2_0 BJT AC 1

7/23/2019 2_0 BJT AC 1

http://slidepdf.com/reader/full/20-bjt-ac-1 1/51

7/23/2019 2_0 BJT AC 1


Pemodelan AC dari suatu rangkaian diperoleh dengan :

1. Hilangkan semua sumber DC, ganti dengan short circuit

2. Ganti kapasitor dengan short circuit

3. Gambar rangkaian pengganti

Remember pada rangkaian RC seri dg sumber V, maka arus

yang mengalir pada kapasitor (Xc) adalah i=v/R2+Xc2,

sehingga bila Xc<< R, maka arus akan mengalir (sepertiasumsi short circuit

Pada dasarnya transitor bukan saja digunakan untuk proses input yang linier

saja (DC), namun perlu (punya kemampuan) analisis (proses) sinyal non-

linier (AC) kebanyakan bagian dari transmitter/reciever dari suatu pancaran

gelombang terdiri dari komponen transistor

Berarti harus ada piranti tambahan (kapasitor) yang akan meloloskan

sinyal AC (bypass/coupling capacitor)

Sedangkan untuk melihat sinyal keluaran dari transitor digunakan

beberapa cara dengan Common Emitter (CE), Common Collector

(CC) dan Common Base (CB)

ingat pelajaran tentang RLC, maka masalah impedansi adalah halpenting untuk dianalisis juga, penguatan yang dapat dilakukan oleh

transistor dan berbagai permasalahan bila transistor di-cascade

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VS

VCC

C1

C2

C3

+

-

Vo

RS

Vi

+

-

RE

RC

R1

R2

VS

+

-

Vo

RS

Vi

+

-

RC

R1

R2

•I/p coupling

capacitor s/c

• Large values

• Block DC and

pass AC signal

• Bypasscapacitor s/c

•Large values

DC supply “0”

potential

Voltage-divider configuration

under AC analysis

Penggambaran ulang

konfigurasi diatas setelah

peniadaan sumber DC dan

menggunakan kapasitor

sebagai short/close

komponen

• O/p coupling

capacitor s/c

• Large values

• Block DC and

pass AC signal

Ilustrasi-1

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VS

RS

R2R1R

c

Transistor small-

signal ac

equivalent cct

Vo

Zi

Ii

Zo

Io

Vi

++

- -

B

E

C

Redrawn for small-signal AC analysis

Modeling of BJ T begin

HERE!

VS

+

-

VoRS

Vi

+

-

RCR1

R2

Ilustrasi-2 lanjutan

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DC and AC Analysis -- Application of Superposition

• DC analysis:

Find the DC equivalent circuit by replacing all capacitors by opencircuits and inductors (if any) by short circuits.

Find the DC Q-point from the equivalent circuit by using theappropriate large-signal transistor model.

• AC analysis:

Find the AC equivalent circuit by replacing all capacitors by short

circuits, inductors (if any) by open circuits, dc voltage sources byground connections and dc current sources by open circuits.

Replace the transistor by its small-signal model (to bedeveloped).

Use this equivalent circuit to analyze the AC characteristics of theamplifier.

Combine the results of dc and ac analysis (superposition) toyield the total voltages and currents in the circuit.

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Lihat Malvino (3rd edition) halaman 177 gambar 7-4/Slide selanjutnya. Kalau

pada analisa DC (kuliah2 sebelum ini) pad penggambaran garis beban

terdapat nilai Q yang bervariasi pada kurva Ic vs Vcc berupa garis lurus saja

Namun pada gambar 7-4 d tsb digambarkan untuk yang beroperasi disekitar

Q dg bentuk sinyal sinus luaran pada VBE akan bervariasi berupa sinus juga (ingat pada kuliah sebelumnya kita hanya memiliki harga pasti (konstan)

dari VBE

Hal tersebut akan terjadi bila sinyal yang masuk ke transistor adalah AC,

sehingga berdasar gambar ketiga diatas, maka diode tergantikan oleh

tahanan di emiter (r’e)

+VBE

+VBE

r’e

IC IC

Pendekatan Diode untuk

Analisa DC

Pendekatan Diode untuk

Analisa AC

Tahanan bulk

dari emiter

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where r’ e = ac emitter resistance

I E = the dc emitter current, found as V E / R E for example.

