Bj Transistors

8/3/2019 Bj Transistors

1/12

5 Bipolar Junction Transistors5.1. IntroductionThe bipolar junction transisior (BlT), or simpl), transistot is a 3-terminal devicethat can be used as an amplifier or as a switch. The internal operation of thcdevice is moderately complex, but r{'e only nced to consider the extcmal"terminal charactedstics" o{ the device. Our aim is to develop a mathcmaticaland model-based description of the transistor so that we can describe itsbehaviour in a circuil and design circuits that use the transistor.5.1.1. Symbols

cl

f -'tE

Fundamental EquationSince the transistor has 3 terminals. there are several combinations of currentsand voltages that we could rneasure. For our purposes there are two significantcharactcdstics: one involving the base terminal (the input characteristic) and theother involving the collcctor terminal (the output characteristjc).

Base A

"NPN" transistor "PNP" transistorThe terms NpN and pNI'refer to the intemal construction o{ the hansistor. Thedifferences between NtN and PNP devices are largely superficial so wc willconcentrate on developing an understanding of the NPN device. It will then be aneasy matter to apply that knowledge to PNP devices.5.2. Terminal CharacteristicsThe terminal characteristics of a device describe the electrical behaviour at itsterminals without reference to the intemal mechanisms driving that behaviour.We treat the device as a "black box" and develop an understanding of itsoperation based solely on how it interacts with other comPonents, and ho$ itreacts to external electrical stimuli.

i^pJ u,* r {--'tCrtr"rro, r,,*,5

E

"':1i,,"\1,Voltage & Current Conventions

49


2/12

50 BiPolar luncuon Transistors

5.2.1. Input CharacteristicTo investigate the input characteristic we set uP the test circuit shown bclow Thevoltage source VBB is varied and the basc current is and base-emitter voltage .rrrare mcasured. The input charactedstic shows the dePendence of i, on our, and isshown below.

6

10-a

lnput Characteristic Test Circuita.2 0.4 0.6uo,. V

Input Characteristic (Typical)0

From the graph we see that for ?sr < 0.5 V no curent flows. For osE i 0.7 V anappreciable current flows and we say that the base-emitter junction is/oru'aldbiased1').

5.2,2. Output CharacteristicWe can explore the output charactedstic using the test circuit shown below. For agiven base-circuit voltage Vss we vary the collector-circuit voltage ycc arrdrecord both the collector current i. and the collector-emitter voltage o.E. Thismeasurement procedure is then repeated for another setting of yBr

Output Characteristic Test Circuit

IThevoltages quoted here and elscshere in this doament arc for transistors constructed usingSiljcon s the scmiconductor-by far ihe most common matenal tsed for thc manufacture ofelcctronic devices.

Eleclrica and Digital Systems 2008


3/12

B polar Junction TransistorsThe output characterishc is shown below. From the graph we see that for ir andr,..- "large enough," i. is independent of .i.r and is instead compietelydetermined by the value of iF.

0.8 --

- l:yi*:,0.6 ".a1 i."'I :

0

-!

,.btP,{a. lttf** L\

Electrical and Digitat Systems 2008Or. Mark Andrews

16Output Charaderistic (Typicat)

l0

5.2.3. Regions of OperationCarefui examination of the input and output characteristics show that there are 3distinct regions in which the transistor can operate. These regions are called theactive regiorl the safuration region and the cutoffregion. The regions are showtrin the diagram betow.

Region


4/12

52 Bipolar lunction Transistors

5.2.3.1. Cutoff Region\\rhen o,. drops below 0 5 V, is and i. droP to zero The transistor is then in thecutoff region'1.

conditions when Cutoffu,. < 0.5Vio =0

5.2,3.2. Saturation RegionIf ?,sF (or ii ) increases to the Point where r)!. > 0 then the transistor enters thesaturation regionr{.Conditions when SaturatedlBF =07\J0r=05Vocr = 0.2V = ocr:o.rrr

5.2,3.3. Active Regionlf ?uE > 0.5 and we maintain urc < 0 then the transistor is in the activc regionConditions when Active

i,= f i,/t is called the .r,'/eni 84i, of the transistor' Typically, p is in the range 50 to300.

rrThefolowinserpressionsareusedsynonymously:,,theiransistorisincutof',,,,rhetransistoriscutoff" and "ihe transisior h off "" The following expressions are used slnonlmosly: "the iransistor is in saturation"' "thctransistoris satated" and "lhe tr sistoris on "Electrical and Digital svstems 2008


5/12

Bipolar Junction Transisto

5.3. Transistor ModelsTo do calculations involving thc transistor, or design circuits that use ttansistors,we need simple models that desc be the operation of the transistor.

