BIPOLAR JUNCTION TRANSISTOR - Delta Univ · 2015. 12. 24. · 1- 2 • BASIC TRANSISTOR OPERATION...
Transcript of BIPOLAR JUNCTION TRANSISTOR - Delta Univ · 2015. 12. 24. · 1- 2 • BASIC TRANSISTOR OPERATION...
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Electronics
BIPOLAR JUNCTION TRANSISTOR
by
Prof. Fahmy El-Khouly
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CHAPTER THREE
BIPOLAR JUNCTION TRANSISTORS
1-1 • TRANSISTOR CONSTRUCTION
constructed with three doped semiconductor regions separated by twop-n junctions
heavily doped emittermoderately doped collector lightly doped base
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p-n-p
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(b) p-n-p
Standard bipolar junction transistor (BJT) symbols
E
(a) n-p-n
The term bipolar refers to the use of both holes and electrons as carriers in the transistor structure
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1- 2 • BASIC TRANSISTOR OPERATION
In order for the transistor to operate properly as an amplifier, the two p-n junctions must be correctly biased with external dc voltages
The operation of the p-n-p is the same as for the n-p-n except that the roles of the electrons and holes, the bias voltage polarities, and the current directions are all reverse
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(a)n-p-n (b)p-n-p
FIGURE(1- 3 ) Forward-reverse bias of a bipolar transistor.
the base-emitter (BE) junction is forward-biased
and the base-collector (BC) junction is reverse-
biased.
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FIGURE-(1-4)
illustration of
BJT action.
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1-3 Transistor Currents
IE=IC+IBAs mentioned before, IB is very small compared to IE or IC.
The capital-letter subscripts indicate dc values.
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1-4 • TRANSISTOR CHARACTERISTICS AND PARAMETERS
(a) n-p-n
When a transistor is
connected to dc bias
voltages, Vbb
forward-biases the
base-emitter junction,
and Vcc reverse-
biases the base-
collector junction.
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the dc current gain of a transistor (βDC)
βDC =IC / IB
Typical values of βdc range from less than 20 to 200 or higher, βdc is usually designated as an equivalent hybrid parameter, hFE, on transistor data sheets
hFE = βDC
The ratio of the dc collector current Ic to the dc emitter current IE is the dc alpha
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αdc =Ic / IE
Typically, values of αDC range from 0.95 to 0.99 or greater, but αDC is always less than 1
Relationship of βDC and αDC :--
IE / IC = IC / IC + IB / IC
1 / αDC = 1+ 1 /βDC
αDC = { βDC / ( 1 + βDC ) }
βDC = { αDC / (1 - αDC ) }
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Current and Voltage Analysis
.IB: dc base current
IE: dc emitter current
IC: dc collector current
vBE: dc voltage at base
with respect to emitter
VCB. dc voltage at
collector with respect to
base
VCE - dc voltage at
collector with respect to
emitter
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VBB : forward-biases the base-emitter junction VCC reverse-biases the base collector junctionVBE=0.7volts
VRB = VBB– VBE
IB * Rb = VBB– VBE
IB =( VBB– VBE ) / RB
VCE = VCC– IC RC
VCB = VCE– VBE
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1– 6 Collector Characteristic Curves
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FIGURE ( 1- 9 ) Collector characteristic curves
In linear region IC = βDC IB
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1 - 7 Cutoff
Cutoff: Collector leakage current (ICEO) is extremely small and is usually neglected. Base-emitter and base-collector junctions are reverse-biased
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1-8 Saturation
As IB increases due to increasing VBB, IC also
increases and VCE decreases due to the
increased voltage drop across RC- When the
transistor reaches saturation, IC can increase
no further regardless of further increase in IB.
Base-emitter and base-collector junctions are
forward-biased
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1-9 DC Load Line
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EXAMPLE
Determine whether or not the transistor in Figure
(1-15 ) is in saturation.
Assume VCE Sat=0.2V.
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Solution First, determine Ic(sat)
IC(sat) = (VCC – VCE (sat.) ) / RC
= (10V-0.2V ) / Ik Ω
= 9.8 mANow, see if IB is large enough to produce IC(sat)
IB(sat) =(VBB – VBE) / RB
=(3 V - 0.7 V) /10 k Ω= 0.23 mA= 0.23 mA
IC = βdc ib = (50) (0.23 mA) = 11.5 mA
Therefore, the transistor is saturated
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1-11 Maximum Transistor Ratings
*The transistor has limitations on its operation.
