Electronic Materials - Problem sheet solution.pdf

MATERIALS SCIENCE TUTORIAL: Electronic Materials (ND Broom) Problem 1.

Calculate the conductivity of pure copper at 300 and -50 °C given that its resistivity at room temperature (20°C) is 1.67 x 10 -8Ω.m and α = 0.0068/ °C. Cu

Assume there is no residual component of resistance. (note: ρ =ρo + ρo α ΔT, where ρo is the resistivity at a given reference temperature.)

Problem 2. [Note: σ = nq(μe + μh); σ = σoexp(-(Ea-Ev)/k T); T); σ = σoexp(-Eg/2k T); T);

T)] σ = σoexp(-(Ec-Ed)/k T) Choose the appropriate relationships from those given above and the data from the table shown below to calculate the conductivity of pure germanium at 323, 360, 425, 473 and 500°K. Plot your data of  σ vs T and explain any important limitation the curve tells us about intrinsic semiconductors. Note: Boltzmann's constant k = 8.63 x 10 -5 eV/° K. Electronic charge q = 1.6 x 10-19 C. PROPERTIES OF SOME COMMON SEMICONDUCTORS AT ROOM TEMPERATURE (300K)

Energy Gap Electron mobility Hole mobility Carrier Density 2 μe[m / (V.s)] μh[m2 / (V.s) Material Eg (eV) ne (= nh) (m-3)  __________________________________  __________________ ___________________________________ ____________________________________  _________________   _  Si 1.107 0.140 0.038 14 x 10 15 Ge 0.66 0.364 0.190 23 x 10 18 CdS 2.59 0.034 0.0018 GaAs 1.47 0.720 0.020 1.4 x10 12 InSb 0.17 8.00 0.045 13.5 x 10 21

Problem 3

Write down the expression for the conductivity of an intrinsic semiconductor. Manipulate this expression to show how the energy band gap for such a semiconductor can be determined graphically from conductivity vs temperature data.

Problem 4.

You are required to characterise a new type of pure (i.e. undoped) semiconductor  material. (a). If its conductivity at 20°C is 250/ Ω.m and at 100°C is 1100/ Ω.m, what is its band gap, Eg? (b). Calculate its conductivity at 50°C.

Worked solutions Problem (1)

Use ρ = ρo + ρoαΔT

∴

When ρ

=

resistivity at given reference temperature i.e. 20°c (in this problem)

ρ300°c

= =

1.67 x 10-8+ 1.67x10 -8 x 0.0068 x (300-20) 4.88 x 10-8 Ω m

Conductivity σ = 1/ ρ

∴

=

σ300°c

ρ -50°c

−50° c

4.88x10 −8

=

0.205x 108 Ω-1 m-1

= = =

1.67 x 10-8 + 1.67x10-8 x 0.0068 x(-50-20) 1.67 x 10-8 (1-0.48) 0.868 x 10 -8 Ω m

=

σ  

1

=

1 0.868 x10 −8 1.15x 108 Ω-1m-1

Problem (2)

Conductivity σ

=

&

=

σ

nq (µe +µh) - Eg σ   e 0 2kt

(1) (2)

since we are considering a pure or intrinsic semi conductor.

From data table Eg for Ge Electron mobility µe = Hole mobility µh = & charge q on electron =

= 0.66ev 0.364 m2/V.sec 0.190 m2/V.sec 0.16x10-18Coulombs

Since the semiconductor is undoped the number of electrons that enter the conduction  band equals the number of holes left in the valence band ∴

nĕ = nh = n = 23x10 18 /m3

Using eqn (1) σ 300K 

= =

23 x 1018 x 0.16 x 10 -18(0.364+0.190 2.04/ Ω m

We can now use this value of  σ300.k to obtain value of equation constant σo If we re-arrange eqn (2) + Eg = σo σ   exp 2k T

i.e

σo

= = =

σ300K  exp

+ Eg

2k T

300 K 

2.04 x exp(0.66/(2x8.62x10 -5x300) 7.1x 105 Ω-1 m-1

(2) cont…/  Now obtain conductivity σ at remaining temperatures of 323K, 360K, 473.K & 500K  using eqn (2) Since we now have σo Eg & T − 0.66 σ 323° K  = 7.1 x 105x exp 5 2 × 8.62 × 10− × 323 = 5.07 / Ω. m − 0.66 σ 360° K  = 7.1x10 5x exp 5 2 × 8.62 × 10 − × 360 = 17.2/ Ω. m − 0.66 = 7.31x105 x e σ 473° K  5 2 × 8.62 × 10 − × 473 = 217/ Ω. m − 0.66 σ 500° K  = 7.31 x105 x e 5 2 × 8.62 × 10− × 500 = 335/ Ω. M  Now plot data as shown below in friendly “low-tech” format:-

Problem (3 ) Holes and electrons are the charge carriers in an intrinsic semi conductor  ∴ use the equation

σ

=

σo exp

=

lnσo -

− Eg

2kT 

Take natural log of both sides ln σ

 Eg

2kT 

which can be expressed in the form of:y = C + mx 1 − Eg where y = ln σ, x = and gradient m = T  2k  and intercept on y axis is ln σ0 1  Now plot ln σ vs T  In σ gradient (y/x) =

1 T 

Energy gap E g

− Eg

2k 

(.K)

= - [gradient x 2 k ]

]

Problem (4) Use σ =

σo exp [-Eg/2kT]

(a)

at T at T

= =

20°C (293K) 100°C (373K)

∴

250

=

σo exp [

&

1100

=

σo exp [

(2) divided by (1) gives :1100 = 250 ln

i.e

1100 250

1.482

=

= =

Eg

= =

σ = 250/ Ω m σ = 1100/ Ω m

− Eg

2 × k × 293 − Eg

2 × k × 373

σ  

0

exp

σ  

0

 Eg

[

2k   Eg

(1)

]

(2)

 Eg

[

1

2k  293

1

2k  293  Eg

]

−

1 373

−

1 373

]

]

[3.413 - 2.681] x 0.732x10 -3

2k  1.482 × 2 × 8.62x10 -5 0.732 × 10 −3 0.35eV

4(b) Using value of Eg from (a) and knowing that σ = 250/ Ω at T = 293K we can now solve for σo − 0.35 250 = ] ∴ σo exp [ 2 × k × 293 + 0.35 and σo = 250xexp [ ] 2 × 8.62 ×10 −5 × 293 = 250xexp [6.93] = 250x1022 = 2.56x105 /Ω m We can now obtain the conductivity at 50oC (323K) − 0.35 = 2.56x105xexp [ ] σ323 2 × 8.62 ×10 −5 × 323 = 2.56x105 x exp [-6.286] = 2.56x105 x 186x10 -3 = 477 / Ω m