Celso José Faa de Aaújo, M.c.
he deal dode: (a) dode ccut symbol; (b) - chaactestc; (c) equalent ccut n the eese decton; (d) equalent ccut n the fowad decton. odes
(a) Rectfe ccut. (b) nput waefom. (c) Equalent ccut when (d) Equalent ccut when 1 0 (e) Output waefom. odes
odes Examples (deal ode) OE LOGC GAE BAERY-LOA CRCU ( snusodal) he cuent flew to battey though dode at 20% of nput sgnal peod whose mean alue s 100mA. (a) pp =? (b)r=? (c) (Peak) =?
odes mplfed physcal stuctue of the juncton dode.
(a) he pn juncton wth no appled oltage (open-ccuted temnals). (b) he potental dstbuton along an axs pependcula to the juncton. odes
(a) (b) (a) he pn juncton excted by a constant-cuent souce n the eese decton. o aod beakdown, s kept smalle than s. Note that the depleton laye wdens and the bae oltage nceases by olts, whch appeas between the temnals as a eese oltage. (b) he pn juncton excted by a constant-cuent souce supplyng a cuent n the fowad decton. he depleton laye naows and the bae oltage deceases by olts, whch appeas as an extenal oltage n the fowad decton. odes
odes he dode - elatonshp wth some scales expanded and othes compessed n ode to eeal detals.
n s e 1 s n e fo = hemal oltage he Fowad-Bas Regon = atuaton Cuent s a constant fo a gen dode at a gen tempeatue. t s also called cale n 0.5 ln k Cuent due to fact that s dectly popotonal to the coss-sectonal aea of de dode. s of the ode of 10 15 A. q 2.3 n log k = Boltzmann s constant = 1.38x10 23 Joules/Keln = the absolute tempeatue n Keln q = the magntude of electonc chage = 1.60x10 19 Coulomb (20 o )=25.2m 25m 1 n 2 n=1 standad ntegated-ccut and n=2 dscete twotemnal components. odes
40 35 30 25 20 15 10 5 4.5 n s e 1 5 4 0-5 -0.1 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 odes 3.5 2.5 1.5 0.5-0.5 0.5 0.525 0.55 0.575 0.6 0.625 0.65 0.675 0.7 0.725 0.75 Gaph of - chaactestc of a p-n juncton dode =1x10-15 A, n=1 and =25m 3 2 ode Cuent (ma) 12 11 10 9 8 7 6 5 4 3 2 1 0-1 -2-0.2-0.15-0.1-0.05 0 0.05 0.1 ode oltage () 1 ode Cuent (x10 15 A) ode Cuent (ma) ode oltage () 0 ode oltage ()
10 1 10 0 10-1 10-2 10-3 10-4 10-5 10-6 60m/ecade of Cuent n=1 ode Cuent (A) 10-7 10-8 10-9 10-10 120m/ecade of Cuent n=2 10-11 0.46 0.52 0.58 0.64 0.7 0.76 0.82 0.88 0.94 1 ode oltage () = 1x10-15 A = 25m ode - chaactestc on semlog scale odes ln 2 n 2.3 log 2 n 2.3 2.3 n n 25m; 25m; n n 1 2 0.0575 0.115 60m 120m 2 1 n ln 2 1 2.3 n log 2 1 2 1
n n e e n 1 ode 1(0.7 1mA) ode 2 1 2(0.7 1A) 1000 1 2 10 1 e 3 e 0.7 0.7 25m 25m 6.9x10 6.9x10 Aea( ode2) 1000 x Aea( ode 1 ) 12 x 10-13 13 A 16 A 1 10 8 0.8 6 4 ode 2 2 0 ode Cuent (A) 0.6 0.4 ode Cuent (A) 0.2-2 0-4 ode 1-0.2 0.5 0.55 0.6 0.65 0.7 0.75 0.8 0.85 0.9 ode oltage () n = 1 = 25m -6-0.2-0.15-0.1-0.05 0 0.05 0.1 0.15 0.2 ode oltage () n = 1 = 25m odes Example: Aea x cale Cuent
Fowad Bas Regon 1.6 x 10-3 ode (0.7 -- 1mA; n = 1 and =20 o ) 1.5 1.4 1.3 1.2 =21 o 1.1 1 0.9 =20 o ode Cuent (A) 0.8 0.7 0.6 0.69 0.692 0.694 0.696 0.698 0.7 0.702 0.704 0.706 0.708 0.71 ode oltage () Geneal ode 2m o C empeatue Effect odes Reese Bas Regon doubles fo eey 10 o C n tempeatue.
