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finalSol

# finalSol - ECE 3150 Final Exam Solution Spring 2009 Name...

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Unformatted text preview: ECE 3150 Final Exam Solution Spring 2009 Name: Student ID: The temperature is 300K with V T = kT/q =26mV, while the semiconductor is silicon with E gap = 1.1eV, and the intrinsic concentration n i = 10 10 cm-3 . np = n i 2 can be used to estimate the minority concentration. The minority diffusion length as: n n n D L τ = , where D n = μ n kT/q is the diffusivity and τ n = τ p =1 μ s is the minority recombination lifetime. μ n = 800 cm 2 /Vs; μ p = 400 cm 2 /Vs. Diode equations: W d = x n + x p bi D A si V N N q + = 1 1 2 ε ε , N A x p = N D x n , = 2 ln i D A T bi n N N V V ε =8.85 × 10-14 F/cm, ε si =11.7 is the relative dielectric constant of silicon, ε ox =3.9, q =1.6 × 10-19 coul is the elemental charge, k the Boltzmann constant, and T the temperature. For a bipolar junction diode, I D = I (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 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 ’ ) ( 29 ( 29 ( 29 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 saturation region, the quasi-static circuit model can be approximated as: For nMOSFET below the threshold voltage, ( 29 T th GS T DS V V V κ V V th D e e I L W I / / ) 1 ( 2--- = and ox si C C m + 2245 ≡ 1 1 κ , where I th is the current at V GS =V th in the EKV model. The small signal parameters can be approximated as: g m = κ I D /V T , r o = V A /I D and A vo = κ V A /V T . 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+ A vo R sig ) ||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 ) 2245 1 A vo = g m r o = 2V A /V OV V A is the early voltage. +- v in r o g m v in v out +- D G S S For a CS amplifier with a R L and C L load and an input resistance of R sig , the time constant τ can be estimated by R sig (C gs + (1- A v )C gd ) + (R L ||r o ) C L . The corner frequency is f c = 1/2 πτ . 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 ....
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finalSol - ECE 3150 Final Exam Solution Spring 2009 Name...

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