Bplar-Junctn (BJT) transstrs References: Barbw (Chapter 7), Hayes & Hrwtz (pp 84-4), Rzzn (Chapters 8 & 9) A bplar junctn transstr s frmed by jnng three sectns f semcnductrs wth alternately dfferent dpngs. The mddle sectn (base) s narrw and ne f the ther tw regns (emtter) s healy dped. Tw arants f BJT are pssble: NPN and PNP. C NPN Transstr C PNP Transstr B n p C B B C B p n C B B C n E E p E E E Crcut Symbls E Crcut Symbls We wll fcus n NPN BJTs. Operatn f a PNP transstr s analgus t that f a NPN transstr except that the rle f majrty charge carres reersed. In NPN transstrs, electrn flw s dmnant whle PNP transstrs rely mstly n the flw f hles. Therefre, t zerth rder, NPN and PNP transstrs behae smlarly except the sgn f current and ltages are reersed..e., PNP = NPN! In practce, NPN transstrs are much mre ppular than PNP transstrs because electrns me faster n a semcnductr. As a results, a NPN transstr has a faster respnse tme cmpared t a PNP transstr. At the frst glance, a BJT lks lke ddes placed back t back. Indeed ths s the case f we apply ltage t nly tw f the three termnals, lettng the thrd termnal flat. Ths s als the way that we check f a transstr s wrkng: use an hm-meter t ensure bth ddes are n wrkng cndtns. (One shuld als check the resstance between CE termnals and read a ary hgh resstance as ne may hae a burn thrugh the base cnnectng cllectr and emtter.) The behar f the BJT s dfferent, hweer, when ltage surces are attached t bth BE and CE termnals. The BE junctn acts lke a dde. When ths junctn s frward based, electrns flw frm emtter t the base (and a small current f hles frm base t emtter). The base regn s narrw and when a ltage s appled between cllectr and emtter, mst f the electrns that were flwng frm emtter t base, crss the narrw base regn and are cllected at the cllectr regn. S whle the BC junctn s reersed based, a large current can flw thrugh that regn and BC junctn des nt act as a dde. The amunt f the current that crsses frm emtter t cllectr regn depends strngly n the ltage appled t the BE junctn, BE. (It als depends weakly n ltage appled ECE60L Lecture Ntes, Sprng 004 55
between cllectr and emtter, CE.) As such, small changes n BE r B cntrls a much larger cllectr current C. Nte that the transstr des nt generate C. It acts as a ale cntrllng the current that can flw thrugh t. The surce f current (and pwer) s the pwer supply that feeds the CE termnals. A BJT has three termnals. Sx parameters; C, B, E, CE, BE, and CB ; defne the state f the transstr. Hweer, because BJT has three termnals, KL and KCL shuld hld fr these termnals,.e., E = C B BC = BE CE B _ CB BE C CE E Thus, nly fur f these 6 parameters are ndependent parameters. The relatnshp amng these fur parameters represents the characterstcs f the BJT, usually shwn as B s BE and C s CE graphs. The abe graphs shw seeral characterstcs f BJT. Frst, the BE junctn acts lkes a dde. Secndly, BJT has three man states: cut-ff, acte-lnear, and saturatn. A descrptn f these regns are gen belw. Lastly, The transstr can be damaged f () a large pste ltage s appled acrss the CE junctn (breakdwn regn), r () prduct f C CE exceed pwer handlng f the transstr, r (3) a large reerse ltage s appled between any tw termnals. Seeral mdels aalable fr a BJT. These are typcally dded nt tw general categres: large-sgnal mdels that apply t the entre range f alues f current and ltages, and small-sgnal mdels that apply t AC sgnals wth small ampltudes. Lw-frequency and hgh-frequency mdels