Question

Question 2 A metal-semiconductor junction has barrier potential height of 1.265 V. The semiconductor is uniformly doped with 1015 cm-3 Phosphorus and the other parameters are as listed below. a) Derive the electric field distribution, E as a function distance, x at thermal equilibrium. The metal-semiconductor interface is defined as x=0. State the boundary condition used. b) Derive the potential distribution, V as a function of distance, x under thermal equilibrium. Determine the potentials at the metal-semiconductor interface (x=0) and at x=W where W is the space-charge width. c) Determine W at equilibrium. Parameters: Temperature is 300K. Semiconductor effective density of states in conduction band = 2.8x1019 cm3 Semiconductor dielectric constant = 11.7 Richardson's constant = 120 A/K?cm2 Thermal voltage = 0.0259 V Permeability of free space = 8.85x10-14 F/cm Electronic charge = 1.9x10-19 C

Answer 1

ANSWER :
0 Given that 15cm 3 , = 300 K. cm3 Thermal voltage potential beight : 1-265V uniformly doped with phosphorous = 10 Temperatwe (+) = 300k Semi conductor effective density of states = 2,8X10 dielectric constant (&) = 1107 Richardson's constant = 120 Alkeem? 0.0259 V con 26 min Ex 8,85 x10 14 Flemi charge a (c) Electric Held dis tribution Ce); Here x is at thermal equilibrium, from law, the electric field a function distance, E(X) = (x2-x) OLXLXd. es - 1.9x10-19 distribution as gauss - qald Xd LT ex=0 dele) de where GE. JA TO / Xd. P(x)=qNj 02x2yd. f(x) =0 id cT ad Emare -qNdxd Es.

(x) = charge density d (6(x)) l Es: de ed AX)= -5 £ .dx X oad E(X) = -f and and .de Es & € (x)= q nd gold leg.x) where ६ zero -ad Eg Egi di-electric constant of the conductor & space charge width [xd=w] - We also assume the electric field distribution of outside the depletion region. E (x=0) qN ld Cxd-o). and CXd ) Eg total change in the depletion layer, Na uniformly doped region (b) Potential distribution [@ce)] we know that V.6? =¢ [gradient of the potential] where Que (or)

v=6=_ Seoda N = oft of Nd (xs-x)dx. N =+ and १ Sexy-x) dx Ve and I 2 ہے 2 2. Es $ = v= fald 2 2 [ 244 from the above equalia lion XLO $(x) = 0 8(x) = and [xy - ( My _ x 1? OLULA [@g=w 2 E5 Cre) andade 2 Es & dex [&d=w > The total potential difference across the semiconductors is equal to the built in potential. The boundary condition provides the following relation between the semi-conductor potential at the surface, the applied voltage and the depletion qNd&d 2 layer width is. 0,-va--$(x=0) 2 Es

where & = PB- it don Nv Na 11 'Space charge width exd) is. reda LES (8,- Va) and. Cryaw] el W= 2&g (deva) [ Na=0) 11 and at equilibrium, "W5 2850 qnd 2(11.7) (0) = 8-93x105m qald W = .8.93810m -> - - Veena ฝง 3 ) 1015 cm 3 C&B= 1265 v sv = 1-265 - (26x10.5) Pr (? 2.8x10 19 cm3 Ø = 1-265 -0-2662 8,2 0,733 v Electric field distribution [E6x)) = Ele)--and -LXg-x). Xe Es 2 E560) fald

2011: 1) (0.733) x 8-85X.16-14 (19x10-19) (2015) 8.938 X10 0-5 6: = 0.8938 x 16 Xd = we 0.8938 mm €(2) こ (119x10-19) (1015) (8.938X165) 11-7X8185 x 1614 x=0 1640679 16481 vlcm. E(X) / a potential equal! P(x = xd ) = faldtd 2 2&si = (1.9x10-19) (1015) (8.938x10-52 2x 11.7x8785X10-14 -0.7329 v هاو $(x)) = 0.7329 v