d) If the energy gap of the InP semiconductor at T 4.5 x 10-4 eV/K and B = 327 K. Find the energy gap of it at room OK is 1.425 eV, a = %3D %3D temperature.
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- The plot of the E field with respect to x in a metal semiconductor junction at 300K is shown in the figure. The semiconductor is Si and E(0)=-2x10* V/cm and xo-0.2x10 cm. What is the semiconductor type and built in potential value. レEo) O a. n type 200 mV O b. n type 800 mv Oc p type 400 mV Od. p type 200 mV O e. n type 400 mV Of. p type 800 mva) If the electron concentration increase along the x-axis of a conductor as shown in equation below: n=3-1030x2+2- 102x+1026 And D=1.2 x 10-4 m/s. Find the diffusion current at x =Smm? b) Find the minimum electron concentration nmin in semiconductor?Consider the semiconductor crystal at 300 K. Where n, 1.8 x 10^6 cm. a- In a sample containing only 2.5x10^15 cm* ionized donors, where is the Fermi level (E-E;) in e.v ? b- In a sample containing 2x10^15 cm* ionized donors and 9 x 10^14 cm ionized acceptors, where is the Fermi level (Er-E;) in e.v ?
- B/ A new semiconductor material is to be n-type and doped with 6x10¹ cm³ donor atoms. Assume complete ionization and assume N,= 0. The effective density of states functions are N, =1.&r 10¹ cm and N,= 1.2x10 cm³ at 7-300 K. A special semiconductor device fabricated with this material requires that the electron concentration be no greater than 6.08x10¹ cm³ at 7= 400 K. What is the minimum band gap energy required in this new material?Energy (eV) d) If the energy gap of the InP semiconductor at T = OK is 1.425 eV, α = 4.5 x 10-4 eV/K and ß = 327 K. Find the energy gap of it at room temperature. 2.8 2.7- 2.6- 2.5- 2.4- 2.3- 2.2- 2.1- 2.0- 1.9- 1.8- 1.7 1.6- 1.5- 1.4- 1.3- C₂A₁B 1-X Г L X 1.2+ 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 AB X CBQ 2/ If the electron density of a pure semiconductor at a temperature of 17 C is m3/1016, and when the temperature increases by ten times, the electron density becomes m3/1019. If impurities of arsenic are added to one end of this material, the concentration of the majority charge carriers becomes m3/1023, and impurities of boron are added to the other end, so that the concentration of the majority charge carriers becomes m3/1021, thus forming a p-n junction with a contact area of 10-7 m2. Calculate what I am at 17.C 1- Fermi position at each end 2- Energy gap in ev 3- The ratio of the current of holes to the current of electrons through the junction if you know that the mobility of electrons is m/Vs 0.5 and the mobility of holes is m/Vs 0.25 and the length of the minority electrons is 0.4 mm and the length of the minority holes is 0.3 mm 4- Density of carriers for each party (majority and minority) 5- The effort of the divider 6- The junction current at an amplitude of 0.4 7- The…
- An intrinsic semiconductor has n=2x1016/m was doped with 1018 /m acceptor impurities. If the mobility of the electrons is 0.13m²/y.s and the mobility of holes is 0.04m/y.s, determine the conductivity, and what is the percentage ratio of the electrons and holes conductivity?Q1/ At 300K, the intrinsic concentration of Ge is 2.5 x 1019 m-3. Given thet m2 and 0.18 V. Sec m2 the mobility of electron and hole are 0.38 respectively. V. Sec Find the conductivity of pure Ge semiconductor.Question 2: а. Find the conductivity of an intrinsic semiconductor which have the following values: µe = (0 + 0.25) m²/V.s; µp = (0+ 0.035) m²/V.s; n¡ = (0 + 1.55) × 10*m³.
- Silicon is doped with phosphorus atoms (column V of Mendeleev table) with a concentration of 1018 cm-3 a- What is, at 27 °C, the electron density in doped Si. Use this result to derive the hole density. Which type of semiconductor is obtained? b- Calculate, at 27 °C, the position of the Fermi level EF and plot the band diagram.(e) Intrinsic silicon has effective densities of states in the conduction band and the valence band of 3.2 × 10¹⁹ cm−³ and 1.8 × 10¹⁹ cm-³, respectively. If the band gap is 1.12 eV, what is the concentration of intrinsic charge carriers in silicon at 300 K? A. 9.46 x 10⁹ m-³ 9.46 x 10⁹ cm-³ 0 m-3 2.40 x 1019 cm-3 B. C. D.Example 1: Find the hole concentration in N-type semiconductor when the donor concentration is 2 x102cm-3 and intrinsic value of silicon material at T = 300° K is 1.25 x10 cm-3. ст %3D ст