-
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
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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.