BJT CnnnncrERrsrrcs nNo Panemrrens ; 169

Transistor ilC ftlodel\bu can view the unsaturated BJT as a device with a current input and a dependent cur-rent source in the output circuit, as shown in Figure 4-7 for an npn.The input circuit is aforward-biased diode through which there is base current. The output circuit is a de-pendent current source (diamond-shaped element) with a value that is dependent on thebase current, Is, and equal to Bocls.Recall that independent current source symbolshave a circular shape.

FIGURE 4_7Collector ldeal dc model of an npn transistor.

BIT Cinauit AnalysisConsider the basic transistor bias circuit configuration in Figure 4-8. Three transistor dcJurrents and three dc voltages can be identified.

1s: dc base current

1s: dc emitter cuffent

16: dc collector current

VsB: dc voltage at base with respect to emitterVs: dc voltage at collector with respect to baseYsB: dc voltage at collector with respect to emitter

FIGURE 4-8Transistor currents and voltages.

The base-bias voltage source, Vss, forward-biases the base-emitter junction, and the--ollector-bias voltage source, V66, reverse-biases the base-collector junction. When thebase-emitter junction is forward-biased, it is like a forward-biased diode and has a nominaltorward voltage drop ol

Vsn 3 0'7 V-\lthough in an actual transistor VsB can be as high as 0.9 V and is dependent on cuffent,u'e will use 0.7 V throughout this text in order to simplify the analysis of the basic concepts.Keep in mind that the characteristic of the base-emitter junction is the same as a normalJiode curve like the one in Figure 1-28.

Since the emitter is at ground (0 V), by Kirchhoff's voltage law, the voltage across Rs is

Eq*;rtion 4-3

ICIB

Emitter

Rc

RB vcar!

Va":Vss-VsB

170 . BrpoLAR JuNcroN Tnarusrsrons

Iquation 4-4

fquation 4-5

lquation 4*6

EXAMPLE 4-2

Also, by Ohm's law,

Substituting for Va, yields

Vp, : lsRs

lsRs: VsB -

Vss

Solving for 1s,

- Vnr

- Vsrts= R"

The voltage at the collector with respect to the grounded emitter is

Vqs: Vgq -

Vp.

Since the drop across R6. isVa. : 16R6

the voltage at the collector with respect to the emitter can be written as

Vcs=Vgg-IaRgwhere 16 : Bocls.

The voltage across the reverse-biased collector-base junction isVCs=Vco-Vw,

Determine Is, Ig, IB, VsB, VcB, and Vss in the circuit of Figure 4-9 . The transistor hasa Bp6 : 150.

FIGU RE 4_9

vcc10v

7se5V

S*J*fir:lr From Equation 4-3, Vsa = 0.7 V. Calculate the base, collector, and emitter currentsas follows:

, _Vss - Vse_ 5V - 0.7V _ Irn..aIB -

-

IJUUnR3 10 koIc: Focls: (150X430pA) : 64.5mAIr: Ic+ IB:64.5mA + 43OP'A:64.9mA

Solve for V6p, and V6s.

Vca:Vcc- lcRc:10V -

(64.5mA)(100O): 10V -

6.45Y:3'55VVcB: Vca- Vsa:3'55V

- 0'7V:2'85V

Since the collector is at a higher voltage than the base, the collector-basejunction isreverse-biased.

ffrf#t#{J Fxthlwx Determine Is, Is, IB,Rs- 220 Q, Vss :

BJT CnnnncrERrsrcs aNo PenamgrEns . 171

VgB, and Vss in Figure 4-9 for the following values: Ra : 22kQ,6Y Vcc: gVandFoc:90.

Collector Characteristic CurvesUsingacircuitlikethatshorryninFigure4-10(a), asetofcollectorcharacteristiccurvescanbe generated that show how the collector current, 16, varies with the collector-to-emittertoltage, Vge, for specified values of base current, Is. Notice in the circuit diagram that bothI'ss and Vggare variable sources of voltage.

Assume that Vss is set to produce a certain value of Is and V66 is zero. For this condi-tion, both the base-emitterjunction and the base-collectorjunction are forward-biased be-cause the base is at approximately 0.7 V while the emitter and the collector are at 0 V. Thebase current is through the base-emitter junction because of the low impedance path to

(a) Circuit

vcp

Open the Multisim fileE04-02 in the Examples folder on your CD-ROM. Measureeach current and voltage and compate with the calculated values.

