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Introduction to concepts in dc bias

circuit design of the amplifier

(BJT CE amplifier design-Voltagedivider bias )



Outline•

Concept of faithful amplification

•Identification of dc circuits

• What is Q point?

•How to measure and locate Q point?

• Interpret location of Q point



Concept of faithful amplification

• The major goal of amplifier is to amplify

applied input signal faithfully.

• When is applied to the amplifier

then different types of output waveforms are possible .

Let us find out what are the possibilities



Faithful amplification-key specification

• The waveforms shown below are possible

outputs of the amplifier

• Can you identify which of the above waveform

represents faithful amplification of appliedinput signal( )?

• (click on the waveform to select answer)



Try again

Look at upper part of waveform it is

clipped which shows distortion in

output waveform and output is not

faithful amplification of input signal



• Look at lower part of

waveform ,it is clipped which

shows distortion in output

waveform and output is not

faithful amplification of input

signal

Try again



Look at both lower and upper

part of waveform ,it is clippedwhich shows distortion in

output waveform and output is

not faithful amplification ofinput signal

Try again



• You are right! This is perfect sine wave

with no clippings at extremities and known

as faithful amplification

NextGo back to

question



Essential components to decide

faithful amplification



For BJT amplifier circuit faithful amplification

can be predicted from

a)Amplitude of input signal and dc bias

circuit

b)Dc bias circuit and ac circuit

c)Ac circuit and amplitude of input signal



• This combination of ac and dc circuit helps to

decide only one parameter required to predict

faithful amplification and thus not sufficient.

Try again



• For this combination, amplitude of input signal

is partially useful to decide faithful

amplification but not complete specification

and ac circuit alone is not sufficient ,it needs

dc circuit to decide other parameter required to

predict faithful amplification.

• Try again



• You are right !

• Faithful amplification of the amplifier can be

predicted from dc bias circuit and input signal

amplitude.

Go back

to

question

Next



Identification of dc circuits

We will study effect of dc bias circuit on amplifier design

Let us find what is dc bias circuit



• For the given CE amplifier circuit identify which is

appropriate dc circuit?

VccVcc

Select option by clicking on the circuit below



• In the selected circuit capacitors are present

,but for dc condition you need to open circuit

capacitors and only resistors in the circuit will

remain.

Try again



• You are right! . For dc condition all capacitors

open circuited and only resistors in the circuit

will remain. The circuit for dc conditions is

Vcc

Go back to

questionNext



Dc circuit parameter

Let us find what are important parameters of dc circuit to

decide faithful amplification



Q . Which of the following do you think is useful to

predict faithful amplification?

(click the box to select option)

Value of βdc

Q point location

Resistor values



• The values are helpful in calculating current

and voltage values of the dc circuit but are

not useful to predict faithful amplification

Try again



• You are right!

Faithful amplification depends on

location of Q point

NextGo back to

question



How to locate Q point?

•

To locate Q point which of the followingcharacteristics are important?

• (Select options by clicking the buttons)

Only a only b only c a&b both a & c both

b&c both

c)Input characteristicsa)Output characteristics b)Load Line

Ic= Vcc/R c+ R E

Vcc



• You are right partially. One of the important

characteristics to locate Q point is output characteristics

,but not sufficient to locate Q point.

Select other options



Input characteristics are useful to locate Q

point partially (i.e. value of IBQ)but are notuseful to find other values of Q point(VCEQ,ICQ )

Try again



• You are right partially. One of the

important characteristics to locate Q point

is load line ,but not sufficient to locate Q

point.

Select other options



• This combination cannot fix Q point since ,it

is only set of characteristics .

Try again



• Load line is useful and with input

characteristics it can locate only IBQ value

.Other values(VCEQ and ICQ can not be

calculated.

Try again



• You are right! you need both output

characteristics and load line to locate Q point

Go back

to

question Next



•1.To locate Q point we need to drawOutput characteristics

• Output characteristics – This is a set of

characteristics . For each fixed value of basecurrent(IB) the relation between collector to emittervoltage(VCE) and collector current(IC) is plotted .

