a) A silicon wafer is uniformly doped p-type with NA=10¹³/cm³. At T=0K, what are the equilibrium hole and electron concentrations?
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![a) A silicon wafer is uniformly doped p-type with NA=10¹³/cm³. At T=0K, what are the equilibrium
hole and electron concentrations?](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F2e521a1e-f4ee-4c02-af71-8dfc2cc184d3%2F5a67ab64-2ec1-4285-bc8b-e11e799dabbc%2Fb9534ia_processed.png&w=3840&q=75)
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- How is a solid-state diode tested? Explain.Consider an extrinsic semiconductor. Given: /n = 6550 cm² /V sec Mp = 400 cm²/V sec ni = 5.6 x 1012/cm³ Find out the hole and electron concentration in the semiconductor when conductivity is minimum (in cm-³).The applied electric field in p-type silicon is E=10V/cm. The semiconductor conductivity is 1.5(Ω-cm)-1and cross-sectional area is 10-5cm2. Determine the driftcurrent. (b) A drift current density of 120A/cm2is established in n-type silicon with an applied electric field of 18V/cm. If the electron and hole mobilities are µn =1250 cm2/V-s and µp =450 cm2/V-s, respectively, determine the required doping concentration.
- Consider a silicon P- N step junction diode with Nd = 1018 cm-3 and Na = 5 × 1015 cm-3 . Assume T=300K. Calculate the capacitance when it is under reverse biased at 1.5V. Assume a cross sectional area of 1um2. If you want to make the capacitance decrease by factor of 3 what should be the width of the depletion layer?This about PN Junctions of semiconductors.The linear electron and hole concentration profiles in a 4 um wide region of silicon material is shown in the figure below. The silicon material is subjected to electron injection from the left and hole injection from the right as shown in the figure. Assume that the cross-sectional area of the material ?=1 ??2, electron mobility ??=1312.75 ??2/? ?, and hole mobility ??=463.33 ??2/? ?. Find the total current (to one decimal place) flowing through the material. ?=300?, ?=1.6×10−19 ?, ?=8.62×10−5 ??/?, ?=1.38×10−23 ?/?, and 1 ??=1.6×10−19 ?. Explain how depletion and diffusion capacitances differ
- Please Help ASAP!!!! Just do A-CThe figure given below represents the energy-band diagram in the p-type semiconductor of a MOS capacitor, indicating surface potential as 0.254 V and the space charge width is 0.30 µm. Then the acceptor doping concentration is x 10¹5 cm-3. (Assume Esi = 11.7 &&0 8.85 x 10-14 F/cm). Os = 0.254 V oxide Xd 0.30 μm p-type E Efi EF MM EvA silicon diode is in connected to a DC voltage source with Forward biased, the net currentflowing through the diode is (25mA) where the applied voltage across the terminals of thediode is (820mV). Determine diode temperature, if Is "dark saturation current", the diodeleakage current density in the absence of light is 3.4 × 10−10 A
- A silicon diode has a forward voltage drop of 1.2V for a forward DC current of 100mA. It has a reverse current of 1μA for a reverse voltage of 10V. Calculate the: a. Bulk resistance b. Reverse resistance c. AC resistance at forward DC current of 2.5mACompute the value of DC resistance and AC resistance of a Germanium junction diode at 250C with reverse saturation current, I0 = 25µA and at an applied voltage of 0.2V across the diode.Here are some statements about a p-n junction diode. Some are TRUE and some are FALSE. i. Applying a negative bias to the p-side and a positive bias to the n-side allows a forward current flow. ii. When the p-n junction is under reverse bias, the Fermi level is continuous across the junction. iii. The forward bias current is made up of holes from the p-side and electrons from the n-side flowing across the junction. iv. Under reverse bias, you can get minority electrons flowing from the p-side to the n-side of the junction. v. To get a large built-in voltage, you need to heavily dope the p-side and n-side of the junction. vi. The built-in voltage (or contact potential) of a p-n junction is typically twice the value of the band-gap of the semiconductor. Which of the following statements is correct: (i) and (ii) are both FALSE (ii) and (v) are both FALSE (iii) and (vi) are both FALSE
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