5.6 Small-Signal Operation and Models

Transcription

1 3/16/2011 secton 5_6 Small Sgnal Operaton and Models 1/2 5.6 Small-Sgnal Operaton and Models Readng Assgnment: Now let s examne how we use BJTs to construct amplfers! The frst mportant desgn rule s that the BJT must be based to the acte mode. HO: BJT GAIN AND THE ATIVE REGION For a BJT amplfer, we fnd that eery current and eery oltage has two components: the D (.e., bas) component a alue carefully selected and desgned by a EE, and the small-sgnal component, whch s the A sgnal we are attemptng to amplfy (e.g., audo, deo, etc.). HO:D AND SMALL-SIGNAL OMPONENTS There are two extremely mportant crcut elements n small-sgnal amplfer desgn: the apactor of Unusual Sze (OUS) and the Inductor of Unusual Sze (IOUS). These deces are just realzable approxmatons of the Unfathomably Large apactor (UL) and the Unfathomably Large Inductor (ULI). These deces hae radcally dfferent propertes when consderng D and small-sgnal components! HO:D AND A IMPEDANE OF REATIVE ELEMENTS Jm Stles The Un. of Kansas Dept. of EES

2 3/16/2011 secton 5_6 Small Sgnal Operaton and Models 2/2 It turns out that separatng BJT currents and oltages nto D and small-sgnal components s problematc! HO:THE SMALL-SIGNAL IRUIT EQUATIONS But, we can approxmately determne the small-sgnal components f we use the small-sgnal approxmaton. HO: A SMALL-SIGNAL ANALYSIS OF HUMAN GROWTH HO: A SMALL-SIGNAL ANALYSIS OF A BJT Let s do an example to llustrate the small-sgnal approxmaton. EXAMPLE: SMALL-SIGNAL BJT APPROXIMATIONS There are seeral small-sgnal parameters that can be extracted from a small-sgnal analyss of a BJT. HO: BJT SMALL-SIGNAL PARAMETERS HO: THE SMALL-SIGNAL EQUATION MATRIX Let s do an example! EXAMPLE: ALULATING THE SMALL-SIGNAL GAIN Jm Stles The Un. of Kansas Dept. of EES

3 3/16/2011 BJT Gan and the Acte Regon 1/4 BJT Amplfer Gan and the Acte Regon onsder ths smple BJT crcut: Q: Oh, goody you re gong to waste my tme wth another of these pontless academc problems. Why can t you dscuss a crcut that actually does somethng? R V O I R B A: Actually, ths crcut s a fundamental electronc dece! To see what ths crcut does, plot the output oltage O as a functon of the nput I. V O BJT n cutoff BJT n acte mode BJT n saturaton I V

4 3/16/2011 BJT Gan and the Acte Regon 2/4 Note that: I O Mode 0 V utoff V 0 Saturaton Why, ths dece s not useless at all! It s clearly a: Dgtal deces made wth BJTs typcally work n ether the cutoff of saturaton regons. So, what good s the BJT Acte Mode?? Sr, t appears to me that the acte regon s just a useless BJT mode between cutoff and saturaton.

5 3/16/2011 BJT Gan and the Acte Regon 3/4 Actually, we wll fnd that the acte mode s extremely useful! To see why, take the derate of the aboe crcut s transfer functon (.e., dv dv ): O I V O V dv dv O I V I We note that n cutoff and saturaton: V whle n the acte mode: dv O dv I 0 dvo 1 dv >> I Q: I e got better thngs to do than lsten to some egghead professor mumble about derates. Are these results een remotely mportant?

6 3/16/2011 BJT Gan and the Acte Regon 4/4 A: Snce n cutoff and saturaton dv dv = 0, a small change n nput oltage V I wll result n almost no change n output oltage V O. ontrast ths wth the acte regon, where dv dv >> 1. Ths means that a small change n nput oltage V I results n a large change n the output oltagev O! O I O I I see. A small oltage change results n a bg oltage change t s oltage gan! The acte mode turns out to be excellent. Whereas the mportant BJT regons for dgtal deces are saturaton and cutoff, bpolar juncton transstors n lnear (.e., analog) deces are typcally based to the acte regon. Ths s especally true for BJT amplfer. Almost all of the transstors n EES 412 wll be n the acte regon ths s where we get amplfer gan!

7 3/16/2011 D and Small Sgnal omponents 1/5 D and Small-Sgnal omponents Note that we hae used D sources n all of our example crcuts thus far. We hae done ths just to smplfy the analyss generally speakng, realstc (.e., useful) juncton dode crcuts wll hae sources that are tme-aryng! The result wll be oltages and currents n the crcut that wll t ). lkewse ary wth tme (e.g., t ( ) and ( ) For example, we can express the forward bas juncton dode equaton as: D( t) nvt t = I e D ( ) Although source oltages S ( ) t can be any S general functon of tme, we wll fnd that often, n realstc and useful electronc crcuts, that the source can be decomposed nto two separate components the D component V S, and the small-sgnal component ( t ). I.E.: s s t or currents ( ) ( t ) = V ( t ) S S s Jm Stles The Un. of Kansas Dept. of EES

8 3/16/2011 D and Small Sgnal omponents 2/5 Let s look at each of these components nddually: * The D component V S s exactly what you would expect the D component of source S( t )! Note ths D alue s not a functon of tme (otherwse t would not be D!) and therefore s expressed as a constant ( e.g., V = 12. 3V ). S Mathematcally, ths D alue s the tme-aeraged alue of ( ) S t : 1 T V S = S( t) dt T where T s the tme duraton of functon S ( t ). 0 * As the notaton ndcates, the small-sgnal component s( t ) s a functon of tme! Moreoer, we can see that ths sgnal s an A sgnal, that s, ts tme-aeraged alue s zero! I.E.: 1 T T 0 ( t) dt = 0 s Ths sgnal ( t ) s also referred to as the small-sgnal component. s * The total sgnal ( t ) s the sum of the D and small sgnal S components. Therefore, t s nether a D nor an A sgnal! Jm Stles The Un. of Kansas Dept. of EES

9 3/16/2011 D and Small Sgnal omponents 3/5 Pay attenton to the notaton we hae used here. We wll use ths notaton for the remander of the course! * D alues are denoted as upper-case arables (e.g., V S, I R, or V D ). * Tme-aryng sgnals are denoted as lower-case arables (e.g., ( t), ( t), ( t )). S r D Also, * A sgnals (.e., zero tme aerage) are denoted wth lower-case subscrpts (e.g., ( t), ( t), ( t )). * Sgnals that are not A (.e., they hae a nonzero D component!) are denoted wth upper-case subscrpts (e.g., ( t), I, ( t), V ). S D R D s d r Note we should neer use arables of the form V, I, V. Do you see why?? e b Q: You say that we wll often fnd sources wth both components a D and small-sgnal component. Why s that? What s the sgnfcance or physcal reason for each component? Jm Stles The Un. of Kansas Dept. of EES