5.3.1. Cutoff ModelCi+ =0I |Rr < 0.5 V

i.=0B-ot=0

5.3.2. Saturation Modelt -. l,"lFosc = 0.5 Vo..=0.2V=r..,r,r,

'rli- =0Etl

E5.3.3. Active Model

ia

-r.( 44 lbtJ,J.'--" Y-y-l8v(q.) f t"rflr rulBt!.'J",

pa-'l-"I t'u--=shu-'ny 4t ^+"* "ft, I z2 \'6 ,,7


6/12

56 Eipolar lunction Transistors5,4. The Transistor as an AmplifierWhen used as an amplifier, the transistor must be operated in the active region.In the previous section we saw that the bansistor biasing resistors determine, inpalt, the region in which the transistor operates. ln fact, the biasing resistorsperform the dual role of maintaining operation in the active region anddetermining thc gain of the circuit.5.4, 1. Common-Emitter AmplifierConsider the transistor circuit shown in the figure below. There are a Iewchanges compared to the previous circuits we've looked at. In particular, we nowhave an input voltaBe o,,, connected in series with the base biasing voltage %s,and an emitter resistor Rr.

y.c =10vv.. = 1vvcc R! = 10 koR.=25kOR. = 1ko/3 = 100'oBE=07V

uin

"l' = 0.1sin 2dof VNotes1. Recall that wc must have irlL > 0.5 V otherwise the transistor will be cutoff.The purpose of yB is to ensurc that VBB +.,i" keeps the transistor active,even when . ir al il\ mo\l negdtive.2. The purpose of RF is to stabilise thelliagainst va ations in / and to

provide thermal stability through negatjve fccdback's.

rr Ntave fccdback and dlemal stabilily are curentll' beyond the scope of this course bur anapproximate nal) sis is as follows. There is an erha, bur very smatt, anent ;cs, rhat flows fronthe collector ;,rto the base when r,. < 0. This cuent is unavoidablc becuse of ihe physicsdescribing the device's intenl opcration. i.jo is amplificd by l, justlike ta, and is highjytemperatre dependenl.lf i, in.reases be.ause ofa tcmperat!.e increasc (via t.!.)then ?linoea*s, causing ?$ to decrcase, therebv redu.ing td and hence opposinS ihe increa* in t. _ Thisis callcd r{diire./,t d}r.k d is d exlrcmely important md powrtut concepi in etectricalengineering, nd in enginecring systems in general Do vou need ro know this for the cxam? No.Electrical and Diqital systems 2008

1AL


7/12

Eipolar lunction Transistors5.4,1.1. AnalysisFor the input loop (containing the base terminal) we ha.ve-l%i + -!r,,, t! RB asF-i. Rt=0 and since iE=(B+1) i, we find' V"P t 'a'" R,. {/- rr &Summing the voltages around the output loop we must haveV.. - i. R. - r,. = 0, a1d since i. = / . i, we get zJc = V.t I . iBR. .

57

12 ,- *-/ ^ P"-a.&* L I

We can now calculate the ouqut voltaSe r.examples are shown below. for any input voltae ?," . Someir = 2.703 pA'L)c = 3.243 V

t',,, = -0.1ViB =1.802 pAr,. = 5.495 V

From these values we can see that for an input va ation ofAtr,,, =0 ( 0.1)=+0.1V the output varies by La( =3.243-5.495= 2.252V.Thisrepresent. a voltage gain of C = ^J 2 2-5,2 22.52 . The gain is negalive" ^.r . 0.1because as the input voltage /ed ccs, the output voltage ir?c,'eases (and vice versa;.see the figures below to see how the output changes for a sinusoidal input.

= sv

F."l ,, V,vrt*r c)vs*(fr)su= rv= B"ttBr=#,- 4.L k

o.1

o.l!0

1

oo

6

5.

