# Chapter 5 Bipolar Junction Transistors by dyz36301

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```									         Chapter 5
Bipolar Junction Transistors
Chapter Goals

• Explore the physical structure of bipolar transistor
• Study terminal characteristics of BJT.
• Explore differences between npn and pnp transistors.
• Develop the Transport Model for bipolar devices.
• Define four operation regions of the BJT.
• Explore model simplifications for the forward active
region.
• Understand the origin and modeling of the Early effect.
• Present a PSPICE model for the bipolar transistor. Discuss
bipolar current sources and the current mirror.
Physical Structure

• The BJT consists of 3 alternating layers
of n- and p-type semiconductor called
emitter (E), base (B) and collector (C).
• The majority of current enters collector,
crosses the base region and exits through
the emitter. A small current also enters
the base terminal, crosses the base-
emitter junction and exits through the
emitter.
• Carrier transport in the active base
region directly beneath the heavily
doped (n+) emitter dominates the i-v
characteristics of the BJT.
Transport Model for the npn Transistor
• Base-emitter voltage vBE and
base-collector voltage vBC
determine the currents in the
transistor and are said to be
positive when they forward-bias
their respective pn junctions.
• The terminal currents are the
• The narrow width of the base           collector current(iC ), the base
region causes a coupling between       current (iB) and the emitter
the two back to back pn junctions.     current (iE).
• The emitter injects electrons into   • The primary difference between
base region; almost all of them
the BJT and the FET is that iB is
travel across narrow base and are
removed by collector.                  significant, while iG = 0.
npn Transistor: Forward Characteristics
Base current is given by
i         I  v   

 

i  F  S exp BE  1
   
B              V  
   
F     F        T  
   


20   500 is forward current gain
F
Emitter current is given by
                  v  I   
i i i  S exp BE  1
    
E C B        V  


F      T   
Forward transport current is
       

v   


             0.95              F 1.0

i  i  I exp BE  1
                            F        1
C F     S   V  

 T  
                                F
IS is saturation current        In this forward active operation region,
1018 A  I  10 9 A                              i          i
S                                        C        C 
i     F    i     F
VT = kT/q =0.025 V at room temperature               B          E
npn Transistor: Reverse Characteristics
0    20 is reverse current gain
R

Base currents in forward and reverse modes
are different due to asymmetric doping levels
in the emitter and collector regions.

Emitter current is given by
I   v      
Reverse transport current is    i   S exp BC  1
          

v   



          C        V


T   

i  i  I exp BC  1
   
R
R    E   S   V  
                         
T  

       
0    R  0.95
Base current is given by             R  1
i I     v      
R
i  R  S exp BC  1
          
B         V  
R  R       T   
npn Transistor: Complete Transport
Model Equations for Any Bias

v       v     I      v                
i  I exp BE   exp BC   S  exp BC             1
                
                                             
C   S    V 

T 
  V
T   R 
            V              
                       T              


v        v     I      v               
i  I exp 

BE   exp BC   S  exp BE


           
 1

E   S    V 

T 
   V  
T          V             
         
F     T              

I  v       I 
     v                
i  S exp BE  1  S  exp BC             1
      
                                
B       V    

T  
      V              
F               
R     T               

The first term in both the emitter and collector current expressions gives
the current transported completely across the base region.
Symmetry exists between base-emitter and base-collector voltages in
establishing the dominant current in the bipolar transistor.
pnp Transistor: Operation

• The voltages vEB and vCB are positive when they forward bias
their respective pn junctions.
• Collector current and base current exit the transistor terminals
and emitter current enters the device.
pnp Transistor: Forward Characteristics

Base current is given by:

i I     v      
i  F  S exp EB  1
          
B         V  
F  F       T   

Emitter current is given by:
Forward transport current is:                   
       v

  
1     
i  i  i  I 1
exp EB  1
   
                E C B       S  
       V  
   
i  i  I exp
v  
EB  1


  F      T  



   
C F     S   V  

  T  
    
pnp Transistor: Reverse Characteristics

Base current is given by:

i  I     v      
i  F  S exp CB  1
          
B         V  
R  R       T   

Emitter current is given by:
Reverse transport current is:            
        v

   
1     
i   I 1

exp CB  1
   
                C     S 
       V  
   

i  i  I exp
v  
CB  1


   R     
T  
    
    
R    E   S   V  

   T  
    
pnp Transistor: Complete Transport
Model Equations for Any Bias

v        v      I      v     

i  I exp EB   exp CB  

  

        S  exp CB  1
             
C   S   V 
T 
    V
T   R 
           V  
          
     T    


v        v      I      v     

i  I exp EB   exp CB  

  

        S  exp EB  1
             
E   S   V 
T 
    V
T   F 
           V  
          
     T    


I    v       I 
       v     
i  S exp EB  1  S  exp CB  1
       
                       
B       V    

T  
        V  
F             
R     T    
Circuit Representation for Transport
Models

In the npn transistor (expressions analogous for the pnp transistors), total
current traversing the base is modeled by a current source given by:
v       v   
         

i  i  i  I exp BE   exp BC 
               
T F R       S    V 
         V  
T        T 
Diode currents correspond directly to the 2 components of base current.

