finalSol06 - ECE 315 Final Exam Solution Fall 2006 Name...

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1 ECE 315 Final Exam Solution Fall 2006 Name: ___________________________________ Student net ID: _______________________ When you are asked to explain a question briefly, use only 1-3 short sentences . Any wrong information you put down can be subject to point deduction, whether it is relevant to the question. The default temperature is 300K, and the default semiconductor is silicon with the bandgap of 1.1eV. The Debye length is defined as D B si D N q T k L 2 0 ε ε = , and the minority diffusion length as: n n n D L τ = , where D n is the diffusivity, τ n is the minority recombination lifetime, k B the Boltzmann constant, and T the temperature . Diode equations: W d = x n + x p bi D A si V N N q + = 1 1 2 0 ε ε , N A x p = N D x n , 2 i D A bi n N N q kT V = , I D = I 0 (exp(V D /V T ) – 1) , with + = D i p p A i n n N n L D N n L D qA I 2 2 0 and V T = k B T/q . The minority injection level at the edge of the depletion region can be approximated by p n = p n0 × exp( V D /V T ), which is valid for both forward and reverse biases. 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: (notice that C ox μ ch = k n ) ( ) ( ) ( ) saturation 2 2 linear 2 2 2 ' 2 2 ' 2 th GS Dsat DS OV n th GS ch ox D th GS Dsat DS DS DS th GS n DS DS th GS ch ox D V V V V V k L W V V C L W I V V V V V V V V k L W V V V V C L W I = = = = < = = μ μ And in the above- V th saturation region, the quasi-static small-signal model can be approximated as: For V GS < V th , ( ) ( ) T th GS th D V V V I I / exp κ , where I th is the current at V GS =V th in EKV. The single-stage amplifiers with the MOSFET in the saturation region or BJT in the active region: CS CS with R S CG CD R in 1/g m + R L /A vo R out ( R L load) (r o ||R L ) (r o ||R L )+A vo R S (r o ||R L ) 1/g m || (r o ||R L ) A v ( R L load) g m (r o ||R L ) g m (r o ||R L )/(1+g m R S ) g m (r o ||R L ) 1 MOSFET Diff pair operation range: OV id OV V v V 2 2 max , < < , and the small signal transconductance gain at small v id can be evaluated from: = 2 id OVcm d v V I i . g mn v gs D A o OV D OV ch ox m I V r V I V C L W g = = = 2 μ V A is the early voltage.
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