25mVe

E

r I

AC Emitter Resistance

BE e

E

V r

I

I E

V BE V BE V BE

Q2

I E

I E

Q1

Beberapa literatur memberikan nilai 26 mV

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The determination of ac beta.

I B

I C

Q

I B

I C

C cac

B b

I i

I i

hFE = dc beta

hfe = ac beta

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AC Model of A BJT

npn pnp

B C

E

B C

E

b

c

e

i b

i c

= aci b

r' e

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Standarisasi notasi AC dan

DC

IE, IC, IB ie, ic, ib

VE, VC, VB ve,vc,vb

VBE, VCE, VCB vbe, vce, vcb

dc, , r’e hFE, hfe, hrb

DC AC

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Fig 9.12a Example 9.2. Analisa Output Untuk

Sinyal AC dan DC (Cshort)

V CC

Q1

R E

R C

R 1

R 2

R L

C C 1

C C 2

C B

GND

Q1

R E

R C

R 1

R 2

R L

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Fig 9.12c-d Example 9.2.

penyederhanaan rangkaian

Q1

R C

R 1

R 2

R L

Q1

R 1||R

2

R C ||R L

(c)

(d)

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Voltage Gain DC of CE

Lcccc

Lccout

R Rr wherer i

R Riv

..

)(

out

in

c C C

ve e e

v i r r A

v i r r

Q1

R 1||R

2

R C ||R

Li b

i c

= i b

r' e

v in

v out

Q1

R 1||R

2

R C ||R

L

in e ev i r

25mVe

E

r I

“-” fase out berbeda 180’ input

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V CC

Q1

R E

R C

R 1

R 2

R L

C C 1

C C 2

C B

GND

Q1

R E

R C

R 1

R 2

R L

Q1

R C

R 1 R 2

R L

Q1

R 1||R

2

R C ||R

L

Q1

R 1||R 2

R C ||R

Li b

i c

= i b

r' e

v in

v out

Q1

R 1||R 2

R C ||R

L

out

in

c C C v

e e e

v i r r A

v i r r

out

where

c C L

c C C C L

v i R R

i r r R R

in e ev i r

1) AC circuit

2) AC Circuit

2.a)

2.b)

3) AC Circuit

3.a)

3.b)

Voltage Gain CE

Summary

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Perhatikan Malvino p.173 gbr. 9.2.b

-vbe+(ic/).rb+ie(r’e+re)0 didapat ie

/' rBer rE

vbbie

vc=ie.rC

ve=ie.rE

vb=ie(r’e+rE)

Penguat Tegangan ACCara 1 A=(vc/vb)

Cara 2 A=(rC/rE+r’e)

Singkatnya

Analias Dc untuk

mendapatkan IE

Analias AC untuk

liat gain dan bentuk

sinyal output

Penghubung AC &

DC melalui,

r’e=25mV/IE

C t h 1 ( t d b )

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+ 30 V

100

R C

10kR 1

20 k

R 2

10k

R L10 k

~

Vbb

10k

10k

20k 10k

10 k

~100

r’e 5k

~100

Contoh-1 (penguat dg common base)+ 30 V

10,1 k

R C 10k20 k

10k

Untuk analisa

DC, maka

semua kapasitor

open shg.

Menjadi gbr. (2)

Untuk dapat r’e diperlukan IB

(DC), maka dg voltage

devider Diperoleh IB, 10 V,

shg dg asumsi VBE≈0, maka

VE=10 V dan IE=10/10,1Katau sekitar 1 mA, shg

didapat r’e=(25 mV/IE) atausama dengan 25 Ω

Seri antara 100 Ω

dan 10 KΩ

1

2

3Untuk analisa AC, maka

kapasitor gbr (1) semua

short dan VCC = 0 /' rBer rE

vbbie

4

vc=ie.rC

ve=ie.rE

vb=ie(r’e+rE)

Cara 1 A=(vc/vb) Cara 2

A=(rC/rE+r’e)

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Vi

Vo

RC

C1

C2

VCC

R1

RER2

R1

RER2

Redrawn the input side of the network

Thevenin

VCC

R1

R2RTH

Determining RTH

R1

R2VCC VR2

+

-

ETH

+

-

Determining ETH

RTH

RE

Inserting the Theveninequivalent cct.