*maximum ratings are given for collector-to-base voltage, collector-to-emitter voltage, emitter-to-base voltage, collector current, and power dissipation.
*If VCE is maximum, Ic can be calculated as
*The product of VCE and Ic must not exceed the maximum power dissipation.
IC= (P D(Max)) / VCE
* If IC is maximum, VCE can be calculated by rearranging Equation
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VCE =( P D(Max)) / Ic
Assume P D (max) is 500 mW, VCE max is 20 V, and Ic (max)
is 50 mA
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EXAMPLE ( 6 )
The transistor in Figure (1-18 ) has the following
maximum ratings: PD (max) = 800 mW, VCE (max)
= 15 V, and IC (max) = 100 mA. Determine the
maximum value to which VCC can be adjusted
without exceeding a rating. Which rating would be
exceeded first?
ICVCEPD(Max)
100 mA
50 mA
33 mA
25 mA
5 V
10 V
15 V
20 V
500 mV
500 mV
500 mV
500 mV
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Solution First, find IB so that you can determine IC.
IB = (VBB - VBE ) / RB = ( 5 –0.7 ) / 22k Ω = 195 uA
IC = βdc *IB = (100) * (195 uA) = 19.5 mAIC is much less than IC ( max) and will not change with VCC It is determined only by IB and βdc. The voltage drop across RC is
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VRC = IC * RC = (19.5 mA) * ( 1 k Ω) = 19.5 V .
Now you can determine the value of VCC when
vCE = VCE (max) = 15 V.
VRC=VCC – VCE
VCC(max) = VCE(Max) + VRC = 15 V + 19.5 V
= 34.5 V
PD = VCE(max) * Ic = (15 V) * (19.5 mA) = 293 mW
Since PD (Max). is 800 mW, it is not exceeded when
VCC = 34.5 V. So, VCE (max) = 15 V is the limiting
rating in this case. If the base current is removed
causing the transistor to turn off, VCE(max) will be
exceeded first because the entire supply voltage,
VCC, will be dropped across the transistor.
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1-18 • THE TRANSISTOR AS A SWITCH
(a) Cutoff— open switch
(b) Saturation — closed switch
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1-19 Conditions in Cutoff
As mentioned before, a transistor is
in the cutoff region when the base-
emitter junction is not forward-biased.
Neglecting leakage current, all of the
currents are zero, and vce is equal to Vcc
VCE cutoff = VCC
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1-20 Conditions in Saturation
when the base-emitter junction is forward-biased and there is enough base current to produce a maximum collector current, the transistor is saturated
IC sat= ( VCC – VCE sat ) /Rc
Since VCE sat is very small compared to VCC it can
usually be neglected.
The minimum value of base current needed to
produce saturation is
IB min = Ic sat / βdc
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IB should be significantly greater than IB min to
keep the transistor well into saturation
EXAMPLE
For the transistor circuit in Figure (1-23), what is
vCE when Vin = 0 V?.
What minimum value of IB is required to saturate
this transistor if βdc is 200 Neglect vCE SAT .
Calculate the maximum value of RB when Vin = 5
V.
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FIGURE )1-23)Vcc=10V,Rc =1k Ω
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Solution
When Vin = 0 V, the transistor is in cutoff
(acts like an open switch) and
VCE=Vcc=10 V
Since Vce sat is neglected (assumed to be
0 V),
Vcc = 10 V
iCE sa t = vcc / Rc
= 10 / 1K Ω =10 mA
iB min = Ic sat / βdc =(10 mA/200) =50 uA
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When the transistor is on, vBE= 0.7 V. The voltage
across RB is :-
VRB =VIN-VBE = 5V-0.7V=4.3V
Calculate the maximum value of RB needed to
allow a minimum IB of 50 uA by Ohm's law as
follows:
Rb max = ( VRB / IB(min) ) = ( 4.3 v/ 50 uA ) =
86 k Ω
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A Simple Application of a Transistor Switch
transistor is in cutoff and, since there is no collector current, the LED does not emit light
When the square wave goes to its high level, the transistor saturates. This forward-biases the LED, and the resulting collector current through the LED causes it to emit light
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CHAPTER FOUR
TRANSISTOR BIASING CIRCUITS
2-2 DC Bias
The dc operating point is often referred to as the Q-point
(quiescent point منطقة السكون).
