T=300K# x 2.2 4 O 1.2 100 µm DD S D - nanoHUBWeek13HW... · ECE305# # Spring2015# ECE/305# # 3#...

ECE 305 Spring 2015

ECE-‐305 Spring 2015 1

ECE 305 Homework SOLUTIONS: Week 13

Mark Lundstrom Purdue University

1) Measured IV data for an n-‐channel MOSFET are shown below. Relevant parameters

for this MOSFET are:

T = 300 K Oxide thickness: x0 = 2.2 nm Relative dielectric constant: KO = 4 Power supply voltage: VDD = 1.2 V Series resistances: RS = RD = 100 Ω− µm Actual channel length = 85 nm Assume that the MOSFET is 1 micrometer wide, W = 10−4 cm .

ECE 305 Spring 2015


HW Week 13 Solutions (continued)

Answer the following questions: 1a) From the so-‐called “on-‐current”, ID VGS =VDD ,VDS =VDD( ) = 1.2 V , estimate the

average velocity of electrons in the channel at the source end of the channel. (Note: You will need to correct for the effect of the series resistance as discussed in HW12.)

Solution:

Qn(0) = Cox VGS − ID RS −VT( )

Cox =

KOεo

xo

= 1.6×10−6 F cm2

Qn(0) = 1.6×10−6( ) 1.2−1120×10−6 ×100− 0.26( ) = 1.32×10−6

Qn(0) q 8.3×1012 cm-2

υ(0) =

ION

WQn(0)= 1120×10−6

10−4 ×1.32×10−6 = 8.5×106 cm/s

υ(0) = 8.5×106 cm/s

1b) From the linear region of operation, estimate the effective mobility of this

MOSFET.

Solution: At the highest gate voltage:

RTOT = RCH + RS + RD = 340Ω− µm

RCH = RTOT − RS + RD = 340− 200( )Ω− µm RCH = 140Ω− µm Because the width is given as 1 micrometer, RCH = 140 Ω− µm .

Assume very small VDS with a small ID , so the intrinsic gate voltage is about the applied gate voltage.

υ(0) =

ION

WQn(0)

ID =

WµnCox

L′VGS −VT( ) ′VDS

ECE 305 Spring 2015



RCH = 1WµnCox

LVGS −VT( )

µn =

LRCHWCox VGS −VT( ) =

LRCHW Qn

Qn = −Cox VGS −VT( ) = −1.6×10−6 1.2− 0.28( ) = 1.47 ×10−6

µn =

LRCHW Qn

= 8.5×10−7

140×10−4 ×1.47 ×10−6 = 412 cm2/V-‐s

µn = 412 cm2 /V-s

2) Assume the square law theory of the MOSFET:

ID =

WµnCox

LVG −VT( )VD −

VD2

2⎡

⎣⎢

⎤

⎦⎥ 0 ≤VD <VDsat VG ≥VT

ID =

WµnCox

2LVG −VT( )2

VD ≥VDsat VG ≥VT

2a) Derive an expression for the electric field vs. position along the channel in the

linear region of operation.

Solution: The drain current is a drift current:

ID y( ) =WµnQn y( )E y y( ) , where in the square law theory:

Qn y( ) = −Cox VG −VT −V y( )⎡⎣ ⎤⎦ . Equate this to the expression for the “linear region” current:

ID y( ) =WµnQn y( )E y y( ) = ID =

WµnCox

LVG −VT( )VD −

VD2

2⎡

⎣⎢

⎤

⎦⎥

E y y( ) = − 1

L

VG −VT( )VD −VD

2

2⎡

⎣⎢

⎤

⎦⎥

VG −VT −V y( )⎡⎣ ⎤⎦

ECE 305 Spring 2015


HW Week 13 Solutions (continued) For very small drain voltages, we can ignore VD

2 2 , and Qn y( ) ≈ −Cox VG −VT( ) , so

E y y( ) ≈ −

VD

LVD <<VG −VT which is just what we would expect for a resistor.

2b) Derive an expression for the electric field vs. position along the channel at

VD =VDsat .

