Chapter 5 Solutions

Chapter 5 Solutions Prob. 5.1 Find the time that it takes to grow the first 200nm, the next 300nm, and the final 400nm. Draw and calculate step heights after reoxidation. Time for first 200nm = 0.13 hours from Appendix VI at 1000°C Time for next 300nm = 0.6 hours for 500nm – 0.13 hours for 200nm = 0.47 hours Time for final 400nm = 1.8 hours for 900nm – 0.6 hours for 500 nm = 1.2 hours Oxidation Time After Etch = 6.0 hours for 2000nm – 1.4 hours for 900nm at 1100°C = 4.6 hours Oxide Growth Inside Window = 1700nm Step in Si = 264 nm

T an his th d wo sa eir is p rk w le co ro is ill o u vi pr de f a rse de ot st ny s d s ec ro p an o te y ar d le d th t o a ly by e s in f th se for Un te is ss th ite gr w in e ity o g us d S of rk ( stu e o tat th inc de f i es e lu nt ns co w d le tr p or in a uc y k g rn to rig an on in rs h d th g. in t la is e D t w no W iss ea s t p or em ch in er ld m W ina g itt id tio ed e n . We or b)

Step in oxide = 564 nm

564nm

1120nm

504nm 0.44 · 900=396nm 264nm

556nm

952nm

880nm

Si

1144nm

748nm

Original Si surface

Prob. 5.2 Plot the distributions for B diffused into Si (Nd= 5 ⋅1016

3 ⋅10-14 cms ) with (a) constant source No= 5 ⋅1020 2

No= 5 ⋅1013

1 cm3

1 cm3

1 cm3

) at 1000°C for 30min (D=

and (b) limited source

on the surface prior to diffusion.

The Gaussian distribution differs from Equation 4-44 because all atoms are assumed to diffuse into the sample (i.e. there is no diffusion in the –x direction). s D ⋅ t = 3 ⋅10-14 cms ⋅ 30min ⋅ 60 min = 5.4 ⋅10-10cm2 = 0.0735µm 2


⎛ ⎞ x ⎛ x ⎞ (a) N = No ⋅ erfc ⎜ ⎟ = No ⋅ erfc ⎜ 0.147µm ⎟ ⎝ 2 D ⋅ t ⎠ ⎝ ⎠

xj=0.4μm

(b) N =

Ns π ⋅ D⋅t

⋅e

x ⎞ ⎟ ⎝ 2 D⋅t ⎠

- ⎛⎜

2

⎛

x

⎞

2

- ⎜ ⎟ Ns ⋅ e ⎝ 0.147µm ⎠ = 0.1302µm

xj=0.3μm

Prob. 5.3

For the unlimited source in Problem 5.2, calculate the time to achieve a junction depth of 1 micron.

Use N o from Appendix VII and D from Appendix VIII. ⎛ 1µm ⎛ x ⎞ 21 1 ⎜ ⋅ N = N o ⋅ erfc ⎜ = 10 erfc ⎟ cm3 ⎜ 2 3 ⋅10-14 cm2 ⋅ t ⎝ 2 D ⋅ t ⎠ s ⎝ ⎛ ⎞ 2 ⋅1016 1 3 1µm cm ⎟ = erfc ⎜ = 2 ⋅10-5 ⎜ 2 3 ⋅10-14 cm2 ⋅ t ⎟ 1021 cm1 3 s ⎝ ⎠ 1µm 10-4 cm = = 3.0 2 2 2 3 ⋅10-14 cms ⋅ t 2 3 ⋅10-14 cms ⋅ t t = 9260s = 2 hrs 34 minutes 20 seconds

⎞ ⎟ = 2 ⋅1016 ⎟ ⎠

1 cm3

Prob. 5.4 Find the implant parameters for an As implant into Si with the peak at the interface.

R p = 0.1µm → Energy = 180keV from Appendix IX Straggle = ΔR p = 0.035µm

φ

Ion Distribution from Equation 5-1a = N(x) =

φ

φ

N peak = 5 ⋅1019

1 cm3

φ = 5 ⋅1019

⋅ 8.77 ⋅10-6cm = 4.39 ⋅1014

1 cm3

=

2π ⋅ ΔR p

=

-6

2π ⋅ 3.5 ⋅10 cm

=

⋅e

1 ⎛ x - R p ⋅⎜ 2 ⎜⎝ ΔR p

φ

8.77 ⋅10-6 cm

1 cm 2 1 cm 2


I⋅t I ⋅ 20s = = 4.39 ⋅1014 -19 q ⋅ A 1.6 ⋅10 C ⋅ 200cm 2 Beam Current = I = 0.7 mA

φ=

2π ⋅ ΔR p

-

⎞ ⎟ ⎟ ⎠

2

Prob. 5.5 Calculate and plot the P distribution.

Energy = 200keV → R p = 0.255µm, ΔR p = 0.0837µm from Appendix IX Dose = φ = 2.1 ⋅1014

1 cm 2

From Equation 5-1a, N(x) = Peak N =

2.1⋅1014

1 cm 2 -6

φ 2π ⋅ ΔR p = 1019

⋅e

-

1 ⎛ x - R p ⋅⎜ 2 ⎜⎝ ΔR p

⎞ ⎟ ⎟ ⎠

2

=

2.1⋅1014

2π ⋅ 8.37 ⋅10 cm

1 cm3

2π ⋅ 8.37 ⋅10 cm Let y be the distance (µm) on either side of R p . 1 cm3

⋅ e-71.37⋅y

2


N(R p ± y) = 1019

1 cm 2 -6

⋅e

-

1 ⎛ x - 0.255µm ⎞ ⋅⎜ ⎟ 2 ⎝ 0.1µm ⎠

2

Prob. 5.6 In patterning the structure shown in the question, design the mask aligner optics in terms of numerical aperture of the lens and the wavelength of the source.

lmin =

0.8 ⋅ λ = 1µm NA

DOF =

λ 2 ⋅ ( NA )

2

= 2µm

⎞ 1 ⎛ 1µm ⋅ NA ⎞ ⎛ ⎟ = 2µm → NA = 0.3125 λ = 0.39µm ⎜ ⎟ ⋅ ⎜⎜ 2 ⎝ 0.8 ⎠ ⎝ 2 ⋅ ( NA ) ⎟⎠

W : xn


Prob. 5.7 In a p+-n Si junction, the n-side has a donor concentration of 1016 cm-3. If ni=1010 cm-3, relative dielectric constant εr= 12, calculate the depletion width at a reverse bias of 100V? What is the electric field at the mid-point of the depletion region on the n side? (Hint: Remember that P+ means very heavily doped!)

