DEPARTMENT OF

ELECTRICAL ENGINEERING

Bipolar Junction Transistors (BJTs)

February 11, 2025

© Schultz, M. E., & Grob, B. (2015). Grob’s basic electronics. Dubuque, IA:
McGraw-Hill.
© Sedra, A. S., & Smith, K. C. (1998). Microelectronic circuits. New York: Oxford
University Press.
© Theraja, B. L., & Theraja, A. K. (2005). A textbook of electrical technology: In
S.I. system of units. New Delhi: S. Chand and Co.
© Malvino, A. P., & Bates, D. J. (2016). Electronic principles. New York:
McGraw-Hill Education.
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 DEPARTMENT OF
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Introduction

The first junction transistor was invented by William Schockley

in 1951.

Transistor invention has led to many other semiconductor

inventions, including the integrated circuits (ICs),...

...a small device that contains thousands of miniaturised

transistors.

Because of IC s, modern computers and other electronic

miracles have become possible.

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 DEPARTMENT OF
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The unbiased BJT

A BJT is a three terminal, current controlled device made up

of three alternating layers of doped semiconductor materials.

It may be used to control the flow of current. But, it can also:

▶ amplify voltage, current or power, and
▶ be used as an electronic switch

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The word bipolar is an abbreviation for “two polarities”,...

...that is, charge carriers of both polarities, i.e, both electrons and
holes participate in the current-conduction process in a BJT.

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Construction of BJT

The following two types of semiconductor transistors may be

constructed;

1. npn-transistor.
2. pnp-transistor.

The construction and schematic symbols for the npn and pnp
transistors are shown in Figures 1, 2, 3 and 4.

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 DEPARTMENT OF
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Figure 1: Construction symbols. (a) Figure 2: Schematic symbols. (a)

npn-transistor. (b) pnp-transistor. npn-transistor. (b) pnp-transistor.

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Figure 3: A simplified structure of the npn-transistor.

Figure 4: A simplified structure of the pnp-transistor.

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In both npn and pnp BJTs, the emitter region is heavily doped,
and its job is to emit or inject current carriers into the base.

For an npn transistor, the n-type emitter injects free electrons

into the base,...

...while the p-type emitter for a pnp-transistor, injects holes into

the base.

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The base region is very thin (narrow) and lightly doped.

Most of the current carriers injected into the base cross over
into the collector and...

...very little flow out through the base lead.

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The collector region is moderately doped and is the largest of

all three regions...

...because it must dissipate more heat than the emitter or base

regions.

Its function is to collect or attract current carriers injected

into the base region.

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Both npn and pnp transistors consist of the following two

pn-junctions,

1. The emitter-base junction (EB-junction) and

2. The collector-base junction (CB-junction).

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Current carriers and junction potential

In npn-transistors, the majority current carriers are free

electrons in the emitter and collector,...

...whereas the majority carriers are holes in the base

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In pnp-transistors the majority carriers are holes in the emitter

and collector,...

...and the majority carriers are free electrons in the base

For silicon BJT, the barrier potential for both EB and

CB-junctions equals approx. 0.7 V

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Biased BJT

The term bias1 is defined as a control voltage or current.

For a transistor to function properly as an amplifier,...

...a bias - an external dc supply voltage (or voltages) must be

applied to produce the desired IC .

1
Biasing is the application of fixed dc supply to a terminal of an electronic
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In Figures 5 and 6, two external voltage sources shown as

batteries, are used to establish the required bias conditions.

A BJT is said to be operating the active mode/region when its

EBJ is forward biased and its CBJ is reversed biased.

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Figure 5: Current flow in an npn-transistor biased to operate in the active

mode. (The reverse current components due to drift of thermally
generated minority carriers are not shown.)
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The VBE causes the p-type base to be higher in potential

than the n-type emitter,...

...thus forward-biasing the EB-junction.

VCB causes the n-type collector to be at a higher potential

than the p-type base, thus reverse-biasing the CB-junction.

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The forward-biasing voltage VBE causes the majority carriers

at the emitter to be injected into the base.