Execses (1) 1. Consde a slcon wth n=1.5. Fnd the change n oltage f cuent changes fom 0.1mA to 10mA. Ans. 172.5m 2. A slcon juncton dode wth n=1 has =0.7 at =1mA. Fnd the oltage dop at =0.1mA and =10mA. Ans. 0.64; 0.76 3. Usng the fact that a slcon dode has =10-14 A at 25 o C and that nceases by 15% pe o C se n tempeatue, fnd the alue of at 125 o C. Ans. 1.17x10-8 A 4. he dode n the ccut of fgue below s a lage, hgh-cuent dece whose eese leakage s easonably ndependent of oltage. f =1 at 20 o C, fnd the alue of at 40 o C and at 0 o C. Ans. 4; 0.25 odes
Execses (2) 1. Fo the ccut shown below, both dodes ae dentcal, conductng 10mA at 0,7 and 100mA at 0,8. Fnd the alue of R, fo whch =50m. 2. n the ccut shown below, both dodes hae n = 1, but 1 has ten tmes the juncton aea of 2. What alue of esults fo 1 of 1mA? o obtan a alue fo of 50m, what cuent 1 s needed? 3. A juncton ode s opeated n a ccut n whch t s suppled wth a constant cuent. What s the effect on the fowad oltage of the dode f an dentcal dode s connected n paallel? Assume n= 1. odes
odes Gaphcal Analyss
teate Analyss etemne the cuent and the dode oltage fo the ccut below wth =5 and R=1k. Assume that the dode has a cuent of 1mA at a oltage of 0.7 and that ts oltage dop changes by 0.1 fo eey decade change n cuent. Recuent Equaton: 2 0.7 0.1 log 1m o begn the teaton, we assume that =0.7 5 0.7 ' 4.3mA ' 0.7 0.1 log 1k 0.763 1k 4.3m 1m 5 '' 4.237mA '' 0.7 0.1 log ''' 5 0.7627 1k 0.7627 4.2373mA 4.2373mA ''' 0.7 0.1 log 0.763 4.237m 1m m 4.2373 1m 0.7627 0.7627 odes
Pecewse-lnea Model of the dode fowad chaactestc and ts equalent ccut epesentaton. odes
Example of Pecewse-lnea model of the dode fowad chaactestc. etemne the cuent and the dode oltage fo the ccut below wth =5 and R=1k, utlzng the pecewse-lnea model whose paametes ae gen n fgue below. O = 0.65 =20 5 1k O 0.65 20 4.26mA 0.65 4.26m 20 0.735 odes
he constant-oltage-dop model of the dode fowad chaactestc and ts equalent ccut epesentaton. etemne the cuent and the dode oltage fo the ccut below wth =5 and R=1k, utlzng the constant-oltage-dop model ( O = 0.7). 5 0.7 1k 4.3mA odes
Execses (3) 1. Fo the ccuts n Fg. below fnd and fo the case =5 and R=10K. Assume that the dode has a oltage of 0.7 at 1mA cuent that oltage changes by 0.1/decade of cuent change. Use (a) teaton (b) the pecewse-lnea model wth =20 and (c) the constant-oltage-dop model wth =0.7. Ans. (a) 0.434mA, 0.63; (b) 0.431mA, 0.689; (c) 0.43mA, 0.7. 2. Consde a dode that s 100 tmes as lage (n juncton aea) as a dode whose chaactestcs ae O =0.65 and =20 how would the model paametes O and change? Ans. O does not change; deceases by a facto of 100 to 0.2. 3. esgn the ccut n fgue below to pode an output oltage of 2.4. Assume that the dodes aalable hae 0.7 dop at 1mA and that =0.1/decade change n cuent. Ans. 760. odes
eelopment of the dode small-sgnal model. Note that the numecal alues shown ae fo a dode wth n = 2. d s a small-sgnal esstance o ncemental esstance d 1 e d d d n d n e e d n d n n n odes d e d d d n 1 d n d e d n Equalent ccut model fo the dode fo small changes aound bas pont Q. he ncemental esstance d s the nese of the slope of the tangent at Q, and 0 s the ntecept of the tangent on the axs.