1

⎡ 2ε ( V0 -V ) ⎛ N a +N d ⎞ ⎤ 2 W = ⎢ ⎜ ⎟ ⎥ q ⎝ N a N d ⎠ ⎥⎦ ⎣⎢ V = -VR VR = 100 V; V0 = VR ⇒ VR + V0 : VR ε = εrε0 ;

ε r = 12

ε 0 = 8.85 ×10−14 F/cm

⎡ 2 ×12 × 8.85 ×10−14 F/cm × (100 V ) ⎛ 1 ⎞ ⎤ ∴ x n = ⎢ ⎜ 16 −19 -3 ⎟ ⎥ 1.6 ×10 C ⎝ 10 cm ⎠ ⎦ ⎣ = (1.33 ×10−7 )

1

2

cm = 3.64 ×10 −4 cm = 3.64 µm

1

2

q N a ( x-x n ) ε ε0 = - q Nd x n ε ∴ Electric field at the middle of depletion region on n-side is

ε=

x q 1.6 ×10−19 C 3.64 ×10−4 cm 16 -3 × × = - Nd n = 10 cm ε 2 12 × 8.85 ×10−14 F/cm 2 5 = -2.74 ×10 V/cm


Prob. 5.8 A semiconductor with a bandgap of 0.8 eV and an intrinsic carrier concentration of 1012 cm-3, is doped with 1018 cm-3 donors on the left half, and 1017 cm-3 acceptors on the right half. Draw the equilibrium band diagram, Calculate the junction potential and the position of the Fermi level, and indicate them on the band diagram. Suppose an electron at the conduction band edge on the p-side goes over the n-side without scattering. Assuming parabolic bandstructure, calculate its wavevector there. The effective mass of the carriers is 0.2mo.

n = nie

Eip -E F

E F -Ein kT

p = nie

(

)

1018cm -3 = 1012cm -3 e

E F -Ein kT

kT

(

)

1017 cm -3 = 1012cm -3 e

( )

Eip -E F kT

( )

E F -E in = ( 0.026 eV ) ln 106

E ip -E F = ( 0.026 eV ) ln 105

E F -E in = 0.36 eV

E ip -E F = 0.3 eV

⇒ V0 = 0.36 eV + 0.3 eV = 0.66 eV An electron going over to the n-side from the p-side without scattering has K.E. = 0.66 eV =

ћ 2k 2 2m*

( 0.66 ×1.6 ×10 J ) × 2 × (0.2 × 9.11×10 (1.054 ×10 J ⋅ s ) −19

⇒ k = 1.86 ×109 m -1

−34

kg

)

2


⇒ k=

−31

Prob. 5.9 An abrupt Si junction (area=0.0001 cm2) has the following parameters. n-side p-side Nd = 5X 1017cm-3 Na = 1017cm-3 Draw and label the band diagram, and calculate the difference between the Fermi level and the intrinsic Fermi level on both sides. Calculate the built-in potential at the junction in equilibrium and the depletion width. What is the total number of exposed acceptors in the depletion region?

For n-Si: N d = 5 ×1017 cm -3 ⎛ N ⎞ ⎛ 5 × 1017 cm-3 ⎞ E F -E i = ln ⎜ d ⎟ ⇒ E F -E i = ( 0.026 eV ) × ln ⎜ = 0.45 eV 10 -3 ⎟ kT ⎝ 1.5 × 10 cm ⎠ ⎝ n i ⎠ For p-Si: N a = 1×1017 cm -3

∴

⎛ N a ⎞ ⎛ 1×1017 cm -3 ⎞ E F -E i = ln ⎜ = 0.41 eV ∴ ⎟ ⇒ E F -E i = ( 0.026 eV ) × ln ⎜ 10 -3 ⎟ kT ⎝ 1.5 ×10 cm ⎠ ⎝ n i ⎠ ⇒ Vbi = 0.45 eV + 0.41 eV = 0.86 eV The depletion width, W =

⇒W=

(

2εSi Vbi e

⎛ 1 1 ⎞ + ⎜ ⎟ ⎝ N a N d ⎠

)

2 × 13 × 8.86 ×10−14 F/cm × ( 0.86 V ) ⎛ 1 1 ⎞ + ⎜ 17 -3 17 -3 ⎟ −19 1.6 ×10 C 5 ×10 cm ⎠ ⎝ 1×10 cm T an his th d wo sa eir is p rk w le co ro is ill o u vi pr de f a rse de ot st ny s d s ec ro p an o te y ar d le d th t o a ly by e s in f th se for Un te is ss th ite gr w in e ity o g us d S of rk ( stu e o tat th inc de f i es e lu nt ns co w d le tr p or in a uc y k g rn to rig an on in rs h d th g. in t la is e D t w no W iss ea s t p or em ch in er ld m W ina g itt id tio ed e n . We or b)

= 1.16 ×10−5 cm N a x p = N d x n ,where x p and x n are the depletion layer widths in the p, and n regions ⇒ 1×1017 cm-3 × x p = 5 ×1017 cm -3 × x n ⇒ x p = 5x n x p + x n = W ⇒ 6x n = W

⇒ x n = 1.93 ×10-6 cm ; x p = 9.67 ×10-6 cm

(

) (

) (

No. of exposed acceptors = N a x p A = 1×1017 cm -3 × 9.67 ×10-6 cm × 10−4 cm 2

)

= 9.67 × 107

Prob. 5.10 A p-n junction diode has a doping concentration of 1016 cm-3 on the p- side, and is very highly doped on the n-side. The intrinsic carrier concentration is 109 cm-3, bandgap is 2 eV and εr=15. Sketch the band diagram for a reverse bias of 2V, and calculate the values of band edges with respect to the quasi-Fermi levels far from the junction. Calculate the depletion charge per cm2 on the n-side. If an electron at the conduction band edge on the p-side goes over to the n-side without scattering, calculate its velocity. Electron and hole effective masses = 0.4m0.

N a = 1016cm -3 N d → high p = Na = n ie φ F,n =

⎛ qφ F,p ⎜⎜ ⎝ kT

⎞ ⎟⎟ ⎠

16

-3

(

9

⇒ 10 cm = 10 cm

-3

)⋅e

⎛ φ F,p ⎞ ⎜⎜ ⎟⎟ ⎝ 0.026 eV ⎠

⇒ φ F,p = 0.42 eV

Eg

= 1 eV 2 Barrier height Δφ = φ F,n + φ F,p = 1.42 eV N a x p = N d x n , where x p , x n are the depletion layer widths on the p, and n sides

⇒ N d x n ≈ N a W, since x p + x n = W and x p ? x n 2εS ( Δφ+Vr )

In equilibrium, W =

qN a

(

) C ) × (10

2 × 15 × 8.86 ×10−14 F/cm × (1.42 V + 2 V )

=

(1.6 ×10

−19

16

cm-3

)

(


= 7.54 ×10−5 cm = 0.75 µm

) (

)

⇒ N d x n = 7.54 ×10−5 cm × 1016cm -3 = 7.54 ×1011 cm -2

Energy of electron from E c on n-side, E = Δφ+Vr = 1.42 eV + 2 eV ⇒v=

2E = m*

(

2 × 3.42 ×1.6 × 10−19 J 0.4 × 9.11×10

−31

kg

) = 1.73 ×10

6

m/s

Prob. 5.11 In a p-n junction, the n-side doping is five times the p-side doping. The intrinsic carrier concentration =1011 cm-3 and bandgap is 2eV at 100 C. If the built-in junction potential is 0.65V, what is the doping on the p- side? If the relative dielectric constant of this semiconductor is 10, what is the depletion capacitance at 2 V reverse bias for a diode of cross-sectional area of 0.5 cm2? Draw a qualitatively correct sketch of the band diagram and label the depletion widths and voltage drops for this bias.

