2010 Semiconductor Devices

Midterm Exam #1 [Solution]

Pf. Sung June Kim

08-Oct-2010

1. The lattice constant or unit cell side-length is [eqn: a`=`5.43` ANGSTROM] in Si at room temperature. Calculate (a) the number of atoms per unit cell (2 pts) and (b) the volume density of atoms per cm3(3 pts), respectively. and explain. Si의 단위 셀의 한 모서리의 길이(Lattice constant)가 실온에서 [eqn: a`=`5.43` ANGSTROM]라 할 때, (a) 단위 셀에 포함된 원자의 개수와 (b) cm3당 원자의 체적밀도를 구하고  각각의 경우에 대한 설명을 하시오.

sol)

(a) Since there are eight Si atoms per unit cell (2 pts) and (b) the volume of the unit cell is [eqn: a ^{3}], it follows that there are [eqn: {8} over {a ^{3}}] (1 pts) almost exactly [eqn: 5` TIMES `10 ^{22}] atoms/cm3 (2 pts)

2. A Si sample is doped with 1014 boron atoms per cm3. Si가 [eqn: 10 ^{14} /cm ^{3}]의 붕소로 도핑되어 있다 하자.
Note that [eqn: E _{G}](Si) = 1.08 eV, [eqn: m _{p} */m _{n} *]= 0.71 at 470K and [eqn: E _{i} =( {3} over {4} )kT`ln(m _{p} */m _{n} *)]. (10 pts).

Figure Prob. 2(a). Intrinsic carrier concentration - T(K).

(a) What are the carrier concentrations in the Si sample at 300K? (4 pts)
300K에서의 carrier concentration들은 어떻게 되는 지 계산하고 설명하시오.

sol)

(a) Boron in Si is an acceptor (class III). Thus [eqn: N _{A} =10 ^{14} /cm ^{3}]. At 300K, [eqn: n _{i} =1.00 TIMES 10 ^{10} /cm ^{3}] and the given [eqn: N _{A}] is clearly much greater than [eqn: n _{i}]. Moreover, since the [eqn: N _{D}] doping was omitted from the problem statement, we infer [eqn: N _{D} << N _{A}]. [eqn: p SIMEQ N _{A} `,`n SIMEQ n _{i} ^{2} /N _{A}] (when [eqn: N _{A} >> n _{i}] and [eqn: N _{A} >> N _{D}] , nondegenerate, total ionization) This equation may be used to calculate the carrier concentrations (4 pts).

∴ [eqn: p=N _{A} =10 ^{14} /cm ^{3} `;`n=n _{i} ^{2} /N _{A} =10 ^{6} /cm ^{3}]

(b) Calculate the energy levels of following and draw them in the energy diagram of figure prob. 2(b) at 470K.; [eqn: E _{i}] , [eqn: E _{F}] , [eqn: {1} over {2} (E _{c} +E _{v} )]. (6 pts) 470K에서 [eqn: E _{i}] , [eqn: E _{F}] , [eqn: {1} over {2} (E _{c} +E _{v} )]은 어디에 위치하게 되는지 계산하고, 그 위치는 어떠한지 에너지 다이어그램(Figure Prob 2(b))에 그려보시오.

sol)

(b) The Si ample is heated to 470K. With [eqn: m _{p} */m _{n} *]=0.71 and [eqn: kT=0.0405`eV] at 470K,

[eqn: ( {3} over {4} )kT`ln(m _{p} */m _{n} *)=-0.0104`eV].           ∴[eqn: E _{i}] is deduced to be located 0.0104 eV below midgap.

Because [eqn: N _{A}] is comparable to [eqn: n _{i}] at 470K, equation (2.37) [eqn: E _{F} -E _{i} `=`kT`ln( {n} over {n _{i}} )=-kT`ln( {p} over {n _{i}} )] (2 pts)

must be used to compute the positioning of [eqn: E _{F}] . Specially, with [eqn: n _{i} =10 ^{14} /cm ^{3}] and [eqn: p=1.62 TIMES 10 ^{14} /cm ^{3}],

∴ [eqn: E _{i} -E _{F} =kT`ln(p/n _{i} )`=`0.0405`ln(1.62 TIMES 10 ^{14} /10 ^{14} )`=`0.0195`eV] (2 pts)

Here [eqn: E _{F}] is only slightly moved from [eqn: E _{i}] as pictured below (2 pts).

3. The carrier distributions as a function of energy in the conduction and valence bands were noted to have peaks at energies very close to the band edges. 에너지에 대한 함수로 표현되는 carrier의 분포는 conduction 밴드와 valence 밴드각각의 밴드 끝 부분 근처에서 최대값을 갖는 peak점을 보이는 형태를 갖게 된다.

