A crash-course in transistor circuits

Transcription

1 A crash-cours in transistor circuits Patrick R. LClair July 19, 2011 Contnts 1 Transistors Basic charactristics Notation Modling transistor bhavior Simpl viw: considr gain Simplst viw: forgt gain Typical transistor circuits Switch Currnt sourc Transistors Th transistor is our first xampl of an activ componnt which can produc an output signal with mor powr than its input signal. This dos not violat consrvation of nrgy, th additional powr simply coms from th transistor s powr supply, th main point is that th transistor has a sort of fdback which will allow it to ffctivly add powr to a givn signal. It is in fact not rally th powr gain that w ar most intrstd in, but this fdback proprty, th thing which maks a transistor an activ componnt. 1.1 Basic charactristics W will not dwll on how on maks a transistor, or much of th thory bhind thir opration. W wish only to hav basic opration knowldg of how to us thm. With that in mind, w should first spll out th basic proprtis of a transistor. A transistor is ssntially a valv in th simplst sns (not a pump). It dos not forc currnt to flow, it rathr prmits it to flow to a controllabl dgr whil th rst of th circuit surrounding th transistor tris to forc currnt through it. 1

2 1.2 Notation 2 A transistor has thr trminals (rathr than two lik a rsistor or capacitor), labld th mittr, th bas, and th collctor. Roughly spaking, th collctor collcts powr from th powr sourc, dlivring it to th mittr trminal which mits currnt. Th bas is th control trminal: applying a voltag to th bas can ithr opn or clos th connction btwn th collctor and mittr, modulating th currnt flow. In trms of a (rough) hydrodynamic analogy, th collctor corrsponds to a prssurizd watr sourc, th mittr th drain, and th bas a valv to control th flow rat. A crucial aspct is that this valv is snsitiv - vry small currnts into th bas can opn and clos th valv to allow much largr currnts to flow btwn collctor and mittr. In this way, amplification of powr (normally calld gain ) can b achivd - an input signal to th bas can produc a much largr signal at th mittr. 1.2 Notation In th sort of circuits w will discuss, all potntial diffrncs ar masurd rlativ to th circuit s ground point, which dfins V = 0. Thus, it is nough to rfr to th potntial V at any point in th circuit, knowing that V is spcifid rlativ to ground. Subscripts on voltags rfr to th transistor trminals: V = mittr voltag rlativ to ground (1) V b = bas voltag rlativ to ground (2) V c = collctor voltag rlativ to ground (3) Th sam is don for currnts, whr th convntion is that positiv currnts flow toward lowr potntial (i.., closr to ground): I = currnt into/out of mittr (4) I b = currnt into/out of bas (5) I c = currnt into/out of bas (6) Two non-idntical subscripts ar usd to rfr to potntial diffrncs in shorthand, for xampl: V bc V b V c = bas-collctor potntial diffrnc (7) V c V c V = collctor-mittr potntial diffrnc (8) V b V b V = bas-mittr potntial diffrnc (9)

3 1.3 Modling transistor bhavior 3 Two idntical subscripts ar shorthand for a powr supply voltag bing fd in (usually to th collctor): V cc = collctor supply voltag (10) 1.3 Modling transistor bhavior Transistor coms in two flavors, npn and pnp. W will focus on th npn varity, but in most cass pnp transistors bhav th sam way, but with opposit polarity. Thir circuit symbols ar drawn lik this: c c b b npn pnp From this point on, w will worry only about npn transistors. Transistors of th npn typ follow four gnral ruls, which w will simplify vn furthr to com up with a rough working dscription of transistor circuits. 1. Th collctor potntial is highr than th mittr potntial, V c >V. 2. Th bas-mittr and bas-collctor circuits look lik diods. Normally, th bas-mittr diod is conducting, but th bas-collctor diod is rvrs-biasd. b c npn 3. Thr ar maximum valus of I c, I b, and V c that, if xcdd, will usually dstroy th transistor. Additionally, thr ar othr limits such as powr dissipation (I c V c ), V b, tmpratur, tc., that must b managd for propr function of your circuit. 4. If th prcding ruls ar obyd, th collctor currnt I c is roughly proportional to th bas currnt I b : I c βi b (11) whr β, th currnt gain, is typically around 100.