V0 =

kT ⎛ N a ( 5N a ) ⎞ ln ⎜ ⎟ = 0.65 V q ⎝ n i 2 ⎠

2 ⎛ 373 K ⎞ ⎡ 11 ⎤ = ( 0.026 eV ) ⎜ ⎟ ⎣⎢ln ( 5 ) + 2 ln ( N a ) − ln 10 ⎦⎥ ⎝ 300 K ⎠ 1 ⇒ ln ( N a ) = ( 20.11 + 49.05 ) = 34.58 2 ⇒ N a = 1.04 ×1015 cm -3

)


(

⇒ N d = 5N a = 5.2 ×1015 cm -3 W=

2εS ( Δφ+Vr ) ⎛ N a +N d ⎞ ⎜ ⎟ q ⎝ N a N d ⎠

(

)

2 × 10 × 8.85 ×10−14 F/cm ( 0.65 V + 2 V ) ⎛ 1.04 + 5.2 ⎞ = ⎜ −19 15 -3 ⎟ 1.6 ×10 C ⎝ 1.04 × 5.2 × 10 cm ⎠ = 1.84 µm

(

)

10 × 8.85 ×10−14 F/cm × 0.5 cm 2 εA Cd = = = 2.4 nF W 1.84 ×10−4 cm

Prob. 5.12 a) Calculate contact potential V0, in a Si p-n junction at 300 K.

V0 =

kT N a N d ln q n i2

V0 = 0.0259 ln

Na Nd (1.5 ⋅ 1010 ) 2

b) Plot E0 vs. Nd


Prob. 5.13 Find the electron diffusion and drift currents at xn in a p+-n junction.

I p (x n ) = q ⋅ A ⋅

Dp Lp

⋅ pn ⋅ e

I = I p (x n = 0) = q ⋅ A ⋅

q ⋅V kT

Dp Lp

⋅e

-

xn Lp

for V ?

kT q

q ⋅V

⋅ p n ⋅ e kT

Assuming space charge neutrality, the excess hole distribution is equal to the excess electron distribution δn(x n ) = δp(x n ) I n (x n )diff

q ⋅V - Lx n Dn dδp = q ⋅ A ⋅ Dn ⋅ = - q⋅A⋅ ⋅ p n ⋅ e kT ⋅ e p dx n Lp

Prob. 5.14

pn Lp

q ⋅V

- Lx n ⎞ - Lx n ⎤ ⎡ ⎛ ⋅ ⎢ D n ⋅ ⎜1 - e p ⎟ + D n ⋅ e p ⎥ ⋅ e kT ⎝ ⎠ ⎣ ⎦


I n (x n )drift = I - I n (x n )diff - I p (x n ) = q ⋅ A ⋅

A Si junction has N a = 1017

and N d = 1016 cm1 3 . Find (a)EF, Vo, and band diagram and (b) compare this value of Vo to that from Equation 5-8.

(a) Eip - E F = kT ⋅ ln

pp ni

1 cm3

= 0.0259eV ⋅ ln

1017

1 cm3 10 1 cm3

1.5 ⋅10

= 0.407eV

1016 cm1 3 nn E F - Ein = kT ⋅ ln = 0.0259eV ⋅ ln = 0.347eV ni 1.50.754eV ⋅1010 cm1 3 q ⋅ Vo = 0.407eV + 0.347eV = 0.754eV 0.407eV 0.347eV

1017 cm1 3 ⋅1016 cm1 3 Na Nd (b) q ⋅ Vo = kT ⋅ ln = 0.0259eV ⋅ ln = 0.754eV 2 n i2 1.5 ⋅1010 1 3

(

cm

)

V0 =

V0 = 0.0259V ln

x n0 =

x p0 =

ε 0

= W = 0.333 µm Nd 1+ Na

W = 0.83 nm Na 1+ Nd

Q + = −Q − = q A x n0 N d = 0.107 nC

−q Nd x n0 = −5.1⋅104 V/cm ∈Si


Prob. 5.15

Find V0, xn0, xp0, Q+, and

ε

4 ⋅1018

O

(1.5 ⋅10

in Si at 300 K.

N N kT ln a 2 d q ni 1 cm3 10

⋅1016 1 cm3

)

1 cm3 2

V0 = 0.8498 V

W=

2 ∈Si V0 ⎛ 1 1 ⎞ + ⎜ ⎟ q ⎝ N a N d ⎠ W = 0.334 µm

Prob. 5.16 Describe the effect on the hole diffusion current of doubling the p+ doping. The depletion edge and electron diffusion current on the p+ side may be ignored and

L p = D p ⋅ τ p = 20 cms ⋅ 50 ⋅10-9s = 10-3cm = 10µm 2

x

⎞ - L n i2 ⎛ qV δp = ⋅ ⎜ e kT -1⎟ ⋅ e p N d ⎝ ⎠

(

)

2

x 2µm 1010 cm1 3 ⎛ 0.6 ⎞ - 10µm ⎞ - Lp d(δp) 1 n i2 ⎛ qV 1 0.026 kT =- ⋅ ⋅ ⎜ e -1⎟ ⋅ e = - -3 ⋅ ⋅ ⎜ e -1⎟ ⋅ e = -8.6 ⋅1016 16 1 dx L p N d ⎝ 10 cm 10 ⎠ ⎝ ⎠ cm3 d(δp) 2 = 1.609 ⋅10-19C ⋅ 20 cms ⋅ 8.6 ⋅1016 cm1 4 = 0.277 cmA2 J p (diffusion) = - q ⋅ D p ⋅ dx

1 cm 4

Prob. 5.17


Since this is independent of the p+ doping, there will be no change.

For the Si p+-n junction, find I for Vf=0.5V. I = q⋅A⋅

Dp Lp -19

q ⋅V

-3

2

I = 1.6 ⋅10 C ⋅10 cm ⋅

Prob. 5.18

Dp

⋅ p n ⋅ e kT = q ⋅ A ⋅

D p ⋅τ p

10 cms

2

10 cms ⋅10−6 s 2

⋅

⋅

n i2 qkT⋅V ⋅e nn

(1.5 ⋅1010

1 cm3 16 1 cm3

5 ⋅10

)2

0.5eV

⋅ e 0.0259eV = 0.55µA

(a) Why is Cs negligible in reverse bias?

