prelimSol - ECE 315 Prelim Exam Solution Fall 2005 Name: _...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
ECE 315 Prelim Exam Solution Fall 2005 Name: ___________________________________ Student net ID: _______________________ For a semiconductor resistor with length of l and cross section of A , the resistance can be estimated as ( 29 p n qp qn A l A l R μ ρ + = = , where n and p are the electron and hole concentration, and n and p are the electron and hole mobility. Diode equations in case: W d = x n + x p bi D A si V N N q + = 1 1 2 0 ε , N A x p = N D x n , = 2 ln i D A bi n N N q kT V , 0 =8.85 × 10 -14 F/cm, si =11.7 is the relative dielectric constant of silicon, q =1.6 × 10 -19 coul is the elemental charge, k B the Boltzmann constant, and T the temperature (default=300K). The thermal voltage k B T/q at room temperature is 26mV. For nMOSFET with the threshold voltage V th , the drain current I D in the linear and saturation regions above threshold ( V GS > V th ) are: ( 29 ( 29 saturation 2 linear 2 2 2 th GS Dsat DS th GS ch ox D th GS Dsat DS DS DS th GS ch ox D V V V V V V C L W I V V V V V V V V C L W I - = - = - = < - - = And in the saturation region, the quasi-static circuit model can be approximated as: 1. For a semiconductor wire as shown below with a cross section of 2 μ m × 1 μ m and length of 100 μ m contains a p-n junction in the middle with N A =10 16 cm -3 and N D =10 16 cm -3 . Assume n i =10 10 cm -3 and n =1,000cm 2 /Vs and p =400cm 2 /Vs. Calculate the wire resistance in Ohm. Notice that a real number is required, instead of just formula. Show your derivation and brief justification. ( Hint: the depletion region has majority carriers much less than the doping level. ) (15 pts) g mn v GS N A =10 16 cm -3 , N D =0 N D =10 16 cm -3 , N A =0 1 m 1 m 1 m 100 m 1
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
The majority carrier is the main component in drift current (or Ohm’s law resistivity), but we need to substrate out the area that is in the depletion region. m W V mV n N N q kT V i D A bi μ 43 . 0 72 . 0 10 2 10 6 . 1 10 85 . 8 7 . 11 2 72 . 0 12 60 ln 16 19 14 2 = × × × × = = × = = - - Since N A =N D , W is equally distributed on the two sides, which makes the two resistors in parallel have an effective area of 1 μ m by (1 - 0.43/2) μ m. The resistance is then: ( 29 = + × × × × × × = - - - - k R 570 400 1000 10 10 6 . 1 1 10 784 . 0 10 10 100 16 19 4 4 4 . 2.
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

Page1 / 8

prelimSol - ECE 315 Prelim Exam Solution Fall 2005 Name: _...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online