4 1.4 Simpl viw: considr gain 4 It is th currnt gain β that allows a small currnt into th bas connction to produc a larg currnt at th collctor and mittr. As a bit of warning, w nvr simply rly on th currnt gain β. It is gnrally quit variabl, vn for transistors of th sam typ and modl, and thus only approximatly spcifid. Our circuit dsigns will nvr rly on β having a particular valu, as it would mak thm unprdictabl (and thrfor bad). Point 2 is xtrmly important for undrstanding whr currnt flows in a transistor circuit. It also allows us a simpl way to tst if a transistor is good. For small applid voltags to only two trminals, Th bas-mittr (b) connctions should bhav lik a diod and conduct on way only. Th bas-collctor (b) connctions should bhav lik a diod and conduct on way only. Th collctor-mittr (b) connctions should not conduct ithr way. Th last charactristic is tru bcaus if th bas has zro potntial, th valv btwn collctor and mittr is closd. Or, if th bas is unconnctd, thn I b =0, and thus I c =0 (q. 11) no mattr what potntial is applid btwn collctor and mittr. Points 1 and 4 ar th kys to dsigning som simpl transistor circuits, and thy can b put a bit mor succinctly: 1. V c is gratr than V by at last a fw tnths of a volt (so currnt flows from c to ) 2. Th surrounding circuit mad such that V b is about 0.6 V highr than V, i.., V b V 0.6. W shall procd with two viws of transistor opration: on simpl, th othr vn simplr. 1.4 Simpl viw: considr gain In th simpl viw, w look at th transistor as a currnt amplifir or a currnt-controlld valv, and xplicitly mak us of th fact that th collctor currnt is qual to th bas currnt tims th gain, I c =βi b. Looking at th thr trminals of th transistor, w can apply consrvation of charg (currnt). b Ib c Ic I Flowing in to th transistor w hav I c +I b, which must qual I :

5 1.4 Simpl viw: considr gain 5 I = I c + I b = βi b + I b = (1 + β) I b (12) Thus, th currnt flowing out of th mittr trminal is a factor 1+β largr than th control currnt flowing into th bas. This allows us to tak a small currnt input at th bas and produc a larg currnt output, but at ssntially th sam voltag. Th xtra currnt gaind coms from th powr supply. This lts us mak a dvic known as a followr: Vb in, small I b +Vcc supply c R V out larg I It is calld a followr bcaus th mittr output voltag (V ) follows th input bas voltag (V b ) (i.., thy vary in th sam way), but th output has a much largr currnt. Rcall that our simplifid ruls statd that V b V 0.6 V, so th mittr voltag will do whatvr th bas voltag dos, just 0.6 V lowr. A small voltag at th bas V b producs only a small currnt I b, maning th input has a rlativly larg rsistanc. Th currnt flowing out of th mittr, and thus through th load R is much largr: I =(1 + β)i b (kp in mind that th V output voltag is a voltmtr connction, it draws no currnt). W v takn an input signal with a givn voltag and small currnt (low powr) and transformd it into an output signal with similar voltag and much mor currnt (high powr) a simpl amplifir of sorts. If w think of this in trms of rsistanc, th followr has anothr function. Th mittr output has a currnt largr by a factor 1+β, but th mittr voltag is roughly th sam as th bas voltag. Almost th sam voltag, highr currnt, this mans a rlativly low rsistanc. This is on of th main functions of a followr: th output voltag is narly th sam as th input voltag, but th currnt is much highr, and th rsistanc much lowr. This is anothr on of th main functions of a followr, transforming th output circuit from high to low rsistanc, important whn conncting diffrnt circuits togthr.

6 1.5 Simplst viw: forgt gain Simplst viw: forgt gain Th prcding is a vry usful circuit, but th dsign could b clarr yt. If w only want to worry about making rough sklton circuits, figuring w can flsh thm out and fancy thm up latr on, w can simplify th transistor ruls vn furthr. W forgt about gain ntirly, and just focus on two points: 1. call V b =V b V =0.6 V, and dsign th surrounding circuit to mak this tru 2. call I c =I Th scond point is clarr if w rmmbr I = I c + I b basd on consrvation of charg. If w prsum th bas currnt to b much smallr than I or I c (it is at last a factor β 100 lowr than I ) and nglct it, thn I I c. Now look at our followr circuit abov again. Lt s say th bas voltag changs by som amount V b. Sinc th mittr voltag is V =V b 0.6, its chang is th sam: V = V b. Th rsistor R thn ss a voltag chang of V, sinc it is connctd btwn mittr and ground, and this producs a currnt chang in th rsistor of I = V R = V b R (13) Th mittr currnt is dictatd by th choic of th valu of R and th voltag chang at th bas. So what? Compar this with th currnt chang at th bas itslf, a factor 1 + β smallr: I b = I 1 + β = V b R (1 + β) Thus th followr givs us th sam chang in voltag at th mittr that was applid to th bas, but with a much largr currnt. Lt s say w want to hav I =1 ma through th rsistor R =1 kω. This mans w nd an mittr voltag of V =I R =1 V. That mans th bas has to hav a voltag 0.6 V highr, or V b =1.6 V. This is accomplishd by making a connction to th supply voltag V cc through rsistors, but mor on that blow. (14) 2 Typical transistor circuits 2.1 Switch Lt s mak a somwhat mor practical circuit. Considr th diagram blow, and lt s say th load is a light bulb. What dos this circuit do? Lt s follow th ruls. Whn th mchanical switch is opn, thr is no bas currnt. No bas currnt mans no mittr currnt and no collctor currnt, sinc I c =βi b and I c I. That mans no currnt through th light bulb, it stays dark. Onc w clos th switch, w now hav a bas currnt, sinc th bas is connctd to th 10 V supply. Th mittr is connctd to ground, maning V = 0, so that mans th bas riss to

