At what temperature will intrinsic silicon become a conductor? Assume that un = 350 cm²/Vs and up = 60 cm²/Vs. The properties of silicon are as follows: B = 1.08 x 1031 K-3 cm-6, k = 8.62 x 10-5 eV/K, and EG = 1.12 eV. Hint: o≥1000(92-cm) typing format please for a conductor
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- High density 6w density to low density low bavyier VoltaJe Higharier, Voltad to high low doPed density atoms Diffusion Curment from Conductivity of Semi conductor In crease with FoYward biased P-N junctionA simple p*n junction is designed to work as IMPATT diode. The doping concentrations in the p* layer is 1019 cm-3 while the doping in the n-layer is 0.7 x1016 Calculate the peak electric field if the breakdown voltage is 80 V and the dielectric constant is 11.9. Express your answer in the unit of kV/cm. cm-3Consider an p-type silicon for which the dopant concentration ND = 1018/cm3. Find the electron and hole concentrations at T = 350 K. electron = 1018/cm³ & holes= 2.25 × 102/cm³ electron = 2.25 x 102/cm³ & holes= 2.25 x 104/cm3 electron = 17.22 × 104 /cm³ & holes= 1018/cm³ 3 electron 1018/cm³ & holes= 3 %3D 1018/cm3
- A gallium arsenide semiconductor at T = 300 K is homogeneously doped with Nd= 5 X 1016 cm-3 and N₂ = 0. Calculate the a drift current for an applied E field of 20 V/cm Jdrf = 1360 A/cm² Jdrf = 2432 A/cm² drf = 624 A/cm² Jdrf=512 A/cm²a. If you do not go completely around the loop when applying Kirchhoff’s voltage law, then the algebraic sum of the voltages cannot be determined the algebraic sum of the voltages will always be positive the algebraic sum of the voltages will always be negative the algebraic sum is the voltage between the start and finish points b. A p -type semiconductor is a semiconductor doped with impurity atoms whose electron valence is +4 pentavalent impurity atoms trivalent impurity atomsAn abrupt silicon pn junction at zero bias has dopant concentrations of Nd = 1 X 1017 cm-3 cm³ and Na and N₂ = 5 X 1016 cm¯³ at T = 300K. Determine the peak electric field for this junction for a reverse voltage of 2 V. Emax = 3.88 X 105 V/cm Emax = 1.35 X 105 V/cm Emax 1.70 X 105 V/cm O Emax = 3.21 X 105 V/cm =
- The circle is below(for D1-1N4148 is 1N4148 Diode). and also there have a table(. Note that the first 3 rows of measurement are obtained when the diode is reverse-biased) Vout_dc (V) VD (V) ID (μA) 0.51 -0.51 0.0 0.23 -0.20 0.0 0.10 -0.08 0.0 0.00 0.00 0.0 0.05 0.04 0.0 0.11 0.09 0.0 0.17 0.15 0.1 0.19 0.20 0.1 0.27 0.25 0.4 0.32 0.30 1.6 0.36 0.34 3.1 0.42 0.40 8.2 0.46 0.42 14.3 0.52 0.47 28.0 Questions: 1) plot the diode’s I-V characteristics on a linear scale (ID on the y-axis and VD on the x-axis) 2) [RP4] 3) [RP5]Find the barrier voltage in a pn junction at T=300k assuming VT= 25.9 mV, NA=1018/cm3, ND=1017/cm3 0.0764 V 0.874 V 1.754 VPROBLEM: The diode (blue line) should be at -10V not -8V SOFTWARE USED: LTspice XVII---Figure 1 is my work, and figure 2 is the correct sample. Please tell me what I did wrong. Thank you :))---Below is the .asc filehttps://www.mediafire.com/file/zxauoh86zi79nfu/EXPERIMENT_01.asc/file
- the non-inductive resistance of kelvin double bridge are set at 100ON each, the standard resistance is set at 0.10020 and the low resistance 0.0000122 was very low. determine the unknown resistanceDiodes Are Connected In Series To Share A Total... 21 Question • Two diodes are connected in series to share a total reverse voltage of VD=D 5 KV. Reverse leakage currents of the diodes are ID1=30 mA and Ip2=35 mA. - Find Vp1 and Vp2 for R,=R2=100 k2 - Find R, and R2 for VD1=VD2 R1 VD1 Hint: Is=L,+IRj= I,2+IR2 VD R2 VD28. Liquids with solid impurities a) Have higher dielectrie strength b) Of large size have higher dielectric strength c) Have lower dielectric strength as compared to pure liquids d) None of the above 9. Peak to peak ripple is defined as a) the difference between average de voltage and peak value b) the difference between maximum and minimum de voltage c) the difference between maximum ac and average dc voltages d) the difference between ac (rms) and average de voltages 10. In a Cockroft-Walton circuit, input voltage 100 kV load current 25 mA, supply frequency 100 Hz, each capacitor 10 nF. The optimum no. of stages for maximum output voltage is a) 1 b) 2 c) 10 d) 35