Most current carriers injected into the base from the emitter
flow out through the collector lead,...

...while only few carriers flow out through the base lead.

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Figure 6: Current flow in a pnp-transistor biased to operate in the active

mode.
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This happens for these two reason in a pnp BJT ;

1. Only a few holes are available for recombination in the

base.

2. The positive CB voltage (for an npn BJT) attracts nearly

all the free electrons in the p-type base over to the collector
side before they can recombine with holes in the base.

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Only a small CB voltage is needed to create an electric field

strong enough to collect almost all free electrons injected into the
base.

Too large CB voltage can exceed the breakdown voltage and

destroy the transistor.

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Graph of IC versus VCE for different IB values

It is possible to connect a BJT as shown in Figure 7.

Two external voltage sources are used to establish the required

bias conditions for active-mode operation.

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Figure 7: The circuit of a Common-Emitter connection.

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In the circuit,

▶ VBB is the base supply voltage used to forward bias the

base-emitter junction,

▶ RB is the base resistor, providing the desired value of

current IB , while

▶ VCC is the collector supply voltage, that provides the

reverse bias voltage required for the CB-junction.

Note from Figure 7 that

VCE = VCB + VBE

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Operating regions of a transistor circuit

Depending on the bias condition (forward or reverse) of the

EB-junction and CB-junction,...

...different modes of operation of the BJT are obtained, as

shown in Table 1.

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Table 1: BJT Modes of Operation

Mode EB-junction CB-junction

Cut-off Reverse Reverse
Active Forward Reverse
Saturation Forward Forward

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The base supply voltage, VBB can be adjusted to provide a wide

range of IB and IC .

Using the circuit in Figure 7, it is possible to plot a graph of IC

versus VCE for different IB values as shown in in Figure 8

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Figure 8: Graph of IC versus VCE for different IB values from a

Common-Emitter connection.

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Cut-off region
▶ When VBB < 0.7 V , very little (ideally zero) base current
flows because the EBJ is not forward biased.

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▶ With EBJ reversed biased, the CBJ will be reversed biased ∵

IE or the currents IB and IC must pass thro the EBJ.

▶ The region below the IB = 0 curve is called the cut-off

region ∵ only a small collector current, IC flows.

▶ A transistor is said to be cut off when its collector current,

IC = 0.

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Figure 9: Operating regions of a transistor circuit.

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Saturation region

▶ When VBB > 0.7 V and VCB is not sufficient to reversed

biased the CBJ...

...or to attract all the charge carriers injected into the

base, the CBJ will obey Ohms law.

▶ When VCE increases from zero, IC increases linearly.

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▶ This produces the early rising part of the curve, where

VCE is between 0V and a few tenths of a volt.

▶ The sloping part of the curve is called the saturation line.

The region between the line VCE = 0 and the saturation

line is called the saturation region.

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In the region, an n-type collector has insufficient positive

bias voltage to collect all the free electrons injected into
the base.

Also, the IB is larger than normal and βDC is smaller than

normal.

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Active region

▶ When EBJ is forward biased and VCB is sufficient to

reversed bias the CBJ...

...the VCB will attract all the charge carriers injected into
the base.

▶ Since the quantity of charge carriers are fairly constant

without carrier regeneration, IC will be fairly constant.

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The collector curves are ∴ nearly horizontal in the active

region of a transistor.

When a transistor is operating in the active region, the IC is

greater than the IB by a factor, β.

The active mode of operation is the one used if the transistor is

to operate as an amplifier.

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The collector circuit acts like a current source in the active

region.

VCC can be varied from few tenths of voltage to several volts

without having effect on IC ,...

...provided VCB breakdown voltage rating of the transistor is

not exceeded.

Thus, IC is solely controlled by IB not VCC .

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Breakdown region

▶ When the CB voltage is too large, the CB diode breaks down,

causing a large, undesired collector current to flow.

▶ This is the breakdown region. This area of operation should

always be avoided.

