Homework 4 - College of Engineering | UMass Amherst · Solution....

Homework 4

1. Problem. Find the resistivity ρ (in ohm-cm) for a piece of Si doped with both acceptors (NA = 1019 cm−3)and donors (ND = 1016 cm−3). Since the electron and hole mobilities depend on the concentration of thedopants, use the following emprirical expressions to evaluate them:

µn = 232 +1180

1 +(NA+ND8×1016

)0.9, (1)

µp = 48 +447

1 +(NA+ND1.3×1016

)0.76, (2)

where the mobilities are expressed in cm2/V s and NA and ND in cm−3.Solution. Subtituting for the values of NA and ND in the given equations,

µn = 247.06 cm2/Vs ,

and

µp = 50.84 cm2/Vs .

Since σ = eNDµn + eNAµp and ρ = 1/σ, we have

ρ = 1.22 × 10−2

Ωcm .

2. Problem. Consider a sample of p-type Si doped with NA = 1018 cm−3 and ND = 0. Over a length of 1 µmthe electron concentration drops from 1016 cm−3 to 1013 cm−3. Using Eq. (1) above, calculate the currentdensity due to diffusion alone.

ECE609 Spring 2010 1

Solution. The electron current due to diffusion alone is:

Jdiff = eDndn

dx.

Since from Einstein relation eDn = kbTµn, assuming dn/dx ≈ 1016 cm−3/10−4 cm and using µn fromthe equation given in the first problem, we have µn = 342.18 cm2/Vs and Jdiff = 141.65 A/cm2.

3. Problem. A p-n Si junction is formed between a n-type region with ND = 2 × 1018 cm−3 and a p-typeregion with NA = 5 × 1016 cm−3. Find:(a) the width of the depletion region on the n side, xn0 (in µm),(b) the width of the depletion region on the p side, xp0 (in µm),(c) the built-in potential Vbi (in eV),(d) the fraction of Vbi which drops over the n-side of the junction,(e) the fraction of Vbi which drops over the p-side of the junction.Finally, using the results just obtained, plot (to scale as accurately as you can) the band diagram of the junctionat equilibrium.Solution. (a) From

W =

[2εkBT

e2ln

(NAND

n2i

) (1

NA+

1

ND

)]1/2

,

we get W = 0.154 µm and

xn0 =WNA

NA + ND= 0.00375 µm .

(b) Similarly,

xp0 =WND

NA + ND= 0.15 µm .


(c) The built-in potential is given by:

Vbi =kBT

eln

(NAND

n2i

)= 0.892 V .

(d) The potential drop on the n side of the junction is:

Vbi,n =Vbixn0

W= 0.021 V .

(d) On the other end, the potential drop on the p side of the junction is:

Vbi,p =Vbixp0

W= 0.871 V .

4. Problem. A Si p-n junction has a saturated reverse current Is = 10−14 A at 300 K. Determine the forwardbias required to get a current of (a) 50 µA and (b) 2.50 mA.Solution. (a) From the expression for the diode current

I = Is

[exp

(eV

kBT

)− 1

]

and from the fact that Is = 10−14 A we have V = 0.5784 V when I = 50 µA. (b) Similarly, for I = 2.5 mAwe have V = 0.6797 V.

5. Problem. The saturated reverse current of a GaAs p-n junction is 5 × 10−21 A. Calculate the current undera forward bias of (a) 0.8 V and (b) 1.2 V.Solution. (a) As in the previous problem, we have I = 0.129 µA for V = 0.8 V and (b) I = 0.66 A for V =1.2 V.


6. Problem. Derive Eq. (419) of the Lecture Notes, Part 2.Solution. We require:

J(x = ln) = J = J

1

Mn+

∫ ln

−lpdx αp exp

[−∫ x

−lpdx

′(αn − αp)

]exp

[∫ ln

−lpdx

′(αn − αp)

]

(3)Now let’s multiply both sides of this equation by:

1

Jexp

[−∫ ln

−lpdx′(αn − αp)

],

so that Eq. (3) becomes:

exp

[−∫ ln


]=

1

Mn+

∫ ln

−lpdx αp exp

[−∫ x


], (4)

or1

Mn= exp

[−∫ ln


]−∫ ln

−lpdx αp exp

[−∫ x


]. (5)

Now let’s add and substract αn in the integrand of the last term:

1

Mn= exp

[−∫ ln


]+

∫ ln

−lpdx (αn − αp) exp

[−∫ x


]−

−∫ ln

−lpdx αn exp

[−∫ x


]. (6)


The second term on the right-hand side can be integrated immediately, since in general:

∫ b

adx f(x) exp

(−∫ x

adx

′f(x

′)

)= − exp

(−∫ x

adx

′f(x

′)

)∣∣∣∣b

a= 1− exp

(−∫ b

adx

′f(x

′)

).

So, Eq. (6) becomes:

1

Mn= exp

[−∫ ln


]+ 1 − exp

[−∫ ln


]−

−∫ ln

−lpdx αn exp

[−∫ x

−lpdx

′(αn − αp)

]= 1 −

∫ ln

−lpdx αn exp

[−∫ x

−lpdx

′(αn − αp)

],

(7)so that

1 − 1

Mn=

∫ ln

−lpdx αn exp

[−∫ x

−lpdx

′(αn − αp)

], (8)

which is Eq. (419) of the Lecture Notes, Part 2.Note that the result above is completely general. We have not made use of the assumption αn = αp, nor ofany arbitrary assumption of the type

exp

[∫ x

−lpdx(αn − αp)

]= exp

[(αn − αp)(x + lp)

],

which is only valid if αn and αp do not depend on x, and, finally, it is valid also far from breakdown, so it does

not require any condition of the sort∫ ln−lp

dxα = 1.

