Mark Lundstrom 10/11/2012 Homework: Week 2 Solutions Nanoscale Transistors Mark Lundstrom, nanoHUB-­‐U Fall 2012 There are four HW questions below. Homework is graded in multiple choice format – no partial credit. Doing the homework before the solutions and tutorials are posted is a great way to test your understanding of the material covered. 1) Assume a silicon MOS capacitor at room temperature biased in depletion with ψ S = 0.3 V and N A = 5 × 1017 cm-­‐3, and compute the charge, QD , in the depletion region in C/cm2. The depletion charge in C/cm2 is closest to: a) -­‐0.1 microCoulombs/cm2 b) -­‐0.2 microCoulombs/cm2 c) -­‐1.0 microCoulombs/cm2 d) -­‐2.0 microCoulombs/cm2 e) -­‐10 microCoulombs/cm2 In addition, but not a part of the assignment, knowing QD , you should also be able to easily obtain the width of the depletion region and the electric field at the surface of the silicon. You should also be able to deduce the electron concentration in the bulk and the electron concentration at the surface. (Assume ni = 1010 cm-­‐3.) Problem 1) Solution: QS = QD = −qN AWD = − 2qN Aε Siψ S QD = − 2 × 1.6 × 10 −19 × ( 5 × 1017 ) × (11.7 × 8.854 × 10 −14 ) 0.3 = − 0.22 × 10 −6 C/cm 2
The additional (not required) questions are solved as follows. QD
2.2 × 10 −7
WD =
=
= 28 nm −qN A 1.6 × 10 −19 × 5 × 1017
−QD
ε SiE S = −QD → E S =
= 2.1× 10 5 V/cm ε Si
n0 ( bulk ) =
ni2
= 2 × 10 2 cm-­‐3 NA
n0 ( surface ) = n0 ( bulk ) eqψ S
kBTL
= 200 × e0.3 0.026 = 2.1× 10 7 cm-­‐3 Nanoscale Transistors 1 nanoHUB-­‐U Fall 2012 Mark Lundstrom 10/11/2012 Homework: Week 2 Nanoscale Transistors (continued) Assume a silicon MOS capacitor at room temperature biased in depletion with ψ S = 0.3 V and N A = 5 × 1017 cm-­‐3 (i.e. problem 1). Assume that the gate oxide thickness is t ox = 2 nm with κ ox = 4 and compute the gate voltage. (You may assume that the metal semiconductor workfunction difference is zero.) 2) Note: A typical metal-­‐semiconductor workfunction difference is -­‐1 eV, which would shift your answer by -­‐1V. a) 0.2 V b) 0.4 V c) 0.6 V d) 0.8V e) 1.0 V Problem 2) Solution: Q (ψ )
VG = − D S + ψ S Cox
Cox =
κ ox ε 0 4 × 8.854 × 10 −14
=
= 1.77 × 10 −6 F/cm2 −7
t ox
2 × 10
VG =
0.22 × 10 −6
+ 0.3 = 0.42 1.77 × 10 −6
3a) This problem uses the Matlab (Octave) script attached at the end of this assignment. It is preloaded on the nanoHUB tool, OCTAViEw, which you can access from the class home page. Alternatively, you if you have direct access to Matlab or Octave, you can run the script attached at the end of the assignment. Edit the script so that it describes a MOSFET with channel doping of N A = 1018 cm-­‐3 and an oxide thickness of t ox = 1.2 nm. Assume a metal-­‐semiconductor workfunction difference of -­‐1 V and no charge at the oxide-­‐Si interface. Also assume TL = 300 K and ni = 1010 cm-­‐3. We often assume that above threshold, the surface potential is equal to 2ψ B . Is this true? Run the script and determine the surface potential under on-­‐state conditions when Qn q = 1013 cm-­‐2. You may assume that well above threshold, QS ≈ Qn . How does the actual surface potential at Qn q = 1013 cm-­‐2 compare to 2ψ B ? Nanoscale Transistors 2 nanoHUB-­‐U Fall 2012 Mark Lundstrom 10/11/2012 a) ψ S ≈ 1.7ψ B V b) ψ S ≈ 2.3ψ B V c) ψ S ≈ 2.6ψ B V d) ψ S ≈ 2.9ψ B V e) ψ S ≈ 3.2ψ B V HINTS for using OCTAViEw: You can change the vertical and horizontal axes on the plots by clicking in them. The min and max can be set, and either log or linear can be selected. Select the square box on the right hand side to restore the initial settings. Place the pointer over a line on the plot to read (x, y). Problem 3a) solution: Select the second plot (|Qs/q| vs. psi_S) and read off the surface potential for |Qs/q| = 1e13 as shown below: Note that for the given doping, ψ B = 0.48 V; the surface potential greater than this by 0.3ψ B = 0.14 V, which is 5.5 in units of kBTL q . Nanoscale Transistors 3 nanoHUB-­‐U Fall 2012 Mark Lundstrom 10/11/2012 3b) We often assume that the surface potential is constant above threshold. Run the same script as in prob. 3a) and estimate dψ S dVG well above threshold. Use this estimate to estimate the semiconductor capacitance well above threshold. a) CS ≈ 0.7Cox b) CS ≈ 1.4Cox c) CS ≈ 7Cox d) CS ≈ 19Cox e) CS ≈ 68Cox Problem 3b) solution: Use the first plot (psi/psi_B vs. V_G) and determine the slope, d (ψ S ψ B ) dVG in the region near V_G = 1.0 (on-­‐current conditions). As shown in the screen shot below, we can read off two points from the curve and get: psi/psi_B = 2.37 V_G = 1.11 psi/psi_B = 2.35 V_G = 0.93 d (ψ S ψ B ) Δ (ψ S ψ B ) ⎛ 2.37 − 2.35 ⎞ 0.02