10 3/16/2011 D and Small Sgnal omponents 4/5 A1: Frst, the D component s typcally just a D bas. It s a known alue, selected and determned by the desgn engneer. It carres or relates no nformaton the only reason t exsts s to make the electronc deces work the way we want! A2: onersely, the small sgnal component s typcally unknown! It s the sgnal that we are often attemptng to process n some manner (e.g., amplfy, flter, ntegrate). The sgnal tself represents nformaton such as audo, deo, or data. Sometmes, howeer, ths small, A, unknown sgnal represents not nformaton but nose! opyrght 2009 Lon Precson. Nose s a random, unknown sgnal that n fact masks and corrupts nformaton. Our job as desgners s to suppress t, or otherwse mnmze t deleterous effects. Jm Stles The Un. of Kansas Dept. of EES

11 3/16/2011 D and Small Sgnal omponents 5/5 * Ths nose may be changng ery rapdly wth tme (e.g., MHz), or may be changng ery slowly (e.g., mhz). * Rapdly changng nose s generally thermal nose, whereas slowly aryng nose s typcally due to slowly aryng enronmental condtons, such as temperature. Note that n addton to (or perhaps because of) the source oltage ( t ) hang both a D bas and small-sgnal component, S all the currents and oltages (e.g., ( ), ( ) R t D t ) wthn our crcuts wll lkewse hae both a D bas and small-sgnal component! For example, the juncton dode oltage mght hae the form: It s hopefully edent that: ( D t ) = co. s ωt D VD = 066. V and ( t) = 0001cos. ω t d t Jm Stles The Un. of Kansas Dept. of EES

12 3/16/2011 D and A Impedance of Reacte Elements 1/6 D and A Impedance of Reacte Elements Now, recall from EES 211 the complex mpedances of our basc crcut elements: ZR = R Z = 1 jω ZL = jωl For a D sgnal ( ω = 0), we fnd that: ZR = R Z 1 = lm = ω 0 jω Z = j(0) L = 0 L Thus, at D we know that: * a capactor acts as an open crcut (I =0). * an nductor acts as a short crcut (V L = 0). Jm Stles The Un. of Kansas Dept. of EES

13 3/16/2011 D and A Impedance of Reacte Elements 2/6 Now, let s consder two mportant cases: 1. A capactor whose capactance s unfathomably large. 2. An nductor whose nductance L s unfathomably large. 1. The Unfathomably Large apactor In ths case, we consder a capactor whose capactance s fnte, but ery, ery, ery large. For D sgnals ( ω = 0), ths dece acts stll acts lke an open crcut. Howeer, now consder the A sgnal case (e.g., a small sgnal), where ω 0. The mpedance of an unfathomably large capactor s: Zero mpedance! Z 1 = lm = 0 jω An unfathomably large capactor acts lke an A short. Jm Stles The Un. of Kansas Dept. of EES

14 3/16/2011 D and A Impedance of Reacte Elements 3/6 Qute a trck! The unfathomably large capactance acts lke an open to D sgnals, but lkewse acts lke a short to A (small) sgnals! ( t) = 0 c I = 0 = lm Q: I fal to see the releance of ths analyss at ths juncture. After all, unfathomably large capactors do not exst, and are mpossble to make (beng unfathomable and all). A: True enough! Howeer, we can make ery bg (but fathomably large) capactors. Bg capactors wll not act as a perfect A short crcut, but wll exhbt an mpedance of ery small magntude (e.g., a few Ohms), proded that the A sgnal frequency s suffcently large. In ths way, a ery large capactor acts as an approxmate A short, and as a perfect D open. Jm Stles The Un. of Kansas Dept. of EES

15 3/16/2011 D and A Impedance of Reacte Elements 4/6 We call these large capactors D blockng capactors, as they allow no D current to flow through them, whle allowng A current to flow nearly unmpeded! Q: But you just sad ths s true proded that the A sgnal frequency s suffcently large. Just how large does the sgnal frequency ω need to be? A: Say we desre the A mpedance of our capactor to hae a magntude of less than ten Ohms: Z < 10 Rearrangng, we fnd that ths wll occur f the frequency ω s: 10 > Z 1 10 > ω 1 ω > 10 For example, a 50 µf capactor wll exhbt an mpedance whose magntude s less than 10 Ohms for all A sgnal frequences aboe 320 Hz. Jm Stles The Un. of Kansas Dept. of EES

16 3/16/2011 D and A Impedance of Reacte Elements 5/6 Lkewse, almost all A sgnals n modern electroncs wll operate n a spectrum much hgher than 320 Hz. Thus, a 50 µf blockng capactor wll approxmately act as an A short and (precsely) act as a D open. 2. The Unfathomably Large Inductor Smlarly, we can consder an unfathomably large nductor. In addton to a D mpedance of zero (a D short), we fnd for the A case (where ω 0): Z = lm jωl = L L In other words, an unfathomably large nductor acts lke an A open crcut! V = 0 ( t ) = 0 L = lm L L The unfathomably large nductor acts lke an short to D sgnals, but lkewse acts lke an open to A (small) sgnals! Jm Stles The Un. of Kansas Dept. of EES

17 3/16/2011 D and A Impedance of Reacte Elements 6/6 As before, an unfathomably large nductor s mpossble to buld. Howeer, a ery large nductor wll typcally exhbt a ery large A mpedance for all but the lowest of sgnal frequences ω. We call these large nductors A chokes (also known RF chokes), as they act as a perfect short to D sgnals, yet so effectely mpede A sgnals (wth suffcently hgh frequency) that they act approxmately as an A open crcut. For example, f we desre an A choke wth an mpedance magntude greater than 100 kω, we fnd that: Z L > 10 5 ωl > 10 5 ω > 5 10 L Thus, an A choke of 50 mh would exhbt an mpedance magntude of greater than 100 kω for all sgnal frequences greater than 320 khz. Note that ths s stll a farly low sgnal frequency for many modern electronc applcatons, and thus ths nductor would be an adequate A choke. Note howeer, that buldng and A choke for audo sgnals (20 Hz to 20 khz) s typcally ery dffcult! Jm Stles The Un. of Kansas Dept. of EES

18 3/28/2011 The Small Sgnal rcut Equatons 1/4 The Small-Sgnal rcut Equatons Now let s agan consder ths crcut, where we assume the BJT s n the acte mode: V R O I B R B BE The four equatons descrbng ths crcut are: 1) R = 0 (KVL) I B B BE 2) = β (BJT) B 3) = V R (KVL) O 4) BE VT = (BJT) I e s

19 3/28/2011 The Small Sgnal rcut Equatons 2/4 Now, we assume that each current and oltage has both a smallsgnal and D component. Wrtng each equaton explctly n terms of these components, we fnd that the four crcut equatons become: (1) ( VI ) RB ( IB b) ( VBE be) = 0 ( V R I V ) ( R ) = 0 I B B BE B b be (2) I c = β ( IB b) I = β I β c B b (3) VO o = V R( I c) V = ( V R I ) R O o c (4) I = I e c s ( VBE be ) V BE be VT VT I = I e e c s V T Note that each equaton s really two equatons! 1. The sum of the D components on one sde of the equal sgn must equal the sum of the D components on the other. 2. The sum of the small-sgnal components on one sde of the equal sgn must equal the sum of the small-sgnal components on the other. Ths result can greatly smplfy our quest to determne the small-sgnal amplfer parameters!