7. Problem. (a). Plot the Zener tunneling current JZener given by Eq. (425) of the Lecture Notes as a functionof applied bias Va usingm

∗ = 0.32 mel = 0.32×9.1 10−31 kg and EG = 1.1 eV (the gap of Si). Assume


for the field F the approximate value F ≈ Fmax, where the maximum field in the junction, Fmax, is given byEq. (364). Assume NA = 1017cm−3 and ND = 1019cm−3 to estimate the width lp and ln of the depletionregions.(b). Repeat the calculation, but now using EG = 0.64 eV (the gap of Ge) and m∗ = 0.22 mel.(c). Repeat the calculations in (a) and (b), but now assuming higher doping, NA = 1019cm−3 andND = 1021cm−3. What happens? Why?(d). Considering that the current VLSI technolgy requires increasingly higher doping, can you foresee anyproblems with the possible use of Ge (instead of Si) when the width of the depletion regions shrinks?Solution. Higher doping enhances tunneling. Ge exhibits a much larger Zener tunneling current than Si becauseof its smaller band-gap. This drawback will become more severe as the doping concentrations increase.The figure below shows the result:


–2 0 2 4 6 8 1010–40

10–20

100

1020

high doping

low doping

Ge

Ge

Si

Si

–Vapplied (V)

J Zen

er (

A/c

m2 )

Note that the current is nonzero only under reverse bias such that Va < −(EG − Vbi) or under forward biasVa > EG + Vbi. For a bias of smaller magnitude, |Va|, tunneling cannot occurr because there are no finalstates available. The plot show only the current in reverse bias, since this matters in devices. (Under forwardbias the diode current is very large anyway and tunneling is not a concern.)NOTE: The most important dependence of the Zener tunneling current on the applied reverse bias Va is via thedepletion width lp appearing in the expression for the field Fmax = eNAlp/εs. Many of you have forgottenthis crucial dependence.

8. Problem. The current trends in VLSI technology also demand reduced applied biases (i.e., a smaller Va).Without doing any calculation, do you think that the strictest breakdown limitations will be due to impactionization or to Zener tunneling?


Solution. Zener tunneling will dominate, since the energy threshold for impact-ionization is ∼ EG, while forZener tunneling is EG − eVbi, where Vbi is the built-in potential of the drain/body junction in a MOSFET.

9. Problem. Using Eqns. (450) and (451), plot the total charge at the GaAs-side of the GaAs-AlxGa1−xAsheterojunction as a function of the interface potential ψi = eψi/(kBT ). Indicate clearly the accumulation,depletion, inversion, and strong inversion regions. Refer to the discussion on page 135 of the Notes for help.Solution. The equations you were told to use account only for the charge of the majority carriers (electrons)and donors. They do not account for minority carriers (holes), so that it is impossible to obtain the charge ininversion (weak or strong), unless the hole charge-density is added. The charge density of the minority carriersis included in the equations for the charge at the interface of an MOS capacitor. So, one could use a similarequation, remembering that in that case we considered a p-type substrate, while here we are dealing with ann-type substrate, so the plot should be ‘flipped’ (as in the ‘mirror image’ ψi → −ψi).

10. Problem. Derive Eq. (457) from Eq. (458).Solution. The potential energy is

Vim(z) = − e2

16πεsz− eFzz . (9)

The maximum of the potential energy is given by

0 =dVim(z)

dz=

e2

16πεsz2− eFz . (10)

The value of z at which this occurs is thus:

z0 =

(e

16πεsFz

)1/2

. (11)


Inserting this into Eq. (9) we get:

Vim(z0) = − e2

16πεsz0− eFzz0 = − e

(eFz

4πεs

)1/2

, (12)

so that the barier-lowering (in V) will be ∆ΦB = [eFz/(4πεs)]1/2.

11. Problem. Using the derivation of Eq. (480) as a guide, derive Eq. (482).Solution. The expression given in class, Eq. (475) of the Notes, Part 2, for the tunneling current from the metalto the semiconductor is:

jMS =em∗

M

2π2h3

∫ ∞

0dE E fM(E) [1 − fS(E)] exp

−2

∫ zt

0dz κ(z)

. (13)

Approximating the potential barrier with φ(z) ≈ φ′B − Fzz, we have for the tunneling distance zt =

(eφ′B − E)/(eFz) and:

∫ zt

0dz κz(z) =

(2m∗S)1/2

h

∫ zt

0dz (eφ

′B − eFz − E)

1/2. (14)

Using the new integration variable y = eφ′B − eFz − E we get:

∫ zt

0dz κz(z) =

(2m∗S)1/2

ehFz

∫ eφ′B−E

0dy y

1/2=

2(2m∗S)1/2

3ehFz(eφ

′B − E)

3/2, (15)

so that

jMS =em∗

M

2π2h3

∫ ∞

0dE E fM(E) [1 − fS(E)] exp

−4(2m∗

S)1/2

3ehFz(eφ′

B − E)3/2

, (16)

which is Eq. (482).


Homework 4 - College of Engineering | UMass Amherst · Solution....

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