als exst (hgh-frequency mdels accunt fr capactance f each junctn). Obusly, the smpler the mdel, the easer the crcut calculatns are. Mre cmplex mdels descrbe the behar f a BJT mre accurately but analytcal calculatns becme dffcult. PSpce prgram uses a hgh-frequency, Eber-Ms large-sgnal mdel whch s a qute accurate representatn f BJT. Fr analytcal calculatns here, we wll dscuss a smple lw-frequency, large-sgnal mdel (belw) and a lw-frequency, small-sgnal mdel n the cntext f BJT amplfers later. ECE60L Lecture Ntes, Sprng 004 56
A Smple, Lw-frequency, Large Sgnal Mdel fr BJT: As the BE junctn acts lke a dde, a smple pece-wse lnear mdel can be used : BE Junctn ON: BE = γ, and B > 0 BE Junctn OFF: BE < γ, and B = 0 where γ s the frward bas ltage ( γ 0.7 fr S semcnductrs). When the BE junctn s reersed-based, transstr s OFF as n charge carrers enter the base and me t the cllectr. The ltage appled between cllectr and emtter has nt effect. Ths regn s called the cut-ff regn: Cut-Off: BE < γ, B = 0, C E 0 Snce the cllectr and emtter currents are ery small fr any CE, the effecte resstance between cllectr and emtter s ery large (00 s f MΩ) makng the transstr behae as an pen crcut n the cut-ff regn. When the BE junctn s frward-based, transstr s ON. The behar f the transstr, hweer, depends n hw much ltage s appled between cllectr and emtter. If CE > γ, the BE junctn s frward based whle BC junctn s reersed-based and transstr s n acte-lnear regn. In ths regn, C scales lnearly wth B and transstr acts as an amplfer. Acte-Lnear: BE = γ, B > 0, C B = β cnstant, CE γ If CE < γ, bth BE and BC junctns are frward based. Ths regn s called the saturatn regn. As CE s small whle C can be substantal, the effecte resstance between cllectr and emtter n saturatn regn s small and the BJT acts as a clsedcrcut. Saturatn: BE = γ, B > 0, C B < β, CE sat Our mdel specfes CE sat, the saturatn ltage. In realty n the saturatn regn 0 < CE < γ. As we are manly nterested n the alue f the cllectr current n ths regn, CE s set t a alue n the mddle f ts range n ur smple mdel: CE sat 0.5 γ. Typcally a alue f sat 0. 0.3 s used fr S semcnductrs. ECE60L Lecture Ntes, Sprng 004 57
The abe smple, large-sgnal mdel s shwn belw. A cmparsn f ths smple mdel wth the real BJT characterstcs demnstrates the degree f apprxmatn used. B C Saturatn BJT ON Acte Lnear BJT OFF γ BE sat CE Cut Off Hw t Sle BJT Crcuts: The state f a BJT s nt knwn befre we sle the crcut, s we d nt knw whch mdel t use: cut-ff, acte-lnear, r saturatn. T sle BJT crcuts, we need assume that BJT s n a partcular state, use BJT mdel fr that state t sle the crcut and check the aldty f ur assumptns by checkng the nequaltes n the mdel fr that state. A frmal prcedure wll be: ) Wrte dwn a KL ncludng the BE junctn (call t BE-KL). ) Wrte dwn a KL ncludng CE termnals (call t CE-KL). 3) Assume BJT s n cut-ff (ths s the smplest case). Set B = 0. Calculate BE frm BE-KL. 3a) If BE < γ, then BJT s n cut-ff, B = 0 and BE s what yu just calculated. Set C = E = 0, and calculate CE frm CE-KL. Yu are dne. 3b) If BE > γ, then BJT s nt n cut-ff. Set BE = γ. Sle abe KL t fnd B. Yu shuld get B > 0. 4) Assume that BJT s n acte lnear regn. Let E C = β B. Calculate CE frm CE-KL. 4a) If CE > γ, then BJT s n acte-lnear regn. Yu are dne. 4b) If CE < γ, then BJT s nt n acte-lnear regn. It s n saturatn. Let CE = sat and cmpute C frm CE-KL. Yu shuld fnd that C < β B. Yu are dne. ECE60L Lecture Ntes, Sprng 004 58
