Lecture 8 bjt_1

Lecture 8:

BIPOLAR JUNCTION
TRANSISTORS
Code: Semester II
EEE2213 2010/2011

BJT STRUCTURE
Basic structure of the bipolar junction transistor (BJT) determines its
operating characteristics.
Constructed with 3 doped semiconductor regions called emitter, base,
and collector, which separated by two pn junctions.

2 types of BJT;
(2)npn: Two n regions separated by a p region
(3)pnp: Two p regions separated by an n region.

BIPOLAR:
refers to the use
of both holes &
electrons as
current carriers
in the transistor
structure.

Base-emitter junction: the pn junction joining the base region & the
emitter region.
Base-collector junction: the pn junction joining the base region & the
collector region.

A wire lead connects to each of the 3 regions. These leads labeled as;
E: emitter
B: base
C: collector

BASE REGION: lightly doped, & very thin

EMITTER REGION: heavily doped

COLLECTOR REGION: moderately doped

BASIC BJT OPERATION
For a BJT to operate properly as an amplifier, the two pn junctions
must be correctly biased with external dc voltages.

Figure: shows a bias arrangement for npn BJTs for operation as an
amplifier.

In both cases, BE junction is forward-biased & the BC junction is
reverse-biased.  called forward-reverse bias.

Look at this one circuit as two separate
circuits, the base-emitter(left side) circuit and
the collector-emitter(right side) circuit. Note
that the emitter leg serves as a conductor for
both circuits. The amount of current flow in
the base-emitter circuit controls the amount of
current that flows in the collector circuit.
Small changes in base-emitter current
yields a large change in collector-current.

The heavily doped n-type emitter region has a very high density of
conduction-band (free) electrons.

These free electrons easily diffuse through the forward-based BE
junction into the lightly doped & very thin p-type base region
(indicated by wide arrow).

The base has a low density of holes, which are the majority carriers
(represented by the white circles).

A small percentage of the total number of free electrons injected into
the base region recombine with holes & move as valence electrons
through the base region & into the emitter region as hole current
(indicated by red arrows).

BJT operation showing electron flow.

When the electrons that have recombined with holes as valence
electrons leave the crystalline structure of the base, they become free
electrons in the metallic base lead & produce the external base
current.

Most of the free electrons that have entered the base do not recombine
with holes because the base is very thin.

As the free electrons move toward the reverse-biased BC junction,
they are swept across into the collector region by the attraction of the
positive collector supply voltage.
The free electrons move through the collector region, into the external
circuit, & then return into the emitter region along with the base
current.

The emitter current is slightly greater than the collector current
because of the small base current that splits off from the total current
injected into the base region from the emitter.

Transistor Currents
The directions of the currents in both npn and pnp transistors and their
schematic symbol are shown in Figure (a) and (b). Arrow on the emitter
of the transistor symbols points in the direction of conventional
current. These diagrams show that the emitter current (IE) is the sum of
the collector current (IC) and the base current (IB), expressed as follows:
I E = IC + I B

BJT CHARACTERISTICS &
PARAMETERS
Figure shows the proper bias
arrangement for npn
transistor for active
operation as an amplifier.
Notice that the base-emitter
(BE) junction is forward-
biased by VBB and the base-
collector (BC) junction is
reverse-biased by VCC. The dc
current gain of a transistor is
the ratio of the dc collector
current (IC) to the dc base The ratio of the dc collector current (IC)
current (IB), and called dc beta to the dc emitter current (IE) is the dc
(β DC). alpha. α DC = IC/IE

Ex 4-1 Determine βDC and IE for a transistor where IB = 50 μA and IC = 3.65 mA.

Ex 4-1 Determine βDC and IE for a transistor where IB = 50 μA and IC = 3.65 mA.

I C 3.65mA
β DC = = = 73
IB 50 µA
IE = IC + IB = 3.65 mA + 50 μA = 3.70 mA

I C 3.65mA
α DC = = = 0.986
I E 3.70mA

Analysis of this transistor circuit to predict the dc voltages and currents
requires use of Ohm’s law, Kirchhoff’s voltage law and the beta for the
transistor;
Application of these laws begins with the base circuit to determine the
amount of base current. Using Kichhoff’s voltage law, subtract the VBE
=0.7 V, and the remaining voltage is dropped across RB .

Thus, VRB = VBB - VBE.

Determining the current for the base with this information is a matter of
applying of Ohm’s law. VRB/RB = IB

The collector current is
determined by
multiplying the base
current by beta.

Thus, IC= βDC * IB

What we ultimately
determine by use of
Kirchhoff’s voltage law
for series circuits is that,
in the base circuit, VBB is
distributed across the
base-emitter junction
and RB in the base
circuit. In the collector
circuit we determine that
VCC is distributed
proportionally across
RC and the
transistor(VCE).