▶ It is assumed that the breakdown region will not occur when

the circuit is designed properly.

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Switching applications (e.g., logic circuits) utilise both the

cut-off mode and the saturation mode.

As the name implies, no current flows in the cut-off mode because

both junctions are reverse biased.

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Current flow

The forward bias on the EBJ will cause current to flow across
this junction.

The current will consist of two components:

1. Electrons injected from the emitter into the base, and

2. Holes injected from the base into the emitter.

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The current that flows across the EB-junction will constitute

the emitter current, iE , as indicated in Figure 5.

Following the usual conventions, the direction of iE is in the

direction of the positive-charge flow (hole current) and...

...opposite to the direction of the negative-charge flow

(electron current).

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iE is equal to the sum of the hole current and the electron

current.

However, since the electron component is much larger, than

the hole component,...

...iE will be dominated by the electron component in an npn

transistor.

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These electrons will be minority carriers in the p-type base

region.

Because the base is usually very thin, the excess minority carrier
concentration in the base in the steady state...

...will have an almost-straight-line profile, as indicated by the

solid straight line in Figure 10.

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Figure 10: Profiles of minority-carrier concentrations in the base and in

the emitter of an npn transistor operating in the active mode: vBE > 0
and vCB ≥ 0
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The electron concentration will be highest denoted by (np (0))

at the emitter side and lowest (zero) at the collector side.

As in the case of any forward-biased pn-junction, the concentration

np (0) will be proportional to e vBE /VT

np (0) = np0 e vBE /VT (1)

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where
▶ np0 is the thermal-equilibrium value of the minority carrier
concentration in the base region,

▶ vBE is the forward base-emitter bias voltage, and

▶ VT is the thermal voltage, which is approximately 25mV at

room temperature.

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The tapered carrier concentration profile in Figure 10 occurs ∵ the

electrons injected into the base diffuse through the base
region toward the collector.

This causes some electron diffusion current.

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This electron diffusion current In is directly proportional to

the slope of the straight-line concentration profile,

dnp (X )
In = AE qDn
dX
  (2)
np (0)
= AE qDn −
W

where
▶ AE is the cross-sectional area of the EB-junction,
▶ q is the magnitude of the electron charge,
▶ Dn is the electron diffusivity in the base, and
▶ W is the effective width of the base.

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Some of the electrons that are diffusing through the base region
will combine with holes, which are the majority carriers in the base.

However, since the base is usually very thin and lightly

doped,...

...the proportion of electrons “lost” through this

recombination process will be quite small.

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The recombination in the base region causes the excess minority

carrier concentration profile to deviate...

...from a straight line to the slightly concave shape indicated

by the broken line in Figure 10.

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The collector current

From the description above, we see that most of the diffusing

electrons will reach the boundary of the CB depletion region.

These electrons will ∴ get “collected” to constitute the iC .

Thus, iC = In which will yield a negative value for iC ,...

...indicating that iC flows in the negative direction of the

x-axis (i.e., from right to left).

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Since we will take this to be the positive direction of iC , we

can drop the negative sign in (2).

Doing this and substituting for np (0) from (1), we can thus
express the collector current iC as

iC = IS e vBE /VT (3)

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where the saturation current IS is given by

AE qDn np0
IS =
W
Substituting np0 = ni2 /NA where
▶ ni is the intrinsic carrier density and
▶ NA is the doping concentration of the acceptor atom in
the base,
we can express IS as

AE qDn ni2
IS = (4)
NA W

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An important observation to make here is that the magnitude of

iC is independent of vCB .

That is, as long as the collector is +ive w.r.t the base, the
electrons that reach the collector side of the base region will
be swept into the collector and register as collector current.

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Typically IS is in the range of 10−12 A to 10−18 A (depending on

the size of the device).

Notice in (4), that IS is proportional to ni2 , as a result, it is a

strong function of temperature, approx. doubling for every
5◦ C rise in temperature.

Since IS is directly proportional to the junction area (i.e., the

device size), it will also be referred to as the scale current.

