Copy of Teorie Electronic A Rezolvare

8/14/2019 Copy of Teorie Electronic A Rezolvare

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1.Distributia Fermicaa)Fermi Dirac distribution for electrons. - 001_Mendeleev Table.ppt Pag. 6-9

Fig. ..1 pct.

Definitions of probabilities .. 1pct

This is the probability to have an electron on energy level E1 and E2 (occupied states) and no electronson E3 and E4 (unoccupied states)

This is the probability to have an electron on energy level E3 and E4 (occupied states) and no electronson E1 and E2 (unoccupied states)Both probabilities must be equal in the case of thermal equilibrium, therefore we can write the followingequality:

( ) ( ) ( ) ( )

=

1

P

11

P

11

P

11

P

1

2143 EEEE2 pct.

E E E E1 2 3 4+ = + 1 pct

( )kT

EEeF

e

E+

=1

1f

..2 pct.

Coments on the shape of F-D distribution ..2 pct

Bonus 1 pct

Total .10 pct

( ) ( ) ( )( ) ( )( )4321 EP1EP1EPEP

( )( ) ( )( ) ( ) ( )4321 EPEPEP1EP1

( ) ( ) 2/1k2/3*

n3kEm2

h

4EN

=


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.b)Fermi Dirac for holes.

E

E

E

E

1

2

4

3

This is the probability to have an electron on energy level E1 and E2 (occupied states) and no electronson E3 and E4 (unoccupied states)

This is the probability to have an electron on energy level E3 and E4 (occupied states) and no electrons

on E1 and E2 (unoccupied states)

If the condition of energy conservation

probability to have an empty state is :

( ) ( ) ( )( ) ( )( )4321 EP1EP1EPEP

( )( ) ( )( ) ( ) ( )4321 EPEPEP1EP1

( ) ( ) ( ) ( )

=

1

EP

11

EP

11

EP

11

EP

1

2143

E E E E1 2 3 4+ = +

( ) ( )

kT

EEep F

e

EfEf

+

==

1

11

( ) ( ) 2/1k2/3*

p3kpEm2

h

4EN

=


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2.Densitatea purtatorilor de sarcina intr-un semiconductor intrinsicThe density of charge carriers in intrinsic semiconductors. - 002_The Density of Charge Carriers in anIntrinsic.ppt -> Pag. 1-7

se dau ( )kT

EEe F

e

E

+=

1

1f i ( ) ( ) 2/1

2/3*

32

4knk Em

hEN

=

Barem de corectare:E E Ek c= .. 1 pct

+

==CE

ee dEENEnn )()(f .1 pct

( ) kTEE

kT

EEe

F

Fe

e

E

+

=

1

1f ..1 pct

e e eE E E E

kT

E E

kT

E E

kTF c c c F c

+

= .1 pct

2

kTEEx c= ..1 pct

( )( ) dxexkTe

h

mn xkT

EE

nFc

22

0

2/3

3

2/3*

224

=

.2 pct

( )

==

000

2

22

222 dxexedxex xxx .1 pct

( )kT

EE

CkT

EE

nFcFc

eNeh

kTmn

==3

2/3*22

1 pct

din oficiu ..1 pct

Total 10 pctIn an intrinsic semiconductor, the density of the two types of charge carriers is the same

n=p

3.P conductor


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Se d: ( )kT

EEe F

e

E

+=

1

1f i kT

EE

V

vF

eNp

=

Barem de corectare:

i comentarii .3pct

f E

e

A E E

kTA F

=

+

( )1

1

...1pct

N N f E N eA A A A

E E

kTA F

= ( )..1pct

p N e NV

E E

kTA

F v

= =

.1pct

N e N eV

E EkT

A

E EkT

F V A F = .1pct

EE E kT N

NF

A v V

Ap=

++

2 2ln i comentarii...2pct

4.N conductor


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Hole (p)

Quasi-free electron (n)

