ECE-­‐606 1)
Purdue University SOLUTIONS: ECE 606 Homework Week 12 Mark Lundstrom Purdue University (revised 4/18/13) The energy band diagram for an MOS capacitor is sketched below. Assume T = 300K and an oxide thickness of x0 = 1.1 nm. Answer the following questions using the delta-­‐
depletion approximation as needed. (Note that E F = Ei at the oxide-­‐silicon interface.) (This problem is similar to prob. 16.7, Pierret SDF). 1a) Sketch the electrostatic potential vs. position inside the semiconductor. Solution: ECE-­‐606 1 Spring 2013 ECE-­‐606 Purdue University 1b) Roughly sketch the electric field vs. position inside the oxide and semiconductor. Solution: Note: We assume that there is no charge in the oxide and no charge at the oxide-­‐Si interface. 1c) Do equilibrium conditions apply inside the semiconductor? Explain Solution: YES. The oxide insures that no current flows, so the metal and semiconductor are two separate systems in equilibrium with possibly different Fermi levels. (Note: We assume that light is not shining on the semiconductor.) 1d) Roughly sketch the hole concentration vs. position inside the semiconductor. ECE-­‐606 Solution: 2 Spring 2013 ECE-­‐606 Purdue University HW Week 12 continued 1e) What is the hole concentration in the bulk? Solution: p ( x → ∞ ) = ni e( Ei −EF ) kBT = 1010 e0.51/0.026 = 3.3 × 1018 cm-­‐3 p ( x → ∞ ) = 3.3 × 1018 cm -3 1f) What is the hole concentration at the surface? Solution: p ( x = 0 ) = ni e( Ei −EF ) kBT = 1010 e0 = 1 × 1010 p ( x = 0 ) = 1010 cm -3 1g) What is the surface potential? Solution: φS = φ ( x = 0 ) − φ ( x → ∞ ) = 0.51 V ECE-­‐606 3 Spring 2013 ECE-­‐606 Purdue University HW Week 12 continued 1h) What is the gate voltage? Solution: The Fermi level in the metal aligns with the Fermi level in the semiconductor, so the gate voltage must be zero. (Note: The fact that there is a volt drop across the oxide and the semiconductor with VG = 0 indicates that there is a workfunction difference between the metal and the semiconductor.) 1i) What is the voltage drop across the oxide? Solution: The electric field at the surface of the semiconductor is given by eqn. (16.27) in SDF as: 2qN AφS
κ Sε0
ES =
We find the electric field in the oxide from: κ oxE ox = κ SE S so E ox =
κS
κ ox
2qN AφS
κ Sε0
The volt drop across the oxide is: Δφox = x0E ox = x0
κS
κ ox
2qN AφS
κ Sε0
alternatively we can write this as: Δφox =
Q (φ )
x0
2qκ S ε 0 N AφS = − D S κ ox ε 0
Cox
where QD (φS ) is the depletion charge in C/cm2 in the semiconductor and Cox is the oxide capacitance in F/cm2 ECE-­‐606 4 Spring 2013 ECE-­‐606 Purdue University HW Week 12 continued Putting in numbers: Cox =
κ ox ε 0 3.9 × 8.854 × 10−14
=
= 3.6 × 10−6 F/cm2 x0
1.1× 10−7
We found the doping density in 1e) and the surface potential in 1g), so QD = − 2qκ S ε 0 N AφS = − 2 × 1.6 × 10−19 × 11.8 × 8.854 × 10−14 × 3.3× 1018 × 0.51 QD = −7.5 × 10−7 C/cm2 Δφox = −
QD (φS )
Cox
=
7.5 × 10−7
= 0.21 3.6 × 10−6
Δφox = 0.21 V 2)
Assume an MOS capacitor on a p-­‐type Si substrate with the following parameters: NA = 2.7 x 1018 cm-­‐3 for the bulk doping oxide thickness: x0 = 1.1 nm QF = 0.0 (no charge at the oxide-­‐Si interface) κ ox = 4 T = 300K VG = 1.0V Also assume that the structure is ideal with no metal-­‐semiconductor workfunction difference. Determine the following quantities by analytical calculations (assume VG = 1.0V). You should use the delta-­‐depletion approximation for these calculations. 2a) The flatband voltage, VFB Solution: VFB = 0 because there is no workfunction difference and no charge at the interface. 2b) The surface potential, φS Solution: VG = φS + Δφox = φS −
ECE-­‐606 QS (φS )
Cox
= φS +
5 2qκ S ε 0 N A
Cox
φS Spring 2013 ECE-­‐606 Purdue University HW Week 12 continued VG = φS + β φS β=
2qκ S ε 0 N A
Cox
= 0.295 φS + β φS − VG = 0 is a quadratic equation for φS φS =
− β ± β 2 + 4VG
2