20 3/28/2011 The Small Sgnal rcut Equatons 3/4 You see, all we need to do s determne four smallsgnal equatons, and we can then sole for the four small-sgnal alues,,,! b c be o From (1) we fnd that the D equaton s: V R I V = I B B BE 0 whle the small-sgnal equaton from 1) s: R = B b be 0 Smlarly, from equaton (2) we get these equatons: I = β I (D) B c = β (small sgnal) b And from equaton (3): V = V R I (D) O o = R (small-sgnal) c

21 3/28/2011 The Small Sgnal rcut Equatons 4/4 Fnally, from equaton (4) we, um, get, er just what the heck do we get? (4) I = I e c s ( VBE be ) V BE be VT VT I = I e e c s V T???? Q: Jeepers! Just what are the D and small-sgnal components of: BE be VT VT I e e??? s V A: Precsely speakng, we cannot express the aboe expresson as the sum of a D and small-sgnal component. Yet, we must determne a fourth small-sgnal equaton n order to determne the four small sgnal alues,,,! b c Howeer, we can approxmate the aboe expresson as the sum of D and small-sgnal components. To accomplsh ths, we must apply the small-sgnal approxmaton (essentally a Taylor seres approx.). be o We wll fnd that the small-sgnal approxmaton prodes an accurate small-sgnal equaton for expressons such (4). We wll lkewse fnd that ths approxmate equaton s accurate f the small-sgnal oltage be s, well, small!

22 3/16/2011 A Small-Sgnal Analyss of Human Growth 1/6 A Small-Sgnal Analyss of Human Growth Say the aerage heght h of a human (n nches) s related to hs/her age t n months by ths equaton: ht ( ) = x10 (45 t12) nches 65 nches ( ) h t Those awkward adolescent years! We shrnk when we age! t 70 years Say that we want to calculate the aerage heght of a human at an age of t =58, 59, 59.5, 60, 60.5, 61, and 62 months. Whew! Let me get out my calculator!

23 3/16/2011 A Small-Sgnal Analyss of Human Growth 2/6 ht ( = 580). = nches ht ( = 590). = nches ht ( = 595). = nches ht ( = 60. 0) = nches ht ( = 60. 5) = nches ht ( = 61. 0) = nches ht ( = 62. 0) = nches Q: Wow, ths was hard. Isn t there an easer way to calculate these alues? A: Yes, there s! We can make a small-sgnal approxmaton. For a small-sgnal approxmaton, we smply need to calculate two alues. Frst: ht ( ) = ht ( = 60) = nches t = 60 In other words, the aerage heght of a human at 60 months (.e., 5 years) s nches. Lkewse, we calculate the tme derate of ht ( ), and then ealuate the result at 60 months: dht ( ) = ( 2059x (45 t 12) 5.75 ) = dt t = 60 = x10 ( ) = nches/month t 60

24 3/16/2011 A Small-Sgnal Analyss of Human Growth 3/6 In other words, the aerage 5 year old grows at a rate of 0.34 nches/month! Now let s agan consder the earler problem. If we know that an aerage 5-year old s nches tall, and grows at a rate of 0.34 nches/month, then at 5 years and one month (.e., 61 months), the lttle bugger wll approxmately be: (0.34)(1) = nches ompare ths to the exact alue of nches a ery accurate approxmaton. We can lkewse approxmate the aerage heght of a 59-month old (.e., 5 years mnus one month): (0.34)( 1) = nches or of a 62-month old (.e., 5 years plus two months): (0.34)(2) = nches Note agan the accuracy of these approxmatons! For ths approxmaton, let us defne tme t =60 as the ealuaton pont, or bas pont T : T ealuaton pont

25 3/16/2011 A Small-Sgnal Analyss of Human Growth 4/6 We can then defne: t = t T In ths example then, T = 60 months, and the alues of t range from 2 to 2 months. For example, t = 59 months can be expressed as t = T t, where T = 60 months and t = 1 month. We can therefore wrte our approxmaton as: ht ( ) dht ( ) ht ( ) t t= T dt t= T For the example where T =60 months we fnd: dht ( ) ht ( ) ht ( ) t t = 60 dt = t t = 60 Ths approxmaton s not accurate, howeer, f t s large. For example, we can determne from the exact equaton that the aerage heght of a forty-year old human s: ht= ( 480) = 65 nches or about 5 feet 5 nches.

26 3/16/2011 A Small-Sgnal Analyss of Human Growth 5/6 Howeer, f we were to use our approxmaton to determne the aerage heght of a 40-year old ( t = t T = = 420 ), we would fnd: ht ( ) (420) = nches The approxmaton says that the aerage 40-year old human s oer 15 feet tall! Where exactly do I fnd these dad-gum humans? The reason that the aboe approxmaton prodes an naccurate answer s because t s based on the assumpton that humans grow at a rate of 0.34 nches/month. Ths s true for 5-year olds, but not for 40-year olds (unless, of course, you are referrng to ther wastlnes)! We thus refer to the approxmaton functon as a small-sgnal approxmaton, as t s ald only for tmes that are slghtly dfferent from the nomnal (ealuaton) tme T (.e., t s small).

27 3/16/2011 A Small-Sgnal Analyss of Human Growth 6/6 If we wsh to hae an approxmate functon for the growth of humans who are near the age of forty, we would need to construct a new approxmaton: dht ( ) ht ( ) ht ( ) t t = 480 dt t = x 10 6 =.. t Note that forty-year old humans hae stopped growng! The mathematcally astute wll recognze the small-sgnal model as a frst-order Taylor Seres approxmaton!

28 3/28/2011 A Small Sgnal Analyss of a BJT lecture 1/12 A Small-Sgnal Analyss of a BJT The collector current of a BJT s related to ts base-emtter oltage BE as: = I e S BE V T BE Jm Stles The Un. of Kansas Dept. of EES

29 3/28/2011 A Small Sgnal Analyss of a BJT lecture 2/12 One messy result Say the current and oltage hae both D.. ( I, V ) and small-sgnal (, ) components: BE c be and ( t) = I ( t) c ( t) = V ( t) BE BE be Therefore, the total collector current s: ( t) = I e S I ( t) = I e c S BE ( t) V T VBE be( t) V T Jm Stles The Un. of Kansas Dept. of EES

30 3/28/2011 A Small Sgnal Analyss of a BJT lecture 3/12 Apply the Small-Sgnal Approxmaton Q: Ykes! The exponental term s ery messy. Is there some way to approxmate t? A: Yes! The collector current c s a functon of base emtter oltage BE. Let s perform a small-sgnal analyss to determne an approxmate relatonshp between c and BE. Note that the alue of ( t) = V ( t) s always ery close to the D.. BE BE be oltage for all tme t (snce ( t ) s ery small). be We therefore wll use ths D.. oltage as the ealuaton pont (.e., bas pont) for our small-sgnal analyss. Jm Stles The Un. of Kansas Dept. of EES