(a) (b) (c) (d) he analyss of the ccut n (a), whch contans both dc and sgnal quanttes, can be pefomed by eplacng the dode wth the model of Fg. (a), as shown n (b). hs allows sepaatng the dc analyss [the ccut n (c)] fom the sgnal analyss [the ccut n (d)]. odes
Example of usng small-sgnal model (1) Consde the ccut shown n fg. below fo the case R=10K. he powe supply + has a C alue of 10 on whch s supemposed a 60Hz snusod of 1 peak ampltude (powe supply pple). Calculate both the C oltage of the dode and the ampltude of the sne-wae sgnal appeang acoss t. Assume the dode to hae a 0.7 dop at 1mA cuent and n=2. C analyss (we assume 10 0.7 0.93mA 10K 53.8Ω d AC analyss d d n (peak (peak) model s to 2 25m 0.93m peak) 5.35m p justfed. 2 d d R,snce d 0.7 2 ) 53.8 53.8 10k n the 10.7m small pp sgnal odes
odes Example of usng small-sgnal model (2) Consde the ccut shown n fg. below. A stng of thee dodes s used to pode a constant oltage of about 2.1. We want to calculate the pecentage change n ths egulated oltage caused by (a) 10% change n the powe-supply oltage and (b) connecton of a 1K load esstance. Assume n = 2. (a) C analyss 10 2.1 7.9mA 1K n 2 25m d 6.3Ω 7.9m 3 d 18.9Ω AC analyss 18.9 o 2 2 R 18.9 1k 18.5m o 0.9% o snce the d small p 6.2 m p sgnalmodel n 37.1m s pp justfed. O O (b) Fo all thee dodes the small-sgnal model we can use the paametes O =1.95 and d = 18.9 obtaned fom (a) analyss. 1 1 k //18.9 k //18.9 1 k 2.061 o 10 2.1 1k//1k 1k//1k 18.9 39m 1.95 2.061
Execses (4) 1. Fnd the alue of the dode small-sgnal esstance at bas cuent of 0.1, 1, and 10mA. Assume n=1. Ans. 250; 25; 2.5. 2. Fo a dode that conducts 1mA at a fowad oltage dop of 0.7 and whose n=1, fnd the equaton of the staght-lne tangent at =1mA. Ans. = (1/25)( -0.675). 3. Consde a dode wth n=2 based at 1mA. Fnd the change n cuent as a esult of changng the oltage by (a) 20m; (b) 10m; (c) -5m (d) +5m (e)+10m; (f) +20m. n each case, do the calculatons () usng the small-sgnal model and () usng the exponental model. Ans. (a) 0.40, -0.33mA; (b) 0.20, -0.18mA; (c) 0.10, -0.10mA; (d) +0.10, +0.11mA; (e) +0.20, +0.22mA; (f) +0.40, +0.49mA. 4. esgn the ccut of fg. below so that O =3 when L = 0, and O changes by 40m pe 1mA of load cuent. Fnd the alue of R and the juncton aea of each dode (assume all fou dodes ae dentcal) elate to a dode wth 0.7 dop at 1mA cuent. Assume n=1. Ans. R= 4.8k; 0.34. odes
Opeaton n the Reese Beakdown Regon-Zene odes. Model fo the zene dode. Ccut symbol fo a zene dode. he dode - chaactestc wth the beakdown egon shown n some detal. odes
Lne Load R mn Z mn ZO L z max odes esgn of the hunt Regulato egulaton egulaton O O L un un m/ m/ma Z mn O R R z ZO Lne Regulaton Load Regulaton z z R z z R ( //R) z ( //R) z L
Zeneode -Example he 6.8 zene dode n the ccut of fg. below s specfed to hae Z = 6.8 at Z = 5mA, z = 20, and ZK = 0.2mA. he supply oltage + s nomnally 10 but can ay by 1. (a) Fnd O wth no load and wth + at ts nomnal alue. (b) Fnd the change n O esultng fom the 1 change n +. (c) Fnd the change n O esultng fom connectng a load esstance R L = 2k. (d) Fnd the alue of O when R L = 0.5k. (e) What s the mnmum alue of R L fo whch the dode stll opeates n the beakdown egon? ZO O (a) (b) O O 6.8 20 5m 6.7 z z z O z // R 10 R R 6.83 O // R R L L R 10 R R // 6.76 (c) z z z R 6.7 O 38.5mA (d) // R R L L z 6.7 R 68m (e) L mn R O L 0.5k 0.5k 0.5k ( ZO ZK z ZO R ZK z ) ZK 1526 10 5 odes
odes Block agam of a C Powe upply.