(take positive sign) Putting in numbers, we find: φS = 0.74 V Is this greater than 2φ F ? ⎛ 2.7 × 1018 ⎞
k BT ⎛ N A ⎞
φF =
ln ⎜
⎟ = 0.026 × ln ⎜⎝ 1010 ⎟⎠ = 0.505 q
⎝ ni ⎠
2φ F = 1.01 V No, so our use of the depletion approximation for QS (φS ) in the first equation is justified, and φS = 0.74 V 2c) The electric field in the oxide, E OX Solution: Since there is no metal-­‐semiconductor workfunction difference, the voltage on the gate is just 1V (no built-­‐in voltage to worry about) and the voltage at the oxide-­‐Si interface is 0.77V, so E ox =
VG − φS
1− 0.74
=
= 2.4 × 106 V cm −7
x0
1.1× 10
Alternatively, we could do it another way that would work even when there is a metal-­‐semiconductor workfunction difference. Gauss’s Law gives: κ ox ε 0E ox = −QS = −QB = 2qκ S ε 0 N AφS ECE-­‐606 6 Spring 2013 ECE-­‐606 Purdue University HW Week 12 continued QB = − 2qκ S ε 0 N AφS = − 2 × 1.6 × 10−19 × 11.8 × 8.854 × 10−14 × 2.7 × 1018 × 0.74 QB = −8.17 × 10−7 C/cm2 E ox = −
QB
8.17 × 10−7
=
= 2.35 × 106 κ ox ε 0 4 × 8.854 × 10−14
E ox = 2.3× 106 V/cm
(slight round-­‐off error gives different answers)
2d) The electric field in the silicon at the surface, E S Solution: At the oxide-­‐Si interface, we have: κ ox ε 0E ox = κ Siε 0E S ES =
κ ox
3.9
E ox =
2.4 × 106 = 7.9 × 105 κ Si
11.8
(
)
E S = 7.9 × 105 V/cm 2e) The depletion region depth, WD Solution: WD =
2κ S ε 0
2 × 11.8 × 8.854 × 10−14
φS =
× 0.77 qN A
1.6 × 10−19 × 2.7 × 1018
WD = 1.93× 10−6 cm WD = 19.3 nm 2f) The charge in the silicon, QS in C/cm2 Solution: From the solution to 2c): QS = −8.34 × 10−7 C/cm2 QS = −8.34 × 10−7 C/cm 2
HW Week 12 continued ECE-­‐606 7 Spring 2013 ECE-­‐606 Purdue University HW Week 12 continued 2g) The charge on the gate, QG in C/cm2 Solution: Charge balance dictates that it must be equal and opposite to the charge in the semiconductor (there is no charge at the oxode-­‐Si interface). QG = −QS = +8.34 × 10−7 C/cm 2 2h) The voltage drop across the oxide Solution: VG = Δφox + φS Δφox = VG − φS = 1− 0.77 = 0.23 V Δφox = 0.23 V 2i) The threshold voltage for this MOS capacitor Solution: VT = 2φ F −
QD ( 2φ F )
Cox
2qκ S ε 0 N A (1.01)
= 1.01+
Cox
VT = 1.01+ 0.304 = 1.314 V VT = 1.314 V 3) The gate electrode of an MOS capacitor is often a heavily doped layer of polycrystalline silicon with the Fermi level located at E F ≈ EC for an n+ polysilicon gate and E F ≈ EV for a p+ polysilicon gate. Sketch the following four equilibrium energy band diagrams: 3a) An n-­‐type Si substrate with an n+ polysilicon gate Solution: (separated) ECE-­‐606 8 Spring 2013 ECE-­‐606 Purdue University HW Week 12 continued A small electron transfer will occur from the gate to the semiconductor Note: There is a little depletion in the polysilicon gate. 3b) An n-­‐type Si substrate with a p+ polysilicon gate Solution: (separated) ECE-­‐606 9 Spring 2013 ECE-­‐606 Purdue University HW Week 12 continued Electrons will transfer from the semiconductor to the gate. Note: We again see some depletion in the polysilicon gate. 3c) A p-­‐type Si substrate with an n+ polysilicon gate Solution: (separated) ECE-­‐606 10 Spring 2013 ECE-­‐606 Purdue University HW Week 12 continued Electrons will transfer from the gate to the semiconductor. Note: The polysilicon gate shows some depletion here too. 3d) A p-­‐type Si substrate with a p+ polysilicon gate Solution: (separated) ECE-­‐606 11 Spring 2013 ECE-­‐606 Purdue University HW Week 12 continued A small electron transfer will occur from the semiconductor to the gate. Note: Polysilicon gate also depleted a little. 4) Consider an MOS capacitor with a gate oxide 1.2 nm thick. The Si substrate doping is N A = 1018 cm-­‐3. The gate voltage is selected so that the sheet density of electrons in the inversion layer is nS = 1013 cm-­‐2. Assume room temperature and that Qi = qnS ≈ 2κ Si ε 0 kBT ni2 N A e+ qφs /2 kBT . Answer the following questions. ECE-­‐606 12 Spring 2013 ECE-­‐606 Purdue University HW Week 12 Continued 4a) What is the surface potential? Compare it with 2φ F . Solution: qnS
