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## Q. 4.9

Consider a rectangular cross-sectional semiconductor bar with length $l =12$ mm, width w = 5 mm and thickness d = 4 mm which is placed in the coordinate system as shown in Fig. 4.12. The positive and negative terminals of a battery of voltage Vd = 5 V are connected between the two cross-sectional surfaces of the bar at x = 0 and x $l$ respectively. A magnetic field B = 6 × 103 Gauss is applied perpendicular to the bar along the +Z-direction.

(a)   If the bar is of n-type semiconductor with electron concentration $n_{n}$ = 2.5 ×  1015 cm−3, and the current flowing through the bar is $I$ = 10 mA, determine the magnitude and polarity of the Hall voltage between the terminal 1 and 2 (see 4.12). Also find the value of the Hall coefficient.

(b)   Repeat part (a) for a p-type semiconductor bar with hole concentration $p_{p}$ = 2.5 × 1015 cm−3 .

(c)   If Hall voltage is $V_{H}$,  derive an expression for the mobility of majority carrier in terms of $V_{H}$ , applied voltage $V_{d}$ and dimensions of the semiconductor bar.

(d)  If the Hall voltage $V_{H}$ of the terminal 2 is − 6 mV measured with respect to the terminal 1 and the current passing through the bar is 10 mA, determine Hall coefficient, carrier concentration and mobility of the majority carrier of the semiconductor. ## Verified Solution

(a) Since the semiconductor is of n-type with $n_{n}=2.5\times 10^{15} cm^{-3}$ , the charge density $\rho$  is given by

$\rho = -e n_{n}=-(1.60\times 10^{-19}C)\times 2.5\times 10^{15}cm^{-3}$

$= -4.0\times 10^{-4} C/cm^3$

The negative sign is used to denote that the type of majority carrier involved in the conduction of current in the semiconductor is electron.

Substituting the values of $I=10\times 10^{-3}A$ , $B=6\times 10^{3}$ Gauss  = $(6\times 10^3)\times (10^{-8} Wb/cm^2)=6\times 10^{-5} Wb/cm^2$ , $\rho =4.0\times 10^{-4} C/cm^2$ and $w = 5\times 10^{-1} cm$ in Eq. (4.58); the magnitude of the Hall voltage is given by

$V_{H}=\frac{BI}{\rho w}$

$=\frac{(6\times 10^{-5} Wb/cm^2)\times (10\times 10^{-3}A)}{(4.0\times 10^{-4} cm^3)\times (5\times 10^{-1} cm)}$

$=3.0\times 10^{-3} V = 3.0 mV$

Since the direction of applied electric field is in the +X direction, the velocity of the electron v must be in the –X direction. Thus, the direction of deflection of the electrons can be determined as follows.

Let the magnetic field and velocity of electrons in the bar be expessed as

B = B $\widehat{z}$ and ve  = $-v \widehat{x}$

where $\widehat{x}$ and $\widehat{z}$ are the unit vectors along the +X and +Z directions. Thus, the force acting on an elecron is given by

Fe =e(B $\times$ ve)= eBv ($– \widehat{z}\times \widehat{x}$) = eBv ($-\widehat{y} )$

where  $\widehat{y}$ is the unit vector in the +Y direction. Since the force is acting on the electron in the −Y direction, the polarity of the Hall voltage at terminal 1 is negative with respective to the terminal 2.

Using Eq. (4.59), the Hall coefficient can be obtained as

$R_{H}=\frac{1}{\rho } =\frac{1}{4.0\times 10^{-4} C/cm^{-3}}=-2.5\times 10^3 cm^3/C$

The negative sign in the value of Hall coefficient indicates that the sample used for Hall measurement is an n-type semiconductor.

(b)  In this case the charge density is

$\rho = ep_{p}=(1.60\times 10^{-19} C)\times 2.5\times 10^{15} cm^{-3}$

$=4.0\times 10^{-4} C/cm^3$

Following the similar method as in part (a), the Hall voltage and Hall coefficient are given by

$V_{H}=3 mV$ and $R_{H}=\frac{1}{\rho }=2.5\times 10^3 cm^3/C$

Since the hole moves in the direction of the applied electric field (i.e. opposite to that of electron), the force acting on the hole can be given by (see Sec.1.11)

Fh =e(v $\times$ B)= eBv ($– \widehat{z}\times \widehat{x}$) = eBv ($-\widehat{y} )$

where v = v $\widehat{x}$ is the velocity of hole. Thus, it is observed that holes are also deflected in the −Y direction as electron and hence the terminal 1 will be positive with respect to the terminal 2.

(c)   Since the applid electric field  $\varepsilon _{x}=\frac{V_{d}}{l}$  is in the +X direction, the current in the bar can be expressed as

$I= (\sigma \varepsilon _{x})wd= \frac{\rho \mu V_d wd}{l}$

Thus, the mobility of majority carrier in the given semicondctor is given by

$\mu =\frac{Il}{\rho wdV_{d}}$

Substituting  $I=\frac{\rho wV_{H}}{B}$  from Eq. (4.58) in the above equation, the mobility can be given by

$\mu = \left\lgroup\frac{l}{\rho wdV_{d}} \right\rgroup \left\lgroup\frac{\rho wV_{H}}{B} \right\rgroup$

or

$\mu =\frac{l}{d} \left\lgroup\frac{V_{H}}{BV_{d}} \right\rgroup$

Note that if the polarity of $V_{H}$ is negative at terminal 1, the above equation will represent the mobility of the electron in the semiconductor. On the other hand, if the terminal 2 is negative with respect to terminal 1, then the equation represents the mobility of holes.

(d)   Since the terminal 2 negative with respect to the terminal 1, we may say from part (c) that the majority carrier in the semiconductor is Using Eqs (4.58) and (4.59), we may obtain the Hall coefficient as

$R_{H}=\frac{1}{\rho }=\frac{V_{H}w}{IB} =\frac{(6\times 10^{-3}V)\times (0.5 cm)}{(10\times 10^{-3} A)\times (6\times 10^{-5} Wb/cm^2)}=5\times 10^3 cm^3/C$

Hence the concentration of holes in the semiconductor is given by

$p_{p}=\frac{1}{eR_{H}}=\frac{1}{(1.60\times 10^{-19} C)\times (5\times 10^3 cm^3 /C)}=1.25 \times 10^{15} cm^{-3}$

From part (c), the mobility of the holes can be obtained as

$\mu _{p} =\frac{1}{d}\left\lgroup\frac{V_{H}}{BV_{d}} \right\rgroup =\frac{(1.2cm)\times (6\times 10^{-3} V)}{(0.4cm)\times (6\times 10^{-5} Wb/cm^2)\times (5 V)}=60 cm^2 /V sec$