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The base current

The base current, iB , will be proportional to e vBE /VT and is
expressed as a fraction of the iC as follows:
iC
iB = (5)
β
That is
IS vBE /VT
iB = e (6)
β
where β is a transistor parameter.

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For modern npn-transistors, β is in the range 50 to 200, but it

can be as high as 1000 for special devices.

For reasons that will soon become clear, the parameter β is called
the common-emitter current gain.

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To obtain a high β, which is highly desirable since β represents

a gain parameter,...

...the base should be thin (W small) and lightly doped and the
emitter heavily doped (making NA /ND small).

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The above description indicates that the value of β is highly

influenced by two factors:
▶ the width of the base region, W , and

▶ the relative dopings2 of the base region and the emitter

region, NA /ND

For modern integrated circuit fabrication technologies, W is in the

nanometre range.

2
NA is doping concentration of the acceptor atom while ND is doping concen-
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The emitter current

Since the current that enters a transistor must leave it, it can
be seen from Figure 5 that;

iE = iC + iB (7)

Using (5) and (7) gives

β+1
iE = iC (8)
β

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That is,
β + 1 vBE /VT
iE = IS e (9)
β

Alternatively, we can express (8) in the form

iC = αiE (10)

where the constant α is related to β by

β
α= (11)
β+1

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Finally, we can use (11) to express β in terms of α , that is,

α
β= (12)
1−α
It can be seen from (11) that α is a constant (for a particular
transistor) that is less than but very close to unity.

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The variable α describes how closely the emitter and collector

currents are in a common base circuit.

It is expressed as;
IC
α=
IE
The ratio of dc collector current IC to the dc emitter current IE is
denoted αdc

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There is also an αac for a BJT which is the ratio of change in

the ac collector current iC to the change in the ac emitter
current iE .

In most cases, the α ≈ 0.99 or greater. The thinner and more

lightly doped the base, the closer alpha is to one, or unity.

Equation (12) reveals an important fact: Small changes in α

correspond to very large changes in β.

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This observation manifests itself physically, since transistors of

the same type may have widely different values of β.

For reasons that will soon become apparent, α is called the

common-base current gain.

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Recapitulation and examples

We have seen that the emitter current, IE is the sum of collector

current, IC and base current, IB . Thus;

IE = IB + IC
IC = IE − IB
IB = IE − IC

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Example 1

A transistor has the following currents: IB = 20mA and

IC = 4.98A. Calculate IE .

IE = IB + IC
= 20 mA + 4.98 A
= 0.02 A + 4.98 A
=5A

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Example 2

A transistor has the following currents: IE = 100mA,

IB = 1.96mA. Calculate IC .

IC = IE − IB
= 100 mA − 1.96 mA
= 98.04 mA

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Example 3

A transistor has the following currents: IE = 50mA, IC = 49mA.

Calculate IB

IB = IE − IC
= 50 mA − 49 mA
= 1 mA

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Example 4

A transistor has the following currents: IE = 15mA, IB = 60µA.

Calculate αdc

IC = IE − IB
= 15 mA − 60 µA
= 15 mA − 0.06 µA
= 14.94 mA
IC
αdc =
IE
14.94 mA
=
15 mA
= 0.996
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Transistor biasing

There are basically three possible bipolar junction transistor

configurations3

1. Common-base - has voltage gain but no current gain

2. Common-emitter - has both current and voltage gain

3. Common-collector - has current gain but no voltage gain

3 Ony ang
Ways of connecting oS.Ob
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The common-base (CB) configuration

Figure 11: (a) The common-base npn-transistor biasing. (b) Currents in

a transistor.
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In this configuration, IE is the input current and IC is the

output current.

The input signal is applied between the emitter and base

whereas output is taken out from the collector and base.

The ratio of the collector current to the emitter current is

called dc alpha (αdc ) of a transistor.

IC
∴ αdc =
IE

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The emitter-base junction must be forward-biased, while the

collector-base junction must be reverse-biased

It is called a common-base connection ∵ the base lead is

common to both the input and output sides of the circuit.

