Homework 5

1. Problem. Derive Eq. (171) of the Lecture Notes, Part 2, expressing the density of a system of electronsconfined in two dimensions as a function of Fermi level.Solution. The derivation has been shown in class: The density in 2D can be obtained as we have done in 3D(see page 41, Lecture Notes, Part 1):

n2D =2

(2π)2

∫dkfFD(k) =

2

(2π)2

∫ 2π

0dφ

∫ ∞

0dkk

1

1 + exp[E(k) − EF )/(kBT )]. (1)

Performing the trivial angular integration and changing integration variable from k to E = h̄2k2/(2m∗) weget:

n2D =m∗

πh̄2

∫ ∞

0

dE

1 + exp[(E(k) − EF )/(kBT )]. (2)

Now let’s multiply numerator and denominator of the integrand by exp[−(E(k)−EF )/(kBT )], and settingy = (E − EF )/(kBT ):

n2D = −m∗kBTπh̄2

∫ ∞

−y0dy

e−y

1 + e−y, (3)

where y0 = EF/(kBT ). The integral is just ln(1 + e−y)|∞−y0 = − ln(1 + ey0), so that:

n2D =m∗kBTπh̄2

ln[1 + exp[EF/(kBT )] . (4)

2. Problem. Consider a MOSFET fabricated 20 years ago with the following parameters:

• Gate oxide thickness tox = 200 nm• Substrate doping concentration (p-type) NA = 1.7 × 1015 cm−3

• Gate material: n++ (poly)Si

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• Charge in the oxide Qox = 1010 charges/cm2

• Intrisic carrier concentration in Si at 300 K ni = 1.45 × 1010 cm−3.

a. Calculate the threshold voltage.b. What happens to the threshold voltage if the oxide gets contaminated during fabrication so that Qox

increases by a factor of 5?Consider now a MOSFET fabricated 5 years ago with the following parameters:

• Gate oxide thickness tox = 20 nm• Substrate doping concentration (p-type) NA = 8.7 × 1016 cm−3

• Gate material: n++ (poly)Si• Charge in the oxide Qox = 1010 charges/cm2

• Intrisic carrier concentration in Si at 300 K ni = 1.45 × 1010 cm−3.

c. Calculate the threshold voltage.d. What happens to the threshold voltage if the oxide gets contaminated during fabrication so that Qox

increases by a factor of 5?e. Compare the results of b and d and explain the cause of the different behavior exhibited by the ‘old’ and‘new’ devices.Solution. When Qox = 1010 cm−2 the threshold voltages are:

VTH,old = 0.723 V and VTH,new = 0.723 V . (5)

When Qox = 5 ×1010 cm−2 the threshold voltages are:

VTH,old = 0.352 V and VTH,new = 0.686 V . (6)

The new devices is less affected by oxide charges thanks to the larger Cox.Note: The equation

Φms ≈ kBT

[ln

(NA,sub

ni

)+ ln

(ND,gate

ni

)], (7)

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which is valid only in the non-degenerate limit (observation we have already made in Homework 1!), whichis clearly not the case for the heavily-doped gate. Instead, one should assume that the Fermi level in the gateis approximately lined-up with the bottom of the conduction band. Therefore the term kBT ln(ND,gate/ni)should be replaced by ∼ EGap/(2e).

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3. Problem. Consider a MOSFET with the following parameters:

• Gate oxide thickness tox = 20 nm• Substrate doping concentration (p-type) NA = 1.0 × 1017 cm−3

• Gate material: n++ (poly)Si• Charge in the oxide Qox = 0 charges/cm2

• Intrisic carrier concentration in Si at 300 K ni = 1.45 × 1010 cm−3• Channel width W = 2 µm and channel length L = 1 µm• Electron mobility µn = 600 cm2/Vs• Zero substrate bias.

On a single graph plot the drain current ID as a function of the drain voltage VD with VD ranging from 1.0to 5.0 V. Use the following equations:a. The complete model, Eq. (13), page 174 of the Lecture Notesb. The simplified model, Eq. (22)c. The simplified model, Eq. (22), with n=1.Solution. My apologies: The gate oxide thickness for this problem should have been 20 nm, NOT 200 nm asI had erroneously typed. Using such a large (unrealistic) thickness with the large substrate doping would haveresulted in a huge, unrealistic threshold voltage, as you found. So, asking for the drain current at a gate voltageso much below threshold made no sense. If you have correctly realized this problem, you have been made awarevery directly of the importance of scaling correctly the thickness of the gate oxide. Here is the solution for thecase tox = 20 nm. Considering just the case of VG = VD = 5 V, one should get:

ID,sat =

1.295 × 10−3A (full model)

1.156 × 10−3A (simplified model)

1.832 × 10−3A (simplified, n=1)

. (8)

See the figure for the full I − V characteristics.

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0 1 2 3 4 50.0

0.5

1.0

1.5

2.0Full modelSimplified modelSimplified model with n=1

VDS (V)

I DS (

mA

)

4. Problem. The threshold voltage of an n-channel MOSFET having degenerately doped n+ polysilicon gate is0.7 V. What would the threshold be if the gate material were degenerately doped p+ polysilicon, all otherfabrication parameters being unchanged?Solution. The Fermi level of the n+ poly-Si gate is roughly lined up with the bottom of the conduction band,while it is lined-up approximately with the top of the valence band for a p+ poly-gate. Therefore VTH increasesby about Egap/e to 1.8 V.This quantitative estimate should have been given. Stating quite generally that ‘VTH increases’ is a little tooobvious and too rough.

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5. Problem. Consider a MOSFET with the following parameters:

• Gate oxide thickness tox = 10 nm• W = L = 1 µm• Substrate doping concentration (p-type) NA = 2.8 × 1017 cm−3

• Gate material: n++ (poly)Si• Charge in the oxide Qox = 1010 charges/cm2

• Intrisic carrier concentration in Si at 300 K ni = 1.45 × 1010 cm−3• Channel width W = 2 µm and channel length L = 1 µm• Electron mobility µn = 600 cm2/Vs• VG = VD = 3.3 V• Zero substrate bias.

a. Calculate the drain currentb. Because of unavoidable fabrication parameter fluctuations, tox, NA, and Qox can vary by ± 5%. Eachof these parameters is independent on the others. The increase of some of these parameters will result in alarger current while the increase of some other parameters will result in a lower current. Calculate the maximum‘worst case’ increase and decrease of the current that can result from the fluctuations defined above. Use thesimplified model, Eq. (22).Solution. The nominal current is ID ≈ 464.7µA. The maximum current will be obtained when tox is smallerthen ‘nominal’ value (as the larger Cox will boost the current), when Qox is larger (lower VTH and so largeroverdrive), and when NA is smaller (again, lower threshold, higher overdrive). Clearly, the lower current will beobtained under the opposite condition. Quantitatively, the larger ID, of 538.8 µA, corresponds to tox = 9.5nm, Qox = 1.05 × 1010 cm−2 and NA = 2.66 × 1017 cm−3. The minimum current of 421.3 µAcorresponds to tox = 10.5 nm, Qox = 0.95 × 1010 cm−2 and NA = 2.94 × 1017 cm−3.

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