Tutorial #4 – selected problems

Problem 1 (Problem 3.7 Streetman)Given mn* and mp* from Table 3-1 p 65, calculatethe effective densities of states Nc and Nv for GaAsat 300K (assume mn* and mp* do not vary with temperature). Calculate the intrinsic carrier concentration.Answer:mn*=0.067m mp*=0.48m T=300KThe effective densities of states are given by:

Nm kT

hcmc

*

2-3n 2

24 35 103 2 17( ) ./

Nm kT

hcmv

p*

2-3 2

28 32 103 2 18( ) ./

The intrinsic carrier concentration is given by

n N N ei c v

-E

2kTg

Eg=1.43eV

ni=1.87 x 106 cm-3

Problem 2 For silicon doped to 1014 donors/cm3: Calculate the temperature where the total electron carrier density in this semiconductor is equal to twice the donor concentration. (This gives a good estimate of where the sample changes from “extrinsic” to “intrinsic” behavior.)

Answer: If the material is to remain electrostatically neutral, the sum of the positive charges (holes and ionized donor atoms) must balance the sum of the negative

charges (electrons and ionized acceptor atoms):p0 + Nd

+ = n0 + Na-

Nd+ = 1014 cm-3 Na

- = 0 cm-3

n0 = 2 Nd+ cm-3

p0 = n0 – Nd+ = 1014 (2-1) = 1014 cm-3

ni2 = n0p0

n (T) = n pkT

hm m ei 0 0 2 n

*p* -

E

2kTg

22 3 2 3 4( ) ( )/ /

0422.0)mm(k)2(2

h pneT

4/3*p

*n

3/2

3002kT

E-3/2

g

where h=6.63 x 10-34 J·s k=1.38 x 10-23 J/K = 8.62 x 10-5 eV/K mn

*= 1.1·(9.11 x 10-31 kg) mp

*=0.56·(9.11 x 10-31 kg) Eg = 1.11eV

0422.0eT T

6438.5-3/2

So,

We can solve for T in various ways.One way is to let a calculator solve the equation.

Another way is to solve for T using Excel orQuattro Pro.

T Tpower3/2 exp(-6438.5/T) (Tpower3/2) * exp(-6438.5/T)

300 5196.2 4.779E-10 2.483E-06

400 8000 1.022E-07 0.0008177

500 11180 2.556E-06 0.0285789

510 11517 3.29E-06 0.0378966

513.9 11650 3.621E-06 0.0421863

514 11653 3.63E-06 0.0423016

515 11687 3.719E-06 0.0434697

T=513.9K.

You can start with e.g. T = 300 K. Check lastcolumn to see how close you are to 0.0422.

Repeat for 400 K and 500 K. Increase to 510 K, and then 515 K. Decrease to 514 K. Fine tune by trying 513.9 K. This value gives a product close to 0.0422.

Therefore,

3. Show that for acceptor doping concentration NA

the equilibrium hole concentration p0 is

)4(2

1 220 iAA nNNp

Solution:

=0 n0p0= ni

2

0

2

0 p

nn i

DA NpnN 00

00

2

pp

nN iA

20

20 pnpN iA

020

20

iA npNp

)4(2

1 220 iAA nNNp

negative root not realistic AiA NnN 22 4