The EB junction supply voltage is designated VEE , while the CB

junction supply voltage is designated VCC .

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Figure 12: Voltage polarities and current flow in transistors biased in the
active mode.

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The common-emitter (CE) configuration

Figure 13: Transistor biasing for the common-emitter (CE) connection.

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Figure 13 shows a CE-connection.

VBB provides the forward bias for the BE-junction, and VCC
provides the reverse bias for the CB-junction.

It is called a CE-connection because the emitter lead is common

to both the input and output sides of the circuit.

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Example 5

A transistor has the following currents: IC = 10mA, IB = 50µA.

Calculate βdc
IC
βdc =
IB
10 mA
=
50 µA
= 200

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Example 6

A transistor has βdc = 150 and IB = 75µA. Calculate IC

IC
βdc =
IB
IC = βdc × IB

= 150 × 75 µA

= 11.25 mA

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Example 7

A transistor has βdc = 100. Calculate αdc

βdc
αdc =
1 + βdc
100
= = 0.99
1 + 100

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 DEPARTMENT OF
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Example 8

A transistor has βdc = 0.995. Calculate βdc

αdc
βdc =
1 − αdc
0.995
= = 199
1 − 0.995

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 DEPARTMENT OF
ELECTRICAL ENGINEERING

Transistor biasing techniques

Most common biasing techniques include

1. Base bias,
2. Voltage divider bias, and
3. Emitter bias

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 DEPARTMENT OF
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1. Base bias
a). Base bias with two supplies

▶ VBB – base supply voltage: used to

forward bias the base emitter junction

▶ RB – base resistor: provide the

desired value of current IB

▶ VCC – collector supply voltage:

provides the reverse bias voltage
required by the CB-junction of the
transistor
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 DEPARTMENT OF
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RC – Provides the desired voltage in the collector circuit

VBB − VBE
IB =
RB
VCE = VCC − IC RC

Since the transistor is silicon, VBE equals 0.7 V. Therefore IB is

calculated as
VBB − VBE
IB =
RB

5 V − 0.7 V
=
56 kΩ

= 76.78 µA
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 DEPARTMENT OF
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The collector current, IC , can be calculated next,

IC = βdc IB
= 100 × 76.78 µA
≈ 7.68 mA

With IC known, the collector-emitter voltage, VCE , can be found

as follows

VCE = VCE − IC RC
= 15 V − (7.68 mA × 1 kΩ)
= 15 V − 7.68 V
= 7.32 V

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 DEPARTMENT OF
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b). Base bias with a single supply

In the Figure 14, solve for IB , IC , and VCE and VCB .

Figure 14: Base bias using a single power supply.

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 DEPARTMENT OF
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VCC − VBE 15V − 0.7V

IB = =
RB 180kΩ

= 79.44µA

IC = βdc × IB
= 100 × 79.44µA = 7.94mA

VCE = VCC − IC RC = 7.06V

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 DEPARTMENT OF
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VCC − VBE 15V − 0.7V

IB = =
RB 180kΩ

= 79.44µA

IC = βdc × IB
= 100 × 79.44µA = 7.94mA

VCE = VCC − IC RC = 7.06V

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 DEPARTMENT OF
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dc load line

▶ The dc-load line is a graph

that allows determination
all possible combinations
of IC and VCE for a given
amplifier.

▶ For every value of IC , the

corresponding value of VCE
Figure 15: dc load line showing the
can be determine on the
endpoints IC (sat) and VCE (off ) , as
load line. well as the Q-point values ICQ and
VCEQ .
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 DEPARTMENT OF
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where;
▶ VCEQ represents the quiescent collector-emitter voltage,

▶ ICQ represents the quiescent collector current.

▶ VCE (off ) – This is the collector emitter voltage with IC = 0 for

the cut-off condition.

During cut-off collector-emitter region is seen as an open.