Example : Cmpute the parameters f ths crcut (β = 00). Fllwng the prcedure abe: kω BE-KL: 4 = 40 0 3 B BE CE-KL: = 0 3 C CE, Assume BJT s n cut-ff. Set B = 0 n BE-KL: - 40 kω 4 B BE _ C _ E CE BE-KL: 4 = 40 0 3 B BE BE = 4 > γ = 0.7 S BJT s nt n cut ff and BJT s ON. Set BE = 0.7 and use BE-KL t fnd B. BE-KL: 4 = 40 0 3 B BE B = 4 0.7 40, 000 = 8.5 µa Assume BJT s n acte lnear, Fnd C = β B and use CE-KL t fnd CE : C = β B = 00 B = 8.5 ma CE-KL: =, 000 C CE, CE = 8.5 = 3.75 As CE = 3.75 > γ, the BJT s ndeed n acte-lnear and we hae: BE = 0.7, B = 8.5 µa, E C = 8.5 ma, and CE = 3.75. Example : Cmpute the parameters f ths crcut (β = 00). Fllwng the prcedure abe: kω BE-KL: CE-KL: 4 = 40 0 3 B BE 0 3 E =, 000 C CE, 000 E Assume BJT s n cut-ff. Set B = 0 and E = C = 0 n BE-KL: - 40 kω 4 B BE _ C _ E CE kω BE-KL: 4 = 40 0 3 B BE 0 3 E BE = 4 > 0.7 S BJT s nt n cut ff and BE = 0.7 and B > 0. Here, we cannt fnd B rght away frm BE-KL as t als cntans E. ECE60L Lecture Ntes, Sprng 004 59
Assume BJT s n acte lnear, E C = β B : BE-KL: 4 = 40 0 3 B BE 0 3 β B 4 0.7 = (40 0 3 0 3 0 ) B B = 4 µa E C = β B =.4 ma CE-KL: =, 000 C CE, 000 E, CE = 4.8 = 7. As CE = 7. > γ, the BJT s ndeed n acte-lnear and we hae: BE = 0.7, B = 4 µa, E C =.4 ma, and CE = 7.. Lad lne The peratng pnt f a BJT can be fund graphcally usng the cncept f a lad lne. A lad lne s the relatnshp between C and CE that s mpsed n BJT by the external crcut. Fr a gen alue f B, the C CE characterstcs cure f a BJT s the relatnshp between C and CE as s set by BJT nternals. The ntersectn f the lad lne wth the BJT characterstcs represent a par f C and CE alues whch satsfy bth cndtns and, therefre, s the peratng pnt f the BJT (ften called the Q pnt fr Quescent pnt) The equatn f a lad lne fr a BJT shuld nclude nly C and CE (n ther unknwns). Ths equatn s usually fund by wrtng a KL arund a lp cntanng CE. Fr the example abe, we hae (usng E C ): KL: =, 000 C CE, 000 E, 000 C CE = An example f a lad lne, C CE characterstcs f a BJT, and the Q-pnt s shwn belw. ECE60L Lecture Ntes, Sprng 004 60
BJT Swtches and Lgc Gates The basc element f lgc crcuts s the transstr swtch. A schematc f such a swtch s shwn. When the swtch s pen, C = 0 and =. When the swtch s clsed, = 0 and C = /. In an electrnc crcut, mechancal swtches are nt used. The swtchng actn s perfrmed by a transstr wth an nput ltage swtchng the crcut, as s shwn. When = 0, BJT wll be n cut-ff, C = 0, and = (pen swtch). When s n hgh state, BJT can be n saturatn wth = CE = sat 0. and C = ( sat )/ (clsed swtch). When R c s replaced wth a lad, ths crcut can swtch a lad ON r OFF (LE and mtr dre crcuts f ECE0A Lab). R B B C C The abe BJT crcut s als an nerter r a NOT lgc gate. Let s assume that the lw states are ltages between 0 t 0.5, hgh states ltages are between 4 t 5, and = 5. When the nput ltage s lw ( 0), BJT wll be n cut-ff and = = 5 ( hgh state). When nput ltage s hgh, wth prper chce f R B, BJT wll be n saturatn, and = CE = sat 0. ( lw state). Resstr-Transstr Lgc (RTL) The nerter crcut dscussed abe s a member f RTL famly f lgc gates. Plt f as a functn f s called the transfer characterstcs f the gate. T fnd the transfer characterstcs, we need t fnd fr a range f alues. When < γ, BJT wll be n cut-ff, C = 0 and =. Therefre, fr nput ltages belw certan threshld (dented by IL ), the gate utput s hgh. Fr ur crcut, IL = γ. When exceeds γ, BE junctn wll be frward based and a current B flws nt BJT: B = γ R B As BE junctn s frward based, BJT can be ether n saturatn r acte-lnear. Let s assume BJT s s n saturatn. In that case, = CE = sat and C / B < β. Then: C = sat B > C β = sat β ECE60L Lecture Ntes, Sprng 004 6