7 2.1 Switch 7 +10V mchanical switch 1kΩ Rload=100Ω (.g., bulb) V b = V = 0.6. If th supply is at 10 V abov ground and th bas only 0.6 V abov ground, th othr 9.4 V must b on th 1 kω rsistor. i Th currnt in th 1 kω rsistor is th sam as th bas currnt, sinc it is in sris with th bas, and it must b I b = 9.4 V/1 kω = 9.4 ma. If th gain of our transistor is β =100, th collctor currnt (also th currnt through th bulb) must b I c =βi b =940 ma. Th bulb lights up. Unfortunatly, this is wrong at last th currnt is. Th bulb will light up, but w hav violatd on of our transistor ruls, that th collctor voltag must b highr than th mittr voltag. Th collctor voltag will b 10 V minus th potntial drop on th load rsistor of 100 Ω. If th collctor currnt is 940 ma, this is V c = 10 (0.94 A) (100 Ω) 84 V. First, this is a bit silly sinc th potntial diffrnc on th rsistor is largr than th supply voltag, which can t happn. Scond, this maks th collctor voltag lowr than th mittr voltag, violating on of our main ruls of transistor bhavior. A transistor cannot pull th collctor voltag blow th mittr voltag. With th mittr groundd, that mans that th collctor can hav at maximum a 10 V potntial rlativ to th supply, giving th load rsistor a 10 V potntial diffrnc. What will rally happn in this cas? Closing th switch biass th bas abov ground (positivly), which opns our valv from collctor to mittr. This allows th supply to snd as much currnt as it can through th bulb, with its ntir 10 V across th bulb, and it lights up with a currnt of I c = 10 V/100 Ω = 0.1 A. Onc th supply has droppd 10 V across th rsistor, nothing furthr can happn - th collctor will b at zro volts thn, and can go no lowr. W hav saturatd th transistor by brining th collctor as clos to th mittr as it can go, sourcing maximum currnt to th bulb. Mor gnrally, what w just mad is a transistor switch closing th mchanical switch to bias th bas allows th sourc to apply full powr to th bulb. Rathr than using a mchanical switch to bias th bas, w could imagin that th bas voltag is producd as th output of som snsor (say, a photorsistor or microphon) which will caus th bulb to light up as an indicator whn th i Us consrvation of nrgy: walking from th 10 V sourc to ground via th bas, th sum of th voltag gains and losss must qual what you startd with, 10 V.

8 2.2 Currnt sourc 8 snsor ss a larg nough stimulus. 2.2 Currnt sourc In intro physics, w tll you all about voltag sourcs, and not that thy ar asily mad with lctrochmical clls (i.., battris). You nvr har how to mak a currnt sourc. Why? Bcaus a currnt sourc mans a constant rat of charg flow, implying that th sourc must adjust its powr, dpnding on what it is connctd to, to nsur a crtain flow of charg. A voltag sourc just nds to nsur a potntial diffrnc, which can occur vn without anything connctd. A currnt sourc is thus an activ dvic which rquirs a bit of fdback, a task for which our transistor is idally suitd. Lt s say w want to crat a 1 ma currnt sourc for driving a load of unspcifid rsistanc (say, an LED). W can t do it with a battry, sinc w d hav to know th rsistanc to nsur a crtain currnt. Th simpl circuit blow dos th job: +Vcc Iout 5.6V 1mA Rload (.g., ld) ~5V 5kΩ Th ida is lik this: fix th bas voltag V b. That fixs th mittr voltag V. Th mittr voltag can thn sourc a known currnt by conncting a known rsistanc btwn it and ground. Sinc th collctor and mittr currnts must b (approximatly) th sam, this fixs th collctor currnt, so w just hav to connct our load dvic btwn th collctor and supply and w r don! Lt s fix th bas voltag at V b =5.6 V. That maks th mittr voltag V V b 0.6=5.0 V. With 5 V at th mittr, a 5 kω rsistor connctd btwn mittr and ground producs a currnt of I = 5 V/5 kω = 1 ma. Sinc I c I, that mans that whatvr is connctd btwn th supply voltag and th collctor (within rason) rcivs a currnt of I c =1 ma from th supply. Th only qustions to rsolv ar how to st th bas and mittr voltags? Say w hav V cc = 10 V to b concrt. W can us sris rsistors - a voltag dividr - to st th bas voltag at th dsird 5.6 V.