•

2.Superimpose load line on the outputcharacteristics

• Load Line — The line joining voltageVcc and current Ic= Vcc/R c+ R E

3.Mark IBQ characteristics

IBQ

— Base current of the given circuitunder no signal condition. This is alsoknown as dc bias current. IBQ characteristics are output characteristicsfor IBQ

Output

characteristics

C o l l e c t o r C u r r e n t ( I C )

IB1 =

IB3 =

IB2 =

IB4 =

IB5 =

IB6=

IB8 =

IB7=

Vcc

Ic= Vcc/R c+ R E

IBQ

Q point

The point of intersection of IBQ characteristics and load line is ―Qpoint‖

―Q point‖ is thus point on the load line representing dc bias conditions of the amplifiercircuit

Animation below will show how Q point is located



Find VCEQ and ICQ

Output

characteristics

C o l l e c t o r C u r r e n t ( I C

)

IB1 =

IB3 =

IB2 =

IB4 =

IB5 =

IB6=

IB8 =

IB7=

IBQ

Load line

Q point

ICQ

VCEQ

Collector to emitter voltage under

no signal condition also known

as dc bias voltage(Q point voltage)

Collector current under

no signal condition also known

as dc bias current(Q pointcurrent)



Locate Q point


IB1 =

IB3 =

IB2 =

IB4 =

IB5 =

IB6=

IB8 =

IB7=

IBQ

Q point

. Once we measure values of currents and voltage we can locate Q point by two

methodsIn the first method we fix output characteristics for IBQ(measured value) then

draw load line(by joiningVCC and Ic.) .Fix point as intersection of the two

Ic= Vcc/R c+

R E(4.08mA

)

Vcc=20V



• In the second method we use VCEQ and ICQ values and load

line as shown


IB1 =

IB3 =

IB2

=

IB4 =IB5 =

IB6=

IB8 =

IB7=

Vcc=20V

Ic=

Vcc/R c+

R E=4.08

mA

VCEQ

ICQ Q point



Interpretation of Q point

• Once we locate Q point by any of the two

methods ,we need to interpret the location of

Q point.

• First we will study different regions of load

line which will help us in interpretation of Q

point.

• Remember load line is essential characteristics

to locate and interpret Q point location.



1mA

2mA

3mA

4mA


)

10 µA

20 µA

40 µA

30 µA

Cut off region

In the output

characteristics

the region

below(slashed)

this point is cut

off region(here

both transistor

junctions are

reverse biased)

and transistor is

switched off

In the output characteristics the region

beyond this point (slashed) is saturation region

and transistor saturates(both junctions of

transistor are forward biased) .Output current

is constant and there is no relation between

input and output current

This diagram explains

Important regions of load

line

VBE<0.7V

VCE=0.1V



Interpretation of Q point(contd)

• In the process of interpretation of Q point

• Second important aspect is how to draw

output waveforms for applied input voltage for

given location of Q point.

• In the next slide Q point is already located and

we will observe how can we draw output

waveform?



Why Q point location is important?

• If we change location of Q point will the shape ofoutput waveform change?

• Let us vary location of Q point on the load line and

observe the output waveform



1mA

2mA

3mA

4mA


)

10 µA

20 µA

40 µA

30 µA

Go to

design

tip

Cut off region

Change the

location of Q

point by

selecting

value of IBQ

Click on

the box to

select value

of IBQ



10 µA

20 µA

30 µA

40 µA

1mA

2mA

3mA

4mA

C o l l e c t

o r

C u r r e n

t ( I C

) FIG.1

Q

This is output waveform in

which upper part of waveform

is clipped ,hence does not

represent faithful amplification

Q point is atlower

extremity and

input goes to

cut off region)

output voltage

waveform

output current

waveform

Vary Q point

Input voltage

Cut off region

Base

voltage< 0.7V



10 µA

20 µA

30 µA

40 µA

1mA

2mA

3mA

4mA

Colle

ctor

Curre

nt( IC)

FIG.1

Q

• This is perfect sine wave

with no clippings at

extremities and known as

faithful amplification

Vary Q

point

•

Q pointat the

centre of

load line

output current

waveform

output voltage

waveform

Input voltage

Cut off region



10 µA

20 µA

30 µA

40 µA

1mA

2mA

3mA

4mA

C o l l e c t o r C u

r r e n t ( I C

) FIG.1

Q

This is output waveform in

which only upper portion of

the waveform is present and

lower portion is clipped ,hence

does not represent faithfulam lification

Q point at upper extremity and upper

portion of input voltage saturates

transistor

Vary Q

point output voltage

waveform

output

currentwaveform

Input voltage

Cut off region

VCEQ= 0.1V



10 µA

20 µA

30 µA

40 µA

1mA

2mA

3mA

4mA C o l l e c t o r

C u r r e n t ( I C

)

FIG.1

Q

output current

waveform

output voltagewaveform

Input voltage

Vary Q

point

This is output waveform

in which small part of

lower portion of the

waveform is clipped ,hence

does not represent faithfulamplification

Cut off region

For this location of

Q point upper

portion of the

input just reaches

to saturation



Design tip1

If Q point is located at the extremities then

output waveforms are clipped at upper

extremity or lower extremity.It is important to decide Q point location near

centre to get faithful amplification

Practical value of VCEQ= Supply(VCC)/2(approx.)

You can go back to find

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