finalSol - ECE 315 Final Exam Solution Fall 2007 The...

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1 ECE 315 Final Exam Solution Fall 2007 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 n = p =1 s is the minority recombination lifetime. Diode equations: 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 T bi n N N V V 0 =8.85 × 10 -14 F/cm, si =11.7 is the relative dielectric constant of silicon, 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 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 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 = = = = < = = And in the saturation region, the quasi-static small-signal circuit model can be approximated as: For V GS < V th , if V DS > 3V T , ( ) ( ) 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 model. V T is the thermal voltage at 26mV at room temperature. The small signal parameters can be approximated as: g m = I D /V T , r o = V A /I D and A vo = 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 ||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 For a CS amplifier with a R L and C L load and an input resistance of R sig , the time constant can be estimated by R sig (C gs + (1+A v )C gd ) + (R L ||r o ) C L . g mn v gs D A o OV D OV ch ox m I V r V I V C L W g = = = 2 A vo = g m r o = 2V A /V OV V A is the early voltage.
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2 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 . For the 5-transistor differential amplifier with a current mirror active load, A vo = g mn (r on ||r op ) , similar to a single-stage CS amplifier.
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finalSol - ECE 315 Final Exam Solution Fall 2007 The...

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