"32

Electrical and Digital Systems 200aDr. Mark Andrews

Il6VI

5

. _ vB 'E'-3J;et: 1-a7n ' ,fi" pi.-,"y,a;s- 49.6 r A;. ' i8.2]s"l\

\1. = 1..' l.fa=q\/E "\h: t


8/12

58 Bipolar lunction Transistors5.4.1.2. Expression for cainWe can find an explicit expression for the gain of the circuit by substituting theexpression for is into the equation for r,c.

= v,, /J'i, R.Rs + (/+ 1) Rr RcRB + (l + 1).RF ' B.R,R +(/+1) RF

= ?'a + G'o"'Note that ?co = IJ. when u,,, = 0 .G is lhe cir it ooltage Sdi, and is given by

R, +(/ +1) R.5.4.1,3. Effect of Emitter ResistorFrom the expression for G r"'e can see that it depends on the resistor values (Rc,R, and Rr ) and the transistor cunent gain ( / ). This is undesirable since thetransistor manufacfurers will or y ensure that the current gain is within a fairlybroad rangc; Ior example 50 to 150. If we need to design a circuit with tightlycontlolled gain, it cannot depend too strongly on the transistor parameters.Fortunately the emitter resistor helps us out here.If we choose RB and R, so that (/+l).RF>>RF then

c=

G=.-:9 :

-pP.

p .R,

Furthermore, if / >> 1

Elect.ical and Digltal Systems 2008

(which is nearly always the case) then -4:1P +'l

RF +(/t + 1).Rr/t R"(P +1) Rr

(,, + 1) F_&RIand

which is independent of I ! Thus, provided p is large enough we can make thecircuit voltaBe gain immune to vafiations in the transistor current gain.


9/12

*t''raa4

F?623 'u,azr\^'ap 43Xk

Bipolar lunction Transisto

5,5, The Transistor as a SwitchWhen used as a switch, the transistor is operated in thc saturation and cutoffregions- A pdmary concem is choosin8 thc bias resistors so that thc transistor isdriven into saturation for any value of f .For example, consider the circuit below, wherelassume that u,,, can be either 0or V.. . If the transistor f varies between 50 and 150, how do we best choose Rsso that Q safurates when .r,,, = Y.. ?

59

io\". lRc

Vcr =10vRc = 1ko50


10/12

60 Bipolar lunction Transistors

p" = 4. -uar - 1o-ozv =9.490ka." i, 0.980 mAo find Rr, note that when r';,, = Vcc,Examplei Nor qte implementation. Vcc s is chosen so that when uJi" is Hlctt Q is saturatedand when a,,, is Low, Q is in cutoff.

0V,,

ExamDle: oR qte imDlementation

Rg0. l1000 Vccv,, 0

IExample: $r fl ip-flop implementation

0i H*-"1-l-rL ptest ",A( * .:,7

a. are oFr and Q,is oN.ssume that Qj , Q, andII S goes HIGH then

QL tums ONQr tums oI!, so Q goes HlcHQ. tums oN, so Q goes Low

Electrical nd Digital Systems 2008


11/12

Bipolar luncUon TrnsistorsWhen S refums to a LOW voltage, Qr turns OFI, butstays QN.Similarly, if R goes higb then

is kept LOW because Qr

Q, tums oNQr turns orF, so Q gn". trcrlQ. turns oN, so Q goes Low

5,6. Power DissiptionPower dissipation in a transistor is an important consideration: attempting topass too much current tfuough a device results in overheating and possibledestruction. Thcrefore it is important to understand the power rating oI atransjstor and how power is consumed in a device.

a

w 1,X" 51

I

For the transistor sho'n at right it is clear that theinstantancous power consumed by thc device must bep . isuit t i rr., . However, i . i. solhdt rs.rB, it-,and n e insLad u>e the dppro\[email protected]. Maximum Power RatingAny given transistor has a maximm power rating PM", above which the devicemust not be operated. That is, we must ensure that icz,.r < PM," or ic < P],k, I ocL.This inequality defines the maximum por.Ller c ,"?p that fu her constrains theregion of operation of the transistor.5.5.2. Static Switch Loss

\ u-fla*s tLr"l tl4 "-A"

Documents

Bj Transistors