If an amplifier is not biased with correct dc voltages on the input and
output, it can go into saturation or cutoff when an input signal is
applied.
(a) Linear operation: larger output has same shape as input except it is inverted
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(C) Nonlinear operation: output voltage
limited (clipped) by saturation
dc operating point being too close to saturation.
The dc operating point (Q-point) being too close to cutoff.
(b) Nonlinear operation: output voltage limited (clipped) by
cutoff
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2-3 Graphical Analysis
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VCE = Vcc – Ic * Rc = 10 V - (20 mA)*(220 Ω) =
= 10 V - 4.4 V = 5.6 V .
VCE = 10 V - (30 mA)(220 Ω) = 10 V - 6.6 V = 3.4 V
VCE = 10 V (40 mA)* (220 Ω) = 10 V - 8.8 V = 1.2V
Kirchhoff's voltage law applied around the collector loop gives
Vcc – lc * Rc - VCE = 0
This results in a straight line equation for the load line of the form
y = m x + b as follows:
Ic = - (1 / Rc) VCE + Vcc / Rc
where –(l/Rc ) is the slope and Vcc / Rc is the y-axis intercept point.
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2-6 Waveform Distortion
(a)Transistor driven into saturation
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(B-)Transistor driven into cutoff
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(c) Transistor driven into both saturation and cutoff
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EXAMPLE 2-1
Determine the Q-point in Figure (2-7), and find the maximum
peak value of base current for linear operation. Assume βdc =
200.
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IB = ( vBB - vBE )/ RB = (10V-0.7V) / 47K Ω =
198 uA
IC = βDC * IB=( 200)* (198 uA) = 39.6 mA
vCE. = VCC – IC * RC = 20 V - 13.07 V = 6.93 V
The Q-point is at IC = 39.6 mA and at VCE = 6.93 V. Since IC(cutoff) =0.
you need to know IC(sat) to determine how much variation in collector
current can occur and still maintain linear operation of the transistor.
IC sat = ( VCC / Rc) = (20V / 330 Ω) = 60.6mA
The dc load line is graphically illustrated in Figure (2-8),
showing that before saturation is reached, IC can increase an
amount ideally equal to
IC(sat) – ICQ = 60.6 mA - 39.6 mA =21 mA
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However, IC can decrease by 39.6 mA before cutoff (IC = 0
V) is reached. Therefore, the limiting excursion is 21 mA
because the Q-point is closer to saturation than to cutoff.
The 21 mA is the maximum peak variation of the collector
current. Actually, it would be slightly less in practice
because VcE(sat) is not quite zero. The maximum peak
variation of the base current is determined as follows:
Ib peak = ( Ic max / βDC) = 21 mA / 200 = 105 uA
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2-7 BASE BIAS
IB= (Vcc - VBE )/ RB
VCC – IC* RC- VCE=0
VCE = VCC -IC * RC
IC = βDC {(VCC – VBE ) / RB }
A more practical method is to use VCC as the single bias source, as shown
in Figure
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2-10 • EMITTER BIAS
Although this method
of biasing requires
two separate dc
voltage sources, one
positive and the
other negative, it
does have an
important advantage
as you will learn.
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VEE-IB*RB-VBE-IE*RE=0
Solving for VEE.
VEE = IB * RB + IE * RE + VBE
Since IC = IE and IC = βDC * IB
Substituting for IB,
(IE /βDC)*RB+IE*RE +VBE=VEE
Factoring out IE yields
IE {(RB /βDC) + RE} +VBE=VEE
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Transposing VBE and then solving for IE
IE=( VEE – VBE) / {RE+(RB/ βDC)} ------- (2-4)
Since IC = IE.
IC = ( VEE - VBE ) / {RE+(RB/ βDC)} ---- (2-5)
VE=-VEE + IE*RE ----------------(2-6)
VB= VE +VBE-----------------------(2-7)
VC = VCC – ICRC ----------------- (2-8)
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IC = IE yields
VCE = VCC – IC*RC - (-VEE + IE*RE)
= VCC + VEE – IC(RC + RE)
EXAMPLE (2-3)
Find IE, IC, and VCE in Figure (2-12) for βDC =
100 and VBE = 0.7 V. Draw the dc load line
showing the Q-point.