BJT Circuit Analysis
There are three key dc voltages and three key dc currents to be
considered. Note that these measurements are important for
troubleshooting.

IB: dc base current
IE: dc emitter current
IC: dc collector current
VBE: dc voltage across
base-emitter junction
VCB: dc voltage across
collector-base junction
VCE: dc voltage from
collector to emitter

When the base-emitter junction is forward-biased,
VBE ≅ 0.7 V
VRB = IBRB : by Ohm’s law
IBRB = VBB – VBE : substituting for VRB
IB = (VBB – VBE) / RB : solving for IB
VCE = VCC – VRc : voltage at the collector with respect to the
grounded emitter
VRc = ICRC
VCE = VCC – ICRC : voltage at the
collector with
respect to the emitter

The voltage across the reverse-biased
collector-base junction

Ex 4-2 Determine I , I , I , VB C E , VCE, and VCB in the circuit of Figure. The
BE

transistor has a βDC = 150.

Ex 4-2 Determine I , I , I , VB C E , VCE, and VCB in the circuit of Figure. The
BE

transistor has a βDC = 150.

When the base-emitter junction is forward-biased, IC = βDCIB
= (150)(430 μA)
VBE ≅ 0.7 V
= 64.5 mA
IB = (VBB – VBE) / RB IE = IC + IB
= 64.5 mA + 430 μA
= (5 V – 0.7 V) / 10 kΩ = 430 μA = 64.9 mA
VCE = VCC – ICRC
= 10 V – (64.5 mA)(100 Ω)
= 3.55 V
VCB = VCE – VBE
= 3.55 V – 0.7 V
= 2.85 V
Since the collector is at a higher
voltage than the base, the collector-
base junction is reverse-biased.

Collector Characteristic Curves
Gives a graphical
illustration of the
relationship of collector
current and VCE with
specified amounts of
base current. With
greater increases of VCC ,
VCE continues to increase
until it reaches
breakdown, but the
current remains about the
same in the linear region
from 0.7V to the
breakdown voltage.

Sketch an ideal family of collector curves for the circuit in Figure for IB = 5 μA to 25 μA in 5
μA increment. Assume βDC = 100 and that VCE does not exceed breakdown.

Sketch an ideal family of collector curves for the circuit in Figure for IB = 5 μA to 25 μA in 5
μA increment. Assume βDC = 100 and that VCE does not exceed breakdown.

IC = βDC IB

IB IC
5 μA 0.5 mA
10 μA 1.0 mA
15 μA 1.5 mA
20 μA 2.0 mA
25 μA 2.5 mA

Cutoff
With no IB , the transistor is in the cutoff region and just as the
name implies there is practically no current flow in the
collector part of the circuit. With the transistor in a cutoff state,
the full VCC can be measured across the collector and
emitter(VCE).

Cutoff: Collector leakage current (ICEO) is extremely small and is usually
neglected. Base-emitter and base-collector junctions are reverse-biased.

Saturation
Once VCE reaches its maximum value, the transistor is said to be in
saturation.

Saturation: As IB increases due to increasing VBB, IC also increases and VCE
decreases due to the increased voltage drop across RC. When the transistor reaches
saturation, IC can increase no further regardless of further increase in IB. Base-
emitter and base-collector junctions are forward-biased.

DC Load Line
The dc load line graphically illustrates IC(sat) and cutoff for a transistor.

Active
region of
the
transistor’s
operation.

DC load line on a family of collector characteristic curves illustrating the
cutoff and saturation conditions.

Ex 4-4 Determine whether or not the transistors in Figure is in
saturation. Assume VCE(sat) = 0.2 V.

Ex 4-4 Determine whether or not the transistors in Figure is in
saturation. Assume VCE(sat) = 0.2 V.

First, determine IC(sat)
VCC − VCE ( sat )
I C ( sat ) =
RC
10 V − 0.2 V
= = 9.8 mA
1.0 kΩ
Now, see if IB is large enough to produce IC(sat).

VBB − VBE 3V − 0.7 V 2.3V
IB = = = = 0.23 mA Thus, IC greater than
RB 10kΩ 10 kΩ IC(sat). Therefore, the
I C = β DC I B = (50)(0.23 mA) = 11.5 mA transistor is saturated.

Maximum Transistor Ratings
A transistor has limitations on its operation. The product of VCE
and IC cannot be maximum at the same time. If VCE is maximum,
IC can be calculated as
PD (max)
IC =
VCE

Ex 4-5 A certain transistor is to be operated with VCE = 6 V. If
its maximum power rating is 250 mW, what is the most collector
current that it can handle?
PD (max) 250 mW
IC = = = 41.7 mA
VCE 6V

Ex 4-6 The transistor in Figure has the following maximum ratings: P D(max)= 800
mW, VCE(max) = 15 V, and IC(max) = 100 mA. Determine the maximum value to which VCC
can be adjusted without exceeding a rating. Which rating would be exceeded first?