ETH IE

IB VBE

+

-

21TH R R R

21

CC2

2R TH

R R

VR VE

0R IVR IE EEBETHBTH

givesI1βI Subtitute

BE

ETH

BETH

B

R 1R

VEI

1 2 3

4 5

C t h 1 (C E itt ) T f th b i it t it Th i

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+20 V

hFE

= 200

R E

2.2k

R C

12k

R 1

150k

R 2

20k

R L

50k

Contoh-1 (Common Emitter) Transform the base circuit to its Theveninequivalent.

2

1 2

1 2

20k Ω20V

170k Ω

2.353V

20k Ω 150kΩ

17.65k Ω

th CC

th

RV V

R R

R R R

2.353V 0.7V

( 1) 17.65k Ω 201 2.2kΩ

3.595μA

th BE B

th FE E

V V I

R h R

200 3.595μA 718.9μA

1 201 3.595μA 722.5μA

C FE B

E FE B

I h I

I h I

20V 718.9μA 12kΩ 722.5μA 2.2kΩ

9.784V active

CE CC C C E E V V I R I R

25mV 25mV34.6Ω

722.5μA

12k Ω 50kΩ 9.677kΩ

e

E

C C L

r I

r R R

9.677k Ω279.7

34.6Ω

C v

e

r A

r

1

)

2)

3)

4)

5)

F 1 E l 4 (1)

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Fig 9.13 Example 9.4. (1)

+20 V

hFE

= 200

R E 2.2k

R C

12kR

1

150k

R 220k

R L

50k

Transform the base circuit to itsThevenin equivalent.

2

1 2

1 2

20k Ω20V

170k Ω

2.353V

20k Ω 150kΩ

17.65k Ω

th CC

th

RV V

R R

R R R

2.353V 0.7V

( 1) 17.65k Ω 201 2.2kΩ

3.595μA

th BE B

th FE E

V V I

R h R

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+20 V

hFE

= 200

R E

2.2k

R C

12k

R 1

150k

R 220k

R L

50k

200 3.595μA 718.9μA

1 201 3.595μA 722.5μA

C FE B

E FE B

I h I

I h I

20V 718.9μA 12kΩ 722.5μA 2.2kΩ

9.784V active


25mV 25mV34.6Ω

722.5μA

12k Ω 50kΩ 9.677kΩ

e

E

C C L

r I

r R R

9.677k Ω279.7

34.6Ω

C v

e

r A

r

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Transform the base circuit to itsThevenin equivalent.

2

1 2

1 2

4.7k Ω10V

22.7k Ω

2.070V

4.7k Ω 18kΩ

3.727k Ω

th CC

th

RV V

R R

R R R

2.070V 0.7V

( 1) 3.727k Ω 31 1.2kΩ

33.49μA

th BE B

th FE E

V V I

R h R

+10 V

hFE

= 30

hfe

= 200

R E

1.2k

R C

1.5kR 1

18k

R 24.7k

R L

5k

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30 33.49μA 1.005mA

1 31 33.49μA 1.038mAC FE B

E FE B

I h I I h I

10V 1.005mA 1.5k Ω 1.038mA 1.2kΩ

7.247V active


25mV 25mV24.08Ω

1.038mA

1.5k Ω 5kΩ 1.154kΩ

e

E

C C L

r I

r R R

1.154k Ω47.91

24.08Ω

C v

e

r A

r

+10 V

hFE

= 30

hfe

= 200

R E

1.2k

R C

1.5kR 1

18k

R 24.7k

R L

5k

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Power Gain

p i v A A A

Example 9.7 The amplifier shown in Fig. 9.5 has

values of Av = 45.3 and Ai = 20. Determine thepower gain ( A p) of the amplifier and the outputpower when Pin = 80 mW.

