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A uniformly doped silicon epitaxial npn bipolar transistor is fabricated with a base doping of NB = 3 x 1016 cm-3 and a heavily doped collector region with Nc = 5 x 1017 cm-3. The neutral base width is xB = 0.7 when VBE = VBC = 0.
The VBC (in V) at punch through is (Answer up to one decimal place)
Correct answer is '18.3'. Can you explain this answer?
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A uniformly doped silicon epitaxial npn bipolar transistor is fabrica...
Understanding the Problem
We are given the doping concentrations of the base and collector regions of a silicon npn bipolar transistor. We are also given the neutral base width and the voltages VBE and VBC. We are asked to find the VBC at punch through.
Solution
We can start by using the following equation to find the punch-through voltage:
VPT = (2qNAWB^2)/(εrε0) * ln[(NA*ND)/(ni^2)]
Where:
q = electron charge = 1.6 x 10^-19 C
NA = acceptor concentration in the collector region = 5 x 10^17 cm^-3
NB = donor concentration in the base region = 3 x 10^16 cm^-3
ni = intrinsic carrier concentration in silicon = 1.5 x 10^10 cm^-3
εr = relative permittivity of silicon = 11.9
ε0 = vacuum permittivity = 8.85 x 10^-14 F/cm
WB = neutral base width = 0.7 cm
Plugging in the values, we get:
VPT = (2*1.6*10^-19*5*10^17*0.7^2)/(11.9*8.85*10^-14) * ln[(1.5*10^10*3*10^16)/(1.5*10^20)]
VPT = 18.3 V
Therefore, the VBC at punch through is 18.3 V.
Conclusion
The VBC at punch through for a uniformly doped silicon epitaxial npn bipolar transistor with a base doping of NB = 3 x 10^16 cm^-3 and a heavily doped collector region with Nc = 5 x 10^17 cm^-3, and a neutral base width of xB = 0.7 cm is 18.3 V.
We are given the doping concentrations of the base and collector regions of a silicon npn bipolar transistor. We are also given the neutral base width and the voltages VBE and VBC. We are asked to find the VBC at punch through.
Solution
We can start by using the following equation to find the punch-through voltage:
VPT = (2qNAWB^2)/(εrε0) * ln[(NA*ND)/(ni^2)]
Where:
q = electron charge = 1.6 x 10^-19 C
NA = acceptor concentration in the collector region = 5 x 10^17 cm^-3
NB = donor concentration in the base region = 3 x 10^16 cm^-3
ni = intrinsic carrier concentration in silicon = 1.5 x 10^10 cm^-3
εr = relative permittivity of silicon = 11.9
ε0 = vacuum permittivity = 8.85 x 10^-14 F/cm
WB = neutral base width = 0.7 cm
Plugging in the values, we get:
VPT = (2*1.6*10^-19*5*10^17*0.7^2)/(11.9*8.85*10^-14) * ln[(1.5*10^10*3*10^16)/(1.5*10^20)]
VPT = 18.3 V
Therefore, the VBC at punch through is 18.3 V.
Conclusion
The VBC at punch through for a uniformly doped silicon epitaxial npn bipolar transistor with a base doping of NB = 3 x 10^16 cm^-3 and a heavily doped collector region with Nc = 5 x 10^17 cm^-3, and a neutral base width of xB = 0.7 cm is 18.3 V.
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A uniformly doped silicon epitaxial npn bipolar transistor is fabricated with a base doping of NB = 3 x 1016 cm-3 and a heavily doped collector region with Nc = 5 x 1017 cm-3. The neutral base width is xB = 0.7 when VBE = VBC = 0.The VBC (in V) at punch through is (Answer up to one decimal place)Correct answer is '18.3'. Can you explain this answer?
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A uniformly doped silicon epitaxial npn bipolar transistor is fabricated with a base doping of NB = 3 x 1016 cm-3 and a heavily doped collector region with Nc = 5 x 1017 cm-3. The neutral base width is xB = 0.7 when VBE = VBC = 0.The VBC (in V) at punch through is (Answer up to one decimal place)Correct answer is '18.3'. Can you explain this answer? for Electronics and Communication Engineering (ECE) 2024 is part of Electronics and Communication Engineering (ECE) preparation. The Question and answers have been prepared according to the Electronics and Communication Engineering (ECE) exam syllabus. Information about A uniformly doped silicon epitaxial npn bipolar transistor is fabricated with a base doping of NB = 3 x 1016 cm-3 and a heavily doped collector region with Nc = 5 x 1017 cm-3. The neutral base width is xB = 0.7 when VBE = VBC = 0.The VBC (in V) at punch through is (Answer up to one decimal place)Correct answer is '18.3'. Can you explain this answer? covers all topics & solutions for Electronics and Communication Engineering (ECE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A uniformly doped silicon epitaxial npn bipolar transistor is fabricated with a base doping of NB = 3 x 1016 cm-3 and a heavily doped collector region with Nc = 5 x 1017 cm-3. The neutral base width is xB = 0.7 when VBE = VBC = 0.The VBC (in V) at punch through is (Answer up to one decimal place)Correct answer is '18.3'. Can you explain this answer?.
A uniformly doped silicon epitaxial npn bipolar transistor is fabricated with a base doping of NB = 3 x 1016 cm-3 and a heavily doped collector region with Nc = 5 x 1017 cm-3. The neutral base width is xB = 0.7 when VBE = VBC = 0.The VBC (in V) at punch through is (Answer up to one decimal place)Correct answer is '18.3'. Can you explain this answer? for Electronics and Communication Engineering (ECE) 2024 is part of Electronics and Communication Engineering (ECE) preparation. The Question and answers have been prepared according to the Electronics and Communication Engineering (ECE) exam syllabus. Information about A uniformly doped silicon epitaxial npn bipolar transistor is fabricated with a base doping of NB = 3 x 1016 cm-3 and a heavily doped collector region with Nc = 5 x 1017 cm-3. The neutral base width is xB = 0.7 when VBE = VBC = 0.The VBC (in V) at punch through is (Answer up to one decimal place)Correct answer is '18.3'. Can you explain this answer? covers all topics & solutions for Electronics and Communication Engineering (ECE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A uniformly doped silicon epitaxial npn bipolar transistor is fabricated with a base doping of NB = 3 x 1016 cm-3 and a heavily doped collector region with Nc = 5 x 1017 cm-3. The neutral base width is xB = 0.7 when VBE = VBC = 0.The VBC (in V) at punch through is (Answer up to one decimal place)Correct answer is '18.3'. Can you explain this answer?.
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