31 3/28/2011 A Small Sgnal Analyss of a BJT lecture 4/12 How fast t grows! We frst determne the alue of the collector current when the base emtter oltage BE s equal to the D alue V BE : BE = BE VT VT BE VBE Is e = I e = I = BE = VBE s V Of course, the result s the D.. collector current I. We now determne the change n collector current due to a change n baseemtter oltage (.e., a frst derate), ealuated at the D.. oltage V BE : d d BE BE = V BE S ( S exp BE T ) d I V = d I = V I = V BE BE V T T BE T = V S T e e V V BE BE BE = V BE A V Jm Stles The Un. of Kansas Dept. of EES

32 3/28/2011 A Small Sgnal Analyss of a BJT lecture 5/12 A smple approxmaton Thus, when the base-emtter oltage s equal to the D.. bas oltage V BE, the collector current wll equal the D.. bas current I. Lkewse, ths collector current wll ncrease (decrease) by an amount of BE T ( ) V V IS V e ma for eery 1mV ncrease (decrease) n T BE. Thus, we can easly approxmate the collector current when the base-emtter oltage s equal to alues such as: Respectely, the answers are: BE BE BE BE = VBE 1mV = VBE 3mV = VBE 2mV = V 05mV. BE VT = I ( I V ) e (1) ma S T VT = I ( I V ) e (3) ma S T VT = I ( I V ) e (-2) ma S T VT = I ( I V ) e (-0.5) ma S T V V V V BE BE BE BE where we hae assumed that scale current I S s expressed n ma, and thermal oltage V T s expressed n mv. Jm Stles The Un. of Kansas Dept. of EES

33 3/28/2011 A Small Sgnal Analyss of a BJT lecture 6/12 The small sgnal approxmaton Recall that the small-sgnal oltage be( t ) represents a small change n BE ( t ) from ts nomnal (.e., bas) oltage V BE. For example, we mght fnd that the alue of ( t ) at four dfferent tmes t are: ( t ) = 1 mv be be be be 1 ( t ) = 3mV be ( t ) = 2 mv ( t ) = 0. 5 mv Thus, we can approxmate the collector current usng the small-sgnal approxmaton as: ( ) VBE VT ( t) = I I V e ( t) S T be where of course I VBE VT = I e. S Ths s a ery useful result, as we can now explctly determne an expresson for the small-sgnal current c( t )! Jm Stles The Un. of Kansas Dept. of EES

34 3/28/2011 A Small Sgnal Analyss of a BJT lecture 7/12 The small-sgnal collector current Recall ( t) = I ( t), therefore: c ( ) VBE VT ( t) = I ( t) = I I V e ( t) c S T be Subtractng the D.. current from each sde, we are left wth an expresson for the small-sgnal current c( t ), n terms of the small-sgnal oltage be( t ) : ( ) VBE VT ( t) = I V e ( t) c S T be We can smplfy ths expresson by notng that I VBE VT = I e, resultng n: S and thus: ( S ) I V e T V V S BE T = = I e I V T V V T BE V T I c( t) = be( t) V T Jm Stles The Un. of Kansas Dept. of EES

35 3/28/2011 A Small Sgnal Analyss of a BJT lecture 8/12 Transconductance: A small sgnal parameter We defne the alue I V T as the transconductance g m : g m = I V T A V and thus the small-sgnal equaton smply becomes: ( t) = g ( t) c m be Jm Stles The Un. of Kansas Dept. of EES

36 3/28/2011 A Small Sgnal Analyss of a BJT lecture 9/12 How transstors got ther name Let s now consder for a moment the transconductance g m. The term s short for transfer conductance: conductance because ts unts are amps/olt, and transfer because t relates the collector current to the oltage from base to emtter the collector oltage s not releant (f n acte mode)! Note we can rewrte the small-sgnal equaton as: be( t) 1 = ( t) g c m The alue (1 g m ) can thus be consdered as transfer resstance, the alue descrbng a transfer resstor. Transfer Resstor we can shorten ths term to Transstor (ths s how these deces were named)! Jm Stles The Un. of Kansas Dept. of EES

37 3/28/2011 A Small Sgnal Analyss of a BJT lecture 10/12 We can summarze our results as: Summarzng I = I e S V BE V T D.. Equaton ( t) = g ( t) Small-Sgnal Equaton c m be ( t) = I g ( t) Small-Sgnal Approxmaton m be Note that we know hae two expressons for the total (D.. plus small-sgnal) collector current. The exact expresson: ( t) = I e S VBE be( t) V T and the small-sgnal approxmaton: ( t) = I g ( t) m be Jm Stles The Un. of Kansas Dept. of EES

38 3/28/2011 A Small Sgnal Analyss of a BJT lecture 11/12 Accurate oer a small regon Let s plot these two expressons and see how they compare: Exact Small-sgnal Valdty Regons g m I BE V BE It s edent that the small-sgnal approxmaton s accurate (t prodes nearly the exact alues) only for alues of near the D. bas alue I, and only for alues of near the D. bas alue V. The pont ( V BE, I ) s alternately known as the D.. bas pont, the transstor operatng pont, or the Q-pont. Jm Stles The Un. of Kansas Dept. of EES

39 3/28/2011 A Small Sgnal Analyss of a BJT lecture 12/12 hange the D bas, change the transconductance Note f we change the D.. bas of a transstor crcut, the transstor operatng pont wll change. The small-sgnal model wll lkewse change, so that t prodes accurate results n the regon of ths new operatng pont: Exact Small-sgnal Valdty Regons g m I BE Jm Stles The Un. of Kansas Dept. of EES V BE

40 3/30/2011 Example A Small Sgnal BJT Approxmaton 1/3 Example: Small-Sgnal BJT Approxmatons Say that we wsh to fnd the collector current of a BJT 12 based n the acte mode, wth = 10 A and a base-emtter oltage of: BE IS = cosωt V Q: Easy! Snce: = I e S BE V T we fnd: 0. 6 V ( 0.001cosωt V T T ) ( t) = I e e S rght? A: Although ths answer s defntely correct, t s not ery useful to us as engneers. learly, the base-emtter oltage conssts of a D.. bas term (0.6 V) and a small-sgnal term (0. 001cosω t ). Accordngly, we are nterested n the D.. collector current I and the small-sgnal collector current c. The D.. collector current s obously:

41 3/30/2011 Example A Small Sgnal BJT Approxmaton 2/3 I = I e S = e = 26 ma V T But how do we determne the small-sgnal collector current ( ) c t from: 0. 6 V ( 0.001cosωt V T T ) ( t) = I e e S??? The answer, of course, s to use the small-sgnal approxmaton. We know that: where: ( t) = g ( t) c m be g m I 26mA = = = 106. Ω V 25mV T 1 Therefore, the small-sgnal collector current s approxmately: ( t) = g ( t) c m be ( ωt ) = cos = 1.06cosωt ma and therefore the total collector current s:

42 3/30/2011 Example A Small Sgnal BJT Approxmaton 3/3 Q: Say the D.. bas oltage ncreases from V = 06. Vto V = 07. V. What happens to the BJT collector current? BE BE A: The D.. bas current becomes: V T I = I e = 10 e = 1446 ma!!! S snce the transconductance s now: g m I 1446mA = = = Ω V 25mV T 1 the small-sgnal collector current s: ( t) = g ( t) c m be ( ωt ) = cos = 57.8cosωt ma Qute an ncrease! hangng the transstor operatng pont (.e., the D bas pont) wll typcally make a bg dfference n the small-sgnal result!