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▶ IC (sat) – This is the collector current IC when the transistor
is saturated

▶ At saturation, collector-emitter region is treated like a

short, since VCE = 0. For this condition,

VCC = IC RC
iff RE = 0Ω
IC (sat) = VCC /RC

Check VCE at saturation with Sedra

▶ At saturation, further increases in IB produce no further

increases in IC because VCE = 0.

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 DEPARTMENT OF
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Example 9

In Figure 28–16, solve for IB , IC , and VCE . Also, construct a dc

load line showing the values of IC (sat) , VCE (off ) , ICQ , and VCEQ .

Figure 16: Circuit.

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 DEPARTMENT OF
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VCC − VBE
IB =
RB
12 V − 0.7 V
=
390 kΩ
= 28.97 µA

Next, calculate IC

IC = βdc × IB
= 150 × 28.97 µA
= 4.35 mA

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 DEPARTMENT OF
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VCE is calculated as follows

VCE = VCC − IC RC
= 12 V − (4.35 mA × 1.5 kΩ)
= 12 V − 6.52 V
= 5.48 V

The endpoints for the dc load line are then calculated

VCC 12 V
IC (sat) = = = 8 mA
RC 1.5 kΩ

VCE (off ) = VCC = 12 V

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 DEPARTMENT OF
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Figure 17: dc load line for base-biased transistor circuit.

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 DEPARTMENT OF
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Example 10

Design a base biased npn-transistor circuit with a single supply

having the following parameters;

VCC = 18 V , RC = 2.4 kΩ, βdc = 150

Determine the value of RB that will produce an ICQ of 3.75 mA

and a VCEQ of 9V. Draw and label the dc load line for the circuit
showing all points. What is the power dissipated by the transistor
circuit?
VCC − VBE
RB =
IBQ

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 DEPARTMENT OF
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But;
ICQ 3.75 mA
IBQ = = = 25 µA
βdc 150

18 V − 0.7 V
RB = = 692 kΩ
25 µA

VCC 18 V
IC (sat) = = = 7.5 mA
RC 2.4 kΩ

VCE (off ) = VCC = 18 V

Pdiss = VCEQ × ICQ = 9 V × 3.75 mA = 33.75 mW

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 DEPARTMENT OF
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 DEPARTMENT OF
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Instability of base bias

▶ Base biasing technique provides a very unstable Q-point ∵

the IC and VCE are greatly affected by change in beta β value

▶ If the transistor is replaced with one having a significantly

different value of βDC , the Q-point might shift to a point
located near or at either cut-off or saturation

▶ β also varies with temperature, therefore any change in

Transistor temperature can cause Q-point to shift

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 DEPARTMENT OF
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2. Voltage divider bias

This is a popular way of biasing a transistor.

Advantage lies in its stability. If designed properly, the circuit is

practically not affected by changes in β caused by either transistor
replacement or change in temperature.

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 DEPARTMENT OF
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Example 11

Figure 18: Voltage divider bias.

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 DEPARTMENT OF
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R2
VB = VCC
R1 + R2
5.1 kΩ
= × 15 V
27 kΩ + 5.1 kΩ

= 2.38 V

VE = VB − VBE

= 2.38 V − 0.7 V

= 1.68 V

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 DEPARTMENT OF
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VE 1.68 V
IE = = = 7 mA
RE 240 Ω
Because βdc = 100, we can assume that IC ≈ IE . And so,

IE ≈ IC = 7 mA

VC = VCC − IC RC

= 15 V − (7 mA × 1 kΩ)

= 15 V − 7 V

=8V
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 DEPARTMENT OF
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VCE = VCC − IC (RC + RE )

= 15 V − 7 mA(1 kΩ × 240 Ω)

= 15 V − 8.68 V

= 6.32 V

VCC 15 V
IC (sat) = =
RC + RE 1 kΩ + 240 Ω

= 12.1 mA

VCE (off ) = VCC

= 15 V
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 DEPARTMENT OF
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Figure 19: dc load line for for voltage divider-based transistor circuit.

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 DEPARTMENT OF
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VB , the voltage drop across R2 is the voltage measured from the

base lead to ground.