Ex 4-6 The transistor in Figure has the following maximum ratings: P D(max)= 800
mW, VCE(max) = 15 V, and IC(max) = 100 mA. Determine the maximum value to which VCC

First, find IB so that you can determine IC.
VBB − VBE 5V − 0.7V
IB = = = 195µA
RB 22kΩ
I C = β DC I B = (100)(195µA) = 19.5mA
The voltage drop across RC is.
VRc = ICRC = (19.5 mA)(1.0 kΩ) = 19.5 V
VRc = VCC – VCE when VCE = VCE(max) = 15 V
VCC(max) = VCE(max) + VRc = 15 V + 19.5 V = 34.5 V
PD = VCE(max)IC = (15V)(19.5mA) = 293 mW
VCE(max) will be exceeded first because the entire supply voltage, VCC will be
dropped across the transistor.

Derating PD(max)
P D(max) is usually specified at 25°C.
At higher temperatures, P D(max) is less.
Datasheets often give derating factors for determining P D(max) at
any temperature above 25°C.

Ex 4-7
A certain transistor has a P D(max) of 1 mW at 25°C. The derating
factor is 5 mW/ °C. What is the P D(max) at a temperature of 70°C?

Transistor Datasheet
Refer Figure 4-20 (a partial datasheet for the 2N3904 npn
transistor).
The maximum collector-emitter voltage (VCEO) is 40V.
The CEO subscript indicates that the voltage is measured from
collector to emitter with the base open. VCEO= VCE(MAX)
The maximum collector current is 200 mA.
* Other characteristics can be referred from the datasheet.

A 2N3904 transistor is used in the circuit. Determine the maximum value to which VCC

A 2N3904 transistor is used in the circuit. Determine the maximum value to which VCC

PD(max) = 800 mW
VCE(max) = 15 V
IC(max) = 100 mA.
I B = 195µ A
I C = β DC I B = 19.5mA
VCC(max) = VCE(max) + VRc = 40 V + 19.5 V = 59.5 V

However at the max value of VCE, the power dissipation is

PD = VCE(max)IC = (40V)(19.5mA) = 780 mW
Power Dissipation exceeds the maximum of 645 mW specified on the
datasheet.

THE BJT AS AN AMPLIFIER
Amplification of a relatively
small ac voltage can be had by
placing the ac signal source in
the base circuit.
Recall that small changes in the
base current circuit causes large
changes in collector current
circuit.

The ac emitter current : Ie ≈ Ic = Vb/r’e
The ac collector voltage : Vc = IcRc
Since Ic ≈ Ie, the ac collector voltage : Vc ≈ IeRc
The ratio of Vc to Vb is the ac voltage gain : Av = Vc/Vb
Substituting IeRc for Vc and Ier’e for Vb : Av = Vc/Vb ≈ IcRc/Ier’e
The Ie terms cancel : Av ≈ Rc/r’e

Ex 4-9 Determine the voltage gain and the ac output
voltage in Figure if r’e = 50 Ω.

The voltage gain : Av ≈ Rc/r’e = 1.0 kΩ/50 Ω = 20
The ac output voltage : AvVb = (20)(100 mV) = 2 V

THE BJT AS A SWITCH
A transistor when used as a switch is simply being biased so that it
is in cutoff (switched off) or saturation (switched on). Remember
that the VCE in cutoff is VCC and 0V in saturation.

Conditions in Cutoff & Saturation
A transistor is in the cutoff region when the base-emitter junction is not
forward-biased. All of the current are zero, and VCE is equal to VCC

VCE(cutoff) = VCC

When the base-emitter junction is forward-biased and there is enough base
current to produce a maximum collector current, the transistor is saturated.
The formula for collector saturation current is

VCC − VCE ( sat )
I C ( sat ) =
RC

The minimum value of base current I C ( sat )
I B (min) =
needed to produce saturation is β DC

Ex 4-10 (a) For the transistor circuit in Figure, what is VCE when VIN = 0 V?
(b) What minimum value of IB is required to saturate this transistor if βDC is
200? Neglect VCE(sat).
(c) Calculate the maximum value of RB when VIN = 5 V.