43 3/30/2011 BJT Small Sgnal Parameters lecture 1/5 BJT Small-Sgnal Parameters We know that the followng small-sgnal relatonshps are true for BJTs: = β = g c b c m be Q: What other relatonshp can be dered from these two?? A: Well, one obous relatonshp s determned by equatng the two equatons aboe: β = β = g = g c b m be be b m We can thus defne the small-sgnal parameter r π as: β βv V g I I T T = = m B r π Jm Stles The Un. of Kansas Dept. of EES

44 3/30/2011 BJT Small Sgnal Parameters lecture 2/5 Small-sgnal base resstance Therefore, we can wrte the new BJT small-sgnal equaton: = r be π b The alue r π s commonly thought of as the small-sgnal base resstance. We can lkewse defne a small-sgnal emtter resstance: r e be e We begn wth the small-sgnal equaton c = α e. ombnng ths wth c = g m be, we fnd: α = α = g c e m be be = e g m Jm Stles The Un. of Kansas Dept. of EES

45 3/30/2011 BJT Small Sgnal Parameters lecture 3/5 Small-sgnal emtter resstance We can thus defne the small-sgnal parameter r e as: αv V α = T = T g I I m E r e Therefore, we can wrte another new BJT small-sgnal equaton: be = r e e Note that n addton to β, we now hae three fundamental BJT small-sgnal parameters: I V V gm = r = r = V I I T T π e T B E Jm Stles The Un. of Kansas Dept. of EES

46 3/30/2011 BJT Small Sgnal Parameters lecture 4/5 These results are not ndependent! Snce I = β IB ( I = α I E ), we fnd that these small sgnal alues are not ndependent. If we know two of the four alues β, g, r, r, we can determne all four! m π e g m α β r r π e = = = r r r r e π π e β re r = = ( β 1 π ) r = e g 1 g r m m e α r r π π r = = = e g β 1 1 g r m m π Jm Stles The Un. of Kansas Dept. of EES

47 3/30/2011 BJT Small Sgnal Parameters lecture 5/5 Make sure you can dere these! The results on the preous page are easly determned from the equatons: I V V gm = r = r = V I I T T π e T B E I = I I E B I = β I B I = α I E Make sure you can dere them! Jm Stles The Un. of Kansas Dept. of EES

48 3/30/2011 The Small Sgnal Equaton Matrx lecture 1/2 The Small-Sgnal Equaton Matrx We can summarze our small-sgnal equatons wth the small-sgnal equaton matrx. Note ths matrx relates the small-sgnal BJT parameters be, b, c, and e. olumn Parameters be b c e be 1 r = β 1 π g m g m α re = g m Row Parameters b 1 gm = 1 r β π 1 1 β ( β 1) c g m β 1 α = β β 1 e 1 gm = β 1 r α e 1 β 1 = α β 1 Jm Stles The Un. of Kansas Dept. of EES

49 3/30/2011 The Small Sgnal Equaton Matrx lecture 2/2 Here s how you use ths To use ths matrx, note that the row parameter s equal to the product of the column parameter and the matrx element. For example: 1 b = r π be be b c e be 1 r = β 1 π g m gm α re = g m b 1 gm = 1 r β π 1 1 β ( β 1) c g m β 1 α = β β 1 e 1 gm = β 1 r α e 1 β 1 = α β 1 Jm Stles The Un. of Kansas Dept. of EES

50 3/30/2011 Example alculatng the Small Sgnal Gan 1/2 Example: alculatng the Small-Sgnal Gan For ths crcut, we hae now determned (f BJT s n acte mode), the small-sgnal equatons are: V 1) = R b B be R 2) c = β b O 3) o = R c c I B R B BE 4) g c m be Q: So, can we now determne the small-sgnal open-crcut oltage gan of ths amplfer? I.E.: A o o( t) = ( t) A: Look at the four small-sgnal equatons there are four unknowns (.e.,,,, )! b be ombnng equatons 2) and 4), we get: c o

51 3/30/2011 Example alculatng the Small Sgnal Gan 2/2 β = = r g be b π b m Insertng ths result n equaton 1), we fnd: ( ) = R r B π b Therefore: and snce c = β : b c b = R B B r π π β = R r whch we nsert nto equaton 3): β R = R = o c R r B π Therefore, the small-sgnal gan of ths amplfer s: ( t) β R o A = = o ( t) R r B π Note ths s the small sgnal gan of ths amplfer and ths amplfer only!

52 3/30/2011 The Hybrd P and T Models lecture 1/6 The Hybrd-Π and T Models onsder agan the small-small sgnal equatons for an npn BJT based n the acte mode: be = = g = β = r b c m be b e b c π ( KL) Now, analyze ths crcut: b c r π be - g m = β be b e Jm Stles The Un. of Kansas Dept. of EES

53 3/30/2011 The Hybrd P and T Models lecture 2/6 Do these equatons look famlar? From Ohm s Law: b = r be π b c From KL: = g = β c m be b r π be - g m = β be b And also from KL: = ( KL) e b c e Q: Hey! Aren t these the same three equatons as the npn BJT small-sgnal equatons? A: They are ndeed! Wth respect to the small-sgnal currents and oltages n a crcut (but only small-sgnal oltages and currents), an npn BJT n acte mode mght as well be ths crcut. Jm Stles The Un. of Kansas Dept. of EES

54 3/30/2011 The Hybrd P and T Models lecture 3/6 Two equalent crcuts Thus, ths crcut can be used as an equalent crcut for BJT small-sgnal analyss (but only for small sgnal analyss!). Ths equalent crcut s called the Hybrd-Π model for a BJT based n the acte mode: B b r π be - gm = β be b c b = be r π npn Hybrd-Π Model = g = β c m be b e E = e b c b = eb r π E e = g = β c m eb b = e b c B b r π eb - g = m eb β b c pnp Hybrd-Π Model Jm Stles The Un. of Kansas Dept. of EES

55 3/30/2011 The Hybrd P and T Models lecture 4/6 An alternate equalent crcut Note howeer, that we can alternately express the small-sgnal crcut equatons as: be = = g = β = b e c c m be b e r These equatons lkewse descrbes the T-Model an alternate but equalent model to the Hybrd-Π. e c npn T-Model B b g m be = β b = b e c = g = β c m be b r e be - e = be r e E e Jm Stles The Un. of Kansas Dept. of EES