EC

EV

CB

VB

N e N eC

E E

kTV

E E

kT

C F F V

=

5.Fenomene fizice in semiconductori

ckEEE =

( ) ( ) 2/12/3*3

24

knkEm

hEN

=

EEEvk

= +

==CE

eedEENEnn )()(f

==VE

ppp dEENEpn )()(f

( ) ( ) 2/12/3*3

24

knkEm

hEN

=

( )( ) dEeEEe

h

mn kT

EE

c

E

kT

EE

nc

c

Fc

2/1

3

2/3*

24=

( )kT

EE

C

kT

EE

nFcFc

eNeh

kTmn

==3

2/3*22

kT

EE

VCi

vc

eNNpnn

==2

kT

EEEE

C

VFcvF

eN

N +=

c

vvc

FN

NkTEEE ln

22+

+=


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6.Jonctiunea PN


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Desen densitate de sarcin...................................... 1pct

N L N LA n D p= ................................................................1pct

=S)xL(eN

S)x(nA

E ............................................1pct

)()(

xLeNx

pDE

= ..............................................1pct

)0(LeNLeN pDnA

m ax EE =

=

= ..................................1pct

Desen E(x)............................................................... 1pct

+

+==

p

n

0

L

L

npmaxb

2

)LL(dx)x(V

EE ..... ...................2pct

Desen V(x)............................................................... 1pct

din oficiu ...............................................................................1pctTotal ...................................................................................10pct

7.Jonctiunea PN la echilibru termic(Einstein)Einsteins relations for Diffusion and Mobility coefficients. - 005_PNjunction.ppt Pag. 6


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8.Capacitatile PN

a)de baraj

e

kTD

p

p=

e

kTD

n

n=


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b)de difuzie

9.Tranzistor bipolar(conditii,parametric functionare,curenti)Curenti


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Static working point for CEC.- 007_The Bipolar Junction Transistor (BJT).ppt(pag.9)

IC

IEVBE

.Static working point for CBC. - 007_The Bipolar Junction Transistor (BJT).ppt Pag. 11 si

pagina 8

),();,(21 CEBCCEBBE

VIfIVIfV ==

( )0

1 CBC III ++=

=1

)V,I(fI);V,I(fV CBECCBEBE 21 ==

0CECIII +=


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IC

IBVBE

The CBC connection is the most stable operating configuration of bipolar transistor. This is provided by the factthat we control the output current IC with a current IE , higher than the residual current of the collector and the

current gain()is approximately constant, having values in the range 0.98 - 0.99.

10.Caracteristici statice BC

V f I V I f I V BE E CB C E CB= =1 2( , ); ( , )

IC

IBVBE

11.Caracteristici statice EC

CE characteristics of BT. - 007_The Bipolar Junction Transistor (BJT).ppt Pag. 9

),();,( 21 CEBCCEBBE VIfIVIfV == and comments 2 pct

IC

IEVBE

( )0

1 CBC III ++=

=1

0CECIII +=


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12.Sensibilitatea fata de temperature a BT

Temperature sensitivity of BT.

( )0)()( 000TTa

CC eTITI= plus comentarii 2pct

( )

+=

K

TTTT

0

0 1)( plus comentarii 1pct

V

T mV CBE = 2 2 0. / ..1pct

I

T

I

I

I

T

I

V

V

T

I

T

C C

C

C C

BE

BE C= + + 0

01pct

I

ISC

C

I

0

= .1pct

I

VSC

BE

U= ...1pct

ISC = 1pct

SI

I

I

B

C

=+

1

11pct

13.Modelul de semnal mic-hibrid

Parametrii hibrizi ce se definesc cu relaiile:

oii VhIhV += 1211

oio VhIhI += 2221


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Parametrii "h" se utilizeaz n special n joas frecven unde ei sunt mrimi reale, care nu depind de frecven.Experimental aceti parametri se determin prin msurtori n situaia de gol la intrare (i1 = 0) i scurcircuit la ieire (u2 =0) pentru componenta variabil, ceea ce este foarte uor de realizat la un tranzistor bipolar, deoarece acesta areimpedana de intrare Zintrmic i impedana de ieire Zies mareDin relaia de definiie se observ c parametrii hibrizi au semnificaii fizice diferite: h12, h21 -adimensionali, h11-impedan, h21 - admitana, ceea ce justific denumirea deparametrii hibrizi

14.Conditii in care se poate simplifica un hybrid

Circuitul hybrid simplificat; condiiile de simplificare(03_General Characteristics of anAmplifier 6-8)


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Now, if we take into account the actual typical values for the hybrid parameters in theequation for the input impedance (hre=10-4 ; Ai=hfe=102 ; RL=104-103 ohm ;hie=103 ohm), we will see that the second term in the expression of input impedancecan have values in the range 102-10. In this case we can ignore this term versus thefirst term which has a value 10 to 100 times higher.