Therefre, BJT wll be n saturatn nly f B exceeds the alue gen by the frmula abe. Ths uccrs when becme large enugh: = γ R B B > γ R B sat β = IH Therefre, fr nput ltages larger than the a certan alue ( IH ), the gate utput s lw. Fr alues between these tw lmts, the BE junctn s frward based but the BJT s NOT n saturatn, therefre, t s n acte lnear. In ths case, the utput ltage smthly changes fr ts hgh alue t ts lw alue as s shwn n the plt f transfer characterstcs. Ths range f s a frbdden regn and the gate wuld nt wrk prperly n ths regn. Ths behar can als seen n the plt f the BJT lad lne. Fr small alues f ( B = 0) BJT s n cut-ff. As s ncreased, B s ncreased and the peratng pnt mes t the left and up n the lad lne and enters the acte-lnear regn. When B s rased abe certan lmt, the peratng pnt enters the saturatn regn. A majr drawback f the ths RTL nerter gate s the lmted nput range fr the lw sgnal ( IL ). Our analyss ndcated that IL = γ, that s the gate nput s lw fr ltages between 0 and γ 0.7. Fr ths analyss, we hae been usng a pecewse lnear mdel fr the BE junctn dde. In realty, the BJT wll cme ut f cut-ff (BE junctn wll cnduct) at smaller ltages (0.4 0.5 ). T resle ths shrtcmng, ne can add a resstr between the base and grund (r between base and a negate pwer supply) as s shwn. (Yu hae seen ths crcut n ECE0A, mtr dre crcut.) R B B R T see the mpact f ths resstr, nte that IL s the nput ltage when BJT s just leang the cut-ff regn. At ths pnt, BE = γ, and B s pste but ery small (effectely C ECE60L Lecture Ntes, Sprng 004 6
zer). Ntng that a ltage BE has appeared acrss R, we hae: = BE R = B = BE R IL = = R B BE = BE R B R BE = γ ( R B R Ths alue shuld be cmpared wth IL = γ n the absence f resstr R. It can be seen that fr R B = R, IL s rased frm 0.7 t.4 and fr R B = R, IL s rased t.. R des nt affect IH as B needed t put the BJT n saturatn s typcally seeral tmes larger than. ) RTL NOR Gate By cmbnng tw r mre RTL nerters, ne btans the basc lgc gate crcut f RTL famly, a NOR gate, as s shwn. Mre BJTs can be added fr addtnal nput sgnals. (Yu hae seen n 0B that all hgher leel lgc gates, e.g., flp-flps, can be made by a cmbnatn f NOR gates r NAN gates.) RB RB Exercse: Shw that ths a NOR gate,.e., the gate utput wll be lw as lng as at least ne f the nputs s hgh. RTLs were the frst dgtal lgc crcuts usng transstrs. They were replaced wth ther frms (T, TTL, and ECL) wth the adent f ntegrated crcuts. The majr prblem wth these crcuts are the use f large resstrs that wuld take large space n an IC chp (n tday s chp, resstr alues are lmted t abut 0 kω and capactance t abut 00 pf). Befre we me n t mre mdern gates, we cnsder tw mprtant characterstcs f a dgtal gate. ECE60L Lecture Ntes, Sprng 004 63
Swtchng Tme and Prpagatn elay: Cnsder the nerter gate wth an nput ltage clse t zer (and/r negate). In ths case, the BJT s n cut-ff, C = 0 and the utput f the gate s hgh. Suppse a hgh ltage s appled nstantaneusly t the gate at sme pnt. We expect BJT t enter saturatn wth C = I Csat and utput t drp t the lw state. Hweer, ths des nt ccur nstantaneusly. When the BJT s n cut-ff, BE junctn s reersed based. When a frward ltage s appled t the BE junctn, t takes sme tme fr the BE junctn transtn capactance t charge up. Tme s als requred fr mnrty carres t dffuse acrss the base and enter the cllectr. Ths results n the