56 3/30/2011 The Hybrd P and T Models lecture 5/6 I just couldn t ft the pnp T-model on the preous page E e pnp T-Model B b r e eb - g m be = β b = b e c = g = β c m eb b e = eb r e c Jm Stles The Un. of Kansas Dept. of EES

57 3/30/2011 The Hybrd P and T Models lecture 6/6 So many choces; whch should I use? The Hybrd-Π and the T crcut models are equalent they both wll result n the same correct answer! Therefore, you do not need to worry about whch one to use for a partcular small-sgnal crcut analyss, ether one wll work. Howeer, you wll fnd that a partcular analyss s easer wth one model or the other; a result that s dependent completely on the type of amplfer beng analyzed. For tme beng, use the Hybrd-Π model; later on, we wll dscuss the types of amplfers where the T-model s smplest to use. Jm Stles The Un. of Kansas Dept. of EES

58 3/30/2011 Small Sgnal Output Resstance lecture 1/5 Small-Sgnal Output Resstance Recall that due to the Early effect, the collector current s slghtly dependent on E : E = β 1 B VA where we recall that V A s a BJT dece parameter, called the Early Voltage. Q: How does ths affect the small-sgnal response of the BJT? A: Well, f ( t) = I ( t) and ( t) V ( t) approxmaton: c =, then wth the small-sgnal E E ce E I = β 1 V A E = V E = VE c B ce E E V V E E 1 = β I 1 β I 1 V V V A A A B B ce Jm Stles The Un. of Kansas Dept. of EES

59 3/30/2011 Small Sgnal Output Resstance lecture 2/5 Small-sgnal base resstance Equatng the D components: I V = β I 1 B V E A And equatng the small-sgnal components: V E 1 = β I 1 V V A A c B ce Note that by nsertng the D result, ths expresson can be smplfed to: I I 1 = c ce = ce V V A A Therefore, another small-sgnal equaton s found, one that expresses the small-sgnal response of the Early effect: I = c VA ce Jm Stles The Un. of Kansas Dept. of EES

60 3/30/2011 Small Sgnal Output Resstance lecture 3/5 Small-sgnal base resstance Recall that we defned (n EES 312) the BJT output resstance r o : I V A 1 r o Be careful! Although the Early Voltage V A s a dece parameter, the output resstance r o snce t depends on D collector current I s not a dece parameter! Therefore, the small-sgnal collector current resultng from the Early effect can lkewse be expressed as: c = r ce o Jm Stles The Un. of Kansas Dept. of EES

61 3/30/2011 Small Sgnal Output Resstance lecture 4/5 Small-sgnal base resstance ombnng ths result wth an earler result (.e., c = g m be ), we fnd that the total small-sgnal collector current s: = g = β r ce c m be b ro ce o We can account for ths effect n our small-sgnal crcut models. For example, the Hybrd-Π becomes: B b r π be - g m be = β b ce - r o c = b = be r g π c m be be r o = e b c e E Jm Stles The Un. of Kansas Dept. of EES

62 3/30/2011 Small Sgnal Output Resstance lecture 5/5 Small-sgnal base resstance c And for the T-model: B b g m be = β b be - r e ce - r o E e Often, r o s so large that t can be gnored (cauton: gnorng the output resstance means approxmatng t as an open crcut,.e., r o = ). Jm Stles The Un. of Kansas Dept. of EES

63 4/1/2011 Steps for Small Sgnal Analyss lecture 1/14 BJT Small-Sgnal Analyss Steps omplete each of these steps f you choose to correctly complete a BJT Amplfer small-sgnal analyss. Step 1: omplete a D.. Analyss Turn off all small-sgnal sources, and then complete a crcut analyss wth the remanng D.. sources only. * omplete ths D analyss exactly, precsely, the same way you performed the D analyss n secton 5.4. That s, you assume (the acte mode), enforce, analyze, and check (do not forget to check!). * Note that you enforce and check exactly, precsely the same the same equaltes and nequaltes as dscussed n secton 5.4 (e.g., V = 07. V, V > 0 ). B BE Jm Stles The Un. of Kansas Dept. of EES

64 4/1/2011 Steps for Small Sgnal Analyss lecture 2/14 You must remember ths * Remember, f we turn off a oltage source (e.g., ( t ) = 0), t becomes a short crcut. * Howeer, f we turn off a current source (e.g., ( t ) = 0), t becomes an open crcut! * Small-sgnal amplfers frequently employ apactors of Unusual Szes (OUS), we ll dscuss why later. Remember, the mpedance of a capactor at D s nfnty a D open crcut. Jm Stles The Un. of Kansas Dept. of EES

65 4/1/2011 Steps for Small Sgnal Analyss lecture 3/14 The goals of D analyss and don t forget to HEK The goal of ths D analyss s to determne: 1) One of the D BJT currents ( I, I, I ) for each BJT. B E 2) Ether the oltage V or V for each BJT. B E You do not necessarly need to determne any other D currents or oltages wthn the amplfer crcut! Once you hae found these alues, you can HEK your acte assumpton, and then moe on to step 2. Jm Stles The Un. of Kansas Dept. of EES

66 4/1/2011 Steps for Small Sgnal Analyss lecture 4/14 The D bas terms are requred to determne our small-sgnal parameters Q: I m perplexed. I was eagerly antcpatng the steps for smallsgnal analyss, yet you re sayng we should complete a D analyss. Why are we dong ths why do we care what any of the D oltages and currents are? A: Remember, all of the small-sgnal BJT parameters (e.g., g, r, r, r ) are dependent on D.. alues (e.g., I, I, I ). B E m π e o In other words, we must frst determne the operatng (.e., bas) pont of the transstor n order to determne ts small-sgnal performance! Jm Stles The Un. of Kansas Dept. of EES

67 4/1/2011 Steps for Small Sgnal Analyss lecture 5/14 Now for step 2 Step 2: alculate the small-sgnal crcut parameters for each BJT. Recall that we now understand 4 small-sgnal parameters: I V V V g = r = r = r = T T A m π e o V I I I T B E Q: Ykes! Do we need to calculate all four? A: Typcally no. You need to calculate only the small sgnal parameters requred by the small-sgnal crcut model that you plan to mplement. For example, f you plan to: a) use the Hybrd-Π model, you must determne gm and r π. b) use the T-model, you must determne g and r. m e c) account for the Early effect (n ether model) you must determne r o. Jm Stles The Un. of Kansas Dept. of EES

68 4/1/2011 Steps for Small Sgnal Analyss lecture 6/14 The four Pees Step 3: arefully replace all BJTs wth ther small-sgnal crcut model. Ths step often ges students fts! Howeer, t s actually a ery smple and straght-forward step. It does requre four mportant thngs from the student patence, precson, persstence and professonalsm! Frst, note that a BJT s: A dece wth three termnals, called the base, collector, and emtter. B B - B BE - E - Its behaor s descrbed n terms of currents,, and oltages B E,,. BE B E E E Jm Stles The Un. of Kansas Dept. of EES

