Question

4.Today, most of the components (except for gate dielectrics and some contacts) of electronic microcircuits are based on semiconductor materials, including resistors. With further scaling down of dimensions, ultra-thin semiconductor materials can present undesirably high resistance to current flow. You need to design 1 kS2 resistor made as a 10 nm-long and 100-nm wide Si bar which that will be produce by a shallow ion implantation through a photoresist mask. In order to choose the implantation dose and type of ions (Ast or B+), you need to find resistivity of n- and p-type material which would provide the required resistance, if the thickness of the resistor cannot exceed 5 nm. Estimate the dopant concentration for n- and p-type Si that would be needed to achieve this resistivity using the mobility-carrier concentration relationship (Figure 3.5a in the Pierret text book, also available in lecture slides and any other text on electronic devices). Based on your finding, justify which dopant, As or B, would be the better choice

Answer 1

Resistance (R) = 500 Ohm Area = 10 nm times 100 nm = 10^3 times 10^- 18 m^2 length (l) = 10 times 10^- 9 m R = rho l/A = 1/sigma A [sigma = conducting] sigma = l/RA = 10 times 10^- 9/500 times 10^3 times 10^- 8 = 2 times 10^4 Ohm^- 1 m^- 1 = 200 Ohm^- 1 cm^- 1 for A_5 (pentavalent) dopant rightarrow N_p (n - type) sigma = mu_n N_p q (mu = 1300 cm^2/v sec) N_p = sigma/r_n q = 9.6 times 10^17/cm^3 \r\n for B (trivalent) dopant cone rightarrow NA (P - type) sigma = mu_p N_A q (mu_P = 500 cm^2/v - sec) N_A = 2.5 times 10^18/cm^3 Resistivity rho = 1/sigma = 5 times 10^- 3 Ohm cm As it�s better choice because less dopant cone provide more conductivity