The voltage divider is made up of R1 and R2 . Thus;

 
R2
VB ≈ VCC
R1 + R2

VE = VB − 0.7 V

VE
IE =
RE

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 DEPARTMENT OF
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IC ≈ IE

VC = VCC − IC RC

VCE = VC − VE

= VCC − IC RC − IE RE

= VCC − IC (RC − RE )

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 DEPARTMENT OF
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Figure 20: A circuit.

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 DEPARTMENT OF
ELECTRICAL ENGINEERING

Exercise 1

Using an npn-transistor, sketch a voltage divider bias configuration

with the following parameters: Rb1 = 22kΩ, Rb2 = 10kΩ (with
one terminal of Rb2 connected to ground), RC = 1kΩ, RE = 1kΩ,
VCC = +30V .
From the circuit diagram, determine VB , VE , VC , VCE , IB , IE and
IC

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 DEPARTMENT OF
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Calculating the endpoints for the dc load line

When the transistor is saturated, VCE ≈ 0, for voltage divider

VCC
IC (sat) =
RC + RE

VCE (off) = VCC

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 DEPARTMENT OF
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Exercise 2

Figure 21: A circuit.

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 DEPARTMENT OF
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Exercise 3

For the pnp-transistor shown, calculate the following;

a). VB
b). VE
c). IC
d). VC
e). VCE
f). IC (sat)
g). VC (off )
Figure 22: A circuit.

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 DEPARTMENT OF
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Emitter bias

If both positive and negative power supplies are available,...

...emitter bias gives a solid Q-point that varies very little

with temperature and transistor replacement.

The emitter supply voltage VEE forward biases the emitter-base

junction through the emitter resistor, RE .

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 DEPARTMENT OF
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Figure 23: A circuit.

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 DEPARTMENT OF
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VEE − VBE
IE =
RE
Note that IE is ignored in the equation for calculating IE . A more
accurate formula for calculating IE is,

VEE − VBE
IE = (13)
RB
RE +
βdc
VC = VCC − IC RC (14)

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 DEPARTMENT OF
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Consider the circuit of Figure 24.

Figure 24: A BJT circuit.

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 DEPARTMENT OF
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The circuit is designed to produce a nearly constant current

through the variable RC .

If an ideal VBB = 7 V source is used to establish the base current,

and an RE = 630Ω is available, determine the following:

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 DEPARTMENT OF
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(a) The value of VE and IC , (Assume that βdc is very large).

VBB = VBE + VE

VE = 7V − 0.7V = 6.3V

∵ βdc is very large −→ IC ≈ IE

VE 6.3V
IE = = = 10mA
RE 630Ω

IC ≈ IE = 10mA

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 DEPARTMENT OF
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(b) The range of RC over which the circuit will function properly.
RE is constant, IE ≈ IC is constant, VE must also be
constant. Thus:
VCC = IC RC + VCE + VE

12 − 5.7V = 0.01RC + VCE

6.3V = 0.01RC + VCE

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 DEPARTMENT OF
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When RC = 0Ω
VCE = 6.3V
This seems well within the active region (far from VCC ).

As RC increases, its drop increases, and VCE decreases.

At some value of RC the operation in the active region will

cease.

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 DEPARTMENT OF
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Now when VCE = 0V ,

6.3V = 0.01RC
6.3
RC = = 630Ω
0.01
The circuit will ∴ function as a constant current source as
long as RC is in the range 0 < RC < 630Ω.

When RC exceeds 630Ω, the BJT becomes saturated.

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 DEPARTMENT OF
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If the base resistor RB (not shown in the circuit) is 56 kΩ,

calculate IB

Since the transistor is silicon, VBE equals 0.7 V . Therefore IB is

calculated as
VBB − VBE
IB =
RB

5 V − 0.7 V
=
56 kΩ

= 76.78 µA

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 DEPARTMENT OF
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The Beta Rule

According to this rule, resistance from one part of a transistor

circuit can be referred to another part of the transistor
circuit.

For example, resistance RL in the collector circuit can be

referred to the base circuit and vice versa.