Solution: Begin as in 2a). The drain current is a drift current:

ID y( ) =WµnQn y( )E y y( ) , where in the square law theory:

Qn y( ) = −Cox VG −VT −V y( )⎡⎣ ⎤⎦ Equate this to the expression for the saturation region current:

ID y( ) =WµnQn y( )E y y( ) = ID =

WµnCox

2LVG −VT( )2

VG −VT −V y( )⎡⎣ ⎤⎦E y y( ) = − 1

2LVG −VT( )2

electric field is gradient of voltage, so:

− VG −VT( ) dV

dy+V dV

dy= − 1

2LVG −VT( )2

− VG −VT( )dV

0

V y( )

∫ + V dV0

V y( )

∫ = − 12L

VG −VT( )2dy

0

y

∫

which can be integrated to find:

− VG −VT( )V y( ) + V 2 y( )

2= − 1

2VG −VT( )2 y

L⎛⎝⎜

⎞⎠⎟

or

V 2 y( )− 2 VG −VT( )V y( ) + VG −VT( )2 y

L⎛⎝⎜

⎞⎠⎟= 0

Solve the quadratic equation for V(y):

V y( ) = VG −VT( ) ± VG −VT( ) 1− y / L Choose the minus sign because V 0( ) = 0 and V L( ) =VG −VT (drain end at pinch-‐off).

V y( ) = VG −VT( ) 1− 1− y / L⎡

⎣⎤⎦

ECE 305 Spring 2015



Now we can find the electric field in the channel:

E y( ) = − dV

dy= −

VG −VT( )2L

1− y / L( )−1/2

E y( ) = −

VG −VT( )2L

1− y / L( )−1/2

Examine answer: At y = 0, we get

E 0( ) = −

VG −VT( )2L

As y → L , E y → L( )→−∞ At the pinch-‐off point, the electric field approaches infinity and the inversion layer charge approaches zero. The product of the two is finite. This is a limitation of the simple theory that the velocity saturation model tries to fix.

3) There is often a sheet of “fixed charge” at the oxide-‐semiconductor interface. (By fixed

charge, we mean a charge that does not vary with surface potential, φS .) Assume a 1.5 nm thick oxide with a relative dielectric constant of 4.0. Answer the questions below. 3a) Assume that a fixed charge of QF q = 1011 cm-2 is present. Explain how the fixed

charge shifts the threshold voltage. What is the direction of this shift (e.g. is it a positive or negative voltage shift?).

Solution: The voltage drop across the oxide is changed by an amount:

ΔV = −

QF

Cox

Cox =

KOε0

xo

= 4.0×8.854×10−14

1.5×10−7 = 2.36×10−6

ΔV = −

QF

Cox

= −1.6×10−19 ×1011

2.36×10−6 = −6.8×10−3 V

ΔV = −6.8×10−3 V

This change in voltage drop acriss the oxide shifts VT . The shift is negative. If we are dealing with an N-‐MOSFET, then the threshold voltage will be a little smaller (less positive). If we are dealing with a P-‐MOSFET, the threshold voltage is also a little smaller, which means that the threshold voltage will be a little more negative. Charge at the oxide-‐Si interface reduces the magnitude of VT for N_MOSFETs and increases the magnitude of VT for P-‐MOSFETs.

ECE 305 Spring 2015


HW Week 13 Solutions (continued) Note that this is a fairly large charge by today’s standards. High quality gate oxides can have 10 times less fixed charge. In the late 1970’s, however, oxides were 100 nm thick and the corresponding shifts in voltage were much larger (~0.45 V) for this QF . (Do you see why?)

3b) In HW 1, we saw that the number of atoms per cm2 on a (100) Si surface is

NS = 6.81×1014 cm-2 . Assume each atom has one charged dangling bond, so a

fixed charge of QF q = 6.81×1014 cm-2 is present. How much is the voltage shift now?

Solution: The voltage drop across the oxide is changed by an amount:

ΔV = −

QF

Cox

= 1.6×10−19 × 6.81×1014

2.36×10−6 = −46 V

ΔV = −46 V This is an enormous shift in voltage. MOSFET technology would not be possible without the remarkable ability of SiO2 to passivate (e.g. neutralize) more than 99.9% of the dangling bonds.

