Download Introduction to BJT Amplifier

Introduction to BJT
Amplifier
BJT (Review)
Still remember about BJT?
I E  I B  IC
The emitter current (iE) is the sum of the collector current (iC) and the base
current (iB)
iE  iB  iC
iB << iE and iC
OTHER PRAMETERS &
EQUATIONS?
BJT
 Basic structure and schematic symbol
n
E
p
n
p
E
C
E
C
approximate
equivalents
E
B
C
npn type
C
C
B
B
p
B
B
E
n
C
E
transistor
symbols
B
pnp type
Refresh..
 Common-emitter
current gain, β

Range: 50 < β < 300
 Common-base current
gain, α

Range: always slightly
less than 1
 The current relationship
between these 2
parameters are as
follows:
iE  iC  iB
iC   iB
 iE  (1   )iB
 iC
iE  (1   )




  
iE
iC  
1  
 
But   
1 



 iC  iE
Refresh..
 BJT as amplifying device


B-E junction is forward-biased
B-C junction is reverse-biased
BIASING OF BJT
 Remember…! for normal operation

emitter-base junction is always forwardbiased
AND

collector-base junction is always reversebiased
FORWARD BIASING E/B JUNCTION
REVERSE BIASING C/B JUNCTION
BIASING NPN TRANSISTOR
Common-Emitter Circuit
(a) with an npn transistor
(b) with a pnp transistor
(c) with a pnp transistor biased with a positive voltage source
DC Analysis
- Common-Emitter Circuit
Transistor currentvoltage characteristics
of the common-emitter
circuit
DC Analysis
- Common-Emitter Circuit
Common-emitter circuit with an
npn transistor
Common-emitter dc equivalent
circuit, with piecewise linear
parameters
DC Analysis
- Common-Emitter Circuit
VBB  VBE (on)
IB 
RB
Usually VBE(on) = 0.7 V
I C  I B
VCC  I C RC  VCE
Common-emitter dc equivalent circuit
or V CE VCC  I C RC
Look for calculation examples in Neamen (Chapter 3), Example 3.3 & 3.4
DC Analysis
- Load Line & Modes of Operation
Figure A
Base on Figure A, using KVL
around B-E loop:
VBB  VBE
IB 
 I BQ
RB
Base-emitter junction characteristics
and the input load line
DC Analysis
- Load Line & Modes of Operation
Base on Figure A, 2 end
points of the load line are
found by setting IC = 0
VCC  I C RC  VCE
VCC  I C RC  VCE
So, VCE = VCC = 10 V
When VCE = 0,
IC = VCC/RC = 5 mA
IBQ is the value from the
previous slide = 15 µA
So, ICQ = βIBQ
If β = 200,
ICQ = 3000 µA = 3 mA
V CEQ VCC  I CQ RC
Common- emitter transistor characteristics
and the collector-emitter load line
So, VEQ = 4 V
BJT as an Amplifier
• Amplification of a small ac
voltage by placing the ac
signal source in the base
circuit
• Vin is superimposed on the
DC bias voltage VBB by
connecting them in series
with base resistor RB:
I C   DC I B
• Small changes in the base
current circuit causes large
changes in collector current
circuit