Ex 4-10 (a) For the transistor circuit in Figure, what is VCE when VIN = 0 V?
(b) What minimum value of IB is required to saturate this transistor if βDC is
200? Neglect VCE(sat).
(c) Calculate the maximum value of RB when VIN = 5 V.
(a) When VIN = 0 V
VCE = VCC = 10 V

(b) Since VCE(sat) is neglected,

VCC 10 V
I C ( sat ) = = = 10 mA
RC 1.0 kΩ
I C ( sat ) 10 mA
I B (min) = = = 50 µA
β DC 200

(c) When the transistor is on, VBE ≈ 0.7 V.
VR = VIN – VBE ≈ 5 V – 0.7 V = 4.3 V
B

Calculate the maximum value of RB VRB 4.3V
RB (max) = = = 86 kΩ
I B (min) 50 µA

Transistor Construction
There are two types of transistors:
• pnp
• npn
pnp

The terminals are labeled:
• E - Emitter
• B - Base
• C - Collector

npn

42

Transistor Operation
With the external sources, VEE and VCC, connected as shown:

• The emitter-base junction is forward biased
• The base-collector junction is reverse biased

43

Currents in a Transistor

Emitter current is the sum of the collector and
base currents:

IE = IC + IB

The collector current is comprised of two
currents:
IC = IC + I CO
majority minority

44

Common-Base Configuration

The base is common to both input (emitter–base) and
output (collector–base) of the transistor.

45

Common-Base Amplifier

Input Characteristics

This curve shows the relationship
between of input current (IE) to input
voltage (VBE) for three output voltage
(VCB) levels.

46

Common-Base Amplifier

Output Characteristics
This graph demonstrates
the output current (IC) to
an output voltage (VCB) for
various levels of input
current (IE).

47

Operating Regions

• Active – Operating range of the
amplifier.
• Cutoff – The amplifier is basically
off. There is voltage, but little
current.
• Saturation – The amplifier is full on.
There is current, but little voltage.

48

Approximations

Emitter and collector currents:

I ≅I
C E

Base-emitter voltage:

VBE = 0.7 V (for Silicon)

49

Alpha (α )

Alpha (α ) is the ratio of IC to IE :
IC
αdc =
IE

Ideally: α = 1
In reality: α is between 0.9 and 0.998

Alpha (α ) in the AC mode:
mode
ΔI C
αac =
ΔI E

50

Transistor Amplification

Currents and Voltages: Voltage Gain:
V 200mV VL 50V
I E = Ii = i = = 10mA Av = = = 250
Ri 20Ω Vi 200mV
I ≅I
C E
I ≅ I = 10 mA
L i
V = I R = (10 ma )(5 kΩ) = 50 V
L L

51

Common–Emitter Configuration

The emitter is common to both input
(base-emitter) and output (collector-
emitter).

The input is on the base and the
output is on the collector.

52

Common-Emitter Characteristics

Collector Characteristics Base Characteristics

53

Common-Emitter Amplifier Currents
Ideal Currents

IE = IC + IB IC = α IE

Actual Currents

IC = α IE + ICBO where ICBO = minority collector current

ICBO is usually so small that it can be ignored, except in
high
power transistors and in high temperature
environments.
When IB = 0 µA the transistor is in cutoff, but there is some minority
current flowing called ICEO.
I CBO
I CEO = I B = 0 μA
1− α

54

Beta (β )
β represents the amplification factor of a transistor. (β is
sometimes referred to as hfe, a term used in transistor modeling
calculations)

In DC mode:
IC
βdc =
IB

In AC mode:
∆IC
βac = VCE =constant
∆IB

55

Beta (β )
Determining β from a Graph

(3.2 mA − 2.2 mA)
β AC =
(30 μA − 20 μA)
1 mA
= V = 7.5
10 μA CE
= 100

2.7 mA
β DC = VCE = 7.5
25 µA
= 108

56

Beta (β )

Relationship between amplification factors β and α

β α
α= β=
β+1 α −1

Relationship Between Currents

I C = βI B I E = (β + 1)I B

57

Common–Collector Configuration

The input is on the
base and the output is
on the emitter.

58

Common–Collector Configuration

The characteristics are
similar to those of the
common-emitter
configuration, except the
vertical axis is IE.

59

Operating Limits for Each Configuration

VCE is at maximum and IC is at
minimum (ICmax= ICEO) in the cutoff
region.

IC is at maximum and VCE is at
minimum (VCE max = VCEsat = VCEO) in
the saturation region.

The transistor operates in the active
region between saturation and cutoff.

60

Power Dissipation
Common-base:
PCmax = VCB I C

Common-emitter:

PCmax = VCE I C

Common-collector:

PCmax = VCE I E

61

Transistor Specification Sheet

62

Transistor Specification Sheet

63

Transistor Testing
• Curve Tracer
Provides a graph of the characteristic curves.

• DMM
Some DMMs measure β DC or hFE.

• Ohmmeter

64

Transistor Terminal Identification

65

Lecture 8 bjt_1

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