Download Bipolar Junction Transistors (BTTs)

Bipolar Junction Transistors (BJTs)
It is a three terminal semiconductor device consists of two p-n junctions. The basic
principle involved is to use the voltage between two terminals to control the current flowing
in the third.
There are two types of BJT (npn and pnp)
* The transistor is design to have heavily doped
Emitter and very thin and lightly doped base.
Depending on the bias condition (forward or reverse) of the junctions, different modes of
operation are obtained.
BJT modes of operation:
Mode
Cutof
EBJ
CBJ
Reverse Reverse
Saturation Forward Forward
Active
Application
Switching Applications logic circuit
Forward Reverse Amplifier
pnp transistor in the active mode
1) The forward bias voltage on Emitter –base junction causes current to flow across this
junction. Two components: holes injected from emitter into the base (Ip) and electron injected
from base into the emitter (In).
Ip is much higher than In (base is thin and lightly doped).
2) Most of the holes ( I P ) injected into the base
will diffuse (continue their way) through the
base toward the collector. This is because of that
the base is thin and attracted by the large negative
of the battery (VBC). Some of the holes (1   ) I P
will attracted by the negative of
the battery (VBE ). (i.e. some of the holes will
combine with the base electrons and disappear). These electrons will be substituted by
electrons from the battery)
3) Collector-Base reverse current (ICBo)
Minority carrier base holes (Ipo) and collector electrons (Ino) generates reverse current across the
collector base junction (Ico) or leakage current.
4) Transistor terminal currents:
IE  I p  In
I C  I p  I po  I no
I B  (1   ) I p  I n  I po  I no
Note that
I C  I B  I p  I po  I no  I p  I p  I n  I po  I no  I p  I n  I E
(i.e. The current enters the
transistor should be equal
current leaving it )
Summary pnp transistor in active
Current
Change carrier
IP
Majority holes injected from emitter into base
In
Majority electrons injected from base into emitter
I P
Most of IP that go to collector
(1   ) I P
Some of IP that do not go to collector
Ipo
Minority holes injected from base to collector
Ino
Minority electrons injected from collector to base
5) IB is very small and usually of the order of A 's or nA 's for most practical cases with
 (collector to emitter-current gain)  1  I C  I E by making the deflection area very
small.(Base very thin)
6) Current relations in active mode
I p  I co
IC



   100 (Base collector current gain)
I B (1   ) I p  I n  I co   1
I E  IC  I B
I C  I E
I C  I B
7) The emitter current IE can be expressed as
I E  I SE (eVBE / VT  1) (Forward bias pn
junction
current)
I C  I E  I SC (eVBE / VT  1) ISC: constant
called
saturation current or current scale factor
npn transistor
IE  I p  In
I B  (1   ) I n  I p  I po  I no
I C  I n  I po  I no
conclusion
In the active mode, the forward-bias voltage VBE cause an exponentially related
current iC to flow in the collector terminal. The collector current iC is almost independent of
the value of the collector voltage as long as the collector –base junction remains reversebiased.
In the active mode, the collector terminal behaves as a current source and its value is
determined by VBE .
ic –VCE characteristics
VCE  VBC  VBE
1) when VCE  0  VBC  0.7 (Forward)
IC  0
2) VCE  0.1  VBC  0.6 (Forward)
I C Start appearing
3) VCE  0.2  VBC  0.5 (Forward)
4) VCE  0.3  VBC  0.4 (in between)
5) VCE  0.4  VBC  0.3 (Revere)
Summary
Cutoff
Active
Saturation
Both junction Reverse
BEJ Forward
Both junction Forward
biased
BCJ Reverse
biased
All current practically zero
IC   I B
V BE  0.7
V CE
unknown
I E  IC  I B
Symbol
I B 
IC

V BE  0.7V
V CE  0  0.2V or max 0.3
I E  IC  I B
BJT
For npn transistor
VBE  VB  VE  0.7 to be ON
1) If VBE  0.7  cutoff  I B  I E  I C  0
2) To operate in active region BCJ should be reverse in active
region only: I C  I B ;
I E  (1   ) I B ;
I C  I E
3) To operate in saturation region BCJ should be forward ( I B 
VCE  0.2V only in saturation region
For pnp transistor
VBE and VCE are negative
and in saturation only
VEC  0.2V
IC

)
DC analysis
Example:
Determine the region of operation and the values of
I B , I C and VCE for the circuit shown in the figure
a) RB  300 k
b) RB  150 k
 F  100
Assume the transistor is working
Saturation  VCE  0.2
Check is ( I B 
IC

) ?
10  0.7
 0.031mA
300k
I
10  0.2
IC 
 4.9mA  C  0.049mA  I B
2k

IB 
This means that the assumption is wrong. Assume the transistor is working in active
mode
 IC   I B
but check BCJ is reverse
I B  0.031mA  I C  3.1mA
VC  10  RC I C  10  (2)(3.1)  3.8V
VCE  VC  VE  3.8  0  3.8V
VBC  VB  VC  0.7  3.8  3.1V
 BCJ is reverse which confirms the assumption or since VCE>0.3 the assumption is
verified.
Find the smallest value of RB for which the transistor still operate in the active mode. At
the edge of saturation and active.
VCE  0.3V
10  0.3
 4.85mA
2k
I
I B  C  0.0485mA
IC 

IB 
10  0.7
 0.0485mA
RB
RB 
9.3
 191.8k 
0.0485mA
For the circuit shown find VB, I B , I E , I C
Assume the transistor is working in active mode
 IC   I B
but check BCJ is reverse
20 I B  0.7  5 I E  5
I E  (1   ) I B  53.3I B
286.5 I B  4.3  I B  0.015mA
I E  I C  I B  (1   ) I B  0.8mA
VB  20 I B  0.3V
VE  VB  0.7  1V
VC  5  5I C  1.08V
VBC  VB  VC  0.3  1.08  1.38V  Re verse  Active 
VEC  VE  VC  1  (1.08)  2.08V >0.3V  Active 
Note: β is very large  I B  0 and IE=IC
  52.3
Find all the currents and voltages?   100
Assume the transistor working in the active
 I C  100 I B check BCJ is reverse
10  10 I E  0.7  10 I B  10 I E  10
I E  (   1) I B
I B  9.5 A
I C  0.95mA
I E  0.96mA
VC  10  10 I E  0.5V
VB  VC  10 I B  0.405V
VE  VB  0.7  0.3V
VBC  0.405  0.5  0.1V
VCE  0.5  (0.3)  0.8V  0.3
(Re verse  active)
(active)
Small signal equivalent circuit
Amplifier
IB  IS
q
VBE
kT
(e
 1)
vbe  VBE  v BE
iB  I S
q
vBE
kT
(e
 1)
q
q
 q (VBE vbe) 
vbe
VBE
kT
kT
kT


.e
 ISe
I B  ib  I S e




 IS
IS
q
VBE
kT
e
q
VBE
kT
e
 IB  IS
vbe
ib
 IS
q
vbe )
kT
q
VBE
kT
e
q
vbe
kT
q
VBE
e kT
q
VBE
kT
e
q
vbe
kT
kT / q
1


q
IB
IB.
kT
1
25


1000 I B 40 I B
ib  I S
vbe
ib
(1 
I C  I B
iC  i B
I C  iC   ( I B  i B )  I C  I B
iC  i B
Small signal ac equivalent circuit of the BJT
Transconductance g m 
 ib  
IC
VT
vbe
 40 I B vbe
1
40 I B
 40 I C vbe  g m vbe
g m  40 I C
g m r  40 I C vbe
1
 g m r  
40 I B