ES 330 Electronics II Homework 03 (Fall 2017 Due Wednesday, September 20, 2017)

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Pae1 Nae Soluios ES 330 Elecroics II Hoework 03 (Fall 017 ue Wedesday, Sepeber 0, 017 Proble 1 You are ive a NMOS aplifier wih drai load resisor R = 0 k. The volae (R appeari across resisor R = 1.5 vols wih a applied ae-o-source volae ( of 0.7 vol. Sall-sial A easurees ive a volae ai A v = -10 /. (a Fid he hreshold volae of he N-chael MOSFET. R k A 0 k ; 00 μa/ ; 1.5 ; 0.7 ; 10 R 1 We kow A ko R ad R IR ko R A ko R A 10 ad so O 0.3 1 R k O R 1.5 O O R Therefore, 0.7 0.3 0.40 O (b The process rascoducace paraeer k = 00 A/ ; wha is he MOSFET s ae widh-o-leh raio (i sybols, W/L? Usi k O A 10 1.67 A/ R 0.30 O A k R W W k 1.67 A/ ; 8.33 ' k k L L k ' 0. A/ Proble Suppose you are ive he coo-eier bipolar aplifier as show scheaically below. (Noe: This is Fiure 7.6 (o pae 377 of Sedra ad Sih, 7 h ediio

Pae The power supply volae = +5 vols ad he load resisor R = 1 k. For he rae of collecor bias curres, I = 0.5 A, 1 A,.5 A, 4 A ad 4.5 A, deerie he correspodi collecor-o-eier volaes E ad volae ais A v for each of he collecor curres. Place aswers i he able below. For I = 0.5 A, we caculae I R 0.5 0.5 Av 19.3 or A v 0 T 0.06 0.05 I R 5.0 0.5 4.5 E This calculaio is repeaed ow for each of he collecor curres I. I (A E (vols A v (/ 0.5 A 4.5-19. / 1 A 4.0-38.5 /.5 A.5-96. / 4 A 1.0-153.8 / 4.5 A 0.5-177.1 / NOTE: This able assues TH = 0.06. Proble 3 Bipolar Trasisor Operaio (10 pois The essece of rasisor operaio is ha for chae i v be, represe i by v be, resuls i a chae i collecor curre i c, represeed by i c. The sall-sial approxiaio eas keepi v be sall eouh o allow i c o be liearly relaed o v be by he relaioship, i c = v be. The paraeer is he rascoducace of

Pae3 he rasisor. Whe passi i c hrouh resisor R, a chaae i oupu volae v o is eeraed. (a Usi he expressio, A v = - [I/T]R, where T is he heral volae kt/q = 0.06 vol (o MOSFET hreshold volae, derive a siple expressio for rascoducace. A v I R T We also kow A v v v Therefore, i R R v ou BE I BE / T (b alculae he value of whe I = 0.5 A. 0.5 For I = 0.5 A, we calculae 19.3 A 0.06 Proble 4 Usi Grahical Aalysis You are preseed wih he NPN bipolar rasisor circui show below: I his proble you are o cosruc a raphical drawi of he i v E characerisic of he BJT, wih base curre values of i B = 10 A, 0 A, 30 A, 40 A ad 50 A, o esiae aplifier paraeers. To siplify he proble we iore he Early effec ; eai he oupu resisace is ifiie (i.e., horizoal lies o he i v E characerisic ad ake he BJT s curre ai = 100 a all curre levels. Give: = +5 vols ad R = 1 k; hese wo paraeers allow you o cosruc ad draw he load lie upo he BJT s i v E characerisic curve.

Pae4 (a raw he collecor curre lies o he raph ad he draw he load lie esablished by he collecor load resisor R. (b Esiae he peak-o-peak collecor volae swi resuli fro drivi he base curre i B over he rae of 10 A (iiu o 40 A (axiu. Use he drawi above o esiae his peak-o-peak volae swi. Peak-o-peak v swi = 4 vols 1 vol = 3 vols (c Assui he BJT biased a E = ½, fid he values of I ad IB a his Q-poi (i.e., Q is he quiese bias poi. The bes bias poi Q is idway bewee he oal volae rae of (ha would volae. Thus, E = ½ correspodi o E =.5 vols. Fro he above plo poi Q correspods o I =.5 A. Base curre IB =.5 A/ =.5 A/100 = 5 A. (d Assui he currre value a bias poi Q fro par (c, ive ha BE = 0.700 vol ad RB = 100 k, fid he required value of power supply BB. BB = IB RB + 0.700 = (0.00005(100,000 + 0.7 =.5 + 0.7 = 3. vols

Pae5 Proble 5 Trascoducace of NMOS Trasisor We have a NMOS rasisor wih k = 10 A/. The overvolae O paraeer is se a O = 0. vol so ha he rasisor is i is sauraed ode of operaio. Noe ha k = OX(W/L. (a Wha is he bias drai curre I? I 1 1 10 (0. ko 0. A (b Nex, we superipose a sall volae upo he bias ae volae,, wih apliude v = + 0.0 vol. Fid he correspodi icreeal collecor curre id by evaluai he oal collecor curre i ad he subraci he bias curre I. v v, where v 0.0 v 0. 0.0 = 0. O 1 1 10 (0. kvo 0.4 A i i i I 0.4 0. 0.04 A d (c Repea he calculaio fro par (b bu ow wih v = - 0.0 vol. For v 0.0 ad 0. 0.0 0.18 1 1 10 0.18 kvo 0.16 A i i d O 0. 0.16 0.038 A (d Use he resuls fro pars (b ad (c o esiae he value of he rascoducace of he rasisor. id 0.04 0.038 A 0.08 A A/ v 0.0 ( 0.0 0.04 (e alculae he rascoducace usi Equaio (7.33 i Sedra ad Sih (fro pae 384. [Noe: Equaio (7.33 reads = k O.] opare his resul wih wha you obaied i par (d above. [I oher words, how well do hey aree?] k 10 0. A/ O Yes, hey aree.

Pae6 Proble 6 Three MOSFET Trascoducace Expressios For he NMOS rasisor we kow ha 1 W I OX ; Threshold volae L where i I i ( A ad v I i d The rascoducace is defied o be di dv d (a Show ha you ca express i he for, W OX k L (b Nex, show ha his expressio for ca be i a alerae for, aely, W I k I L OX (c Fially, sill aoher for for is I I ( ( O erive all hree expressios for MOSFET rascoducace. Noe: I his course hese hree equaios are sufficie o deerie i ay proble you will ecouer. Sar wih he equaio for i ad usi v v, 1 1 i k v I id k v 1 i I id k ( ( v v v is very sall (copared o (, ad ( is a er (rea i as a cosa whe aki he derivaive. 1 I k i k v v v did By defiiio,, k( ko dv ( ad d ( sice (.

Pae7 Sari wih he equaio, k ( Reeberi I k( ; ives ( ad subsiui ives k I Fially, for he hird equaio forula for, I I k k, he subsiui ( Usi (, we solve for for k io he forula for ives I I I k I 4 ( ( O I k