≈
=⎜
=
= 0.11 ⎝ 1.11− 0.93 ⎟⎠ 0.18
dVG
ΔVG
so d (ψ S ψ B )
dψ S
≈ 0.11× 0.48 = 0.05 ≈ 0.11 ψ B = 0.48 dVG
dVG
Nanoscale Transistors 4 nanoHUB-­‐U Fall 2012 Mark Lundstrom 10/11/2012 From the two-­‐capacitors in series model discussed in Week 2, Lecture 2, we have dψ S
Cox
1
=
=
dVG Cox + CS 1+ CS Cox
CS
1
1
=
−1 =
− 1 = 19 Cox dψ S dVG
0.05
4) Assume a MOSFET with the following parameters: S = 81mV/dec DIBL = 81 mV/V Cinv = 1.55 × 10 −6 F/cm2 Assume that CG−VS = Cinv . Compute C D−VS and compare it to CG−VS . This gives a feel for how big C D−VS can be in an electrostatically well-­‐designed (a so-­‐called “well-­‐
tempered”) MOSFET. a) C D−VS ≈ 0.02 × CG−VS b) C D−VS ≈ 0.04 × CG−VS c) C D−VS ≈ 0.04 × CG−VS d) C D−VS ≈ 0.06 × CG−VS e) C D−VS ≈ 0.08 × CG−VS Problem 4) Solution: DIBL = C D−VS CG−VS → C D−VS CG−VS = 0.081 End of homework assignment. This assignment contains 4 questions. Nanoscale Transistors 5 nanoHUB-­‐U Fall 2012 Mark Lundstrom 10/11/2012 Appendix: Matlab (Octave) script for problem 3. % Surface potential vs. Gate voltage for MOS Capacitor
% Date: Oct. 24, 2012
% Authors: Xingshu Sun and Mark S. Lundstrom (Purdue
University)
%
% References
% [1] Mark Lundstrom and Xingshu Sun (2012), "Notes on the
Solution of the Poisson-Boltzmann Equation for MOS
Capacitors and MOSFETs, 2nd Ed."
http://nanohub.org/resources/5338.
%Initialize the range of the surface potential
psi = -0.2:0.01:1.2; %[V]
%Specify the physical constants
epsilon_siO2 = 4*8.854*1e-14; %Permittivity of SiO2 [F/cm]
epsilon_si = 11.68*8.854*1e-14; %Permittivity of Si [F/cm]
k_b = 1.380e-23; %Boltzmann constant [J/K]
q = 1.6e-19; %Elementary charge [C]
%Specify the environmental parameters
TL = 300; %Room temperature[K]
ni = 1e10; %Intrinsic semiconductor carrier density [cm-3]
N_A = 1e17; %Acceptor concentration [cm-3]
tox = 2e-7; %The thickness of SiO2 [cm]
V_FB = 0; %The flat-band voltage [V]
%Calculate the gate oxide capacitance
cox = epsilon_siO2/tox; %[F/cm2]
%Calculate the F function (to calculate the total charge)
with respect to different surface potentials. See [1]
eqn. (10)
F_psi1 = exp(-psi*q/k_b/TL) + psi*q/k_b/TL -1;
F_psi2 = ni^2/N_A^2*(exp(psi*q/k_b/TL) - psi*q/k_b/TL -1);
F_psi = sqrt(F_psi1 + F_psi2);
Nanoscale Transistors 6 nanoHUB-­‐U Fall 2012 Mark Lundstrom 10/11/2012 %Initialize the vectors
psi_length = length(psi);
Qs = zeros(psi_length,1);
V_G = zeros(psi_length,1);
psi_B = k_b*TL/q*log(N_A/ni); %the bulk potential
%Calculate the total charges and the corresponding gate
voltages. See [1] eqn. (12)
for i = 1:psi_length
if psi(i) <= 0
Qs(i) = sqrt(2*epsilon_si*k_b*TL*N_A)*F_psi(i);
%Calculate charge when the surface potential is negative
else
Qs(i) = -sqrt(2*epsilon_si*k_b*TL*N_A)*F_psi(i);
%Calculate charge when the surface potential is positive
end
V_G(i) = V_FB+psi(i)-Qs(i)/cox; %Calculate the gate
voltage
end
%Plot surface potential (psi_S) vs. gate voltage
h1=plot(psi/psi_B,V_G,'r','linewidth',3);
set(gca, 'xlim', [-4 13], 'ylim', [-0.3 1.3]);
set(gca,'fontsize',13);
title('psi_S (Surface potential) vs. Gate voltage');
xlabel('Gate voltage (V)');
ylabel('psi_S (Surface potential / bulk potential) (V)');
%Plot semiconductor charge (Qs) vs. surface potential, psi
h2=plot(abs(Qs/q), psi/psi_B,'r','linewidth',3);
set(gca, 'xlim', [-4 13], 'ylim', [-0.3 1.3]);
set(gca,'fontsize',13);
title('psi_S (Surface potential) vs. Gate voltage');
xlabel('surface potential / bulk potential (V)');
ylabel('abs(Qs/q) (V)');
Nanoscale Transistors 7 nanoHUB-­‐U Fall 2012