Esercitazione amp-BJT.pps

8/14/2019 Esercitazione amp-BJT.pps

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1

Coupling and Bypass

Capacitors

AC coupling through capacitors isused to inject ac input signal andextract output signal without

disturbing Q-point

Capacitors provide negligibleimpedance at frequencies of interestand provide open circuits at dc.

C1and C3are large coupling capacitors

or dc blocking capacitors, their

reactance at signal frequency is

negligible.

C2is bypass capacitor, provides lowimpedance path for ac current from

emitter to ground, removingRE

(required for good Q-point stability)

from circuit when ac signals are

considered.


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2

DC and AC Analysis DC analysis:

Find dc equivalent circuit by replacing all capacitors by opencircuits and inductors by short circuits.

Find Q-point from dc equivalent circuit by using appropriate large-signal transistor model.

AC analysis:

Find ac equivalent circuit by replacing all capacitors by shortcircuits, inductors by open circuits, dc voltage sources by groundconnections and dc current sources by open circuits.

Replace transistor by small-signal model

Use small-signal ac equivalent to analyze ac characteristics ofamplifier.

Combine end results of dc and ac analysis to yield total voltagesand currents in the network.


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3

DC Equivalent for BJT Amplifier

All capacitors in original amplifier circuits are

replaced by open circuits, disconnecting vI,

RI, and R3from circuit.


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4

AC Equivalent for BJT Amplifier

k100k3.43

k30k1021

RC

RR

RRB

R


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5

Hybrid-Pi Model of BJT

The hybrid-pi small-signalmodel is the intrinsic low-frequency representation

of the BJT. Small-signal parameters

are controlled by the Q-point and are independentof geometry of BJT

Transconductance:

CI

TV

CI

mg 40

Input resistance:

mg

o

CI

TVo

r

Output resistance:

max)(guadagnoormgf

CIA

V

CI

CEV

AV

or


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6

Equivalent Forms of Small-

Signal Model for BJT

Voltage -controlled current source gmvbecan betransformed into current-controlled currentsource,

Basic relationship ic=ibis useful in both dc and

ac analysis when

BJT is in forward-active region.

bice

v

bici

bi

bi

bevbi

bev

oor

o

or

mg

mg

r


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7

Small-Signal Analysis of

Complete C-E Amplifier: AC

Equivalent Ac equivalent circuit

is constructed by

assuming that allcapacitances havezero impedance atsignal frequency and

dc voltage source isac ground.

Assume that Q-point

is already known.

21RR

BR


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8

Small-Signal Analysis of

Complete C-E Amplifier: Small-

Signal Equivalent

3R

CRorL

R

LRmg

bevov

bvcv

vtA

Terminal voltage gainbetween

base and collector is:

Overall voltage gain from source vi

to output voltage acrossR3is:

r

B

R

I

R

rB

R

LRmgvA

ivbev

vtA

ivbev

bevov

ivov

vA


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9

C-E Amplifier Voltage Gain:

Example Problem:Calculate voltage gain

Given data:F=100, VA=75 V, Q-point is (1.45 mA,

3.41 V), R1= 10 kW, R2= 30 kW,R3= 100 kW, RC=

4.3 kW, RI

= 1kW.

Assumptions:Transistor is in active region, O=

F. Signals are low enough to be considered small

signals.

C

be

Iic

mVv

%20

5


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10

Analysis

dB3.42130

rB

RI

R

rB

R

LRmgvA

mS0.58)mA45.1(4040 C

Img

k72.1mA45.1

)V025.0(100

CI

TVor

k1.54mA45.1

V14.3V75

C

ICE

VAV

or

k83.33 R

CRorL

Rk5.721 RR

BR

mV57.8

)(V)005.0(

rB

R

rB

RI

R

iv

i

BB

B

be

be

vrRR

rRv

mVv

)||(

||

5


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11

Small-Signal Model

Simplification If we assume

rBR

IR

3R

CRormgL

RmgvtAvA

This implies that total signal voltage at input

appears across r.