9 2.2 Currnt sourc 9 +10V R1 R2 W connct th bas to th 10 V supply through rsistor R 1, and th bas to ground through rsistor R 2. Th voltag at th bas will b th sam as th potntial diffrnc across rsistor R 2. Using th ruls for sris rsistors, V b = V R2 = W can solv this for th ratio R 1 /R 2, and w find R 2 R 1 + R 2 V supply = 5.6 V (15) R 1 = V supply 1 = (16) R 2 V b 5.6 Thus, choosing a ratio of 0.8 for th two rsistors sts th dsird bas voltag, which in turn sts th mittr voltag and th mittr and collctor currnts. W don t want much currnt flowing through th bas, so w should choos larg rsistors ( 100 kω rang); picking R 1 = 120 kω and R 2 =150 kω dos th job wll nough. Now w can draw th full currnt sourc circuit. +10V 1mA R1 Rload R2 5kΩ

10 2.2 Currnt sourc 10 This is a prfctly rasonabl 1 ma currnt sourc, taking as much powr as rquird from th 10 V sourc to maintain 1 ma (or until w rach th supply s or transistor s limits... rmmbr w cannot drop mor than 10 V across th load rsistor and kp th collctor abov 0 V) Variant: automatic night light Our nxt circuit is an automatic night light. W want an LED to light up whnvr th ambint light lvl is too low. A photorsistor will hlp, it has a rsistanc in th dark of 50 kω, and only a fw kω undr bright illumination. In gnral, its rsistanc follows a powr-law dpndnc as a function of th incidnt light intnsity, but th main point is that it has a low rsistanc whn in bright light, and a vry high rsistanc in th dark. Th ida, thn, is that w can us our prvious currnt sourc sklton circuit to powr th LED, but us a currnt through th photorsistor to st th bas voltag - high rsistanc (dark) givs a high bas voltag, opning th collctor-mittr connction and lighting th bulb, low rsistanc (light) givs a nar-zro bas voltag, closing th collctor-mittr connction and turning off th LED. Th circuit blow accomplishs this, a sort of hybrid of our transistor switch and currnt sourc circuits. 100kΩ photo R b c +Vin LED 2N Ω -Vin -Vin Bfor calculating anything, lt s think qualitativly about this circuit. Th 100 kω and photorsistor mak a voltag dividr from th supply. If th photorsistor is lit, it has a low rsistanc, and th 100 kω ats up most of th supply voltag. That mans th voltag at th bas is low, th bas currnt is tiny, and so ar th mittr and collctor currnts. Th collctor-mittr channl is closd, and th LED is dark. If th photorsistor is dark, it has a rsistanc comparabl to th 100 kω, so it will tak a significant fraction of th supply voltag. That mans th bas has a significant voltag on it, producing a larg mittr and collctor currnt, and th LED lights. Th collctor-mittr channl is opnd by putting a voltag on th bas, which w achivd by darkning th photorsistor with a currnt running through it. Stting th valu of th rsistor at

11 2.2 Currnt sourc kω sts th illumination thrshold for turning on th LED, so on might b vn mor clvr by making it adjustabl ovr som rang. How about som spcifics? In this cas, w prsum th voltag input V in is about 3.5 V, roughly 2 C cll battris in sris. W want th bas to b nar zro voltag whn th photorsistor is dark, so w mak it a part of a voltag dividr for th bas. Th 100 kω rsistor is chosn to match th photorsistor s dark rsistanc; if it is a bit lowr, th light turns on whn it is vn darkr, if it is a bit smallr, th light turns on at a bit highr illumination thrshold. Why? Whn th photorsistor is in modrat darknss, its rsistanc is 50 kω, so it taks a proportion 50/(50+100)=0.33 of th supply voltag and th 100 kω rsistor th othr That maks th bas voltag about 1.16 V, and thus th mittr voltag only about 1 V. That mans w hav 1 V across th 200 Ω mittr rsistor, so th mittr and collctor currnts ar I I c 5 ma, plnty to light up th LED. Onc th photorsistor is illuminatd, its rsistanc is only a fw kω, lt s say 2 kω for th sak of argumnt. That mans th voltag drop on th photorsistor, th bas voltag, is only 2/(2+100) 0.02 of th supply voltag, or about 0.07 V. That mans ssntially no bas currnt, so no mittr or collctor currnt, and th LED dosn t light. Evn mor to th point, w v saturatd th transistor again - th mittr should b 0.6 V lowr than th 0.07 V bas voltag, but it can t b - that would violat our rul that V b V 0.6, and anyway th mittr is fixd at ground and can t go ngativ. Th mittr will stay at ground (V = 0), and it mans th bas is insufficintly biasd, so th collctor-mittr channl rmains closd, and th LED stays dark.