Notesale: Turn your study into money

Document Preview

Extracts from the notes are below, to see the PDF you'll receive please use the links above

---

Electrical Engineering
Principles & Applications
Chapter 13
Bipolar Junction Transistors
Slide 1

Outcomes

1
...
Analyze simple amplifiers using the loadline technique

Slide 2

Introduction
• A Bipolar Junction Transistor (BJT) is a
three-terminal device
• It is constructed as layers of semiconductor
materials (usually silicon)
• It is used in amplifiers and logic circuits
• BJTs are capable of producing large output
currents
Slide 3

NPN Bipolar Junction Transistors (BJT)
• A layer of p-type material between
two layers on n-type material
• Each PN junction forms a diode but
current in one junction affect the
other
• Three layers are called:
– Emitter
– Base
– Collector

• With the voltages applied, a small
current from the base causes much
larger currents to flow from the
collector to the emitter
Slide 4

Bipolar Junction Transistors (BJT)

rse
eve
r

sed
bia

C

B
for

E
war

d- b

iase

d

Normal amplifier operation:
B-C junction is reverse-biased
B-E junction is forward-biased
Slide 5

Equations of Operation
By KCL:

i E = iC + i B
i
iB = iE −i C = (1 − C )iE
iE

Define: α =

iC
, i
...
i C = αiE
iE

iB = (1 − α )iE
=

(1 − α )

iC

α
0
...
999
Typical value is 0
...
e
...
9 <

< 0
...
e
...
Then applying KVL, we get:

V BB + v in (t ) = R B i B (t ) + v BE (t )

t
npu it
I
cu
Cir

Smaller vin

The DC power-supply voltages VBB and VCC bias the device at an operating
point to amplify the AC input signal vin(t)
An amplified version of the input appears between the collector and ground
Slide 11

Q-Point
• When vin(t) = 0, the operating point is called the
quiescent operating point (Q-point)
• As vin(t) changes, the operating point swings
above and below the Q-point
• Values of iB can be found from the intersection
of the load line with the input characteristic
for each value

Slide 12

Load-Line Analysis of a Common-Emitter BJT
Amplifier (Output Circuit)
t
tp u t
Ou cui
Cir

vCE = 0

VCC = RC iC + vCE
Slide 13

iC = 0

VCC=10V
VBB=1
...
4 V peak,1–kHz
vin(t) = 0
...

Solution:
From the input char
...

ICQ=2
...
8 V peak-to-peak
Output is 4 V
Voltage gain is 5 times (actually -5)
Slide 15

Nonlinear Distortion
Cutoff region: iB and iC are zero
vCE = VCC
Q-point moves to voltage axis
Happens because of the negative peak of vin(t)
Notice the clipping of the +ve peak of vCE

The amplitude of input signal is increased to 1
...
2 V)
happens because ib becomes large enough
operates at a

Slide 16

point above the load line

Nonlinear Distortion
v BC = v BE − v CE
When vCE 0
...
2
Both junctions are forward
biased
vce ≈ 0
...
e
...
e
...
Assume VBE = 0
...

VB = VCC
RB =

R2
= 5V
R1 + R2

1
= R1 R2 = 3
...
2 µA
RB + (β + 1)RE
For

= 100, I C

= β I B = 4
...
72V

For

= 300, IB = 14
...
24 mA, and VCE = 6

---