Osnovi elektronike – Odsek SI Februar...

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Osnovi elektronike – Odsek SI Februar 2017. REŠENJA 1. a) 3 2 5 X r X R R R a R = ⇒ = + = b) 2 0 (3 2 ) 3 2 X U U X U U X X I I U X X I r U X v v v v v R R v v v v R v v R R R R v R a v R - = ⇒ = - + + = ⇒ = + = = + c) Kada pojačanje a →∞ tada je 5 1/ 1/5 r a β β = = ⇒ = Kada je pojačanje 3 a = tada je 3 15 / 8 1 1 3/5 r a a a β = = = + + d) I max min min 2 3 2 2k 3 CC m X CC X m R R v V V R V R V = = + ⇒ = = Ω - 2. 27 , 2 ] V [ 3 1 ] V [ + = G I v v , za V 36 , 1 V 5 , 1 ≤ ≤ - G v (D - ON) ] V [ 2 ] V [ G I v v = , za V 5 , 1 V 36 , 1 ≤ ≤ G v (D - OFF) 3. a) kΩ 4 , 4 2 = R ; kΩ 1 3 = R ; kΩ 5 4 = R . b) 8533,3 = v a c) ∞ → ul R , Ω = k 5 izl R . 4. a) Pošto su kolektorske struje u odnosu 1:10:100, bazne struje u DC analizi utiču sa 10% i mogu da se zanemare. R 1 =7k, R 3 =1.2k, R 2 =12.7k b) Ω = = = = Ω = = = = Ω = = = = = = 0 25 400ms mV 25 / A 10 / k 5 . 2 40ms mV 25 / A 1 / k 25 4ms mV 25 / µA 100 / 4 4 4 3 3 3 2 1 1 2 1 π π π π r m V I g r m V I g r r V I g g T C m T C m T C m m ( 29 ( 29 1700 1 ) 1 ( || || 2 1 3 4 3 4 3 0 4 2 3 3 1 1 2 ≈ + + + - - = ≈ ≈ 43 42 1 4 4 4 3 4 4 4 2 1 R g R g R r R g r R g a m m R m m β π π c) 1 2 π r d) Ω ≈ + 130 1 || 0 2 4 3 β π R r R 5. 1 M -triodna oblast; 2 M -zasićenje; V 025 , 1 = I V .

Transcript of Osnovi elektronike – Odsek SI Februar...

Page 1: Osnovi elektronike – Odsek SI Februar 2017.tnt.etf.bg.ac.rs/~si1oe/Ispiti/Februar_2017_resenja.pdf · v R a v R − = ⇒ = − ... vG ≤1,36 (D - ON) vI [V] =2vG [V] , za 1,5V1,36

Osnovi elektronike – Odsek SI Februar 2017.

REŠENJA

1. a) 3 2 5X rX

RR R a

R= ⇒ = + =

b)

2

0 (3 2 )

3 2

X U UX U

UX X II U

X X

Ir

U X

v v vv v

R Rvv v v R

v vR R R R

v Ra

v R

−= ⇒ =

−+ + = ⇒ = +

= = +

c) Kada pojačanje a → ∞ tada je 5 1/ 1/ 5ra β β= = ⇒ =

Kada je pojačanje 3a = tada je 315 / 8

1 1 3 / 5ra

aaβ

= = =+ +

d) Imax minmin

23 2 2k

3CC m X

CCX

m

R Rv V V R

VRV

= = + ⇒ = = Ω

−

2. 27,2]V[3

1]V[ += GI vv , za V36,1V5,1 ≤≤− Gv (D - ON)

]V[2]V[ GI vv = , za V5,1V36,1 ≤≤ Gv (D - OFF)

3. a) kΩ4,42 =R ; kΩ13 =R ; kΩ54 =R .

b) 8533,3=va

c) ∞→ulR , Ω= k5izlR .

4. a) Pošto su kolektorske struje u odnosu 1:10:100, bazne struje u DC analizi utiču sa 10% i mogu da se zanemare. R1=7k, R3=1.2k, R2=12.7k b)

Ω====

Ω====

Ω======

025

400msmV25/A10/

k5.2

40msmV25/A1/

k25

4msmV25/µA100/

4

44

3

33

21

121

π

π

ππ

r

mVIg

r

mVIg

rr

VIgg

TCm

TCm

TCmm

( ) ( ) 17001

)1(||||2

1

34

343042331

1

2

≈

+

++−

−=

≈≈ 43421

444 3444 21 Rg

RgRrRgrR

ga

m

m

R

mm βππ

c) 12 πr

d) Ω≈++

1301

||0

243 β

π RrR

5. 1M -triodna oblast; 2M -zasićenje; V025,1=IV .

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