For reverse bias of more than a few tenths of a volt, Δpn ≈ -pn. Changes in the reverse bias do not appreciably alter the (negative) excess hole distribution. The primary variation is in the width of the depletion region, giving rise to the junction capacitance. (b) With equal doping, which carrier dominates injection in a GaAs junction? Electron injection dominates since μn?μp. With nn=pp it is clear that a carrier with higher mobility will determine the injection.

Prob. 5.19 (a) Find Cj for V=-10V for a Si p+-n junction 10−2 cm2 in area with N d = 1015

1 cm3

.

On the n side, 0.843 eV

Ei EF

0.555 eV 0.288 eV

Ei

Ec EF Ev

1015 cm1 3 Nd E F - E in = kT ⋅ ln = 0.0259eV ⋅ ln = 0.288eV ni 1.5 ⋅1010 cm1 3 On the p-side, E ip - E F =

1 2

⋅ E g ≅ 0.555eV

q ⋅ Vo = 0.555eV+0.288eV = 0.834eV 1 2

1

-14 F -19 ⋅1015 A ⎛ 2 ⋅ q⋅∈⋅N d ⎞ 10 cm ⎛ 2 ⋅1.6 ⋅10 C ⋅11.8 ⋅ 8.85 ⋅10 cm C j = ⋅ ⎜ ⋅ ⎜ ⎟ = ⎜ 0.834eV - (-10eV) 2 ⎝ Vo - V ⎠ 2 ⎝ 2


-2

1 cm3

⎞ 2 -11 ⎟⎟ = 2.78 ⋅10 F ⎠

(b) What is W just prior to avalanche?

From Figure 5-22, Vbr=300V from Figure 5-22 1

1

F ⎛ 2⋅∈s ⋅Vbr ⎞ 2 ⎛ 2 ⋅11.8 ⋅ 8.85 ⋅10-14 cm ⋅ 300V ⎞ 2 -3 W = ⎜ = ⎜ ⎟ = 1.98 ⋅10 cm = 20µm ⎟ -19 15 1 ⎜ ⎟ 1.6 ⋅10 C ⋅10 cm3 ⎝ q ⋅ N d ⎠ ⎝ ⎠

Prob. 5.20

Show Εo depends on doping of a lightly doped substrate. 1 2

1

-1 2 ⎡ 2 ⋅ q ⋅ V ⎛ 1 ⎡ 2 ⋅ q ⋅ Vo ⎛ N a ⋅ N d ⎞ ⎤ q 1 ⎞ ⎤ o εo = - ⋅ Nd ⋅ x no = - ⎢ ⋅ ⎜ ⋅ ⎜ + ⎟ ⎥ = - ⎢ ⎟ ⎥ ∈ ⎢⎣ ∈ ⎢⎣ ∈ ⎝ N a +Nd ⎠ ⎥⎦ ⎝ Na Nd ⎠ ⎥⎦ the lightly doped side dominates so the doping variation of Vo has only a minor effect 1

ε

o

q ⎡ 2 ⋅ q ⋅ Vo ⋅ Nd ⎤ 2 = - ⋅ Nd ⋅ x no = - ⎢ ⎥ ∈ ∈ ⎣ ⎦

Prob. 5.21 For the abrupt Si n+-p junction, calculate the peak electric field and depletion capacitance for reverse bias. Find the total excess stored electric charge and the electric field far from the depletion region on the p side. Depletion region is mostly on the p-side.

Vtotal = Vr + Vo ; Vr = 100V 2 ∈Si Vtotal ⎛ 1 1 ⎞ + ⎜ ⎟ q ⎝ N a N d ⎠ 1 Since N d is very high, may be neglected. Nd W=

W=

2 ∈Si Vtotal = q ⋅ Na

2 ⋅11.8 ⋅ 8.85 ⋅10-14 cmF ⋅100V = 1.14µm 1.6 ⋅10-19 C ⋅1017 cm1 3

I = 20 ⋅10-3A =

Qn τn


2 ⋅ Vtotal 2 ⋅100V V = = 1.75 ⋅106 cm -4 W 1.14 ⋅10 cm F ⋅ 0.0001cm 2 ∈ A 11.8 ⋅ 8.85 ⋅10-14 cm C j = Si = = 0.916pF W 1.14 ⋅10-4 cm

ε=

Q n = 20 ⋅10-3A ⋅ 0.1 ⋅10-6s = 2 ⋅10-9C

Far from the junction on the p-side, the current is only hole drift. I = 20 ⋅10-3A = q ⋅ A ⋅ p ⋅ µ p ⋅ ε = 1.6 ⋅10-19 C ⋅ 0.0001cm 2 ⋅1017

ε = 62.5

V cm

1 cm3

⋅ 200 cm V⋅s ⋅ ε 2

Prob. 5.22 Find the electron injection efficiency In/I.

(b)

In = I

D np D pn

⎛ D ⎞ ⎛ q⋅V ⎞ D q ⋅ A ⋅ ⎜ p ⋅ p n + n ⋅ n p ⎟ ⋅ ⎜ e kT -1⎟ ⎜ L p ⎟ Ln ⎠ ⎝ ⎠ ⎝

=

In = I

µ np µ pn

and

=

1 DL p 1+ p n ⋅ n Dn Lp n p

pp D pn ⋅ p n µ pn ⋅ p p pn = gives p = p np nn Dn ⋅ n p µn ⋅ nn

I 1 so making n n ? p p (i.e. using n + - p) increases n n L ⋅ µ p ⋅ pp I 1+ p L ⋅ µn ⋅ nn p n n p

Prob. 5.23


(a)

⎛ qkT⋅V ⎞ Dn ⋅ n p ⋅ ⎜ e -1⎟ q⋅A⋅ Ln ⎝ ⎠

Find Ip(xp) when pp=nn.

⎛ D p ⎞ ⎛ qV ⎞ Dn I = q ⋅ A ⋅ ⎜ ⋅p + ⋅ n ⋅ e kT -1⎟ ⎜ L p n L n p ⎟⎟ ⎜ ⎠ ⎝ ⎠ ⎝ is composed of x

- p ⎛ qV ⎞ Dn I n (x p ) = q ⋅ A ⋅ ⋅ n p ⋅ e Ln ⋅ ⎜ e kT -1⎟ Ln ⎝ ⎠

⎡ D ⎛ - x p n ⎢ I p (x p ) = I - I n (x p ) = q ⋅ A ⋅ ⋅ n p ⋅ ⎜1-e Ln ⎜ ⎢ L n ⎝ ⎣ 2 n Since N a =N d , n p =p n = i giving Na ⎡ D I p (x p ) = q ⋅ A ⋅ ⎢ n ⎢ L ⎣ n

x ⎛ - p Ln ⋅ ⎜1 - e ⎜ ⎝

⎤ ⎛ qV ⎞ ⎞ D p ⎟ + ⋅ p ⎥ ⋅ ⎜ e kT -1⎟ ⎟ L p n ⎥ ⎝ ⎠ ⎠ ⎦

⎞ D p ⎤ n 2 ⎛ qV ⎞ ⎥ ⋅ i ⋅ ⎜ e kT - 1⎟ ⎟ + ⎟ L p ⎥ N a ⎝ ⎠ ⎠ ⎦

Prob. 5.24 For the given p-n junction, calculate the built-in potential, the zero-bias space charge width, and the current for a 0.5V forward bias. (a) Calculate built-in potential:

for N a =1015

1 cm3

and N d =1017

Vo = φFP +φFN =

1 cm3

N ⎤ kT ⎡ N a +ln d ⎥ ⋅ ⎢ln q ⎣ n i n i ⎦

⎡ 1015 1017 ⎤ Vo = 0.026V ⋅ ⎢ln +ln ⎥ 10 1.5 ⋅1010 ⎦ ⎣ 1.5 ⋅10 Vo = 0.026V ⋅ [11.1+15.7 ] = 0.70V (b) Calculate total width of space change region

