Transcript Chapter 4 Bipolar Junction Transistor

Chapter 4
Bipolar Junction Transistor
REMEMBER THIS
Current flow in the opposite direction of the electrons flow;
same direction as holes
e
e
e
I
h
h
h
Transistor Structures
 The bipolar junction transistor (BJT) has three
separately doped regions and contains two pn junctions.
 Bipolar transistor is a 3-terminal device.
 Emitter (E)
 Base (B)
 Collector (C)
 The basic transistor principle is that the
voltage between two terminals controls the
current through the third terminal.
 Current in the transistor is due to the flow of both electrons
and holes, hence the name bipolar.
Transistor Structures
 There are two types of bipolar junction transistor: npn and
pnp.
 The npn bipolar transistor contains a thin p-region
between two n-regions.
 The pnp bipolar transistor
contains a thin n-region
sandwiched between two pregions.
3 Regions of Operation
 Active
Operating range of the amplifier.
Base-Emitter Junction forward biased.
Collector-Base Junction reverse biased
 Cutoff
The amplifier is basically off. There is
voltage but little current.
Both junctions reverse biased
 Saturation
The amplifier is full on. There is little
voltage but lots of current.
Both junctions forward biased
OPERATIONS - npn
ACTIVE MODE

The base-emitter (B-E)
junction is forward biased and
the base-collector (C-B)
junction is reverse-biased,.

Since the B-E junction is
forward biased, electrons
from the emitter are injected
across the B-E junction into
the base  IE

Once in the base region, the
electrons are quickly
accelerated through the
base due to the reversebiased C-B region  IC
+
VBE
-
iB
Some electrons, in passing
through the base region,
recombine with majority carrier
holes in the base. This produces
the current IB
TO ILLUSTRATE
C
B
E
VBE
+
•Imagine the marbles as electrons
•A flat base region with gaps where the
marbles may fall/trapped – recombine
•A sloping collector region represents
high electric field in the C-B region
•Hence, when enough energy is given
to the marbles, they will be accelerated
towards to base region with enough
momentum to pass the base and
straight ‘fly’ to the collector
MATHEMATICAL EXPRESSIONS
IC
IB
+
VBE
IE
-
IE = IS [ e VBE / VT -1 ] = IS e VBE / VT
Based on KCL: IE = IC + IB
No. of electrons crossing the base region and then directly into the
collector region is a constant factor  of the no. of electrons exiting the
base region
IC =  IB
No. of electrons reaching the collector region is directly proportional to
the no. of electrons injected or crossing the base region.
IC =  IE
Ideally  = 1, but in reality it is between 0.9
and 0.998.
Based on KCL: IE = IC + IB
IE =  IB + IB = IB(  + 1)
IC =  IB
IC =  IE
IE = IB(  + 1)
Now
With IC =  IB  IB = IC / 
Hence,
IE = [ IC / ] (  + 1)
IC = IE [  /  + 1 ]
Comparing with IC =  IE
=[ /+1]
C
OPERATIONS - pnp
FORWARD ACTIVE MODE

The emitter – base (E- B)
junction is forward biased and
the base-collector (B- C)
junction is reverse-biased,.
IE = IS [ e VEB / VT -1 ] = IS e VEB / VT
**Notice that it is VEB
Based on KCL: IE = IC + IB
IB
IC
B
-
VEB
IE
+
E
pnp Transistor- Active mode
SUMMARY: Circuit Symbols and
Conventions
Based on KCL: IE = IC + IB
npn bipolar transistor simple
block diagram and circuit symbol.
Arrow is on the emitter terminal
that indicates the direction of
emitter current (out of emitter
terminal for the npn device)
pnp bipolar transistor simple
block diagram and circuit symbol.
Arrow is on the emitter terminal
that indicates the direction of
emitter current (into of emitter
terminal for the pnp device)
NPN
PNP
IE = IS [ e VBE / VT ]
IE = IS [ e VEB / VT]
IC =  IB
IC =  IE
IE = IB(  + 1)
=[ /+1]
 = [ / 1 -  ]
Based on KCL: IE = IC + IB
EXAMPLE 4.1
Calculate the collector and emitter currents, given the base current and current gain.
Assume a common-base current gain
and a base current of
.
Also assume that the transistor is biased forward in the forward active mode.
Solution: The common-emitter current gain is
The collector current is
And the emitter current is
Examples
• EXAMPLE 1
• Given IB = 6.0A and
IC=510 A
Determine ,  and IE




EXAMPLE 3
PNP Transistor
 = 60, IC= 0.85mA
Determine , IE and IB
• EXAMPLE 2
• NPN Transistor
• Reverse saturation
current Is = 10-13A with
current gain,  = 90.
Based on VBE = 0.685V,
determine IC , IB and IE
BJT: Current-Voltage
Characteristic
IC versus VCE
Common-Emitter Configuration - npn

The Emitter is common to both input
(base-emitter) and output (collectoremitter).

Since Emitter is grounded, VC = VCE

With decreasing VC (VCE), the junction
B-C will become forward biased too.
 The current IC quickly drops to zero
because electrons are no longer
collected by the collector
Node B
0V
Characteristics of Common-Emitter
- npn
NOTE: VEC for PNP