20 45.3 906 p i v A A A

out in 906 80μW 72.48mW p P A P

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The Effects of Loading

The lower the load resistance is, the lower the voltage gain.

2.4k Ω

96

C C L

C v

e

r R R

r A

r

2k Ω

80

C C L

C v

e

r R R

r Ar

R C

3k

R L

6k

R 1||R

2r' e=25

R C

3k

R L

12k

R 1||R

2r' e=25

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Example 9.8

The load in Fig. 9.16 is open. Calculate theopen-load voltage gain of the circuit.

3k ΩC C r R

3k Ω120 (max. gain)

25Ω

C v

e

r A

r

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INPUT IMPEDANCE

WHY?

For an instrument the ZIN should be

very high (ideally infinity) so it does

not divert any current from the input toitself even if the input has very high

resistance.

e.g. an opamp taking input from a

microelectrode.

Input Circuit Output

Impedance between input

terminals = input

impedancee.g. Microelectrode R=10 Mohm &

therefore Rin=G Ohm!

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Ide DasarnyaOUTPUT IMPEDANCE

Input Circuit Output

Impedance between output terminals= output impedance

WHY?

For an instrument the ZOUT shouldbe very low (ideally zero) so it cansupply output even to very low

resistive loads and not expendmost of it on itself.

e.g. a power opamp driving a motor

A common-emitter (CE) amplifie

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A common emitter (CE) amplifie

capacitors are used for coupling ac without disturbing dc levels

• Vout is biased at middle of V+

• AC Gain: G = - Rc / RE

0

(V+)/20.1

(V+)/2 + G*0.1

(V+)/2 - G*0.1-0.1

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Phase Relationship

The phase relationship between input and output is 180 degrees. The negative

sign used in the voltage gain formulas indicates the inversion.

Ilustrasi-1

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Ilustrasi 1

The input (Vi) is applied to the base and the output (Vo) isfrom the collector.

The Common-Emitter is characterized as having high input

impedance and low output impedance with a high voltage and

current gain.

Removing DC effects of

VCC and Capacitors

Common-Emitter (CE)

Fixed-Bias Configuration

f CIlustrasi-2

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A Practical BJT Amplifier using Coupling

and Bypass Capacitors

AC coupling through capacitors is used

to inject an ac input signal and extract the

ac output signal without disturbing the DC

Q-point. Capacitors provide negligible

impedance at frequencies of interest and

provide open circuits at dc.

In a practical amplifier design, C 1 and C 3

are large coupling capacitors or dc

blocking capacitors, their reactance (XC =

|ZC| = 1/wC ) at signal frequency is

negligible. They are effective open

circuits for the circuit when DC bias isconsidered.

C2 is a bypass capacitor . It provides a

low impedance path for ac current from

emitter to ground. It effectively removes

R E (required for good Q-point stability)from the circuit when ac signals are

considered.

Ilustrasi 2

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Calculating Amp.

Input Impedance

in 1 2 in(base) Z R R Z

Z in(base)

Z in

R 1

R 2

Biasing circuit

V CC

IN(base)DC: 1 FE E

FE E

R h R

h R

in(base)AC: 1 fe e

fe e

Z h r

h r

Rin=(dc+1 )Re

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Example 7-3 (p. 186-188, 3rd malvino)Data, Rin (impedansi input

sebelum kapasitor =1K),R1=10Kk, R2=2,2K, dan Rc=3,6K,

Re=1K sedang beban Rl=1,5K), 150

Pertama tentukan impedansi input yaitu,Zin(base)=.r’e=150.(22,7 Ohm)=3,4K, angka 22,7 dari contoh 7-1

Inpedansi input amplifier zin = R1R2.r’e

zin=10K 2,2K 3,4K=1,18K dari soal sebelumnya (7-2)didapat

unload

Voltage gain adalah -159, sedangkan impedansi output sama dg.Rc, maka

Voltage devider input vin = (1.18/1,18+1) . 1 mV =0,541 mV

Thevenien output voltage vout = A. vin = -159 . (0,541mV)=-

86mVThat is unloaded output

The actual output is what appears across the 1,5K

Vout = (1,5/1,5+3,6). -86mV=-25mV

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Fig 9.17 Example 9.9.