Ycn1.u*y

I Breakdoi,n0.7 v

Aclivc region---------------- regiojlI

(c) Family of16 versus V6B curves for several values oflu(Is1< 182< /B3. etc.)

b) 16 versus V6s curve for one value of 1s

GURE 4-10-:, iector characteristic curves.

172 . BtpoLAR JuNcrroru TneNsrsrons

ground and, therefore, 1g is zero. When both junctions are forward-biased, the transistor isin the saturation region of its operation. Sa{:xs.eag$ulxn is the state of a BJT in which the col-lector current has reached a maximum and is independent of the base cument.

As Vs6 is increased, 76s increases as the collector cunent increases. This is indicatedby the portion ofthe characteristic curve between points A and, B in Figure 4-10(b). 16 in-creases as Vgg is increased becadse V6E remains less than 0.7 V due to the forward-biasedbase-collector junction.

Ideally, when V6B exceeds 0.7 V the base-collectorjunction becomes reverse-biased andthe transistor goes into the active or fifl*rulsr r*g!

BJT CnnnncrERrsrrcs nro Pnnnmerens . 173

16 (mA)

Is=25 PA

Ir--20 pA

Is= 5 PA

FIGURE 4-12

rfats# Fr**Js#? Where would the curve for Islector leakage current?

: 0 appear on the graph in Figure 4-12, neglecting col-i

I1

Cutsff-\s previously mentioned, when 1s : 0, the transistor is in the cutoff region of its operation.This is shown in Figure 4-13 with the base lead open, resulting in a base current of zero.Under this condition, there is a very small amount of collector leakage current, 1gp6, duemainly to thermally produced carriers. Because 1sE6 is extremely small, it will usually beneglected in circuit analysis so that Vce : Vcc.In cutoff, neither the base-emitter nor thecase-collector junctions are forward-biased. The subscript CEO represents collector-to-emitter with the base open.

FIGURE 4_1 3Cutoff: Collector leakage current(lcrd is extremely small and is usu-

,, ally neglected. Base-emifter and

'cc base-collectorjunctionsarereverse-biased.

Saterratimm\\-hen the base-emitter junction becomes forward-biased and the base current is in---reased, the collector current also increases (Ic : Eocls) and V6g decreases as a resultrf more drop across the collector resistor (VcB : Vcc

- 16R6). This is illustrated in

Figure 4-14. When VgB reaches its saturation value, V6s1ru1;, the base-collector junction be-,-omes forward-biased and 1c can increase no furlher even with a continued increase in 1s.-\t the point of saturation, the relation Ic: Bocls is no longer valid. VgB1r.1y for a transis-:or occurs somewhere below the knee of the collector curves, and it is usually only a few:e nths of a volt.

Vcu =

Vc.c

17 4 . BrpoLAR JuNcrtoN Tterustsrons

FIGURE 4-15DC load line on a family of collectorcharacteristic curves illustrating thecutoff and saturation conditions.

FIGURE 4-14Saturation: As Is increases due to in-creasing V66, 16 also increases and V6sdecreases due to the increased volta$edrop across R6. When the transistor Ireaches saturation, Ic can increase nofurther re$ardless of further increase vesin Is. Base-emitter and base-collectorjunctions are forward-biased.

DC load [ineCutoff and saturation can be illustrated in relation to the collector characteristic curvesby the use of a load line. Figure 4-15 shows a dc load line drawn on a family of curvesconnecting the cutoff point and the saturation point. The bottom of the load line is atideal cutoff where Is : 0 and vca: vcc. The top of the load line is at saturation where1c : 1c(su0 and V6B : Vcs(sao. In between cutoff and saturation along the load line is theactive regiion of the transistor's operation. Load line operation is discussed more inChapter 5.

Ic(ruo

EXAMPLE 4*4 Determine whether or not the transistor in Figure 4-16 is in saturation' AssumeVCr(saO :0.2Y.