2 ∈s q

⎛ N +N ⋅ ⎜ a d ⎝ N a N d

⎞ ⎟ ⋅ Vo ⎠


W=

Thermal equilibrium means total potential φT across the P-N junction equals Vo

W=

2 ∈s q ⋅ Na

W=

2 ∈s ⎡ N a ⎤ ⋅ ⎢1+ ⎥ ⋅ Vo q ⋅ N a ⎣ N d ⎦

W=

⎛ N +N ⋅ ⎜ a d ⎝ N a N d

2 ⋅ 8.85 ⋅10-14

(

1.6 ⋅10

W = 1.3 ⋅10-8

cm 2 V

F cm

−19

⎞ ⎟ ⋅ Vo ⎠

⋅11.8

C

)⋅

1

15

10

1 cm3

⎡ 1015 ⎢1 + 17 ⎢⎣ 10

1 cm3 1 cm3

⎤ ⎥ ⋅ 0.70V ⎥⎦

⋅ [1 + .01] ⋅ 0.70V = 9.6 ⋅10-5 cm = 0.96 microns

(c) Forward bias current:

cm 2 cm 2 µ p = 450 τ = 2.5 ⋅10-3s V ⋅s V ⋅s ⎛ kT ⎞ cm 2 cm 2 D n = µ n ⎜ ⋅ 0.026V = 38.9 ⎟ = 1500 V ⋅s s ⎝ q ⎠

µ n = 1500

n i = 1.5 ⋅1010

1 cm3

⎛ kT ⎞ cm 2 cm 2 D p = µ p ⎜ = 450 ⋅ 0.026V = 11.7 ⎟ V ⋅s s ⎝ q ⎠ L n = D n ⋅ τ = 0.31cm

L p = D p ⋅ τ = 0.17cm

⎛ D P D n ⎞ J 0 = q ⋅ n i 2 ⋅ ⎜ + ⎟ ⎝ N d ⋅ L P N a ⋅ L n ⎠

(

)

J 0 = 1.6 ⋅10-19 C ⋅ 1.5 ⋅1010

)(

(

1 cm3

2

) ⋅ (6.88 ⋅10-

18 cm 4 s

⎛ qkT⋅V ⎞ I = A ⋅ J 0 ⋅ ⎜ e -1⎟ = 0.001cm 2 ⋅ 4.5 ⋅10-12 ⎝ ⎠

)(

)

cm 4 s

) = 4.5 ⋅10-

12

C cm 2 ⋅s

0.7V ⎞ ⎡ ⎛⎜ 0.026V ⎤ ⎟ ⎝ ⎠ ⋅ ⎢e -1⎥ = 2.2 ⋅10-3 A ⎣⎢ ⎦⎥


(

C cm 2 ⋅s

+ 1.25 ⋅10-15

Most of the current is carried by electrons because Na is less than Nd. To double the electron current, halve the acceptor doping. Prob. 5.25

Find the total reverse bias junction capacitance and forward bias electric field.

For n + - p in reverse bias, Cj =

A⋅∈s A 2 ⋅ q⋅∈s 25µm 2 = ⋅ ⋅ Na = ⋅ W 2 Vo -V 2

F 2 ⋅1.6 ⋅10-19 C ⋅11.8 ⋅ 8.85 ⋅10-14 cm ⋅1016 ⎛ 1016 1 ⋅1020 1 ⎞ cm3 cm3 ⎟ 0.026V ⋅ ln ⎜ - ( -2V ) ⎜ 1.5 ⋅1010 1 2 ⎟ cm3 ⎝ ⎠

(

)

For n + - p in forward bias, 0.5V ⎞ ⎛ qV ⎞ -9 A ⎛ 0.026 kT − 1 ⎟ = 0.225 cmA2 because only drift current J = J o ⋅ ⎜ e − 1⎟ = 10 cm2 ⋅ ⎜ e ⎝ ⎠ ⎝ ⎠ J = q ⋅ µ p ⋅ N a ⋅ ε in p region far from junction

ε=

-19

0.225 cmA2 2

16

1.6 ⋅10 C ⋅ 250 cm V⋅s ⋅10

1 cm3

= 0.56

V cm

1 cm3

= 4.2 ⋅10-15F

Prob. 5.26 In a p+- n junction with n-doping changed from Nd to 2Nd, describe the changes in junction capacitance, built-in potential, breakdown voltage, and ohmic losses. a) b) c) d)

junction capacitance increases built-in potential increases breakdown voltage decreases ohmic losses decrease

Prob. 5.27 Sketch the equilibrium band diagram with precise values for the bar.

on right side 18

p = 10

16

= 10

1 cm3

1 cm3

⋅e

Ei -E F kT

on left side Na : n i p = Na +

2 i

n = 2 ⋅1016 p 16

p = 2.4 ⋅10

1 cm3

1 cm3

16

= 10

E i - E F = 0.026eV ⋅

1 cm3 1 cm3

⎞ ⎟ = 0.24eV ⎟ ⎠


18 600K ⎛ 10 E i -E F = 0.026eV ⋅ ⋅ ln ⎜ 16 ⎜ 10 300K ⎝

+

1 cm3

10

32

1 cm6

p

⋅e

Ei - E F kT

600K ⋅ ln ( 2.4 ) = 0.046eV 300K

0.046eV

EC Ei 0.24eV EF EV

Prob. 5.28 Plot Ip and In versus distance. Assume that the minority carrier mobilities are the same as the majority carrier mobilities given in Figure 3-23a. 2 2 µ n = 700 cm µ p = 250 cm V⋅s V⋅s 2

cm D n = 0.0259V ⋅ 700 cm V⋅s = 18.13 s

2

2

cm D p = 0.0259V ⋅ 250 cm V⋅s = 6.475 s

2

L n = D n τ n = 18.13 cms ⋅10-6s = 2.54 ⋅10-3cm

L p = D p τ p = 6.475 cms ⋅10-6s = 4.26 ⋅10-3cm

2.25 ⋅1020 cm1 6 n i2 pn = = = 2.25 ⋅103 cm1 3 17 1 Na 10 cm3

n p = p n = 2.25 ⋅103 cm1 3

2

qV kT

Δp n = p n ⋅ e = 2.25 ⋅10

3

1 cm3

⋅e

0.7 0.0259

= 1.23 ⋅1015

xp

2

1 cm3

Δn p = Δp n = 1.23 ⋅1015

xp

Lp


L -3 I n (x p ) = qA n e Ln = 5 ⋅10-5 A ⋅ e 2.54⋅10 cm τn

I p (x n ) = qA

τp

I = 5 ⋅10-5 A + 8.38 ⋅10-5 A = 133.8µA I n (x n ) = I - I p (x n )

