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Bipolar Junction Transistor

(Voltage Gain (Av),

Maximum output / input voltage

value without distortion )

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Course Learning Outcomes (CLO)

CLO1 - Describe theoretically the characteristics andelectrical properties of semiconductors usingappropriate illustration.

CLO2 - Explain correctly basic principles of electronicdevices.

CLO3 - Explain the applications of electronic devices byusing the schematic diagram.

CLO5 - Construct other semiconductor devices circuitry

by using the schematic diagram

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Specific Outcomes 3.2.11 Explain the operations of a common emitter

amplifier circuit with AC input signal.

3.2.12 Compare the input and output waveform interms of amplitude and phase shift.

3.2.13 Determine by calculation the voltage gainbased on the formula: AV = Voutput/ Vinput

3.2.14 Explain with the aid of a load line curve, theoccurrence of amplitude distortion when a veryhigh input signal is given.

3.2.15 Determine the maximum input signal whichproduces undistorted output signal amplitude.

3.2.16Construct and analyze a common emitteramplifier circuit.

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A small-signal amplifier can also be called a voltageamplifier.

Common-emitter amplifiers are one type.

C

BE

Start with an NPN bipolar junction transistor

VCC

Add a power supply

RL

Next, a load resistor

RB

Then a base bias resistor

CC

A coupling capacitor is often required

Connect a signal source

The emitter terminal is grounded and common tothe input and output signal circuits.

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RB RLVCC

CC

C

BE

The output

is phase inverted.

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RBVCC

CC E

When the input signal goes positive:

B

The base current increases.

C

The collector current increases times.

RL

So, RL drops morevoltage and VCE must decrease.

The collector terminal is now less positive.

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RBVCC

CC E

When the input signal goes negative:

B

The base current decreases.

C

The collector current decreases times.

RL

So, RL drops lessvoltage and VCE must increase.

The collector terminal is now more positive.

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350 k

CC EB

C

1 k

14 V

The maximum value of VCE for this circuit is 14 V (VC cut-off).

The maximum value of IC is 14 mA (Ic Saturation).

IC(MAX) =VCC = 14 V

RC 1 k

These are the limits for this circuit.

= 14 mA

VC(MAX) =VCC = 14V

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0 2 4 6 8 10 12 14 16 18

2

4

6

8

10

12

14

VCE in Volts

IC in mA

20 A

0 A

100 A

80 A

60 A

40 A

The load line connects the limits.

SAT.

This end is calledsaturation.

CUTOFFThis end is called cutoff.

LINEAR

The linear region is between the limits.

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0 2 4 6 8 10 12 14 16 18

2

4

6

8

10

12

14

VCE in Volts

IC

in mA

20 A

0 A

100 A

80 A

60 A

40 A

Whats wrong with this Q point?

Good Q-Point(IC 6mA, VC 8V)

NOT Good Q-Point(IC 14mA, VC 0V)

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This is an example of a good Qpoint for linear amplification.

This is an example of nota goodQ point for linear amplification.

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A.C EMITTER OPERATION WHEN RECEIVES

A.C INPUT SIGNAL:

One of the factors might change IB value is a.c. input signal.

A.C. input signal value must not exceed that can cause IB oscillate bigIf a.c. input signal too big, this can because IB oscillates bigger(see dot line on the wave).Thus the value of IC oscillation too bigger and follows by VC. The oscillation of IC and VC are bigger than saturated point of a circuit.At this point, there is no gain value will occur.

So, the waveform that over this point will cutoff.

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A.C. loadlineWhen a.c. signal available, load that are facing at output partsometimes different compared d.c. signal analysis.

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Capacitor will allowed a.c. signal flow through it. So, loadthat are facing at output part is

Collector current value will not drop at d.c. loadline again,so in a.c. analysis it will have a new loadline. It also haveits own saturation point and cut off.

This line known as a.c. loadline.

rL=RC//RL.

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Formula to get a.c saturated point and a.c

shift point:

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Example:

Draw d.c. loadline and a.c. for the circuit below.

Also shows the location of Q-point.

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From this concept, assume equation for voltage gain is :

re input at resistance overall for value is

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Maximum output voltage value without distortion,

Vk(max)(without distortion)(herotan);

We can get the maximum output voltage withoutdistortion by using its loadline graph circuit.

Maximum output voltage without distortionmeans output signal which is oscillate at operation

point with symmetry and without limited/clipped.

We will choose part which is

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We will choose part which issmall peak voltage to getmaximum output voltagewithout distortion .As seen, ifwe take the big peak voltage

value as output voltage, wecan get one of the peaks atoutput voltage is cut off. Atthis state, its not calledmaximum output voltagewithout distortion.

Part 1Part 2

We will choose VP part 1 to get maximumoutput voltage without distortion

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Maximum value input voltage.

If we know the value of maximum outputvoltage without distortion (Vk(max)(without

distortion)), so we can determine the value of

input voltage without distortion

(Vm(max)(without distortion)).

But, the value of voltage gain for the circuitneed to know first.

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Answers..

V35.11VV

mA73.4II

11.35V33.65V-V35R.IVV

mA73.4)06.63)(75(I.I

06.63

555K

V35

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CCQ

CCQ

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BC

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A

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CCt)C(alih)(a.

C

CCt)C(tepu)(a.

==

===

DC load Line

22.7V

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rL

VCQICQ

33.028V83)(4.73)(4.511.35V

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==

5833

3511734

555

VmA

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rLICQVV

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tatepuC

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AC load Line

herotan.tanpa(masukan)(

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kV

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VmVm

26mV867

p22.7Vp

A

VkVm

8675.29

4.58k

re'

rL

Vm

VA

5.294.73mA

25mV

i

25mVre'

=

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====

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Vin max (withoutdistortion)(IC=IE)

Vout max (withoutdistortion)

a

fdc

b

e