Homework 5 1. Problem. Derive Eq. (171) of the Lecture Notes,Part 2,expressing the density of a system of electrons confined 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): n 2D = 2 (2π) 2 dkf FD (k) = 2 2π (2π) 2 dφ 0 0 1 dkk 1+exp[E(k) E F )/(k B T )]. (1) Performing the trivial angular integration and changing integration variable from k to E = h 2 k 2 /(2m ) we get: n 2D = m de π h 2 0 1+exp[(E(k) E F )/(k B T )]. (2) Now let s multiply numerator and denominator of the integrand by exp[ (E(k) E F )/(k B T )],and setting y =(E E F )/(k B T ): n 2D = m k B T π h 2 dy e y, (3) y 0 1+e y where y 0 = E F /(k B T ). The integral is just ln(1 + e y ) y = ln(1 + e y 0),so that: 0 n 2D = m k B T π h 2 ln[1 + exp[e F /(k B T )]. (4) 2. Problem. Consider a MOSFET fabricated 20 years ago with the following parameters: Gate oxide thickness t ox = 200 nm Substrate doping concentration (p-type) N A = 1.7 10 15 cm 3 ECE609 Spring 2010 1
Charge in the oxide Q ox = 10 10 charges/cm 2 Intrisic carrier concentration in Si at 300 K n i = 1.45 10 10 cm 3. a. Calculate the threshold voltage. b. What happens to the threshold voltage if the oxide gets contaminated during fabrication so that Q ox increases by a factor of 5? Consider now a MOSFET fabricated 5 years ago with the following parameters: Gate oxide thickness t ox =20nm Substrate doping concentration (p-type) N A = 8.7 10 16 cm 3 Charge in the oxide Q ox = 10 10 charges/cm 2 Intrisic carrier concentration in Si at 300 K n i = 1.45 10 10 cm 3. c. Calculate the threshold voltage. d. What happens to the threshold voltage if the oxide gets contaminated during fabrication so that Q ox 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 Q ox =10 10 cm 2 the threshold voltages are: V TH,old = 0.723 V and V TH,new = 0.723 V. (5) When Q ox =5 10 10 cm 2 the threshold voltages are: V TH,old = 0.352 V and V TH,new = 0.686 V. (6) The new devices is less affected by oxide charges thanks to the larger C ox. Note: The equation [ ( ) NA,sub k B T ln + ln Φ ms n i ( ND,gate n i )], (7) ECE609 Spring 2010 2
which is valid only in the non-degenerate limit (observation we have already made in Homework 1!),which is clearly not the case for the heavily-doped gate. Instead,one should assume that the Fermi level in the gate is approximately lined-up with the bottom of the conduction band. Therefore the term k B T ln(n D,gate /n i ) should be replaced by E Gap /(2e). ECE609 Spring 2010 3
3. Problem. Consider a MOSFET with the following parameters: Gate oxide thickness t ox = 20 nm Substrate doping concentration (p-type) N A = 1.0 10 17 cm 3 Charge in the oxide Q ox = 0 charges/cm 2 Intrisic carrier concentration in Si at 300 K n i = 1.45 10 10 cm 3 Channel width W =2µm and channel length L =1µm Electron mobility µ n = 600 cm 2 /Vs Zero substrate bias. On a single graph plot the drain current I D as a function of the drain voltage V D with V D ranging from 1.0 to 5.0 V. Use the following equations: a. The complete model,eq. (13),page 174 of the Lecture Notes b. 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 as I had erroneously typed. Using such a large (unrealistic) thickness with the large substrate doping would have resulted in a huge,unrealistic threshold voltage,as you found. So,asking for the drain current at a gate voltage so much below threshold made no sense. If you have correctly realized this problem,you have been made aware very directly of the importance of scaling correctly the thickness of the gate oxide. Here is the solution for the case t ox = 20 nm. Considering just the case of V G = V D = 5 V,one should get: I D,sat = 1.295 10 3 A (full model) 1.156 10 3 A (simplified model) 1.832 10 3 A (simplified,n=1). (8) See the figure for the full I V characteristics. ECE609 Spring 2010 4
2.0 1.5 Full model Simplified model Simplified model with n=1 I DS (ma) 1.0 0.5 0.0 0 1 2 3 4 5 V DS (V) 4. Problem. The threshold voltage of an n-channel MOSFET having degenerately doped n + polysilicon gate is 0.7 V. What would the threshold be if the gate material were degenerately doped p + polysilicon,all other fabrication 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 V TH increases by about E gap /e to 1.8 V. This quantitative estimate should have been given. Stating quite generally that V TH increases is a little too obvious and too rough. ECE609 Spring 2010 5
5. Problem. Consider a MOSFET with the following parameters: Gate oxide thickness t ox =10nm W = L =1µm Substrate doping concentration (p-type) N A = 2.8 10 17 cm 3 Charge in the oxide Q ox = 10 10 charges/cm 2 Intrisic carrier concentration in Si at 300 K n i = 1.45 10 10 cm 3 Channel width W =2µm and channel length L =1µm Electron mobility µ n = 600 cm 2 /Vs V G = V D =3.3V Zero substrate bias. a. Calculate the drain current b. Because of unavoidable fabrication parameter fluctuations, t ox, N A,and Q ox can vary by ± 5%. Each of these parameters is independent on the others. The increase of some of these parameters will result in a larger 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 the simplified model,eq. (22). Solution. The nominal current is I D 464.7µA. The maximum current will be obtained when t ox is smaller then nominal value (as the larger C ox will boost the current),when Q ox is larger (lower V TH and so larger overdrive),and when N A is smaller (again,lower threshold,higher overdrive). Clearly,the lower current will be obtained under the opposite condition. Quantitatively,the larger I D,of 538.8 µa,corresponds to t ox =9.5 nm, Q ox = 1.05 10 10 cm 2 and N A = 2.66 10 17 cm 3. The minimum current of 421.3 µa corresponds to t ox =10.5 nm, Q ox =0.95 10 10 cm 2 and N A =2.94 10 17 cm 3. ECE609 Spring 2010 6