I p (x p ) = I - I n (x p )

e

-

xn Lp

1 cm3

= 8.38 ⋅10-5 A ⋅ e

-

xn 4.26⋅10-3cm

Prob. 5.29 Find the new junction capacitance for the given changes. 1

⎛ N ⎞ 2 C j ∝ ⎜ d ⎟ ⎝ Vr ⎠

C j,original = 10pF 1 2

⎛ 2 ⋅ N d ⎞ C j,new ∝ ⎜ ⎟ = ⎝ 8 ⋅ Vr ⎠

1 2

⎛ N ⎞ ⋅ ⎜ d ⎟ ⎝ Vr ⎠

1 2

C j,new =

1 2

⋅ C j,original = 5pF

Prob. 5.30 Find the minimum width to ensure avalanche breakdown.

Vbr = 300V from Figure 5-22 and Equation 5-23b Vr = V-Vo ; 300V and N a ? N d 1

1

⎡ 2 ⋅11.8 ⋅ 8.85 ⋅10-14 ⋅ 300V ⎤ 2 -3 ⎢ ⎥ = 2 ⋅10 cm = 20µm -19 15 1 ⎢⎣ 1.6 ⋅10 C ⋅10 cm3 ⎥⎦


x no

⎡ 2⋅∈⋅Vr ⎤ 2 ; W = ⎢ ⎥ = ⎣ q ⋅ N d ⎦

Prob. 5.31 Calculate the capacitance and relate to Na.

C=

∈S q ⋅ N a ⋅∈S ⋅A = A⋅ W 2 ( VO +VR )

Vo = 0.55eV+.0259eV ⋅ ln

Na ni

1 1 V +V = 2 ⋅ o R∈s 2 C A q ⋅ Na ⋅ 2 This means the slope of

1 C2

versus VR is

1 so knowing the area and material A ⋅ q ⋅ N a ⋅ ∈2s 2

type allows N a to be found. 1 = 1.197 ⋅10 22 CV2 ⋅ (0.84eV+VR ) 2 C 1 For N a = 1017 cm1 3 , Vo = 0.94eV → = 1.197 ⋅10 22 CV2 ⋅ (0.94eV+VR ) 2 C 1 cm3

, Vo = 0.84eV →


For N a = 1015

Prob. 5.32 Calculate the Debye length for a Si p-n junction on the p-side with Na = 1018 cm-3 and Nd = 1014, 1016, and 1018 cm-3. Compare with the calculated value of W for each case. Find LD and W, leaving Nd as a variable. 1

1

1

⎡ ∈s ⋅kT ⎤ 2 ⎡ ∈S kT ⎤ 2 ⎡11.8 ⋅ 8.85 ⋅10-14 L D = ⎢ 2 ⋅ ⎥ = ⎢ ⎥ = ⎢ 1.6 ⋅10-19C ⎣ q ⋅ N d ⎦ ⎣ q ⋅ N d q ⎦ ⎣ ⎡ 2⋅∈s ⋅kT ⎛ N a ⋅ N d ⋅ ⎜ ln W= ⎢ 2 n i2 ⎢⎣ q ⎝

⎞ ⎛ 1 1 + ⎟ ⋅ ⎜ ⎠ ⎝ N a N d

⎞ ⎤ ⎟ ⎥ ⎠ ⎥⎦

1 2

F cm

⎤ 2 - 12 411 ⋅ 0.0259eV ⎥ N d = 1 N d2 ⎦ 1

⎡⎛ Nd =581 cm1 2 ⋅ ⎢⎜ ln 1 ⎢⎣⎜⎝ 225 cm3

⎞ ⎛ -18 3 1 ⎞ ⎤ 2 ⎟ ⋅ ⎜10 cm + ⎟ ⎥ ⎟ ⎝ N ⎥⎦ d ⎠ ⎠

Plugging the doping values into these equations: LD (cm)

W (cm)

LD / W (%)

1014

4.11 x 10-5

3.01 x 10-4

7.3

1016

4.11 x 10-6

3.27 x 10-5

8.0

1018

4.11 x 10-7

4.93 x 10-6

12.0


Nd (cm-3)

The Debye length varies from 7 to 12 percent of W across this doping range.

Prob. 5.33 For the given symmetric p-n Si junction, find the reverse breakdown voltage.

-x p0 = -W/2

x n0 = W/2

ε0 W 2 εo = q ⋅ Na ⋅ x n0 = q ⋅ Na ⋅ W 2 ∈ ∈ The breakdown voltage plus built-in voltage is: x n0

→ x n0 = x p0 =


Na = Nd

∫ ε ⋅ dx = the area under the triangle

-x p0

W 1 ⎛ 2⋅∈⋅ε o ⎞ ∈⋅ε o2 Vbr +Vo = ε o ⋅ x n0 = ε o ⋅ = ε o ⋅ ⋅ ⎜ ⎟ = 2 2 ⎝ q ⋅ N a ⎠ q ⋅ N a V 2 F ⋅ (5 ⋅105 cm ) 11.8 ⋅ 8.85 ⋅10-14 cm = 16.32V = 17 1 -19 1.6 ⋅10 C ⋅10 cm3


Prob. 5.34

Prob. 5.35 Find the stored charge Qp as a function of time in the n-region if a long p+-n forward bias current is switched from IF1 to IF2 at t = 0.

I I F2

I F1 t Q p (0) = I F1 ⋅ τ p

I F2 =

Q p (t) dQ p + τp dt


Q p (∞) = I F2 ⋅ τ p

Taking the Laplace transform I F2 Q p (s) = + s ⋅ Q p (s) - I F1 ⋅ τ p s τp ⎛ I ⎞ 1 Q p (s) = ⎜ F2 + I F1 ⋅ τ p ⎟ ⎝ s ⎠ s + τ1p

Transforming back to time domain

(

Q p (t) = I F2 ⋅ τ p ⋅ 1- e

-

t τp

) +I

F1

⋅ τp ⋅ e

-

t τp

Prob. 5.36 The forward current in a long p+-n diode is suddenly raised from 0 to I at t = 0. (a) Find and sketch Qp(t).