in(base)

200 24Ω 4.8kΩ fe e

Z h r

in 1 2 in(base)

18k Ω 4.7kΩ 4.8kΩ

2.1k Ω

Z R R Z

+10 V

hFE = 30hfe

= 200

R E

1.2k

R C

1.5kR 1

18k

R 2

4.7k

R L5k

Z in(base)

Z in

Q1

R 2

4.7k

R 1

18k

R C

1.5k

R L

5k

r' e

= 24

hfe = 200

Z in(base)

Z in

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Calculating the Value of Ai

out

in

i

i A

i c

fe

b

ih

i

out out

out

c L C c C L

c C

L

v i R R i r i R

i r i

R

in in in in(base)

in(base)

in

in in

1

b

b fe eb

v i Z i Z

i h r i Z i

Z Z

in

in(base)

C i fe

L

Z r A h

Z R

Q1

R C

R 1

R 2

R L

i c i b

Z in(base)

i out

v out

i in

v in

Z in

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Example 9.10

Calculate the value of Ai for the circuit shown in Fig. 9.17.

in(base) 200 24Ω 4.8kΩ fe e Z h r

in 1 2 in(base) 2.1k Ω Z R R Z

1.15k ΩC C Lr R R

in

in(base)

2.1k Ω 1.15kΩ200

4.8k Ω 5kΩ

20.2

C i fe

L

Z r A h Z R

Q1

R 2

4.7k

R 1

18k

R C

1.5k

R L

5k

r' e

= 24

hfe

= 200

Z in(base)

Z in

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Multistage Amp.

Gain Calculations

1 2 3vT v v v A A A A

1 2 3iT i i i A A A A

pT vT iT A A A

Procedure:

1. Do dc analysis

2. Findr’ e for each stage

3. Find r C for each stage

4. Using r’ e and r C to find Av

for each stage Input impedance of next stage is the load of current stage.

(Z in of next stage is RL of current stage)

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Fig 9.18 Example 9.11. (1)

( ) 1 201 17.4 3.497k Ωin base fe e Z h r

Determine Av of the 1st

stage. Assume that r’ e for the 1st stage is 19.8 and r’ e for the 2nd stage isfound to be 17.4 . Forthe 2nd stage, hfe is 200. 5 6 ( ) 1.329k Ωin in base Z R R Z

The input impedance of the 2

nd

stage:

+15V

C C 1

C C 2

C C 3

R 1

22k

R 2

3.3k

R 4

1kC

B1

R 8

1kC

B2

R 3

5k R 5

15k

R 6

2.5k

R 7

5k

R L

10kQ

1Q

2

hFE

= 150

hfe

= 200

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Fig 9.18 Example 9.11. (2)

+15V

C C 1

C C 2

C C 3

R 1

22k

R 2

3.3k

R 4

1kC

B1

R 8

1kC

B2

R 3

5k R 5

22k

R 6

3.3k

R 7

5k

R L

10kQ

1Q

2

hFE

= 150

hfe

= 200

35k Ω 1.33kΩ=1.05kΩ

C inr R Z

Finally, Av for the 1st stage is found as 1.05k Ω

53.0319.8Ω

C v

e

r A

r

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Example 9.12. (1)

Determine the value of AvT for the amplifier in Figure9.18. r C for the 2nd stage can be found as

7 3.33k ΩC Lr R R

3.33k Ω

191.3817.4Ω

C

ve

r

A r

Av for the 2nd stage is found as

3

1 2 53.03 191.38 10.15 10vT v v A A A

Fi 9 19 Th d CE lifi d

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Fig 9.19 The swamped CE amplifier and

its ac equivalent ckt.

Swamped amplifier is anamplifier that uses a partiallybypassed emitter resistance toincrease ac emitter resistance.