delay tme t d, whch s f the rder f a nansecnd fr a typcal BJT. Befre BJT can enter saturatn, t shuld traerse the acte-lnear regn. The rse tme, t r (n the rder f -0 ns) accunt fr ths transtn. The tme that takes fr the gate t swtch ON s represented by t n. Suppse that the nput ltage t gate s then reduced nstantaneusly t lw state. BJT wll leae saturatn regn and g t cut-ff. Agan, ths nt ccur nstantaneusly. When a BJT s n saturatn, bth BE and BC junctns are frward based and cnductng. As such, an excess mnrty charge s stred n the base. Fr the transstr t leae saturatn and enter acte-lnear (BC junctn t becme reersed based), ths excess charge must be remed. The tme requred fr the remal f excess charge determnes the strage tme, t s (rder f 00 ns). Then, transstr traerses the acte-lnear regn befre enterng cut-ff. Ths accunt fr the fall tme t f (-0 ns). The ttal tme t takes fr the gate t swtch OFF s represented by t ff. As can be seen, BJT swtchng s manly set by the strage tme, t s. Prpagatn delays ntrduced by transstr swtchng tme are mprtant cnstrants n desgnng faster chps. Gate desgns try t mnmze prpagatn delays as much as pssble. Fan-ut: All dgtal lgc crcuts are cnstructed wth crss-cuplng f seeral basc gates (such as NOR r NAN). As such, a basc gate may be attached t seeral ther gates. The maxmum number f gates that can be attached t a dgtal gate s called fan-ut. Obusly, ne wuld lke t hae large fan-ut. ECE60L Lecture Ntes, Sprng 004 64
de-transstr Lgc (TL) The basc gate f TL lgc crcuts s a NAN gate whch s cnstructed by a cmbnatn f a dde AN gate and a BJT nerter gate. de AN Gate: Frst, let s cnsder the dde AN gate as s shwn. T study the behar f the gate we wll cnsder the state f the crcut fr dfferent alues f and (ether 0 r 5 crrespndng t lw and hgh states). T ad the analyss, let s assume = 5 and R A = kω. We nte that by KCL, A = (assumng that there s n current drawn frm the crcut). Case, = = 0: Snce the 5- supply wll tend t frward bas bth and, let s assume that bth ddes are frward based. Thus, = = γ = 0.7 and > 0, > 0. In ths case: = = = 0.7 A = = 5 0.7 = 4.3 ma R A, 000 Current A wll be dded between tw ddes by KCL, each carryng ne half f A (because f symmtery). Thus, = =. ma. Snce dde currents are pste, ur assumptn f bth dde beng frward based s justfed and, therefre, = 0.7. S, when and are lw, and are ON and s lw. Case, = 0, = 5 : Agan, we nte that the 5- supply wll tend t frward bas. Assume s ON: = γ = 0.7 and > 0. Then: = = 0.7 = = 4.3 < γ and wll be OFF ( = 0). Then: A = = 5 0.7 = 4.3 ma R A, 000 = A = 4.3 0 = 4.3 ma Snce > 0, ur assumptn f beng frward based s justfed and, therefre, = 0.7. S, when s lw and s hgh, s ON and s OFF and s lw. ECE60L Lecture Ntes, Sprng 004 65 A R A
Case 3, = 5, = 0 : Because f the symmetry n the crcut, ths s exactly the same as case wth rles f and reersed. S, when s hgh and s lw, s OFF and s ON and s lw. Case 4, = = 5 : Examnng the crcut, t appears that the 5- supply wll NOT be able t frward bas and. Assume and are OFF: = = 0, < γ and < γ. Then: A = = 0 = R A = 5 0 = 5 = = 5 5 = 0 < γ and = = 5 5 = 0 < γ Thus, ur assumptn f bth ddes beng OFF arejustfed. S, when and are hgh, and are OFF and s hgh. Oerall, the utput f ths crcut s hgh nly f bth nputs are hgh (Case 4) and the utput s lw n all ther cases (Cases t 3). Thus, ths s an AN gate. Ths analyss can be easly extended t cases wth three r mre dde nputs. TL NAN Gate: The basc gate f TL lgc crcuts s a NAN gate whch s cnstructed by a cmbnatn f a