I   v        I     v      
i  S exp BE  1  S exp BC  1
                     
B      V  

   

 V  


F      T       R      T   
Operation Regions of the Bipolar
Transistor
Base-emitter junction                Base-collector junction
Reverse Bias                 Forward Bias
Forward Bias        Forward active region          Saturation region
(Normal active region)        (Not same as FET
(Good Amplifier)             saturation region)
(Closed switch)
Reverse Bias           Cutoff region           Reverse-active region
(Open switch)            (Inverse active region)
(Poor amplifier)
i-v Characteristics Bipolar Transistor:
Common-Emitter Output Characteristics

For iB=0, the transistor is cutoff. If iB >0, iC
also increases.

For vCE > vBE, the npn transistor is in the
forward active region, iC = F iB is independent
of vCE..
For vCE< vBE, the transistor is in saturation.

For vCE< 0, the roles of collector and emitter
are reversed.
i-v Characteristics of Bipolar Transistor:
Common-Emitter Transfer Characteristic

This characteristic defines the relation
between collector current and base-emitter
voltage of the transistor.
It is almost identical to the transfer
characteristic of a pn junction diode.
Setting vBC =0 in the collector-current
expression:

v   
   



i  I exp BE  1
   
C   S   V  

 T  
   
Junction Breakdown Voltages

• If reverse voltage across either of the two pn junctions in the transistor
is too large, the corresponding diode will break down.
• The emitter is the most heavily doped region, and the collector is the
most lightly doped region.
• Due to these doping differences, the base-emitter diode has a relatively
low breakdown voltage (3 to 10 V). The collector-base diode is
typically designed to break down at much larger voltages.
• Transistors must therefore be selected in accordance with the possible
reverse voltages in circuit.
Simplified Forward-Active Region
Model
In the forward-active region, the base-emitter junction is forward-biased
and the base-collector junction is reverse-biased. vBE > 0, vBC < 0
If we assume that v        4kT                    4kT
        0.1V v             0.1V
BE       q             BC          q
then the transport model terminal current equations simplify to:
v       I            
vBE 

i  I exp BE  S  I exp




C S      V  
              S      V     
T  R
                     T 

I    
vBE 
   I      I        
vBE 

i  S exp       
S  S exp           
E        V                      V                i  i
 T                        T 
F                 F      F                             C    FE
I              I     I       I                   i  i
vBE                             vBE              FB
i  S exp       
S  S  S exp                      C
B        V                            V

       i  (  1)i
 T                               T          E     F    B
F                  F     R       F
The BJT is often considered a current-controlled current source, although
   fundamental forward active behavior suggests a voltage-controlled current
source.
Simplified Circuit Model for Forward-
Active Region

• Current in the base-emitter diode is amplified by the common-emitter
current gain F and appears at the collector
• The base and collector currents are exponentially related to the base-emitter
voltage.
• The base-emitter diode is often replaced by a constant voltage drop model
(VBE = 0.7 V), since it is forward-biased in the forward-active region.
Simplified Forward-Active Region
Model (Analysis Example)
•   Problem: Find Q-point
•   Given data: F = 50, VBC =VB - VC= -9 V
•   Assumptions: Forward-active region of operation, VBE = 0.7 V
•   Analysis:
V     8200I (V      ) 0
BE         E      EE
8.3V
I           1.01mA
E 8200
I      1.02mA
I  E                19.8A
B  1         51
F
I   I  0.990mA
C    F B
V     (V    )V     V
CE         EE     CC     R
V      9 9 8.3 9.7V
CE
Note: V  I R here.
R E


Biasing for BJT

• The goal of biasing is to establish a known Q-point, which
in turn establishes the initial operating region of transistor.
• In BJT circuits, the Q-point is represented by (VCE, IC) for
the npn transistor or (VEC, IC) for the pnp transistor.
• In general, during circuit analysis, we use a simplified
mathematical relationships derived for the specified
operation region of the transistor.
• The practical biasing circuits used with BJTs are:
– The Four-Resistor Bias network
– The Two-Resistor Bias network
Four-Resistor Bias Network for BJT
R              RR
V    V       1        R   1 2
EQ    CC R  R        EQ R  R
1   2           1 2
V    R   I V     R I      F  75
EQ   EQ B     BE   E E
4 12,000I  0.716,000( 1)I
B               F    B
4V-0.7V
I               2.68A        I   I 201A
B
1.2310 6            C F B
I ( 1)I 204A
E     F   B 