Also referred to as a gain-stabilized amplifier. R

C

r E

R E

C B

R 1

R 2

C 1

C 2 R

L

+V CC

Q1

r E

Q1

r C

R 1//R

2

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Distortion (Bagian dari efek Negatip Swamp)

• The location of the point Q (size of the dc source on input) may

cause an operating point to lie outside of the active range.

Driven to saturationDriven into Cutoff

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Av of Swamped Amp.

C v

e E

r A

r r

r E

Q1

r C

R 1//R

2

r C

R 1//R

2

r E

r' e

i b

i c

= ac i b

Q1

v in

v out

out c C

v

in e e E

v i r A

v i r r

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Fig 9.20 Example 9.13. (1)

R C

1.5k

r E

300

R E 910

C B

R 1

18k

R 2

4.7k

R L

10k

+10V

hFE

= 200hfe

= 150

2

1 2

1 2

4.7k Ω10V 2.070V

22.7k Ω

18k Ω 4.7kΩ 3.727kΩ

th CC

th

RV V

R R

R R R

1

2.070V 0.7V

3.727k Ω 201 1210Ω

5.550μA

200 5.550μA

1.110mA

1 201 5.550μA

1.116mA

th BE B

th FE E E

C FE B

E FE B

V V

I R h r R

I h I

I h I

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Fig 9.20 Example 9.13. (2)

R C

1.5k

r E

300

R E

910C

B

R 1

18k

R 2

4.7k

R L

10k

+10V

hFE

= 200hfe

= 150

10V 1.110mA 1.5k Ω 1.116mA 1210Ω

6.985V (active)

CE CC C C E E E V V I R I r R

25mV 25mV 22.41Ω1.116mA

e

E

r I

1.304k ΩC C Lr R R

1.304k Ω4.046

22.41Ω 300Ω

C v

e E

r A

r r

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Example 9.14

Determine the change in gain for the amplifier in Example9.13 when r’ e doubles in value.

1.304k Ω3.782

44.82Ω 300Ω

C v

e E

r A

r r

4.046 3.782 0.2639v A

Swamping improves the gain stability of a CEamplifier when r E >> r’ e.

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Kuantitas r’e idealnya bernilai (25mV/Ie), akan tetapi karena transistor

lebih sensitip thd kenaikan temperatur (ingat disetiap komponen transistor

seringkali disertai pendingin), sehingga nilai tahanan tsb. pun akan dapat

berubah juga

Perubahan tahanan tersebut kalau melihat beberapa rumusan

sebelumnya akan berpengaruh pada voltage gain, tentunya tak

diharapkan bila terlalu tinggi

Lebih spesifik lagi lihat halaman 189 biasanya untuk mengatasi efek

swamping, maka di emiter ditambah dengan tahan RE sehingga pada

sisi emiter menjadi (r’e+RE), lihat sekali lagi hal 189, bila diberikan

tahanan seperti dimaksud diatas, maka penurunan gain lebih kecil 5%

dengan kenaikan r’e 100%

Sedang untuk gain tanpa RE (lihat barisan dibawah halaman yang sama

dengan kenikan /penurunan voltage gain besar variasi perbedaannya

sangat besar (tidak stabil) seharusnya kenaikan temperatur tidak

mengakibatkan kenaikan gain yang cukup besar

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The Effect of Swamping on Zin

in(base) 1 fe e

fe e

Z h r

h r

in(base) 1 fe e E

fe e E

Z h r r

h r r

b

c

eZ in(base)

r' e

b

c

e

Z in(base)

r' e

r E

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Fig 9.22 Gain stabilization.

Av -r C / r’ e -r C / (r’ e+r E )

Zin(base) hfer’ e hfe(r’ e+r E )

Advantage Higher values of Av than the

swamped amplifier.

Relatively stable Av . Much

smaller distortion.

Disadvantage Relatively unstable values of

Av .

Lower Av than the standard

amplifier.

R E

R E

r E

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Summary

• AC concepts

• Roles of capacitors in amplifiers

• Common-emitter ac equivalent circuit

• Amplifier gain

• Gain and impedance calculations

• Swamped amplifiers

• h-parameters

2_0 BJT AC 1

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