e+ qψ s /2 kBT =
2κ Si ε 0 k BT ni2 N A φS =
⎞
2k BT ⎛
qnS
ln ⎜
⎟ 2
q
⎝ 2κ Si ε 0 k BT ni N A ⎠
Putting in numbers: φS = 1.11 V Now compare this to 2φ F : ⎛ 1018 ⎞
2φ F = 2 × 0.026ln ⎜ 10 ⎟ = 0.958 ⎝ 10 ⎠
φS − 2φ F = 1.11− 0.96 = 0.15 φS − 2φ F = 5.8 ( k BT q ) The actual surface potential under strong inversion is about 6 kBT/q bigger than 2φ F . 4b) How much does the surface potential need to increase to double the inversion layer density? Solution: From: φS =
⎞
2k BT ⎛
qnS
13
ln ⎜
⎟ with nS = 2 × 10 , we find 2
q
⎝ 2κ Si ε 0 k BT ni N A ⎠
φS = 1.14 so ΔφS = 0.03 V ECE-­‐606 13 Spring 2013 ECE-­‐606 Purdue University HW Week 12 continued 4c) How much does the gate voltage need to increase to double the inversion layer density? Solution: ΔVG = ΔφS + Δφox Δφox ≈ −
Cox =
ΔQi
Cox
κ ox ε 0
= 2.88 × 10 −6 xo
Δφox ≈ −
ΔQi
q × 1013
=
= 0.56 V Cox 2.88 × 10 −6
ΔVG = ΔφS + Δφox = 0.03 + 0.56 = 0.59 V ΔVG = 0.59 4d) Explain in words why it is difficult to increase the surface potential for an MOS capacitor above threshold. Solution: It takes only a small amount of additional band bending in the semiconductor to double the inversion layer charge because the inversion layer charge increases exponentially with surface potential. The increased inversion layer charge leads to a large increase in the electric field in the oxide, which increases the volt drop in the oxide and as a result, the gate voltage. So above threshold, it take a large increase in the gate voltage to increase the surface potential a little bit, because most of the increase in gate voltage is lost in the volt drop across the oxide. 5) The body effect coefficient, m = (1+ CS Cox ) , is an important MOS parameter. Typical values are said to be 1 < m < 1.4 , but since m varies with gate bias, we should ask what bias these typical numbers refer to. Consider an MOS capacitor with a gate oxide 1.2 nm thick. The Si substrate doping is N A = 1018 cm-­‐3 and answer the following questions. (Assume room temperature conditions). ECE-­‐606 14 Spring 2013 ECE-­‐606 Purdue University HW Week 12 continued 5a) Compute m for below threshold conditions. Assume that the depletion layer width is W (φS ) = W ( 2φ F ) = WT . Solution: WT =
2κ S ε 0
2φ F qN A
⎛ 1018 ⎞
2φ F = 2 × 0.026ln ⎜ 10 ⎟ = 0.958 ⎝ 10 ⎠
WT =
2κ S ε 0
2φ F = 3.54 × 10−6 cm qN A
CS = C D =
Cox =
κ Si ε 0
= 2.95 × 10 −7 WT
κ ox ε 0
= 2.88 × 10 −6 xo
m = (1+ CS Cox ) = (1+ 0.295 2.88) = 1.10 m = 1.10 below threshold 5b) Compute m for above threshold, strong inversion conditions. Assume that the sheet density of electrons in the inversion layer is nS = 1013 cm-­‐2 and that Qi = qnS ≈ 2κ Si ε 0 kBT ni2 N A e+ qφs /2 kBT . Solution: 2κ Si ε 0 kBT ni2 N A e+ qφs /2 kBT
Qi
dQS
dQi
CS = −
≈−
=
=
dφs
dφs
2kBT / q
2kBT q Qi
1.6 × 10 −19 × 1013
CS =
=
= 30.8 × 10 −6 2kBT q
0.052
m = (1+ CS Cox ) = (1+ 30.8 2.88) = 11.7
m = 11.7 above threshold 5c) Use the results from 5a) and 5b) to explain why the surface potential is easy to change with a gate voltage below threshold but hard to change above threshold. ECE-­‐606 15 Spring 2013 ECE-­‐606 Purdue University HW Week 12 continued Solution: We have two capacitors in series, an oxide capacitance and a semiconductor capacitance. If a gate voltage is applied to the top (oxide capacitance), then the voltage that gets to the semiconductor (the surface potential) is the voltage dropped across the semiconductor capacitance: V
m = (1+ CS Cox ) φS = G m
Below threshold, CS is small, m ≈ 1 , so most of the gate voltage is dropped across the semiconductor. Above threshold, CS is large, m >> 1 , so most of the gate voltage is dropped across the oxide – very little across the semiconductor, so it is hard to change the band bending in the semiconductor. ECE-­‐606 16 Spring 2013