Download Lecture 10 - BJT Introduction

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

Klystron wikipedia, lookup

Regenerative circuit wikipedia, lookup

Valve RF amplifier wikipedia, lookup

Thermal runaway wikipedia, lookup

Amplifier wikipedia, lookup

Resistive opto-isolator wikipedia, lookup

Josephson voltage standard wikipedia, lookup

Power electronics wikipedia, lookup

Molecular scale electronics wikipedia, lookup

Surge protector wikipedia, lookup

Schmitt trigger wikipedia, lookup

Switched-mode power supply wikipedia, lookup

Ohm's law wikipedia, lookup

CMOS wikipedia, lookup

Two-port network wikipedia, lookup

Nanofluidic circuitry wikipedia, lookup

Opto-isolator wikipedia, lookup

Operational amplifier wikipedia, lookup

Current source wikipedia, lookup

Rectiverter wikipedia, lookup

Wilson current mirror wikipedia, lookup

Transistor–transistor logic wikipedia, lookup

Power MOSFET wikipedia, lookup

TRIAC wikipedia, lookup

Current mirror wikipedia, lookup

Transcript
Assume: IZmin = 0.1 IZ(max)
Range of power
supply
Voltage
Regulator
Zener Diode
CHAPTER 3
The load resistor sees a constant voltage regardless of the current
VL = VZ
VRi = VPS - VZ
II = IZ + I L
VC = Vm e –
t / RC
Capacitor
Discharge
Half Wave
Vm TP
Vr =
RC
If the ripple is very small, we
can approximate T’ = Tp where
Tp is the period of the cycle
Vr =
Vm T
RC
′
Full Wave
Vm TP
Vr =
2RC
Vm is the peak value
of the output
voltage
Multiple Diode Circuit
Ripple Voltage, Vr
Vo = Vs - V
Filter
Vo = Vs - 2V
PIV = 2Vspeak - V
Duty
Cycle
Peak and RMS
Vpeak
Vrms =
2
Centertapped
Bridge
PIV = Vspeak - V
Full Wave
Vo = Vs - V
CHAPTER 3
Rectifier
Half Wave
PIV = Vspeak
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 basecollector (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 reverse-biased C-B
region  IC
VBE = V
+
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)
I C = 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
VEB = V
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
pnp bipolar transistor simple
block diagram and circuit symbol.
Arrow is on the emitter terminal
that indicates the direction of
emitter current
NPN
PNP
IE = IS [ e VBE / VT ]
IE = IS [ e VEB / VT]
IC =  IB
IC =  IE
IE = IB(  + 1)
𝜷
𝜶=
𝟏+𝜷
𝜶
𝜷=
𝟏−𝜶
Based on KCL: IE = IC + IB
EXAMPLE
Calculate the collector and emitter currents, given the base current and current gain.
Assume a common-base current gain,  = 0.97 and a base current of iB = 25 µA . 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
Answers:
 = 85
= 0.9884
IE = 516 A
•
•
•
•
EXAMPLE 2
NPN Transistor
Reverse saturation current Is = 10-13A with current gain,  = 90.
Based on VBE = 0.685V, determine IC , IB and IE
Answers:
IE = 10-13 (e 0.685/0.026) = 0.0277 A
IC = (90/91)(0.0277) = 0.0274 A
IB = IE – IC = 0.3 mA
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