V     V     R I R I
CE   CC     C C    E E
              
    R 
 R  E I  4.32V
CE Loop
V
BE Loop
CC  C   C
       
     F 


Q-point is (4.32 V, 201 A)
Four-Resistor Bias Network for BJT
(Check Analysis)
• All calculated currents > 0, VBC = VBE - VCE = 0.7 - 4.32 = - 3.62 V
• Hence, the base-collector junction is reverse-biased and the assumption
of forward-active region operation is correct.
• The load-line for the circuit is:                      
R 

V   V    R  F I 12 38,200I

CE   CC    C   C

 
C
F 


The two points needed to plot the load
       line are (0, 12 V) and (314 A, 0). The
resulting load line is plotted on the
common-emitter output characteristics
for IB= 2.7 A.
The intersection of the corresponding
determines the Q-point.
Four-Resistor Bias Network for BJT:
Design Objectives
• From the BE loop analysis, we know that
VV         V   V
I    EQ    BE      EQ   BE                for REQ (F 1)RE
B R   ( 1)R     ( 1)R
EQ    F     E     F     E
• This will imply that IB << I2 so that I1 = I2 to good approximation in


the base voltage divider. Then the base current doesn’t disturb the
voltage divider action, and the Q-point will be approximately
independent of base voltage divider current.
• Also, VEQ is designed to be large enough that small variations in the
assumed value of VBE won’t have a significant effect on IB.
• Base voltage divider current is limited by choosing I  I / 5
2 C
This ensures that power dissipation in base bias resistors is < 17 % of
the total quiescent power consumed by the circuit, while I2 >> IB.
Four-Resistor Bias Network for BJT:
Design Guidelines
•   Choose I2 = IC/5. This means that (R1+R2) = 5VCC/IC .
•   Let ICRC =IERE = (VCC - VCE)/2. Then RC = (VCC - VCE)/2IC; RE =FRC
•   If REQ<<(F+1)RE, then IERE = VEQ - VBE.
•   Then (VCC - VCE)/2 = VEQ - VBE, or VEQ = (VCC - VCE + VBE)/2.
•   Since VEQ = VCCR1/(R1 +R2) and (R1+R2) = 5VCC/IC,

VCC VCE  2VBE  5VCC  VCC VCE  2VBE 
R1                  
      5              
     2VCC        
  IC         2IC       
• Then R2 = 5VCC/IC - R1.
• Check that REQ<<(F+1)RE. If not, adjust bullets 1 and 2 above.
 Note: In the LabVIEW bias circuit design VI (NPNBias.vi), bullet 1
•
is called the “Base Margin” and bullet 2 is called the “C-E V(oltage)
Drops”.
Problem 5.87 4-R Bias Circuit Design
Two-Resistor Bias Network for BJT:
Example
• Problem: Find the Q-point for the pnp transistor in the 2-resistor bias
circuit shown below.
• Given data: F = 50, VCC = 9 V
• Assumptions: Forward-active region operation with VEB = 0.7 V
• Analysis:
9 V   18,000I 1000(I  I )
EB         B       C B
9 V   18,000I 1000(51)I
EB          B          B
9V 0.7V
I           120A
B 69,000
I  50I  6.01mA
C     B
V    91000(I  I ) 2.87V
EC           C B

      Q-point is : (6.01 mA, 2.87 V)
PNP Transistor Switch Circuit Design
Emitter Current for PNP Switch Design
BJT PSPICE Model
• Besides the capacitances which are
associated with the physical structure,
current iS, capacitance CJS, related to the
large area pn junction that isolates the
collector from the substrate and one
transistor from the next.
• RB is the resistance between external
base contact and intrinsic base region.
• Collector current must pass through RC
on its way to the active region of the
collector-base junction.
• RE models any extrinsic emitter
resistance in the device.
BJT PSPICE Model -- Typical Values