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+10V
-10V
Solution
IE=( VEE – VBE) / {RE+(RB/ βDC)}
= (10V-0.7V)/ (4.7 k Ω +4.7 k Ω /100) = 1.80Ma
IC= IE = l.80 mA
VCE =VCC + VEE – IC*(RC + RE)
= 10 V + 10 V - 1.80 mA*(5.7 k Ω) = 9.74 V
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IC sat ={ VCC - (-VEE)} / (RC+RE) =
={10 V - (-10 V)} / (5.7 k Ω)
=20 V / 5.7 k Ω= 3.51mA
The collector-to-emitter voltage at cutoff is
VCE (cutoff) = VCC - (-VEE) = 10 V - (-10 V) = 20 V
Δ IC(max)= IC sat – IC = 3.51 mA - 1.80 mA= 1.71 mA
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2-13 VOLTAGE-DIVIDER BIAS
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2-14 Input Resistance at the Base
R IN base = (V IN / I IN )
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VIN=VBE + IE*RE
With the assumption that VBE « IE*RE. the equation reduces to
V IN = IE*RE
DC input resistance is RIN = VIN/ IIN.
Now, since
IE = IC = βDC *IBThe input current is the base current:
IIN=IBBy substitution,
RIN (base) = VIN / I IN = (βDC * IB * RE) / IBCanceling the IB terms gives
R IN (base) = βDC* RE ------------------(2-9)
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2-15 Analysis of a Voltage-Divider Bias Circuit
If (βDC* RE ) ≥ 10 R2. then the formula simplifies to
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VE = (VB)- VBE ------------------------- (2-11)
IE= VE / RE ---------------------- (2-12)
IC=IE ----------------------- (2-13)
VC=VCC -IC * RC ------ (2-14)
VCC – IC*RC –IE*RE- VCE= 0
Since IC = IE,
VCE = VCC – IC * RC – IE * RE
= VCC - IC (RC + RE) ------------ (2-15)
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EXAMPLE (2-7)
Determine VCE and IC in Figure (2-22)
if βDC = 100,RC = 1k Ω ,RE = 560 Ω ,R1 =
10K Ω , VCC =10Vand R2 =5.6K Ω.
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Solution
First, determine the dc input resistance at the
base to see if it can be neglected.
RIN (base) = βDC* RE = (100)*(560 Ω) = 56 k Ω
VB = { R 2/ ( R1+ R2)}* VCC
= {5.6k Ω / 15.6k Ω }* 10 V=3.59V
So, VE = VB – V BE =
= 3.59 V-0.7V= 2.89V
And IE = VE / RE = 2.89V / 56
= 5.16 mA
Therefore, IC = IE = 5.16 mA
VCE = VCC – IC (RC + RE)
= 10 V - 5.16 mA*(1.56 kΩ)
= 1.95 V
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2-18 • COLLECTOR-FEEDBACK BIAS
IB = (VC - VBE ) / RB -----------------(2-16)
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Lel's assume that IC » IB- The collector voltage is
VC = VCC - lC*RC
Also, IB = IC / βDC
Substituting for IB and VC in Equation (2-16),
IC / βDC = ( VCC - IC * RC- VBE ) / RB
{(IC*RB )/ βDC} +lC * RC = VCC - VBE.
Then you can solve for IC as follows:
IC { RC + (RB / βDC)} = VCC - VBE.
IC = {VCC - VBE} / { RC + (RB / βDC)} ----
Since the emitter is ground, VCE = VC
VCE = VCC - lC * RC ---------- ( 2-18)
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EXAMPLE (2-10)
Calculate the Q-point values (IC and VCE) for the circuit in
Figure (2-28).RC =10 k ,RB = 100K ,VBE =0.7V, βDC = 100
+10 V
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IC = (VCC- VBE ) / {RC +( RB / βDC) }
= (10 V - 0.7 V) / {lOk Ω +(100k Ω /100) }
= 845 uA
The collector-to-emitter voltage is
VCE = VCC – IC * RC = 10 V - (845 uA)*(10 k Ω) =
1.55 V