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Electronics

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Running head: Electronics 1 Electronics Student’s Name: Institutional Affiliation: Date Electronics 2 1. (12points) The built-in voltage of a silicon pn junctions is 0.675 V at T =300K. The concentration of acceptor atom in the p-region is 10^17cm^-3. Determine the concentration of donor atoms in the n-region Solution where: Vbi -is built in potential in volts =0.675V o Nd - n-type donor atom concentrations =? o Na - n-type acceptor atom concentrations 10^17cm^-3. o ni is the concentration of atoms = ni = 3.87 π‘₯ 1016 T3/2 𝑒 βˆ’(7.02 𝑋 10 ^ 3)/𝑇 = 3.87 π‘₯ 1016 3003/2 𝑒 βˆ’(7.02 𝑋 10 ^ 3)/300 = 1.38 x 10^10 VT at T = 300k is 0.026V therefore the donor atom concentration is given by; Nd = Nd = 𝑛 𝑖^2 π‘π‘Ž 𝑒 (𝑉𝑏𝑖/𝑉𝑇) (1.38 π‘₯ 10^10)^2 10^17 𝑒 (0.675/0.026) Answer = 3.587x 1014cm-3 Electronics 3 SOLUTIONS VPS = 11.5 to 14.2V VZ = 3.9V IZ (min) = 0.1IZ(max) Ri =? P =? Voltage difference is 11.5 -14.2 = 7.6V V = IR Max current is 14.5 - 3.9 = 10.6 To get Ri: IL= 250m A = 10.6 𝑅𝐼 10.6 𝑅𝑖 Ri = 0.04 kκ­₯ The power in the diode P = I2R I= 10.6 = 0.4 265m A P = (265 X 10-3)2 X 0.04Kκ­₯ Answer = 2.8W Electronics 4 SOLUTIONS V out 9V Vref 450 Vin 5V Electronics 5 SOLUTIONS VBE =0.7V VCE =0.2V Ξ²=75 VCE = VCC - ICRC BUT Rc is 620κ­₯ and vcc is 10V 0.2V =10 - 620IC IC=0.016A BUT IC = Ξ² IB Therefore IB = IC Ξ² = 0.016𝐴 75 VB = VE + VBE AND VC = VCC - ICRC = 10-0.016 X 620 = 0.08V Henc ...
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