o (C) ( o max ( O 1/π )/R ) O /2 ω t o (a) Half-wae ectfe. (b) Equalent ccut of the half-wae ectfe wth the dode eplaced wth ts battey-plus-esstance model. (c) tansfe chaactestc of the ectfe ccut. (d) nput and output waefoms, assumng that << R. max ( π 2θ O ) whee P θ ωt o sen 1 O odes
o (C) o max ( ( O 2/π ) )/R O ω t o max ( π 2θ O ) whee P θ 2 ωt o sen O 1 O Full-wae ectfe utlzng a tansfome wth a cente-tapped seconday wndng. (a) Ccut. (b) ansfe chaactestc assumng a constant-oltage-dop model fo the dodes. (c) nput and output waefoms. odes
o (C) o max ( ( 2/π ) 2 O )/R 2 O ω t ( π 2θ ) whee θ ωt o sen odes 2 P o max O he bdge ectfe: (a) ccut and (b) nput and output waefoms. O 1 2 O
L o (C) C max a f L L L p CR o 1 1 π p p 2π 1 2 2 p 1 2 2 p /R ω t 2 P 2 p p oltage and cuent waefoms n the peak ectfe ccut wth CR >>. he dode s assumed deal. (Half Wae) odes
L o (C) C max a 2f L L L p CR o 1 1 π p p 2π 1 2 2 p 1 2 p 2 2 /R ω t P P 2 2 p p p (Cente ap) (Bdge) oltage and cuent waefoms n the peak ectfe ccut wth CR >>. he dode s assumed deal. (Full Wae Rectfe) odes
Execses (5) 1. Consde a bdge-ectfe ccut wth a flte capacto C placed acoss the load essto R, fo the case n whch the tansfome seconday deles a snusod of 12 ms hang 60Hz fequency, and assumng O = 0,8 and a load esstance R= 100. Fnd the alue of C that esults n a pple oltage no lage than 1 pp. What s the C oltage at the output? Fnd the load cuent. Fnd de dodes conducton angle. What s the aeage dode cuent? What s the peak eese oltage acoss each dode? pecfy the dode n tems of ts peak cuent and ts P. Ans. 1281F; 15.4 o (a bette estmate) 14.9; 0.15A; 0.36 ad; 1.45A; 2.74A; 16.2. hus select a dode wth 3.5 to 4 A peak cuent and a 25 P atng (50%). 2. t s equed to use a peak ectfe to desgn a C powe supply that podes an aeage C output oltage of 15 on whch a maxmum of 1 pple s allowed. he ectfe feeds a load of 150. he ectfe s fed fom the lne oltage (120 ms 60Hz) though a tansfome. he dodes aalable hae 0.7 dop when conductng. f the desgne opts fo the half-wae ccut: (a) pecfy the ms oltage that must appea acoss the tansfome seconday. (b) Fnd the equed alue of the flte capacto. (c) Fnd the maxmum eese oltage that wll appea acoss the dode, and specfy the P atng of the dode. (d) Calculate the aeage cuent though the dode dung conducton. (e) Calculate the peak dode cuent. Ans. (a) 11.8 ms ; (b) 889F; (c) 32.7; (d) 2.61A; (e) 1.36 A. odes
odes A AREY OF BAC LMNG CRCU.
Execses (6) 1. Assumng the dodes to be deal, descbe the tansfe chaactestc of the ccuts shown n Fgue below. o=(3/4)+5/4 5 o 6 3-6 o o=(4/7)+15/7 5-6 o= o=(4/5)+3/5 o=(2/3)-2 3 o= 2. Assumng the dodes has a 0.7 dop when conductng and the zene specfed oltage s 8.2 (deal), descbe the tansfe chaactestc of the ccuts shown n Fgue below. 6.75 o=(1/6)+8 9.6 o -9.6 9.6-9.6 o= o=(1/6)-8 odes
he Clamped Capacto o C Restoe O 10 10 4 10-6 t t 10 4 O t t -10-6 10 etemne X to hae an aeage oltage of output A fo the ccut shown n fgue below. Consde the dode deal and the nput sgnal s that shown n fgue below. Ans.: X = -(1/2)A odes
1 -P -2P odes he oltage ouble 1 P t -P O t -2P t
PECAL OE YPE he chottky-bae ode (B) aactos Photododes Lght-Emttng odes (LEs) odes
Photonc eces Lght-emttng dodes (LE) ecombnaton electon (+) hole (-) enegy = E G / h h: Planck s constant h=6.63x10-34 Jsec - heat - lght (E=h) Mateals: 10 GaAs ( nfaed ) GaAs 1-x P x GaAs 0.6 P 0.4 ed lght ( E G 1.8 e ) (ma) 20 1.0 2.0 () B.G. teetman, old tate Electonc eces, 2nd edton, Pentce-Hall, 1980 odes
Photododes (1/2) h > E G p n E R gop G op : optcal geneaton ate Relate numbe of MNORY caes ncease by a lage amount dft cuent nceases. odes dak cuent
Photododes (2/2) PH a,c p = x > 0 b p = x < 0 the dode s a powe souce. No llumnaton a OC Model fo optcally llumnated dode. PH epesents the cuent geneated by absopton of photons n the cnty of the pn juncton. c C b 1 C 0 PH PH n s e OC 0 n ln PH s odes
Photo detecto Optosolato + B + B R R O O PH (a) (b) Basc photo-detecto ccut (a) and model (b) No llumnaton: O = B llumnaton: O = B -R ph LE photodode Lght-to-oltage conete R ph O = Rph odes
ola cells 1.5 1.0 0.5 0.0-0.5 sc Pmax oc Cell Cuent c (A) 3 n = m, = m fo P = P max m m < oc sc Aeage sola enegy (zenth): 100mW/cm 2 Maxmum powe 300 mw -1.0-1 Cell oltage () C emnal chaactestcs fo a pn juncton sola cell 0 1 Effcency of sola cells ange between 10% and 20% odes