Download Bipolar Junction Transistor - Help-A-Bull

Survey
yes no Was this document useful for you?
   Thank you for your participation!

* Your assessment is very important for improving the work of artificial intelligence, which forms the content of this project

Document related concepts
no text concepts found
Transcript
Transistors
Three-terminal devices with three doped silicon
regions and two P-N junctions versus a diode
with two doped regions and one P-N junction
Two basic ways of implementation:
Bipolar Junction Transistor (BJT)
Field-Effect Transistor (FET)
Transistors
•
•
•
•
Many types !
3- terminal devices
Made with semiconductor materials
Used for … amplifier design and as switches !
(but many more … )
• Common types:
– BJT: Bipolar Junction Transistor
– FET: Field Effect Transistor
• MOSFET: Metal Oxide Semiconductor FET
• MISFET: Metal Insulator Semiconductor FET
• CMOS Technology: Complementary Metal Oxide
Semiconductor FETs !
Transistors
Bipolar Junction Transistors
C
C = collector
B = base
E = emitter
n
p
B
n
E
NPN
transistor
Transistors
BJT
Transistors
BJT
Transistors
BJT
PNP
NPN
Typical Transistor Circuit
• Both DC and AC signals
• DC signals for “powering”
up the transistor and
establishing an
“operating point”
• AC signals – what we
want to “process” i.e.
amplify
Common Emitter Amplifier
8
Common Emitter Amplifier
DC Equivalent Circuit
9
Common Emitter Amplifier
AC Equivalent Circuit
10
How Do We Handle Trans ??
• Determine “operating mode” and
replace transistor with appropriate
model (linear!)
Analysis Method
12
Analysis Method
13
a. Since VBB = 0.3V
< 0.7V, transistor
operates
in
CUTOFF region
So IB = IC = 0.
Write
KVL
equation around
the input loop:
VBB = 80k(IB) +
VBE
VBE = 0.3 V
Write
KVL14
equation around
b. Since VBB = 2.7V >
0.7V, transistor is
ON and VBE = 0.7V
Write
KVL
equation around
the input loop:
VBB = 80k(IB) + VBE
IB = (2.7 – 0.7)/80k
= 25 A
Assuming ACTIVE
mode, IC =  IB
> 0.2V, so transistor IS in active region
IC = 2.5 mA
15
c. Since VBB = 6.7V >
0.7V, transistor is
ON and VBE = 0.7V
IC = (VCC – VCE) / 2k = 4.9 mA
Write
KVL
equation around
the input loop:
VBB = 80k(IB) + VBE
IB = (6.7 – 0.7)/80k
= 75 A
Assuming ACTIVE
mode, IC =  IB
< 0.2V, so transistor is in SATURATION region
IC = 7.5 mA
16
Transistors
Field-Effect Transistors
Transistors
Field-Effect Transistors
Transistors
FET
MOSFETs – Circuit Symbols
• n-MOS and p-MOS
“working together”
• “n” and “p” for ntype and p-type
semiconductor
• n-type: negative
charges – electrons
• p-type: positive
charges – “holes”
Transistors
FET
When VG is positive, electrons in
the p-type substrate are attracted
to the oxide–silicon interface, and
form an n-type conduction
channel. The electrical model is
represented by resistors in series.
The transistor is in its ON state.
NMOS
Transistors
FET
When VG = 0, the area
underneath the oxide layer is
still p-type, which forms a
“ back-to-back ” diode with
the n region, as shown in the
electrical representation. The
transistor is in its OFF state.
NMOS
Transistors
FET
PMOS
NMOS vs. PMOS - Operation
NMOS vs PMOS
INPUT
NMOS
INPUT
PMOS
High “1”
ON
High “1”
OFF
Low “0”
OFF
Low “0”
ON
NMOS Inverter
• What happens
when Vin is “high” ?
i.e. logic level “1”
• What happens
when Vin is “low” ?
i.e. logic level “0”
INPUT
High “1”
Low “0”
OUTPUT
CMOS Inverter
CMOS Inverter
CMOS “Gate”
A
B
LO
LO
LO
HI
HI
LO
HI
HI
OUT
CMOS “AND Gate”
A
B
LO
LO
LO
HI
HI
LO
HI
HI
OUT
CMOS “AND Gate”
A
B
LO
LO
LO
HI
HI
LO
HI
HI
OUT
CMOS “AND Gate”
A
B
OUT
LO
LO
LO
LO
HI
LO
HI
LO
LO
HI
HI
HI
CMOS “AND Gate”
A
B
OUT
0
0
0
0
1
0
1
0
0
1
1
1
Related documents