(a) sketch carrier distributions versus Energy of a n-type semiconductor considering [eqn: g _{c} (E)] & [eqn: f(E)].
(5 pts) n-type의 반도체에 대해서 에너지에 대한 함수로 표현되는 carrier의 분포를 [eqn: g _{c} (E)`,`f(E)]를 고려하여 도시하시오.

sol)

(a) (4 pts)

(b) Derive the expression of n as a function of E (5 pts).
에너지에 대한 n 의 값을 구하는 식을 유도하시오([eqn: n(E)])

sol)

(b) The distribution of electrons in the conduction band is given by [eqn: g _{c} (E)f(E)]; the distribution of holes in the valence band is given by [eqn: g _{v} (E)[1-f(E)]]. Working with the electron distribution we note,

[eqn: f(E)= {1} over {1+e ^{(E-E _{F} )/kT}} CONG e ^{-(E-E _{F} )/kT}]… for all [eqn: E` GEQ E _{C}] if the semiconductor is nondegenerate.

Thus, [eqn: g _{c} (E)f(E)`=` {m* _{n} sqrt {2m* _{n} (E-E _{C} )}} over {pi ^{2} hbar ^{3}} e ^{-(E-E _{F} )/kT`} =` kappa (E-E _{C} ) ^{1/2} e ^{-(E-E _{F} )/kT}] (3 pts)

(c) The extrema points of any function are obtained by taking the derivative of the function and setting the derivative equal to zero.

[eqn: {d} over {dE} [g _{c} (E)f(E)]= {kappa } over {2(E-E _{c} ) ^{1/2}} e ^{-(E-E _{F} )/kT} - {kappa } over {kT} (E-E _{C} ) ^{1/2} e ^{-(E-E _{F} )/kT} =0] (5 pts)

Clearly, for electrons in the conduction band

[eqn: {1} over {2 sqrt {E _{peak} -E _{c}}} = {sqrt {E _{peak} -E _{c}}} over {kT} ````]or [eqn: E _{peak`} -E _{c} `=kT/2``````and``E _{peak`} `=`E _{c} ``+`kT/2] (5 pts)

(c) Assuming the semiconductor is nondegenerate, show that the energy at which the carrier distributions peak is [eqn: E _{c} `+`kT/2] and [eqn: E _{V} `-`kT/2] for the conduction and valence bands, respectively (10 pts).

Nondegenerate된 반도체를 생각해 보자. 이 때 carrier의 분포의 peak가 있는 곳의 에너지의 위치가 conduction band에서는 [eqn: E _{c} `+`kT/2], valence band에서는 [eqn: E _{V} `-`kT/2] 에 있는 이유가 무엇인지 식으로 증명하시오.

sol)

(c) The extrema points of any function are obtained by taking the derivative of the function and setting the derivative equal to zero.

[eqn: {d} over {dE} [g _{c} (E)f(E)]= {kappa } over {2(E-E _{c} ) ^{1/2}} e ^{-(E-E _{F} )/kT} - {kappa } over {kT} (E-E _{C} ) ^{1/2} e ^{-(E-E _{F} )/kT} =0] (5 pts)

Clearly, for electrons in the conduction band

[eqn: {1} over {2 sqrt {E _{peak} -E _{c}}} = {sqrt {E _{peak} -E _{c}}} over {kT} ````]or [eqn: E _{peak`} -E _{c} `=kT/2``````and``E _{peak`} `=`E _{c} ``+`kT/2] (5 pts)

4. The equilibrium and steady state conditions before and after illumination of a semiconductor of a semiconductor is characterized by the energy band diagrams shown Figure Prob. 4.( [eqn: T=300K,`n _{i} =10 ^{10} /cm ^{3}] )

아래의 에너지 다이어그램에서 표시한 것과 같은 반도체가 있을 때, illumination의 이전과 이후 모두 평형 상태이면서 동시에 정상상태이라 가정하자.

Figure Prob. 4.