I=

Qp ( t ) dQp ( t ) + τp dt

Qp ( t ) =A+Be

- τt

p

at t = 0, Qp(0) = A + B = 0 at t -> infinity, Qp(∞) = A = I τp thus, B = - I τp - τtp

)


(

Qp ( t ) = I⋅ τp 1- e

(b) Find Δpn(t) and v(t) in the quasi-steady state approximation. ∞

If

Qp = q ⋅ A ⋅ ∫ Δpn ⋅ e

- Lxnp

⋅ dx n = q ⋅ A ⋅ Lp ⋅ Δp n

0

Then

Δp n ( t ) =

Qp ( t )

q ⋅ A ⋅ Lp

Δp n ( t ) = p n ⋅ e Thus

v(t) =

qv kT

(

I ⋅ τp ⋅ 1 - e

=

- τtp

q ⋅ A ⋅ Lp

kT ⎛ Δp n ( t ) ⎞ ⋅ ln ⎜ ⎟ q ⎝ p n ⎠

(

⎛ I ⋅ τ ⋅ 1 - e τ-tp kT ⎜ p v(t) = ⋅ ln ⎜ q ⎜ q ⋅ A ⋅ L p ⋅ p n ⎝

) ⎞⎟⎟ ⎟ ⎠

)

Prob. 5.37 Sketch the voltage across a 1 kΩ resistor in series with a diode (offset 0.4 V, resistance 400 Ω) and a voltage source of 2 sin ωt. An ideal diode is a perfect short (resistance is zero) when the voltage across the diode is greater than the offset voltage. If the voltage is less, the diode is a perfect open (resistance is infinite). Thus,

v s < 0.4, v 0 = 0

(v s − 0.4 ) ⋅ 1kΩ 1kΩ + 400Ω


v s > 0.4, v 0 =


Prob. 5.38

Plot current versus voltage.

I = I p ( x n =0 ) = - q ⋅


Prob. 5.39

Find Qp and I when holes are injected from p+ into a short n-region of length l, if δp varies linearly. l

Q p = q ⋅ A ⋅ ∫ ∂p ⋅ dx n = q ⋅ A ⋅ 0

l ⋅ Δp n 2

A ⋅ D p ⋅ Δp n l

Prob. 5.40 Find the hole distribution and the total current in a narrow-base diode.

d 2 ∂p ( x n ) ∂p ( x n ) = dx 2n L2p ∂p ( x n ) =Ce

- Lx np

+De

xn Lp

Boundary conditions:

When x n = 0, ∂p = Δp n = C + D When x n = l , ∂p = 0 = Ce

- LlP

l

+ De LP

Thus, l

l

e LP - e

- LlP

D = Δp n − C =

-Δp n e l

- LlP

e LP - e


C =

Δp n e LP

- LlP

Plugging C and D back into the solution

a) ∂p ( x n )

⎡ Δp n ⎢e ⎣ = e

l - xn Lp

l Lp

−e

-e

-

- Llp

xn - l Lp

⎤ ⎥⎦

-q ⋅ A ⋅ Dp ⋅ Δp n ⎡ d∂p ( x n ) ⎤ I= ⎢-q ⋅ A ⋅ D p ⋅ = ⎥ dx n ⎦ x →0 Lp ⎣ n ⎡ q ⋅ A ⋅ D p l ⎤ ⎛ qV ⎞ b) I= ⎢ ⋅ p n ⋅ ctnh ⎥ ⋅ ⎜ e kT -1⎟ L p ⎥⎦ ⎝ ⎢⎣ L p ⎠

l

l

⎡ e Lp + e − Lp ⋅ ⎢- l l ⎢⎣ e Lp - e− Lp

⎤ ⎥ ⎥⎦

Prob. 5.41 For the narrow-base diode, find the current components due to recombination in n, and recombination at the ohmic contact. The steady-state charge stored in the excess hole distribution is x

x

n - Ln ⎡ L ⎤ Q p = q ⋅ A ∫ ∂p(x n ) ⋅ dx n = q ⋅ A ∫ ⎢C ⋅ e p + D ⋅ e p ⎥ ⋅ dx n 0 0 ⎣ ⎦

l

l

) ( )

l - l L Q p = q ⋅ A ⋅ L p ⎡⎢-C e Lp -1 +D e p -1 ⎤⎥ ⎣ ⎦

(

l L

Q p = q ⋅ A ⋅ L p ⋅ Δp n ⋅

e p+e l Lp

- Llp

e -e

-2

- Llp

The current due to recombination in n is

qAL p p n ⎡ ⎞ l l ⎤ ⎛ qV = − csch ⎥ ⎜ e kT − 1⎟ ⎢ctnh τp τ p ⎣⎢ Lp L p ⎦⎥ ⎝ ⎠ Lp Dp using = τp Lp Qp


Qp

qAD p p n ⎡ ⎞ l ⎤ ⎛ qV = ⎢ tanh ⎥ ⎜ e kT − 1⎟ τp L p ⎣⎢ L p ⎦⎥ ⎝ ⎠

The current due to recombination at xn = l,

Qp

⎡ qALp pn ⎞ l ⎤ ⎛ qV I− = ⎢ csch ⎥ ⎜ e kT − 1⎟ τ p ⎣⎢ τ p Lp ⎥⎦ ⎝ ⎠

These correspond to the base recombination and collector currents in the p-n-p BJT with VCB = 0 given in Equation 7-20.

Prob. 5.42 If the n region of a graded p+-n has Nd=Gxm, find a)

dε q q = N d = Gx m → ∈ ∈ dx

0

∫ dε =

εo

ε , ε ( x), Q, and C . j

o

W

q G ∫ x mdx ∈ 0

m+1

q W G ∈ m+1 q W m+1 εo = - G ∈ m+1 -ε o =

ε

x

q m b) ∫ dε = G ∫ x dx → ∈ 0 W V

(

)

W

∫ ε ⋅ dx 0


o dV ε=→ - ∫ dV = dx V

qG x m+1 - W m+1 ∈ (m+1)

ε(x) =

W

⎞ qG ⎛ x m+2 - (Vo - V) = - W m+1 ⋅ x ⎟ ⎜ ∈ (m+1) ⎝ m+2 ⎠0

qG ⎛ W m+2 (m+2) ⋅ W m+2 ⎞ qG -(m+1)W m+2 - (Vo - V) = ⋅ ⎜ ⎟ = m+2 m+2 ∈ (m+1) ⎝ m+2 ⎠ ∈ (m+1) Vo - V =

qG ⋅ W m+2 ∈ (m+2)

W

qAGW m+1 c) Q = qA ∫ Gx dx = m+1 0 m

W m+1 = (W

m+1 m+2 m+2

)

m+1

⎛ (V -V)⋅∈⋅(m+2) ⎞ m+2 = ⎜ o ⎟ qG ⎝ ⎠ m+1

qAG ⎛ (Vo -V)⋅∈⋅(m+2) ⎞ m+2 Q= ⋅ ⎜ ⎟ m+1 ⎝ qG ⎠

m+1

d) Q = qAG ⋅ ⎛ (Vo -V)⋅∈⋅(m+2) ⎞ m+2 ⎜ ⎟ m+1 ⎝ qG ⎠ m+1

dQ qAG ⎛ (Vo -V)⋅∈⋅(m+2) ⎞ m+2 = ⋅ ⎜ ⎟ d(Vo -V) m+1 ⎝ qG ⎠

-1

⋅

∈⋅(m+2) qG -1

dQ A⋅∈⋅(m+2) ⎛ (Vo -V)⋅∈⋅(m+2) ⎞ m+2 = ⋅ ⎜ ⎟ d(Vo -V) m+1 qG ⎝ ⎠ 1 m+2 m+2

dQ = A⋅ ∈ d(Vo -V)

(

)

1

⎛ ⎞ m+2 qG ⋅ ⎜ ⎟ ⎝ (Vo -V)⋅∈⋅(m+2) ⎠ 1

Prob. 5.43


⎛ qG⋅∈m+1 ⎞ m+2 dQ = A ⋅ ⎜ ⎟ d(Vo -V) ⎝ (Vo -V) ⋅ (m+2) ⎠

Draw the equilibrium band diagram and explain whether the junction is a Schottky or ohmic contact. Describe how to change the metal work function to switch the contact type.