69 4/1/2011 Steps for Small Sgnal Analyss lecture 7/14 They re both so dfferent not! Now, contrast the BJT wth ts small-sgnal crcut model. A BJT small-sgnal crcut model s: A dece wth three termnals, called the base, collector, and emtter. Its behaor s descrbed n terms of currents,, and oltages,, ce. be cb Exactly the same what a concdence! b c e B b - cb c r π be gm be ce - - E e Jm Stles The Un. of Kansas Dept. of EES

70 4/1/2011 Steps for Small Sgnal Analyss lecture 8/14 Am I makng ths clear? Therefore, replacng a BJT wth ts small-sgnal crcut model s ery smple you smply change the stuff wthn the orange box! Note the parts of the crcut external to the orange box do not change! In other words: 1) eery dece attached to the BJT base s attached n precsely the same way to the base termnal of the crcut model. 2) eery dece attached to the BJT collector s attached n precsely the same way to the collector termnal of the crcut mode 3) eery dece attached to the BJT emtter s attached n precsely the same way to the emtter termnal of the crcut model. 4) eery external oltage or current (e.g.,, o, ) s defned n R precsely the same way both before and after the BJT s replaced wth ts crcut model s (e.g., f the output oltage s the collector oltage n the BJT crcut, then the output oltage s stll the collector oltage n the small-sgnal crcut!). Jm Stles The Un. of Kansas Dept. of EES

71 4/1/2011 Steps for Small Sgnal Analyss lecture 9/14 It s just lke workng n the lab You can thnk of replacng a BJT wth ts small-sgnal crcut model as a laboratory operaton: 1) Dsconnect the red wre (base) of the BJT from the crcut and then solder the red wre (base) of the crcut model to the same pont n the crcut. 2) Dsconnect the blue wre (collector) of the BJT from the crcut and then solder the blue wre (collector) of the crcut model to the same pont n the crcut. 3) Dsconnect the green wre (emtter) of the BJT from the crcut and then solder the green wre (emtter) of the crcut model to the same pont n the crcut. Jm Stles The Un. of Kansas Dept. of EES

72 4/1/2011 Steps for Small Sgnal Analyss lecture 10/14 Ths s superposton turn off the D sources! Step 4: Set all D.. sources to zero. Remember: A zero-oltage D source s a short crcut. A zero-current D source s an open crcut. The schematc n now n front of you s called the small-sgnal crcut. Note that t s mssng two thngs D sources and bpolar juncton transstors! * Note that steps three and four are reersble. You could turn off the D sources frst, and then replace all BJTs wth ther small-sgnal models the resultng small-sgnal crcut wll be the same! * You wll fnd that the small-sgnal crcut schematc can often be greatly smplfed. Jm Stles The Un. of Kansas Dept. of EES

73 4/1/2011 Steps for Small Sgnal Analyss lecture 11/14 Many thngs wll be connected to ground! Once the D oltage sources are turned off, you wll fnd that the termnals of many deces are connected to ground. * Remember, all termnals connected to ground are also connected to each other! For example, f the emtter termnal s connected to ground, and one termnal of a resstor s connected to ground, then that resstor termnal s connected to the emtter! * As a result, you often fnd that resstors n dfferent parts of the crcut are actually connected n parallel, and thus can be combned to smplfy the crcut schematc! * Fnally, note that the A mpedance of a OUS (.e., Z = 1 ω ) s small for all but the lowest frequences ω. If ths mpedance s smaller than the other crcut elements (e.g., < 10Ω), we can ew the mpedance as approxmately zero, and thus replace the large capactor wth a (A) short! Jm Stles The Un. of Kansas Dept. of EES

74 4/1/2011 Steps for Small Sgnal Analyss lecture 12/14 Organze and smplfy or persh! Organzng and smplfyng the small-sgnal crcut wll pay bg rewards n the next step, when we analyze the small-sgnal crcut. Howeer, correctly organzng and smplfyng the small-sgnal crcut requres patence, precson, persstence and professonalsm. Students frequently run nto problems when they try to accomplsh all the goals (.e., replace the BJT wth ts small-sgnal model, turn off D sources, smplfy, organze) n one bg step! Steps 3 and 4 are not rocket scence! Falure to correctly determne the smplfed small-sgnal crcut s almost always attrbutable to an engneer s patence, precson and/or persstence (or, more specfcally, the lack of same). Jm Stles The Un. of Kansas Dept. of EES

75 4/1/2011 Steps for Small Sgnal Analyss lecture 13/14 Ths s a EES 211 problem, and only a 211 problem Step 5: Analyze small-sgnal crcut. We now can analyze the small-sgnal crcut to fnd all small-sgnal oltages and currents. * For small-sgnal amplfers, we typcally attempt to fnd the small-sgnal output oltage o n terms of the small-sgnal nput oltage. From ths result, we can fnd the oltage gan of the amplfer. * Note that ths analyss requres only the knowledge you acqured n EES 211! The small-sgnal crcut wll consst entrely of resstors and (small-sgnal) oltage/current sources. These are precsely the same resstors and sources that you learned about n EES 211. You analyze them n precsely the same way. Jm Stles The Un. of Kansas Dept. of EES

76 4/1/2011 Steps for Small Sgnal Analyss lecture 14/14 Trust me, ths works! * Do not attempt to nsert any BJT knowledge nto your small-sgnal crcut analyss there are no BJTs n a small-sgnal crcut!!!!! * Remember, the BJT crcut model contans all of our BJT small-sgnal knowledge, we do not ndeed must not add any more nformaton to the analyss. You must trust completely the BJT small-crcut model. It wll ge you the correct answer! Trust the BJT smallsgnal model, Luke. Jm Stles The Un. of Kansas Dept. of EES

77 4/1/2011 Example A Small Sgnal Analyss of a BJT Amp 1/10 Example: A Small-Sgnal Analyss of a BJT Amplfer onsder the followng BJT amplfer: 15.0 V R =5 K ( t) = V ( t) O O o R B =5 K β = 100 _ ( t ) 58. V R E =5 K OUS Let s determne ts small-sgnal, open-crcut oltage gan: A o o ( t) = ( t)

78 4/1/2011 Example A Small Sgnal Analyss of a BJT Amp 2/10 To do ths, we must follow each of our fe small-sgnal analyss steps! Step 1: omplete a D.. Analyss The D crcut that we must analyze s: 15.0 V R =5 K I V O R B =5 K β = 100 I B 58. V R E =5 K I E Note what we hae done to the orgnal crcut: 1) We turned off the small-sgnal oltage source ( ( t ) = 0), thus replacng t wth a short crcut. 2) We replaced the capactor wth an open crcut ts D mpedance.