dde AN gate and a BJT nerter gate as s shwn belw (left fgure). Because R B s large, n ICs, ths resstr s usually replaced wth tw ddes. The cmbnatn f the tw ddes and the BE junctn dde leads t a ltage f. fr the nerter t swtch and a IL =.4 fr the NAN gate (Why?). Resstr R s necessary because wthut ths resstr, current B wll be t small and the ltage acrss 3 and 4 wll nt reach 0.7 althugh they are bth frward based (Recall LE drer crcut f ECE0A n whch the LE started t lt fr n abut 0.8 nstead f estmated.4 ). A R A R B B C A R A 3 4 B C R ECE60L Lecture Ntes, Sprng 004 66
TLs were ery ppular n ICs n 60s and early 70s but are replaced wth Transstr- Transstr Lgc (TTL) crcuts. TTL are descrbed later, but as TTLs are eled frm TLs, sme examples f TL crcuts are gen belw. Example: erfy that the TL crcut shwn s a NAN gate. Assume that lw state s 0., hgh state s 5, and BJT β mn = 40. Case : = = 0. It appears that the 5- supply wll frward bas and. Assume and are frward based: = = γ = 0.7 and > 0, > 0. In ths case: 3 4 A 5kΩ 3 4 5 B 5kΩ C kω 3 = = = 0. 0.7 = 0.9 ltage 3 = 0.9 s nt suffcent t frward bas 3 and 4 as 3 = 3 4 BE and we need at least.4 t frward bas the tw ddes. S bth 3 and 4 are OFF and 4 = 0. (Nte that 3 and 4 can be frward based wthut BE junctn beng frward based as lng as the current 4 s small enugh such that ltage drp acrss the 5 kω resstr parallel t BE junctn s smaller than 0.7. In ths case, 5 = 4 and B = 0.) Then: = A = 5 3 5, 000 = 5 0.9 = 0.8 ma 5, 000 And by symmetry, = = 0.5 A = 0.4 ma. Snce bth and are pste, ur assumptn f and beng ON are justfed. Snce 4 = 0, B = 0 and BJT wll be n cut-ff wth C = 0 and = 5. S, n ths case, and are ON, 3 and 4 are OFF, BJT s n cut-ff, and = 5. Case : = 0., = 5 Fllwng arguments f case, assume s ON. Agan, 3 = 0.7 0. = 0.9, and 3 and 4 wll be OFF wth 4 = 0. We fnd that ltage acrss s = 3 = 0.9 5 = 4. and, thus, wll be OFF and = 0. Then: = A = 5 3 5, 000 = 5 0.9 = 0.8 ma 5, 000 and snce > 0, ur assumptn f ON s justfed. Snce 4 = 0, B = 0 and BJT wll be n cut-ff wth C = 0 and = 5. S, n ths case, s ON, s OFF, 3 and 4 are OFF, BJT s n cut-ff, and = 5. ECE60L Lecture Ntes, Sprng 004 67
Case 3: = 5, = 0. Because f the symmetry n the crcut, ths s exactly the same as case wth rles f and reersed. S, n ths case, s OFF, s ON, 3 and 4 are OFF, BJT s n cut-ff, and = 5. Case 4: = = 5 Examnng the crcut, t appears that the 5- supply wll NOT be able t frward bas and. Assume and are OFF: = = 0, < γ and < γ. On the ther hand, t appears that 3 and 4 wll be frward based. Assume 3 and 4 are frward based: 3 = 4 = γ = 0.7 and 4 > 0. Further, assume the BJT s nt n cut-ff BE = γ = 0.7 and B > 0. In ths case: 3 = 3 4 BE = 0.7 0.7 0.7 =. = 3 =. 5 =.9 < γ = 3 =. 5 =.9 < γ Thus, ur assumptn f and beng OFF are justfed. Furthermre: 4 = A = 5 3 5, 000 = 5. = 0.58 ma 5, 000 5 = BE 5, 000 = 0.7 = 0.4 ma 5, 000 B = 4 5 = 0.58 0.4 = 0.44 ma and snce 4 > 0 ur assumptn f 3 and 4 beng ON are justfed and snce B > 0 ur assumptn f BJT nt n cut-ff s justfed. We stll d nt knw f BJT s n acte-lnear r saturatn. Assume BJT s n saturatn: = CE = sat = 0. and C / B < β. Then, assumng n gate s attached t the crcut, we hae C = 5 sat, 000 = 5 0., 000 = 4.8 ma and snce C / B = 4.8/0.44 = < β = 40, ur assumptn f BJT n saturatn s justfed. S, n ths case, and are OFF, 3 and 4 are ON, BJT s n saturatn and = 0.. Oerall, the utput n lw nly f bth nputs are hgh, thus, ths s a NAN gate. Nte: It s nterestng t nte that at the nput f ths gate, the current actually flws ut f the gate. In the example abe, when bth nputs were hgh = = 0, when bth were lw = = 0.4 ma, and when ne nput was lw, e.g., was lw, = 0.8mA. The nput current flwng n (r ut f the gate n ths case) has mplcatns fr the fan-ut capablty f lgc gates as s shwn n the example belw. ECE60L Lecture Ntes, Sprng 004 68
