Whits, EE 320 Lctur 19 Pag 1 of 10 Lctur 19: Common Emittr Amplifir with Emittr Dgnration. W ll continu our discussion of th basic typs of BJT smallnal amplifirs by studying a variant of th CE amplifir that has an additional rsistor addd to th mittr lad: (Fig. 1) (Sdra and Smith, 5 th d.) This is calld mittr dgnration and has th ffct of gratly nhancing th usfulnss of th CE amplifir. W ll calculat similar amplifir quantitis for this circuit as thos in th prvious lctur for th CE amplifir. 2017 Kith W. Whits
Whits, EE 320 Lctur 19 Pag 2 of 10 In contrast to th prvious lctur, w ll us a T small-nal modl (R in sris with r ) for th BJT and w ll also drop r o : it turns out to hav littl ffct hr but complicats th analysis. (Fig. 2) (Sdra and Smith, 5 th d.) Input rsistanc, R in. From this circuit, w s dirctly that th input rsistanc at th bas R ib is dfind as vi Rib ib (1) Notic that hr v i v, unlik th CE amplifir w/o mittr dgnration. Rfrring to th small-nal circuit w s that vi ir R (2) and i ib 1 (3) Substituting ths into (1) givs R 1 r R (4) ib
Whits, EE 320 Lctur 19 Pag 3 of 10 W s from this xprssion that th bas input rsistanc is +1 tims th total rsistanc in th mittr circuit. This is calld th rsistanc rflction rul and applis to th T smallnal BJT modl. [In th prvious lctur, w s in Fig. 7.56(b) that Rib r but r 1r, which obys this rsistanc rflction rul sinc thr is no R in that circuit.] This bas input rsistanc can b much largr than without th mittr rsistanc. That s oftn a good thing. Th dnr can chang R to achiv a dsird input rsistanc [> (+1)r ]. Th total input rsistanc to this CE amplifir with mittr dgnration is thn Rin RB Rib RB 1r R (7.153),(5) Small-nal voltag gain, G v. W ll first calculat th partial voltag gain vo Av (6) vi At th output, vo i RC RL (7) Substituting for i from (2) givs
Whits, EE 320 Lctur 19 Pag 4 of 10 A v R r C R R L (8) Th ovrall small-nal voltag gain G v (from th sourc to th load) is dfind as vo Gv (9) v W can quivalntly writ this voltag gain as vi vo vi Gv Av (10) v v v i with A v givn in (8). By simpl voltag division at th input to th small-nal quivalnt circuit Rin vi v (11) R R in Substituting this rsult and (8) into (10) yilds th final xprssion for th ovrall small-nal voltag gain v RC R o L Rin Gv (7.154),(12) v r R R R 6 in Using R in in (5) and assuming RB Rib thn (12) simplifis to G RC RL v (13) R 1 r R
Whits, EE 320 Lctur 19 Pag 5 of 10 Notic that this gain is actually smallr than th CE amplifir without mittr dgnration [bcaus of th (+1)R trm in th dnominator]. Howvr, bcaus of this trm it can b shown that th gain is lss snsitiv to variations in. Ovrall small-nal currnt gain, G i. By dfinition io Gi (14) i i Using currnt division at th output of th small-nal quivalnt circuit abov RC RC io ic i (15) RC RL RC RL whil at th input RB ib ii (16) RB Rib Substituting i 1ib and (16) into (15) givs RC RB io 1 ii RC RL RB Rib or RR B C Gi (17) R R R R C L B ib Short circuit currnt gain, A is. In th cas of a short circuit load (R L = 0), G i rducs to th short circuit currnt gain: RB Ais (18) R R B ib
Whits, EE 320 Lctur 19 Pag 6 of 10 In th usual circumstanc whn RB Rib, thn Ais (19) which is th sam valu as for a CE amplifir sinc thr is no R in this xprssion. Output rsistanc, R out. Rfrring to th small-nal quivalnt circuit abov and shorting out th input v = 0 Rout RC (20) which is th sam as th CE amplifir (whn ignoring r o ). Exampl N19.1 (Somwhat similar to txt Exampl 7.8). Givn a CE amplifir with mittr dgnration having = 100 and R R L 5 k. Th circuit is biasd as shown: (Fig. 3) (Sdra and Smith, 5 th d.)
Whits, EE 320 Lctur 19 Pag 7 of 10 Th small-nal quivalnt circuit for this CE amplifir with mittr dgnration is th sam as that shown arlir in this lctur: (Fig. 2) (Sdra and Smith, 5 th d.) With I E 1 ma, thn r VT IE 25 mv 1 ma 25. Find th valu of R that givs Ri n 4Rsg i 20 k. From (5) Rin RB Rib, which implis that Rib 25 k. Using (4) Rib Rib 1r R r R 1 Rib 25,000 or R r 25 222.5 1 101 W can choos this R without considring its ffct on I E or th bias bcaus a currnt sourc bias is bing usd. Tricky! If anothr biasing mthod wr usd, this might not b th cas.
Whits, EE 320 Lctur 19 Pag 8 of 10 Dtrmin th output rsistanc. From (20), Rout R C 8 k Comput th ovrall small-nal voltag gain. Using (12) v RC R o L Rin Gv v r R R in R 0.993,080 20,000 Gv 9.86 V/V 25 222.5 20,000 5,000 Dtrmin th opn circuit small-nal voltag gain, G vo. This is th ovrall gain but with an opn circuit load. Hnc, from this dscription w can dfin Gvo Gv R L Using (12) onc again but with R L givs for th CE amplifir with mittr dgnration RR C in Gvo (21) r RRin R For this particular amplifir 0.998,00020,000 G vo 25.6 V/V 25 222.5 20,000 5,000 Not that this is not th sam opn circuit gain A vo usd in th txt. A vo is th partial opn circuit voltag gain A A which using (8) is vo v R L
Whits, EE 320 Lctur 19 Pag 9 of 10 RC Avo (22) r R For this amplifir 0.998,000 A vo 32 V/V 25 222.5 Can you physically xplain why G vo and A vo ar diffrnt valus? Comput th ovrall currnt gain and th short circuit currnt gain. Using (17) RR B C Gi R R R R C L B ib 100100,0008,000 8,000 5,000100,000 25,000 49.2 A/A For th short circuit currnt gain, w us (18) RB 100100,000 Ais 80 A/A R R 100,000 25,000 B ib If v is limitd to 5 mv, what is th maximum valu for v with and without R includd? Find th corrsponding v o. To addrss this qustion, w nd an xprssion rlating v and v. From (11) w know that Rin vi v (11) Rin R whil at th input to th small-nal quivalnt circuit abov
Whits, EE 320 Lctur 19 Pag 10 of 10 v r r R v i (23) Substituting (11) into (23) givs Rin R r R v v (24) R r in With v 5 mv, thn 20,000 5,000 25 R v 5 mv 20,000 25 or v 0.2525 R mv (25) If R = 0, thn from (25) v 6.25 mv is th maximum input nal voltag and bcaus Gv 9.86 V/V, th corrsponding output voltag is vo Gv v 61.6 mv. If R = 222.5, thn from (25) v 61.9 mv is th maximum input nal voltag and th corrsponding output voltag is v 610.1 mv. o This is a dmonstration of yt anothr bnfit of mittr dgnration: th amplifir can handl largr input nals (and hnc potntially largr output voltags) without incurring nonlinar distortion.