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12

Progetto amplificatore piccoli

segnali (carico resistivo)

Generally R3>> RCand R3


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13

C-E Amplifier Input Resistance

Input resistance, the totalresistance looking into the

amplifier at coupling

capacitor C1represents total

resistance presented tosource.

rRRrB

RR

rB

R

21

xixv

in

)(xixv


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14

C-E Amplifier Output Resistance

Output resistance is the total

equivalent resistance looking

into the output of the amplifier

at coupling capacitor C3. Input

source is set to 0 and test

source is applied at output.

CRorC

RR

mg

orCR

x

ixv

out

bevx

vxvxi but vbe=0

As ro>> R

C.


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15

Sample Analysis of C-E

Amplifier Problem:Find voltage gain, input

and output resistances.

Given data:F= 65, VA=50 V

Assumptions:Active-regionoperation, VBE=0.7 V, small signal

operatingconditions.

Analysis: To find the Q-point,dc equivalent circuit is

constructed.

A24566

A24165

A71.3

BI

EI

BI

CI

BI

5)4106.1()1(510 BI

FBEV

BI

V67.3

)4106.1(41055

CEV

EI

CEV

CI

jj

ji i IRV

KVL

S l A l i f C E


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16

Sample Analysis of C-E

Amplifier (contd.)

0.84

in

in)3out

(

RI

R

RRRmg

ivov

vA

S31064.940 C

Img

k64.6

C

IT

Vor

k223

CI

CEVAVor

k23.6

in

r

B

RR

k57.9out

orCRR


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17

CE senza Capacit di bypass


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18

Inverting Amplifiers: Terminal

Voltage Gain

Using test source vbto drive the base

terminal of the transistor, neglecting

ro,

ERL

R

ER

mg

LR

mg

ACEvt

ER

omg

o

L

R

o

ER

or

L

R

o

bvov

vtA

1

)1(/)1(

1oAssuming

Lamplificazione

dipende solo dai

resistori esterni!

LRmg

bevov

bvcv

vtA

Eob Rirv

KVL

)1(


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19

nver ng mp ers: npuResistance and Overall Voltage

Gain

mgro

o

1

Input resistance looking into the base

terminal is given by

)1(

)1(

ERmgrR

CEin

ERor

ibv

RCEin

RCEinBR

IR

RCEinBR

CEin

R

LRo

RCEinBR

IR

RCEinBR

ACEvt

ivbv

ACEvt

ivbv

bvov

ivov

ACEv

Overall voltage gain is

Assuming

Generatore reale! Av

rB

RI

R

rB

R

LRmgvA

iv

bev

vt

A

iv

bev

bev

ov

iv

ov

v

A


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20

Inverting Amplifiers: Output

Resistance

0ixi0i

0ev0)1(

1ev

i)1(ev

o

rth

RE

Ro

ERo

rthR

ev-

i

ButRout= rowhenRE= 0, not infinite.

Now, we also include roin our analysis.

xix

v

outR

rth

RE

RE

RoorR

rthRER

ER

ERr

thR

oro

1out

xii

xiev

ev)ixi(evrvxv

IBTH RRR //

CORTOCIRCUITANDO

IL GENERATORE DI

INGRESSO

neglect ing ro


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21

Inverting Amplifiers: Output

Resistance (contd.) rmgo Assuming and , with

.for

Finite current gain of BJT places an upper limit on size of output

resistance. rappears in parallel withREifRthis neglected. If we letREbe

infinite, maximum value of output resistance is

thR

ERr )( E

Ror

)()(out

)()(1out

ERr

fERrmgorR

ERr

for

ERrmgorR

1)(

E

Rrmg

oroR )1(out

CRorC

RR

xixv

out

Rout modifica Avperch RLdiventa il

parallelo tra il carico

e Rout


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22

Inverting Amplifiers: Current

Gain

oACE

it

ACEit rRR

R

i

rRRRi

i

i

i

iA

rRR

Rii

ii

EoB

B

i

EoB

BiO

i

O

i

Li

EoB

Bi

OL

)1(

)1(

)1(

Esercitazione amp-BJT.pps

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