(a) Determine [eqn: n _{0}] and [eqn: p _{0}], the equilibrium carrier concentrations (5 pts).

carrier concentration인 [eqn: n _{0}]과 [eqn: p _{0}]를 구하여라.

sol) No partial credits

(a)

[eqn: n _{0} =n _{i} `exp( {E _{F} -E _{i}} over {kT} )=(1 TIMES 10 ^{10} )exp( {0.20} over {0.0259} )=2.26 TIMES 10 ^{13} /cm ^{3}]

[eqn: p _{0} =n _{i} `exp( {E _{i} -E _{F}} over {kT} )=(1 TIMES 10 ^{10} )exp( {-0.20} over {0.0259} )=4.43 TIMES 10 ^{5} /cm ^{3}]

(b) Determine [eqn: n] and [eqn: p], under illumination (steady state condition) (5 pts).

illumination이후 정상상태에서의 인 [eqn: n]과 [eqn: p]를 구하여라.

sol) No partial credits

(b)

[eqn: n=n _{i} `exp( {E _{N} -E _{i}} over {kT} )=(1 TIMES 10 ^{10} )exp( {0.22} over {0.0259} )=4.89 TIMES 10 ^{13} /cm ^{3}]

[eqn: p=n _{i} `exp( {E _{i} -E _{P}} over {kT} )=(1 TIMES 10 ^{10} )exp( {0.20} over {0.0259} )=2.28 TIMES 10 ^{13} /cm ^{3}]

(c) Are the values of n , p obtained in (b) is reasonable ? Explain. (5 pts).

(b)번 문제에서의 n과 p의 값이 합당한 가를 설명하시오.

sol) No partial credits

(c)

Yes. Actually [eqn: n`=`n _{0} + TRIANGLE n=n _{0} `+p`=`(2.26+2.28) TIMES 10 ^{13} ` CONG `4.89` TIMES `10 ^{13}]. This is roughly same. so the value of (b) is reasonable.

만약 설명이 동일한데 수치의 차이가 미약하게 나마 존재하여 reasonable 하지 않다고 하였을 경우에도 인정. If the student answered 'No' because there is a little difference between actual value and calculated value (mentioned above), it could be correct.

5. A semi-infinite p-type bar (Figure Prob. 5) is illuminated with light which generates [eqn: G _{L}] electron-hole pairs per cm3-sec uniformly throughout the volume of the semiconductor. Simultaneously, carriers are extracted at x = 0 making [eqn: △n _{p} `=`0] at x = 0. Assuming a steady state condition has been established and [eqn: △n _{p} (x)` << p _{0}] for all x, determine [eqn: △n _{p} (x)](10 pts).

아래의 그림과 같은 한쪽으로 무한한 p-type의 막대가 있다. 이 막대에 [eqn: G _{L}]개/cm3•sec의 전자-홀을 균일하게 생성시키는 빛을 비추었다고 하자. 또, x = 0에서 캐리어가 추출되어 [eqn: △n _{p} `=`0]라고 하자. 정상상태가 되었고 임의의 x에 대하여 [eqn: △n _{p} (x)` << p _{0}]라 할 때에 [eqn: △n _{p} (x)]를 구하시오.

sol)

Since [eqn: TRIANGLE n _{p} (x) << p _{0}], we have low level injection. All conditions for use of the minority carrier diffusion equation are satisfied expect possibly [eqn: ℰ CONG 0], we can write in general

[eqn: {PARTIAL TRIANGLE n _{p}} over {PARTIAL t} =D _{N} {PARTIAL ^{2} TRIANGLE n _{p}} over {PARTIAL x ^{2}} - {TRIANGLE n _{p}} over {tau _{n}} +G _{L}] (2 pts)

Simplifying... [eqn: {Partial TRIANGLE n _{p}} over {Partial t} =0] ... steady state. [eqn: 0=D _{N} {Partial ^{2} TRIANGLE n _{p}} over {Partial x ^{2}} - {TRIANGLE n _{p}} over {tau _{n}} +G _{L}]  ... equation to be solved (2 pts)

Solving...

[eqn: TRIANGLE n _{p} (x)=`G _{L} tau _{n} +Ae ^{-x/L _{N}} +Be ^{x/L _{N}}] ([eqn: L _{N} = sqrt {D _{N} tau _{n}}])

Boundary conditions... (1) [eqn: TRIANGLE n _{p} (0)`=`0] ... given (2) [eqn: TRIANGLE n _{p} ( INF )`=`G _{L} tau _{n}] (2 pts)

The second boundary condition follows from the fact that far from x = 0 the concentration gradient caused by the extraction of carriers at x = 0 must vanish. If [eqn: D _{N} ( {d ^{2} TRIANGLE n _{p}} over {dx ^{2}} )` -> `0.~][eqn: TRIANGLE n _{p} =G _{L} tau _{n}] as deduced from the simplified minority carrier diffusion equation (2 pts) . Thus

B = 0 ... applying boundary condition (2)

[eqn: 0=G _{L} tau _{n} +A] or [eqn: A=-G _{L} tau _{n}]  ... applying boundary condition (1)

and

[eqn: TRIANGLE n _{p} (x)=G _{L} tau _{n} (1-e ^{-x/L _{N}} )] (2 pts)