2010 Semiconductor Devices

Midterm Exam #2 [Solution]

Pf. Sung June Kim

02-Nov-2010

1. Discuss the I-V(current-voltage) relations of followings.
(a) An ideal silicon p+-n junction diode when n doping changes from [eqn: 10 ^{16} cm ^{-3}] to [eqn: 10 ^{18} cm ^{-3}](5 pts).

sol)

For p+-n diodes

[eqn: I _{0} CONG qA {D _{P}} over {L _{P}} {n _{i} ^{2}} over {N _{D}}].        Also,      [eqn: {D _{P}} over {L _{P}} = sqrt {{D _{P}} over {tau _{P}}} = sqrt {{(kT/q) mu _{P}} over {tau _{P}}}].

[eqn: tau _{p1} = tau _{p2}], since the R-G center concentrations are taken to be identical in the two diodes. Moreover, although the semiconductor material is not specified in the problem statement, there is only a small difference between the [eqn: N _{D} =10 ^{16} /cm ^{3}] and [eqn: N _{D} =10 ^{18} /cm ^{3}]. mobilities in most materials. Thus,

[eqn: {I _{o1}} over {I _{o2}} = {N _{D2}} over {N _{D1}} sqrt {{mu _{p1}} over {mu _{p2}}} CONG 100]

The diode 1 current is approximately a hundred times larger than the diode 2 current for all applied voltage.

(b) Germanium and silicon pn junction diodes in comparison(5 pts).

(

sol)

Since Germanium has smaller band-gap than Si, ni(Ge) > ni(Si). Similar to part(a), the I-V characteristics relations of them is like the right figure,

or

sol2)

Since

[eqn: I=I _{DIFF} +I _{R-G} =I _{01} e ^{qV _{A} /kT} +I _{02} e ^{qV _{A} /2kT}],

Wheras [eqn: qAn _{i} W/2 tau _{0} >> I _{0}] in Si diodes at room temperature, the larger [eqn: n _{i}] of Ge typically makes [eqn: I _{0} >qAn _{i} W/2 tau _{0}] in Ge diodes at room temperature. Also, since [eqn: I _{0} PROPTO n _{i} ^{2}] while the reverse bias [eqn: I _{R-G} PROPTO n _{i}], the reverse bias diffusion component of the current will increase at a faster rate with increasing temperature.

(c) A non-ideal pn junction diode with recombination-generation in the depletion region(10 pts).

sol)

In Si diodes at room temperature [eqn: qAn _{i} W/2 tau _{0} >> I _{0}] and the [eqn: I _{R-G}] current dominates at reverse biases and small forward biases. Since the reverse bias [eqn: I _{R-G}]is proportional to W, the reverse current never saturates, but continually increases with increasing reverse bias. The forward bias [eqn: I _{R-G}] varies as [eqn: exp(qV _{A} /2kT)] for [eqn: V _{A} >few`kT/q].

[eqn: I=I _{DIFF} +I _{R-G} =I _{01} e ^{qV _{A} /kT} +I _{02} e ^{qV _{A} /2kT}]

With increasing forward biases the [eqn: I _{DIFF}] component, which in creases more rapidly with voltage, eventually overtakes the [eqn: I _{R-G}] components, leading to the [eqn: q/kT] region on a semi-log plot of the forward-bias characteristics.

2. Consider the "iso-type", p1-p2, step junction shown in Fig. P2. for part (c) and (d).

Fig. P2. iso-type step junction

(a) What is the depletion approximation(3 pts)?

sol) To obtain a quantitative solution for the electrostatic variables, it will be necessary to solve Poisson's equation. The depletion approximation facilitates obtaining closed-form solutions to the equation. Appearing in a myriad of device analyses, the depletion approximation is far and away the most important and most widely encountered of the simplifying approximations used in the modeling of devices.

(1) The carrier concentrations are assumed to be negligible compared to the net doping concentration in a region [eqn: -x _{p} LEQ x LEQ x _{n}] straddling the metallurgical junction.

(2) The charge density outside the depletion region is taken to be identically zero.

(b) Can the depletion approximation be invoked in solving for the electrostatic variables inside the pictured p1-p2 junction? Explain(5 pts).

sol) No. It is true that the minus charge shown on the x < 0 portion of the part (b) [eqn: rho]-plot is caused by a depletion of holes on the higher-doped [eqn: N _{A1}]-side of the junction, leading to net charge associated with the ionized accepters. The plus charge on the [eqn: N _{A2}]-side of the junction, however, cannot be attributed to ionized donors. Theres is no donor doping! The only other source of a positive charge is holes -- There is a hole concentration on the n-side the junction in excess of [eqn: N _{A2}]. Since [eqn: p>N_A2] near the junction, we cannot invoke the depletion approximation.

(c) Make rough sketches of charge density inside the junction(2pts).

sol)

3. Consider a silicon pn junction diode with the doping profile sketched in Fig. P2. (T=300K)

Note:　[eqn: epsilon =(11.7)(8.85 TIMES 10 ^{-14} )], [eqn: kT/q`=`0.0259`V], [eqn: n _{i} =1.5 TIMES 10 ^{10} cm ^{-3}]

Fig. P2.