Saturation Current = 3 e-17 A
Forward current gain = 100
Reverse current gain = 0.5
Forward Early voltage = 75 V
Base resistance = 250 
Collector Resistance = 50 
Emitter Resistance = 1 
Forward transit time = 0.15 ns
Reverse transit time = 15 ns
Minority Carrier Transport in Base
Region
• With a narrow base region, minority carrier density decreases linearly
across the base, and the Saturation Current (NPN) is:
I  qADn bo        i                 where
S      W     N   W
B     AB B
NAB = the doping concentration in the base
ni2 = the intrinsic carrier concentration (1010/cm3)
nbo = ni2 / NAB
Dn = the diffusivity = (kT/q)n
•   Saturation current for the PNP transistor is: I S  qAD p bo         i
W     N    W
B     DB B
• Due to the higher mobility () of electrons compared to holes, the npn
transistor conducts higher current than the pnp for equivalent doping
and applied voltages.
Diffusion Capacitance

• For vBE and hence iC to change, charge stored in the base region must
also change.
• Diffusion capacitance in parallel with the forward-biased base-emitter
diode produces a good model for the change in charge with vBE.

dQ               1 qAnboWB    v

  I
BE   T 
C                            exp    
D dv               V     2        V         F
BE Q  point    T              T  VT

• Since transport current normally represents collector current in the
forward-active region,
I
C  C
D V F
T
Early Effect and Early Voltage
• As reverse-bias across the collector-base junction increases, the width of
the collector-base depletion layer increases and the effective width of base
decreases. This is called “base-width modulation”.
• In a practical BJT, the output characteristics have a positive slope in the
forward-active region, so that collector current is not independent of vCE.
• “Early” effect: When the output characteristics are extrapolated back to
where the iC curves intersect at common point, vCE = -VA (Early voltage),
which lies between 15 V and 150 V.
• Simplified F.A.R. equations, which include the Early effect, are:

v                            I   v
           

  
i   S exp BE
           
     1  CE 

                        B        V


F   FO    V 
                           FO

 T





A 

  
v     v





i  I exp
   BE 

1  CE   I
C S 
   V  V  F B



   T 




A 
BJT Current Mirror
• The collector terminal of a BJT in the
forward-active region mimics the
behavior of a current source.
• Output current is independent of VCC as
long as VCC ≥ 0.8 V. This puts the BJT
in the forward-active region, since VBC ≤
- 0.1 V.
• Q1 and Q2 are assumed to be a
“matched” pair with identical IS, FO, and
VA,.
V   V
I       BB   BE  I  I  I
REF       R       C1 B1 B2
BJT Current Mirror (continued)

V
         V


    I

   V     

I      I exp BE
        1 CE1   2 S exp BE 
REF    S    V

        
       V    
        
    V   
 T

        
          A       FO     T   

V
            
1 CE 2
    V     V         
                V
I   I exp BE  1 CE 2   I                  A
C2   S      V  
V    

REF     V         2

     T          A           1 BE 
V                     V      
A      FO
1 CE2
I              V
MR       O             A       is the "Mirror Ratio".
I         V
REF 1 BE  2
V       
A      FO
With an infinite FO and VA (ideal device), the mirror ratio is unity. Finite

current gain and Early voltage introduce a mismatch between the output
and reference currents of the mirror.
BJT Current Mirror: Example
•   Problem: Find output current for given current mirror
•   Given data: FO = 75, VA = 50 V
•   Assumptions: Forward-active operation region, VBE = 0.7 V
•   Analysis:
V    V
I      BB    BE  12V 0.7V  202A
REF        R         56k
1 12
I  MR  I      (202A)      75  223A
O         REF          1 0.7  2
75 50


VBE =
6.7333e-01
IC2 =
5.3317e-04
IC21 =
5.3317e-04
BJT Current Mirror: Altering the Mirror Ratio
A
I I    E          where ISO is the saturation current of a BJT
S   SO A
with one unit of emitter area: AE =1(A). The
actual dimensions of A are technology-
dependent.
The Mirror Ratio of a BJT current mirror can be changed by simply
changing the relative sizes of the emitters in the transistors. For the
“ideal” case, the Mirror Ratio is determined only by the ratio of the
two emitter areas.
V
1 CE 2
V                          A
I  n.I              A                     n  E2
O      REF   V         2                     A
1 BE                              E1
V      
A      FO
BJT Current Mirror: Output Resistance
• A current source using BJTs doesn’t have an output current that is
completely independent of the terminal voltage across it, due to the
finite value of Early voltage. The current source seems to have a
resistive component in series with it.

V      v              V    v
1  CE2 ce2            1  CE o
V                    V
i i  I            A     I             A
O C2 REF    V         2     REF V          2
1 BE                1 BE 
V                   V      
A      FO            A      FO

              1
               1


io
                        I       V
Ro    


        C2        A
vo
             

V V
    I

     Q pt   


A CE


O

• Ro is defined as the “small signal” output resistance of the current

mirror.

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