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A silicon npn bipolar transistor has doping concentration of NE= 2 x 1018cm-3, NB =1017cm-3 and NC= 15 x 1016 cm-3. The area is 10-3 cm2 and neutral base width is 1 μm. The transistor is biased in the active region at VBE = 0.5V. The collector current is(DB = 20 cm2/s)a)9 μAb)17μAc)22 μAd)11 μACorrect answer is option 'B'. 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 silicon npn bipolar transistor has doping concentration of NE= 2 x 1018cm-3, NB =1017cm-3 and NC= 15 x 1016 cm-3. The area is 10-3 cm2 and neutral base width is 1 μm. The transistor is biased in the active region at VBE = 0.5V. The collector current is(DB = 20 cm2/s)a)9 μAb)17μAc)22 μAd)11 μACorrect answer is option 'B'. 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 silicon npn bipolar transistor has doping concentration of NE= 2 x 1018cm-3, NB =1017cm-3 and NC= 15 x 1016 cm-3. The area is 10-3 cm2 and neutral base width is 1 μm. The transistor is biased in the active region at VBE = 0.5V. The collector current is(DB = 20 cm2/s)a)9 μAb)17μAc)22 μAd)11 μACorrect answer is option 'B'. Can you explain this answer?.

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Here you can find the meaning of A silicon npn bipolar transistor has doping concentration of NE= 2 x 1018cm-3, NB =1017cm-3 and NC= 15 x 1016 cm-3. The area is 10-3 cm2 and neutral base width is 1 μm. The transistor is biased in the active region at VBE = 0.5V. The collector current is(DB = 20 cm2/s)a)9 μAb)17μAc)22 μAd)11 μACorrect answer is option 'B'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of A silicon npn bipolar transistor has doping concentration of NE= 2 x 1018cm-3, NB =1017cm-3 and NC= 15 x 1016 cm-3. The area is 10-3 cm2 and neutral base width is 1 μm. The transistor is biased in the active region at VBE = 0.5V. The collector current is(DB = 20 cm2/s)a)9 μAb)17μAc)22 μAd)11 μACorrect answer is option 'B'. Can you explain this answer?, a detailed solution for A silicon npn bipolar transistor has doping concentration of NE= 2 x 1018cm-3, NB =1017cm-3 and NC= 15 x 1016 cm-3. The area is 10-3 cm2 and neutral base width is 1 μm. The transistor is biased in the active region at VBE = 0.5V. The collector current is(DB = 20 cm2/s)a)9 μAb)17μAc)22 μAd)11 μACorrect answer is option 'B'. Can you explain this answer? has been provided alongside types of A silicon npn bipolar transistor has doping concentration of NE= 2 x 1018cm-3, NB =1017cm-3 and NC= 15 x 1016 cm-3. The area is 10-3 cm2 and neutral base width is 1 μm. The transistor is biased in the active region at VBE = 0.5V. The collector current is(DB = 20 cm2/s)a)9 μAb)17μAc)22 μAd)11 μACorrect answer is option 'B'. Can you explain this answer? theory, EduRev gives you an ample number of questions to practice A silicon npn bipolar transistor has doping concentration of NE= 2 x 1018cm-3, NB =1017cm-3 and NC= 15 x 1016 cm-3. The area is 10-3 cm2 and neutral base width is 1 μm. The transistor is biased in the active region at VBE = 0.5V. The collector current is(DB = 20 cm2/s)a)9 μAb)17μAc)22 μAd)11 μACorrect answer is option 'B'. Can you explain this answer? tests, examples and also practice Electronics and Communication Engineering (ECE) tests.