Similarly, RE can be referred to the base circuit and,

reciprocally, RB can be referred to the emitter circuit.

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 DEPARTMENT OF
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Since current through RL is IC (= βIB ), hence β-factor comes

into the picture.

Similarly, current through RE is IE which is (1 + β)IB , hence

(1 + β) or approx. β-factor comes into the picture again.

The use of this ‘β-rule’ makes transistor circuit calculations

(especially of IB ) quite quick and easy.

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 DEPARTMENT OF
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The ‘β-rule’ may be stated as follows:

1. When referring RL or RC to the base circuit, multiply it by

β. When referring RB to the collector circuit, divide by β.

2. When referring RE to base circuit, multiply it by (1 + β) or

just β (as a close approximation).

3. Similarly, when referring RB to emitter circuit, divide it by

(1 + β) or β.

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 DEPARTMENT OF
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Before applying this rule, you must remember one very

important point otherwise you are likely to get wrong answers.

That is “only those resistances which lie in the path of the

current being calculated are transferred. Not otherwise”.

This will be demonstrated by solving the following problems.

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 DEPARTMENT OF
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Example 12

Calculate the value of VCE in the collector stabilisation circuit of

Figure 25.

Figure 25: Transistor circuit.

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 DEPARTMENT OF
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We will use β-rule to find IC in the following two ways.

i). First Method

Here, we will transfer RL to the base circuit.

VCC − IB RB − IB RL β − VBE = 0 -following the base path

After transferring RL to the base circuit, the base supply

voltage thus becomes = VCC .

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 DEPARTMENT OF
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Thus,

VCC − VBE 20 − 0.7

IB = = = 9.65 mA
RB + βRL 1000 + 100(10)

IC = βIB = 100 × 9.65 mA = 965 mA = 0.965 A

VCE ≈ VCC –IC RL = 20–(0.965 × 10) = 10.35 V

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 DEPARTMENT OF
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ii). Second Method

Now, we will refer RB to collector circuit.

VCC − IC RL − IC RB /β − VBE = 0 -following the base path

After transferring RB to the collector circuit, the collector

current IC thus becomes base current IB .

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 DEPARTMENT OF
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VCC − VBE 20 − 0.7

IC = = = 965 mA
RL + RB /β 10 + 1000/100

VCE = VCC –IC RL = 10.35 V − as above.

We now consider a case when RE is present and RL does not lie in

the path of IB .

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 DEPARTMENT OF
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Example 13

Calculate the three transistor currents in the circuit of Figure 26.

Figure 26: Transistor circuit.

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 DEPARTMENT OF
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i). First Method

Since RE lies in the path of IB .

Remember that when referring RE to base circuit, multiply

it by (1 + β)

VCC −IB RB − IB RE (1 + β) −VBE = 0 -following the base path

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 DEPARTMENT OF
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VCC − VBE
∴ IB =
RB + (1 + β)RE

10 − 0.7
= = 46.38 µA
100k + (1 + 200)0.5k

Thus,

IC = βIB = 200 × 46.38 µA = 9.28 mA,

IE = IB + IC = 9.32 mA

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 DEPARTMENT OF
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ii). Second Method

We now transfer RB to emitter circuit and find IE

directly.

VCC −VBE − IE RB /(1 + β) −IE RE = 0 -following the base path

VCC − VCE 10 − 0.7

IE = = = 9.32 mA
RE + RB /(1 + β) 0.5 + 100/(1 + 200)

IE 9.32 mA
IB = = = 46.38 µA,
(1 + β) (1 + 200)

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 DEPARTMENT OF
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Example 14

Calculate IE in the circuit of Figure 27.

Figure 27: Transistor circuit.

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 DEPARTMENT OF
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If we neglect VBE , then as seen from the circuit of Figure 27.

VEE
IE =
RE + RB /(1 + β)

10
= = 0.99 mA
10 + 10/101

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 DEPARTMENT OF
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Importance of VCE

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 DEPARTMENT OF
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142/142