(a) What is the built-in voltage([eqn: V _{bi}]) across the junction? Explain (5 pts).

sol)

assume [eqn: x _{p} >x _{0}] then [eqn: V _{bi} =- int _{-x _{n}} ^{x _{p}} {ℰdx} `=` {kT} over {q} int _{n(-x _{n} )} ^{n(x _{p} )} {{dn} over {n}} = {kT} over {q} `ln[ {n(x _{p} )} over {n(-x _{n} )} ]].

[eqn: V _{bi}]relationship is valid for an arbitrary nondegenerate doping profile. Moreover, [eqn: n(x _{p} )=n ^{2} /N _{A}], and with [eqn: x _{p} >x _{0} ,`n(-x _{n} )=N _{D}] for the given junction profile, just like in a standard step junction. Assuming the p- and [eqn: x _{eqalign{p# }} >x _{0}] n-region dopings to be nondegenerate, the standard equation result below,

[eqn: V _{bi} = {kT} over {q} `ln`( {N _{A} N _{D}} over {n _{i} ^{2}} )=(0.0259)ln( {(10 ^{16} )(10 ^{16)}} over {(1.5 TIMES 10 ^{10} ) ^{2}} )=0.6946`[V]]

but in this case [eqn: x`=`( {2k _{s} epsilon _{0}} over {q} {N _{D}} over {N _{A} (N _{D} +N _{A} )} V _{bi} ) ^{1/2} =2.15e ^{-5} cm````,`N _{A} =16/cm ^{3}]

which is smaller then 50um.

So we have to solve [eqn: V _{bi}] using [eqn: N _{A`} =10 ^{14} /cm ^{3}].

[eqn: V _{bi} = {kT} over {q} `ln`( {N _{A} N _{D}} over {n _{i} ^{2}} )=(0.0259)ln( {(10 ^{16} )(10 ^{14)}} over {(1.5 TIMES 10 ^{10} ) ^{2}} )=0.575`[V]]

(b) Calculate electric field (ℰ) (5 pts).

sol)

[eqn: ℰ _{-x _{n} LEQ x<0} = int _{0} ^{ℰ(x)} {{qN _{D}} over {epsilon }} dx= {1.6 TIMES 10 ^{-19}} over {(11.7)(8.85 TIMES 10 ^{-14} )} (1 TIMES 10 ^{16} )x=1.545 TIMES 10 ^{9} x``[V/cm]]

[eqn: ℰ _{0 LEQ x<x _{p}} = int _{ℰ(0)} ^{ℰ(x)} {{qN _{A}} over {epsilon }} dx= {1.6 TIMES 10 ^{-19}} over {(11.7)(8.85 TIMES 10 ^{-14} )} (1 TIMES 10 ^{14} )x=1.545 TIMES 10 ^{7} x``[V/cm]]

(c) Calculate the applied reverse-bias voltage required so that the space charge region extends entirely through the p-region (5 pts).

sol)

⇒  ⇒

(d) Determine the space charge width into the n+-region with the reverse-bias voltage calculated in part(c)(5 pts).

sol)

⇒  ⇒

(e) Calculate the peak electric field for this applied voltage(5 pts).

sol)

⇒

4. An abrupt Si pn junction has the following properties at 300K ([eqn: A=10 ^{-4} cm ^{2}], [eqn: {kT} over {q} `=`0.0259]):

Table P.4

p-side	n-side
[eqn: N _{A} =10 ^{17} cm ^{-3}]	[eqn: N _{D} =10 ^{15} cm ^{-3}]
[eqn: tau _{n} =0.1 mu sec]	[eqn: tau _{p} =10 mu sec]
[eqn: mu _{p} =300cm ^{2} /V`⁃s]	[eqn: mu _{n} =1300cm ^{2} /V`⁃s]
[eqn: mu _{n} =700cm ^{2} /V`⁃s]	[eqn: mu _{p} =450cm ^{2} /V`⁃s]
[eqn: epsilon =(11.8)(8.85 TIMES 10 ^{-14} )]

(a) Calculate the forward current at the forward biased by 0.5V and the reverse bias of -0.5V, respectively(5 pts).

sol)

(b) What is the total depletion capacitance at -4V(5 pts)?

sol)

5. Consider the special silicon [eqn: p ^{+} n] step junction diode pictured in Fig. P5.
Note that [eqn: tau _{p} = INF] for [eqn: 0 LEQ x LEQ x _{b}] and [eqn: tau _{p} =0] for [eqn: x _{b} LEQ x LEQ x _{c}].

Fig. P5.

Develop an expression for the room-temperature I-V characteristic of the diode excluding biases that would cause high-level injection or breakdown. Assume the depletion width (W) never exceeds [eqn: x _{b}] for all biases of interest(10 pts).

sol)