FIGURE 4-16

vns3V

17 6 . BlpoLAR JuNcroN TneNsrsrons

FIGURE 4_18

Maximum power dissipation curveand tabulated values. 60

1c(max; + 56

40

EXAMPLE 4_5

&*mximunu Tnar*sEstea' ffi*tfi mgfls

A BJT, like any other electronic device, has limitations on its operation. These limitationsare stated in the form of maximum ratings and are normally specified on the manufacturer'sdatasheet. Typically, maximum ratings are given for collector-to-base voltage, collector-to-emitter voltage, emitter-to-base \^oltage, collector current, and power dissipation.

The product of V6s, and.16 must not exceed the maximum power dissipation. BothVgs, and 1g cannot be maximum at the same time. If V6B is maximum, 16 can be calcu-lated as

Polmax;Ic:vcp.

If 16 is maximum, VgBcan be calculated by rearranging the previous equation as fol-lows:

Polmax;vrr: kFor any given transistor, a maximum power dissipation curve can be plotted on the col-

lector characteristic curves, as shown in Figure 4-18(a). These values are tabulated inFigure 4-18(b). Assume Polmaxy is 500 mW, VcE(max) is 20 Y and 151*^"1 is 50 mA' Thecurve shows that this particular transistor cannot be operated in the shaded portion of thegraph. 1s1m.r; is the timiting rating between points A and B, Polmaxl is the limiting ratingbetween points B and C, and VCE(*ax) is the limiting rating between points C and D.

PD(-u*) vcp.

Vcr (v)

(b)

A certain traasistor is to be operated with V6B : 6 V. If its maximum power rating is250 mW, what is the most collector current that it can handle?

PDrmax) -

250 mW : 41.7 mA\ciuiiom ln:-L vcP' 6 v

This is the maximum current for this particular value of Vss. The transistor can handlemore collector current if V6s is reduced, as long as Pp1**y and 161-u*;y a.re not exceeded.

5101520lI

VcP(max)

500 mW500 mW500 mW500 mW

5V10v15V20v

100 mA50 mA33 mA25 mA

f{sf*ferJ frr*#Jerr If Pp16a;y : 1 W how much voltage is allowed from collector to emitter if the transis-tor is oPerating with 16 : 100 mA?

1. (mA)

BJT CnanecrERrsrrcs nruo Pennruerens ; 177

EXAMPLE 4_6 The transistor in Figure 4-19 has the following maximum ratings: Polma,y : 800 mW,Vcrlmaxl : 15 Y and 161*u^; : 100 mA. Determine the maximum value to which V6scan be adjusted without exceeding aratitg. Which rating would be exceeded first?

FIGURE 4-19

7en5V

S*fxtit* First, find 1s so that you can determine 16..

vss-veE 5V-0.7Vlp:

-

: l9-5uAu D 22k{lI\BIc: Focls: (100X195 pA) : 19.5 mA

16 is much less than /c(ma*) and ideally will not change with V66. It is determined onlyby 1s and pp6.

The voltage drop across R6 isVp": 1gR6 : (19.5mA)(l.0kO) : 19.5V

Now you can determine the value of V66 when V6E : VCE(max) : 15 V.Vpr: Vqg

- Vgs

So,

VCCl-ur; = VCE(max) * (Vn.: 15V + 19.5V:34.5VVgg car be increased to 34.5 V, under the existing conditions, before V6B1,,,u*; isexceeded. However, at this point it is not known whether or not Pp16u*y has beenexceeded.

PD : VcE(,"*;16 : (15VX19.5mA) : 293mWSince Pp1o,*y is 800 mW it is not exceeded when Vgs: 34.5 V. So, VsBlmrx) : 15 Vis the limiting rating in this case. If the base current is removed causing the transistorto turn off, Ycn(max) will tle exceeded first because the entire supply voltage, V6g, willbe dropped across the transistor.

j

#rl*ged? Fr*i.r}r*t The transistor in Figure 4-19 has the following maximum ratings: Polmaxl : 500 mW, i

t/ycE(max) : ZlY,urid 1"1-*; : 200 mA. Determine the maximum value to which V6s i

can be adjusted without exceeding aratirg. Which rating would be exceeded first? iit

,..'.'.*.. --"'-'. --*j

*ermtimffi F*{m;ax)PD1-ury is usually specified at25'C. For higher temperatures, PD(max) is less. Datasheets of-ten give derating factors for determining PD(max) at any temperature above 25"C. For ex-ample, a derating factor of 2 mWl"C indicates that the maximum power dissipation isreduced 2 mW for each degree Celsius increase in temperature.