That means, from a practical standpoint of view, that in the hybrid model of thetransistor, we can neglect the reverse transfer factor hre.

The simplified circuit shown in Fig.4.9 can be used, too, for the computing of theparameters of any amplifier, independent of the transistor connection type (CE;CB orCC). Lets test that right now.

15.Amplificator in BCAmplificator C.B.C.(04_Common Base Amplifier 1-2)


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16.Amplificator in CC


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17.Amplificator cu EC


17/29

Every amplifier is characterised by:voltage amplification Av , current amplification Ai , input impedance zi and output admittance yo .

In order to be able to calculate these parameters it is necessary to transform the d.c circuit in its a.c.equivalent circuit. Here are two rules to be followed:

every capacitance is a short-circuit in a.c. the d.c. biasing source is a short-circuit to the ground in a.c.

Now, in a.c. circuit, we must replace the hybrid circuit of the tranzistor

R B

R g

vg

R C R L

B C

E E

hi e

zi

zo

ho e

hr e o

v hf e b

i

i n p u t o u t p u

i = ii b

i = io c

18.Amplificator cu rezistenta in emitor


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05_The Differential Amplifier

R Ev g v g

T 1 T 2R B R B

R C

+ VC C

- VC C

R C+ ic 1

+ ie 1

i

=

c

t.

e

- ie 2

+ vb 1

- ic 2- vo 1 + vo 2


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+ v / 2g - vg

+ ie 1 - ie 2

R E

R BR B

M i r r o r p

Then the total a.c. current that flows through the resistor RE is zero. That means the emitters haveconstant potential, as being grounded, from the a.c. standpoint of view.


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19.Amplificator diferential in mod diferential

20.Amplificator diferential in mod comun


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21. Reacia n electronic; definiie; tipuri de reacie;


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22. Avantajele reaciei negative. 07_Feedback Amplifiers 5-8


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23.Reactia pozitiva(Bauchauser)

oL

L

i

LL

i

ov

rz

kz

v

iz

v

vA

+===

( )

321

312

zzz

zzzzL

++

+= ofi vvv ==

31

1

zz

z

+=

( ) ( )312321

211zzzzzzr

zkz

o ++++

=

0321 =++ XXX

1

2

X

Xk=


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24.Oscilator LC

Oscilator Hartley/Collpitz+ V

c c

Rc

R B 1

RB 2

RE

j L 1 j L 2

1 / j C

H a r t l e y o s c i l l a t o r

+ Vc c

Rc

RB 1

RB 2

RE

j L

1 / j C1 1 / j C2

C o l l p i t z o s c i l l a t o r

( )CLLCjjLjL

21

2

21

10

1

+==+

LCC

CCCj

CjjL

+

==

21

21

2

21

10

11


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25.Oscilator RC

Oscilator RC cu punte Wien. Condiia de oscilaie.


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26.Circuite basculante

a)astabil

Circuitul astabil. 08_The Astable (free-running) Multivibrator 1-4

R C 1 R C 2

T ( o n )1 T ( o f f )2

C 2C 1

R B 2

I B 2 ( d e s c )

R B 1

+ VC C

+ VC C- VC C+ VC E (s a t )

R C 1 R C 2

T ( o n )1 T ( o n )2

C 2C 1

R B 2

I B 1 ( in c a r )

R B 1

+ VC C

- VC C+ VC E ( s a t )


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b)monostabilCircuitul monostabil. 08_The Astable (free-running) Multivibrator 5

c)bistabil.Circuitul bistabil. 08_The Astable (free-running) Multivibrator 6-7


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27. Amplificator de putere clas A. 06_Power Amplifiers 4-6

+ VC C

R L

v i

F i g . 5 . 2

R B 1

R B 2

R E

C E

I C

I Q

V QV p e a k

I p e a k

I m a x

I m i n

V

=

V

m

in

s

a

t

V

=

V

m

a

x

C

C

Q A

V C E


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9. Amplificator de putere clas B. 06_Power Amplifiers 7-9I C

V p e a k

I p e a k

I m a x

I m i nV

=

V

m

in

s

a

t

V

=

V

m

a

x

C

C

Q B

V C E

+ VC C

- VC C+

-

T 2

R L

i c 1

i c 2

v i

T 1

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