbox{N_{hole}}&=&10*10^{-12}&=&10^{-11}&\approx&0\end{array}$

Based on the above, the right choice is (D) 0.

## References

[1] GATE Examination Question Papers [Previous Years] from Indian Institute of Technology, Madras http://gate.iitm.ac.in/gateqps/2012/ec.pdf [3]

[2] Solid State Electronic Devices, Ben G Streetman, Sanjay Kumar Banarjee (buy from Amazon.com [1]buy from Flipkart.com [2])

[3] Valence band http://en.wikipedia.org/wiki/Valence_band [4]

[4] Conduction band http://en.wikipedia.org/wiki/Conduction_band [5]

URL to article: http://www.dsplog.com/2012/10/27/gate-2012-ece-q26-electronic-devices/

URLs in this post:

[3] http://gate.iitm.ac.in/gateqps/2012/ec.pdf: http://gate.iitm.ac.in/gateqps/2012/ec.pdf

[4] http://en.wikipedia.org/wiki/Valence_band: http://en.wikipedia.org/wiki/Valence_band

[5] http://en.wikipedia.org/wiki/Conduction_band: http://en.wikipedia.org/wiki/Conduction_band