Example: Fnd the fan-ut f ths NAN TL gate. Assume that lw state s 0., hgh state s 5, and BJT β = 40. Other gates 3 4 A 5kΩ 4 5 B 5kΩ R kω L C The crcut s the same TL NAN gate f preus example and we can use results frm preus example here. N ther NAN gates are attached t the utput f ths gate. Fan-ut s the maxmum alue f N. Snce we want t make sure that ur gate perates prperly under all cndtns, we shuld cnsder the wrst case, when all f the secnd stage gates hae maxmum currents. Fr a NAN TL gate, the maxmum current ccurs when all f the nputs are hgh wth exceptn f ne nput. We fund ths alue t be 0.8 ma (Cases & 3 n the preus example). Therefre, the wrst case s when the nput f all secnd stage gates are lw (fr the frst stage, = 0. ) and each draw a current 0.8 ma (a ttal f L = N 0.8 ma s drawn frm the frst stage gate). Cnsderng the frst stage gate, we had fund that = 0. nly fr Case 4. Fr that case, we fund B = 0.44 ma. Then: R = 5 sat, 000 = 5 0. = 4.8 ma, 000 C = R 0.8N = 4.8 0.8N The frst stage gate perates prperly as lng as the BJT s n saturatn,.e., C < β B 4.8 0.8N < 40 0.44 N < 3.7 As the fan-ut shuld be nteger, the fan-ut fr ths gate s 3. Fan-ut f TL gates can be greatly ncreased by a small mdfcatn. Fan-ut can be ncreased by ncreasng the base current f the BJT. B s, hweer, lmted by the current A (and 4 ). Reducng the alue f R A n the AN dde part f the crcut wll hae ncrease B. Unfrtunately, as ths resstr s reduced, pwer dsspatn n the gate ncreases and the fan-ut capablty decreases dramatcally. ECE60L Lecture Ntes, Sprng 004 69
A smple slutn whch keeps current A small but ncreases B drastcally s t replace dde 3 wth a BJT as s shwn. As can be seen, the TL NAN gate s nw made f 3 stages: ) nput stage (ddes), ) drer stage (frst BJT) and 3) utput stage (nd BJT). A R A 4 B C R Transstr-Transstr Lgc (TTL) A smplfed ersn f an IC-chp NPN transstr s shwn. The dece s fabrcated n a p-type substrate (r bdy) n a ertcal manner by embeddng alternatng layers f N and P-type semcnductrs. By embeddng mre than ne N- type emtter regn, ne can btan a multple-emtter NPN transstr as shwn. The multple-emtter NPN transstrs can be used t replace the nput ddes f a TL NAN gate and arre at a NAN gate entrely made f transstrs, hence Transstr-Transstr Lgc (TTL) gates. A smple TTL gate s shwn wth the multple-emtter BJT replacng the nput ddes. Ths transstr perates n reerse-acte mde,.e., lke a NPN transstr n actelnear mde but wth cllectr and emtter swtched. Operatnally, ths BJT acts as tw ddes back t back as shwn n the crcle at the bttm f the fgure. As such the peratn f ths gate s essentally smlar t the TL NAN gate descrbed abe (nte pstn f drer transstr and 4 dde s swtched). Crcut Symbl Smlar t TL NAN gates, a typcal TTL NAN gate has three stages: ) Input stage (mult-emtter transstr), ) drer stage, and 3) utput stage. Mdern TTL gates bascally hae the same cnfguratn as s shwn wth the exceptn that the utput stage s replaced wth the Ttem- Ple utput stage t ncrease swtchng speed and gate fanut. Fr a detaled descrptn f TTL gate wth Ttem- Ple utput stage, cnsult, Sedra and Smth (pages 75 t 80). A R A B R R C ECE60L Lecture Ntes, Sprng 004 70