Test 5 solution 3610 F2020

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Test 5-EECS 3610 Semiconductor Physics and Devices-Fall 2020 Professor Gerd Grau Total: 24 points: Q1 (14 points) All questions in this test refer to the same MOSCAP device with the following parameters. The silicon is uniformly doped with a concentration of 4*10 16 cm -3 phosphorus dopants. The oxide capacitance is 312 nF/cm 2 . The work function of the metal gate is 0.6 V larger than the work function of the silicon. Calculate the threshold voltage in volts. Sketch the low-frequency capacitance-voltage curve for this MOSCAP. Make sure to label any important points but no need to perform any further calculations. Q2 (4 points) MOSCAPs are fabricated according to the specifications in question 1. After fabrication, the actual threshold voltage is measured to be -0.35V. Assuming that this shift in threshold voltage is due to charges at the dielectric-semiconductor interface, calculate the areal density of these charges in nC/cm 2 . Q3 (6 points) After further investigation, you realize that the charge at the dielectric-semiconductor interface is not due to fixed charge, but rather due to interface traps. Each trap can be occupied by one electron. Traps are neutrally charged when they are empty and charged with the charge of one electron when they are occupied. Assume that trap states are uniformly distributed over the energy of the bandgap. Calculate the number of traps per energy interval in the bandgap (in eV) and per unit area of the gate (in cm 2 ). You can approximate the Fermi function as: f(E<E F ) = 1 f(E>E F ) = 0 Enter your answer as a number with three significant figures multiplied by 10 10 eV -1 cm -2 , e.g. if the correct answer is 1.23*10 11 eV -1 cm -2 , enter 12.3 as the answer. Solution Q1 PMOS: 𝑉 𝑡ℎ = 𝑉 ?? − 2𝜙 ? − √2𝑞𝑁 ? 𝜀 𝑆𝑖 2𝜙 ? ? 𝑜𝑥

𝜙 ? = 𝑘𝑇 𝑞 ln 𝑁 𝐷 ? 𝑖 = 0.026 ∗ ln 4∗10 16 10 10 = 0.395𝑉 𝑉 𝑡ℎ = 0.6 − 2 ∗ 0.395 − √2∗1.602∗10 −19 ∗4∗10 16 ∗11.9∗8.85∗10 −14 ∗2∗0.395 3.12∗10 −7 = −0.521𝑉 Standard MOSCAP curve: General shape Correct orientation (PMOS vs NMOS) Curve goes up to Cox Vth and Vfb labelled Q2 Δ𝑉 𝑡ℎ = − 𝑄 𝑜𝑥 ? 𝑜𝑥 𝑄 ?𝑥 = −(𝑉 𝑡ℎ,𝑓?? − 𝑉 𝑡ℎ,0 ) ∗ 𝐶 ?𝑥 = −(−0.35 − (−0.521)) ∗ 3.12 ∗ 10 −7 = −53.4?𝐶/𝑐? 2 Q3 Q ox is due to electrons trapped in trap states at energies below E F above E V . Define density of trap states as N t . 𝑄 ?𝑥 = −𝑞 ∗ 𝑁 𝑡 ∗ (𝐸 ? − 𝐸 𝑣 ) At threshold (PMOS): 𝐸 ? − 𝐸 𝑣 = ? 𝑔 2 − 𝜙 ? = 1.1 2 − 0.395 = 0.155𝑒𝑉 𝑁 𝑡 = − 𝑄 𝑜𝑥 𝑞∗(? 𝐹 −? 𝑉 ) = 53.4∗10 −9 1.602∗10 −19 ∗0.155 = 2.15 ∗ 10 12 𝑐? −2 𝑒𝑉 −1

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