1018 cm1 3 Na = 4eV+0.55eV+0.0259eV ⋅ ln = 5.02eV φs = χ +0.55eV+kT ⋅ ln ni 1.5 ⋅1010 cm1 3 For this p-type semiconductor, (Φm = 4.6eV) < (Φs = 5.02eV); so, the junction is a Schottky barrier. The junction becomes an ohmic contact at Φm > Φs. The metal work function must be raised to 5.02eV to make this an ohmic contact. Schottky Barrier

Ohmic Contact

4eV

4eV

4.6eV 5.02eV

5.02eV

EFn

5.02eV

EFn EFs Metal

Si

EFs Metal

Si


Prob. 5.44

Use InAs to make an ohmic contact to GaAs.

For further discussion, see Woodall, et al., J. Vac. Sci. Technol. 19. 626(1981).

a) Ei - E F = kT ⋅ ln

b) T an his th d wo sa eir is p rk w le co ro is ill o u vi pr de f a rse de ot st ny s d s ec ro p an o te y ar d le d th t o a ly by e s in f th se for Un te is ss th ite gr w in e ity o g us d S of rk ( stu e o tat th inc de f i es e lu nt ns co w d le tr p or in a uc y k g rn to rig an on in rs h d th g. in t la is e D t w no W iss ea s t p or em ch in er ld m W ina g itt id tio ed e n . We or b)

Prob. 5.45

Draw the equilibrium band diagram (a) and the band diagrams for 0.3V forward bias and 2.0V reverse bias (b).

1017 cm1 3 po = 0.0259eV ⋅ ln = 0.407eV ni 1.5 ⋅1010 cm1 3


Prob. 5.46 We want to make a Schottky diode on one surface of an n-type semiconductor, and an ohmic contact on the other side. The electron affinity is 5 eV, bandgap is 1.5 eV, and the Fermi potential is 0.25 eV? What should be values of work functions of the two metals. (Give your answer as greater than or less than certain values.) Sketch the band diagram of the structure.

Prob. 5.47 A semiconductor heterojunction is made between the following materials, A and B with the following parameters: EG (eV) χ (eV) Doping (cm-3) Length (μm) Ln,p (μm) ni (cm-3) τn,p (μs) A: 2 4 NA=1020 0.5 10 108 10 16 10 B: 1 5 ND=10 0.1 100 10 1 Draw the equilibrium band diagram, marking off the band edge energies and EF, with respect to the vacuum level. Calculate the current density if the junction is forward biased such that the minority concentrations are increased by a factor of 106. (HINT: Use appropriate approximations. There is a lot of extraneous information here; the answer is very simple. Remember that this is a p+-n junction, and the length of the semiconductors is << diffusion length; the minority carrier concentration is zero at the ohmic contacts at both ends of the device.)


(

)

Δp n = p n ×106 = 104 cm -3 ×106 = 1010 cm -3 L p = D p τ p ⇒ D p = L p 2 τ p = 100 cm 2 /s

⎛ 1010 cm -3 ⎞ dp 2 = 1.6 ×10 −19 C × 100 cm 2 /s × ⎜ ⎟ = 16 mA/cm −4 dx 0.1 × 10 cm ⎝ ⎠ δp(x) is linear because length of n-region = L p I ≈ I p = qD p

(

) (

)

Prob. 5.48 A p-n junction diode has a doping concentration of 1017 cm-3 on the p- side, and double that on the n-side. The intrinsic carrier concentration is 1011 cm-3, bandgap is 2 eV and εr=15. Sketch the band diagram in equilibrium, and mark off the values of band edges with respect to the Fermi level and the depletion widths on both sides. Repeat the above for a heterojunction, where the bandgap on the n-side is reduced to 1 eV, and the electron affinity on the n-side is 4 eV. Other parameters are kept the same. Bandoffsets are the same for conduction and valence band across the heterojunction.

⎛ (1017 cm -3 ) × ( 2 ×1017 cm -3 ) ⎞ kT ⎛ N a N d ⎞ ⎜ ⎟ = 0.734 V V0 = ln ⎜ ⎟ = ( 0.0259 V ) × ln ⎜ 2 11 -3 ⎟ q ⎝ n i 2 ⎠ (10 cm ) ⎝ ⎠ 12

⎞ ⎤ ⎟ ⎥ ⎠ ⎥⎦


⎡ 2εV0 ⎛ N a +N d W = ⎢ ⎜ ⎢⎣ q ⎝ N a N d

12

⎡ 2 × (15 × 8.85 ×10−14 F/cm ) × ( 0.734 V ) ⎛ (1017cm -3 ) + ( 2 ×1017cm -3 ) ⎞ ⎤ ⎜ ⎟ ⎥ = ⎢ ⎜ (1017 cm -3 ) × ( 2 ×1017 cm -3 ) ⎟ ⎥ 1.6 ×10−19 C ⎢ ⎝ ⎠ ⎦ ⎣ = 0.135 µm

ϕFp = Eip - E F =

⎛ 1017 cm -3 ⎞ kT ⎛ N a ⎞ ln ⎜ = 0.0259 V ln × ) ⎜ 11 -3 ⎟ = 0.358 V ⎟ ( q ⎝ n i ⎠ ⎝ 10 cm ⎠

p-side: E C - E F = 1 eV + 0.358 eV = 1.358 eV E F - E V = 0.642 eV

ϕFn =

⎛ 2 ×1017 cm -3 ⎞ kT ⎛ N d ⎞ = 0.0259 V ln × ln ⎜ ( ) ⎟ ⎜ ⎟ = 0.376 V 11 -3 q ⎝ n i ⎠ ⎝ 10 cm ⎠

n-side: E C - E F = 1 eV - 0.376 eV = 0.624 eV E F - E V = 1 eV + 0.376 eV = 1.376 eV xp = W

Nd 2 = W = 90 nm Nd + Na 3

xn = W

Na 1 = W = 45 nm Nd + Na 3


Heterojunction:

Chapter 5 Solutions

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