79 15.0 V 4/1/2011 Example A Small Sgnal Analyss of a BJT Amp 3/10 Now we proceed wth the D analyss. We ASSUME that the BJT s n acte mode, and thus ENFORE the equaltes V = 07V. and I = β I. B BE We now begn to ANALYZE the crcut by wrtng the Base- Emtter Leg KVL: 5.8 5I 0.7 5( β 1) I = 0 B B 58. V R B =5 K I B R =5 K R E =5 K I β = 100 I E V O Therefore: 5.1 I = = B 5 5(101) 0.01 ma and thus: I = βi = 1.0 ma B I = I I = E B 1.01 ma Q: Snce we know the D bas currents, we hae all the nformaton we need to determne the small-sgnal parameters. Why don t we proceed drectly to step 2?

80 4/1/2011 Example A Small Sgnal Analyss of a BJT Amp 4/10 A: Because we stll need to HEK our assumpton! To do ths, we must determne ether V V. E or B Note that the ollector oltage s: 15.0 V V = 15 IR = 15 (1.0)5 = V R =5 K I And the Emtter oltage s: V = I R E E E = (1.01)5 = 505. V Therefore, V E s: 58. V R B =5 K I B R E =5 K β = 100 I E V O V = V V E E = = 495. V We now can complete our HEK: I = V = E 10. ma > V > 0.7 Tme to moe on to step 2!

81 4/1/2011 Example A Small Sgnal Analyss of a BJT Amp 5/10 Step 2: each BJT. alculate the small-sgnal crcut parameters for If we use the Hybrd-Π model, we need to determne g and r π : m g m I 10. ma ma = = = 40 V V V T VT V rπ = = = 25. K I 0.01 ma B If we were to use the T-model we would lkewse need to determne the emtter resstance: VT V re = = = Ω I 1.01 ma B The Early oltage V A of ths BJT s unknown, so we wll neglect the Early effect n our analyss. As such, we assume that the output resstance s nfnte r = ). ( o

82 4/1/2011 Example A Small Sgnal Analyss of a BJT Amp 6/10 Step 3: arefully replace all BJTs wth ther small-sgnal crcut model V R =5 K _ (t) R B =5 K B be 25K. 40 be ( t ) = O V ( t) O o 58. V E R E =5 K OUS

83 4/1/2011 Example A Small Sgnal Analyss of a BJT Amp 7/10 Step 4: Set all D.. sources to zero. R =5 K R B =5 K B ( ) o t _ (t) be 25K. 40 be E R E =5 K We lkewse notce that the large capactor (OUS) s an approxmate A short, and thus we can further smplfy the schematc by replacng t wth a short crcut.

84 4/1/2011 Example A Small Sgnal Analyss of a BJT Amp 8/10 R =5 K R B =5 K B ( ) o t _ (t) b be 25K. 40 be c e E We notce that one termnal of the small-sgnal oltage source, the emtter termnal, and one termnal of the collector resstor R are all connected to ground thus they are all collected to each other! We can use ths fact to smplfy the small-sgnal schematc.

85 4/1/2011 Example A Small Sgnal Analyss of a BJT Amp 9/10 R B =5 K B b c ( ) o t _ (t) be r π = 25K. 40 be R =5 K - e E The schematc aboe s the small-sgnal crcut of ths amplfer. We are ready to contnue to step 5! Step 5: Analyze small-sgnal crcut. Ths s just a smple EES 211 problem! The left sde of the crcut prodes the oltage dder equaton: be rπ = R r B π 25. = = 3 a result that relates the nput sgnal to the base-emtter oltage.

86 4/1/2011 Example A Small Sgnal Analyss of a BJT Amp 10/10 R B =5 K B ( ) o t _ (t) be r π = 25K. 40 be R =5 K - E The rght sde of the schematc allows us to determne the output oltage n terms of the base-emtter oltage: = R o c = ( gm ) R be = 40(5) be = 200 be ombnng these two equatons, we fnd: = 200 o = = be The open-crcut, small-sgnal oltage gan of ths amplfer gan s therefore: A o o = = 66. 7

87 4/4/2011 Example Small-Sgnal Input and Output Resstances 1/6 Example: Small-Sgnal Input and Output Resstances onsder agan ths crcut: 15.0 V R =5 K ( t) = V ( t) O O o R B =5 K β = 100 _ ( t ) 58. V R E =5 K OUS Recall we earler determned the open-crcut oltage gan A o of ths amplfer. But, recall also that oltage gan alone s not suffcent to characterze an amplfer we lkewse requre the amplfer s nput and output resstances!

88 4/4/2011 Example Small-Sgnal Input and Output Resstances 2/6 Q: But how do we determne the small-sgnal nput and output resstances of ths BJT amplfer? A: The same way we always hae, only now we apply the procedures to the small-sgnal crcut. Recall that small-sgnal crcut for ths amplfer was determned to be: ( t ) R B =5 K B ( ) o t _ (t) be r π = 25K. 40 be R =5 K - E The nput resstance of an amplfer s defned as: R n = For ths amplfer, t s edent that the nput current s: = R = = 7. 5 B r π and thus the nput resstance of ths amplfer s:

89 4/4/2011 Example Small-Sgnal Input and Output Resstances 3/6 R n = = 75. K Now for the output resstance. Recall that determnng the output resstance s much more complex than determnng the nput resstance. The output resstance of an amplfer s the rato of the amplfer s open-crcut output oltage and ts short-crcut output current: R out = oc o sc o Agan, we determne these alues by analyzng the small-sgnal amplfer crcut. Frst, let s determne the open-crcut output oltage. Ths, of course, s the amplfer output oltage when the output termnal s open-crcuted! ( t ) R B =5 K B oc o ( t ) _ (t) be r π = 25K. 40 be R =5 K - E

90 4/4/2011 Example Small-Sgnal Input and Output Resstances 4/6 It s edent that the output oltage s smply the oltage across the collector resstor R : ( ) ( ) oc = g R = 40 5 = 200 V o m be be be Now, we must determne the short-crcut output current sc o. Ths, of course, s the amplfer output current when the output termnal s short-crcuted! ( t ) R B =5 K B sc o ( t ) _ (t) be - r π = 25K. 40 be R =5 K E It s edent that the short-crcut output current s: sc = g = 40 ma o m be be and therefore the output resstance of ths amplfer s: R out oc 200 be V o = = = 5KΩ sc 40 ma o be Now we know all three of the parameters requred to characterze ths amplfer!

91 4/4/2011 Example Small-Sgnal Input and Output Resstances 5/6 A o = V V R n = 75. KΩ R out = 50. KΩ We can therefore wrte the equalent crcut model for ths amplfer: o 75. K 5.0K 66.7 o Note that the nput resstance of ths amplfer s not partcularly large, and output resstance s not at all small. Ths s not a partcularly good oltage amplfer!

92 4/4/2011 Example Small-Sgnal Input and Output Resstances 6/ V Now, reflect on how far we hae come. We began wth the amplfer crcut: R =5 K ( t) = V ( t) O O o R B =5 K β = 100 _ ( t ) 58. V R E =5 K OUS and now we hae dered ts equalent small-sgnal crcut: 5.0K o _ ( t ) 75. K 66.7 Wth respect to small sgnal nput/output oltages and currents, these two crcuts are precsely the same!