4) The gate electrode of an MOS capacitor is often a heavily doped layer of

polycrystalline silicon with the Fermi level located at EF ≈ EC for an n+ polysilicon gate and EF ≈ EV for a p+ polysilicon gate. These heavily doped semiconductors act almost like metals. Sketch the following four equilibrium energy band diagrams:

4a) An n-‐type Si substrate ( N D = 1017 cm-3 ) with an n+ polysilicon gate. Solution: (first separated, then together)

ECE 305 Spring 2015


HW Week 13 Solutions (continued) A small electron transfer will occur from the gate to the semiconductor. The result is:

Note: The semiconductor is accumulated. There is a little depletion of electrons in the polysilicon gate. This does not happen for a metal because there are so many 4b) An n-‐type Si substrate ( N D = 1017 cm-3 ) with a p+ polysilicon gate.

Solution: (first separated, then together)

Electrons will transfer from the semiconductor to the gate.

ECE 305 Spring 2015



Note: The semiconductor is depleted. We again see some depletion in the polysilicon gate.

4c) A p-‐type Si substrate ( N A = 1017 cm-3 ) with an n+ polysilicon gate.


Electrons will transfer from the gate to the semiconductor.

ECE 305 Spring 2015


HW Week 13 Solutions (continued) Note: The semiconductor is depleted. The polysilicon gate shows some depletion here too.

4d) A p-‐type Si substrate ( N A = 1017 cm-3 ) with a p+ polysilicon gate.


A small electron transfer will occur from the semiconductor to the gate.

Note: The semiconductor is accumulated. The polysilicon gate also depleted a little.

5) Compute the “metal”-‐semiconductor workfunction difference for each of the cases

above.

5a) An n-‐type Si substrate with an n+ polysilicon gate.

Solution: We must find the location of the Fermi level in the semiconductor. Begin with

n0 = N D = NCe EFS−EC( ) kBT , which can be solved for

ECE 305 Spring 2015

ECE-‐305 Spring 2015 10


EC − EFS( )q

=kBTq

lnNC

N D

⎛⎝⎜

⎞⎠⎟ NC = 3.23×1019 cm-3

EC − EFS( )q

=kBTq

lnNC

N D

⎛

⎝⎜⎞

⎠⎟= 0.026ln 323( ) = 0.150 eV

ΦMS = χ − χ + EC − EFS( ){ } = − EC − EFS( ) = −0.150 eV

ΦMS = −0.150 eV

The work function difference is small, as can be seen in the energy band diagram of 4a).

5b) An n-‐type Si substrate with a p+ polysilicon gate.

Solution

We have the same Si substrate, so

EC − EFS( )q

= 0.150 eV

ΦMS = χ + EG − χ + EC − EFS( ){ } = 1.12 − 0.15 eV =0.97 eV

ΦMS = 0.97 eV

The work function difference is close to the bandgap of Si, as can be seen in the energy band diagram of 4b). Note: Since the left contact is a p-‐type semiconductor and the right contact is an n-‐type semiconductor, we can show that the magnitude of the built-‐in potential is given by the PN junction equation:

ΦMS q =

kBTq

lnN AN D

ni2

⎛

⎝⎜⎞

⎠⎟

ECE 305 Spring 2015

ECE-‐305 Spring 2015 11


5c) A p-‐type Si substrate with an n+ polysilicon gate.

Solution:

p0 = N A = NV e EV −EFS( ) kBT , which can be solved for

EFS − EV( )q

=kBTq

lnNV

N A

⎛⎝⎜

⎞⎠⎟ NV = 1.83×1019 cm-3

EFS − EV( )q

=kBTq

lnNV

N A

⎛

⎝⎜⎞

⎠⎟= 0.026ln 183( ) = 0.135 eV

ΦMS = χ − χ + EG − EFS − EV( ){ } = −EG + EFS − EV( ) = −0.985 eV

ΦMS = −0.985 eV

The work function difference is very large in magnitude, as can be seen I the energy band diagram of 4c).

5d) A p-‐type Si substrate with a p+ polysilicon gate.

Solution: The same p-‐type substrate, so:

EFS − EV( )q

= 0.135 eV

ΦMS = χ + EG − χ + EG − EFS − EV( ){ } = EFS − EV( ) = 0.135 eV

ΦMS = 0.135 eV

The work function difference is very small in magnitude, as can be seen in the energy band diagram of 4d).

T=300K# x 2.2 4 O 1.2 100 µm DD S D - nanoHUBWeek13HW... · ECE305# # Spring2015# ECE/305# # 3#...

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