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INTRODUCTION TO ELECTRONICS (21604)

HOMEWORK #7

Assoc.Prof.Dr.İnci Ç

İLES

İZDUE DATE and TIME 2 May 2003 - 12.00

SOLUTIONS:

Problem *4.85:  From the circuit on p. 344 of the book it is very obvious that

. Thus E  Bc   I  I  I mA   =+=1   mA I  I   E C  99,01

=+

= β 

 β . Also V V  I V k  I   E  BE  BC  77,1175100   =++= .

For AC analysis take the T-model in Fig. 4.27

on p. 261 with Ω== 251

m

e g 

r 

)175(

 as shown on the

right. )175(   Ω+=Ω+==   ebembi   r vvv ee   g r i .

100k

C, vo

gm v be

or

αie 

B

r e 

175 Ω

E

+

vi 

-

+v be 

-

10k

Now try to solve for vo  assuming there is nocurrent flow through 100k : bemo   v g k v   ⋅−= 10 ,

thus, 50175

10−=

Ω+==

ei

o

r 

k 

v

v A . With the current

through 100k taken into consideration, the

calculations are a bit more complicated becausethe current through 10k is not -αie = -0,99 ie,

but, ei⋅−=ee

k    ik k 

k r 

+i

  ⋅−Ω+=

10100

10017510 9,0

α .

Subsequently, 45)175(

10 10 −≅Ω+

⋅==

ee

k 

i

o

r i

ik 

v

v A .

Problem **4.95:  This is the bootstrapped follower that has special properties we

will see as we solve the problem.  (a) To analyze DC conditions, we first need to (1)remember that all capacitors are open circuited and then (2) find the Thevenin

equivalent of the two 20k resistors in series connected to 9 V power supply.V V 

k k 

k V  BB 5,49

2020

20=

+=   and k k k  R BB 1020||20   == . From VBB  via RBB  and 10k base

resistor, over BE junction of the transistor via 2k resistor to ground one can write thefollowing loop equation 02)10(   =++++   V k  R I   BE  BB B−   k  I V   E  BB . Using the values provided

and presumed known mA I  B E  77,1)1(   =+= I  A I  B 5,(

⇒k k k 

V V 17

2101)1010

6,05,4=

⋅++

−=   µ β  . Also,

4108,1423)1;   k r r  ee   ≈Ω=+=   β (;1,14   r  I 

V 

 E 

T  =Ω=   Π/70   V mAV 

 I 

V 

 I  g 

T 

 B

T 

C m   ===

  β .

GOOD LUCK!

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(b) To analyze AC conditions, look atthe equivalent circuit on the right. It isobvious that the 4 resistors can betaken account as 2 resistors in series.

To find x

 xi

i

v R   = , we have to calculate the

current i10k through 10k resistor in RBE;ib flows through rП.and .k b x   iii 10+=

 

R i

E, vo

B

R BB 

10k +

v be 

-

2k gm v be

10k 

+

vs 

-

r Π 

C = GND

R BE 

R EC With b

bek    i

k 

r 

k 

vi

101010

Π==  one can calculate

bbe x   ik 

r v

r k i

+=

+=   Π

Π

110

1

10

1.

 EC  Rbe Rbe x   Rivvvv  EC  EC  +=+=  with bem R   v g r k 

 EC   

++=

Π

1

10

1i  resulting in

11111

10

1

1

10

11

k 

r k 

 R g r k 

 R

 EC m

i   =

+

⋅

+++

=

Π

Π . Note that, (1) Ri is much larger than RBE+REC. (2)

Ri>>10k, (3) RBE≈REC. We findi

o

i

i

i

o

o

i

i

o

 s

ov

v

v

k  R

 R

v

v

v

v

v

v

v

v A 917,0

10=

+⋅=⋅==  with obei   vvv   +=  

and . Thus, EC  Ro   vv   = 99,01

10

11

1

10

1

=

+++

++

=+=

Π

Π

m EC 

m EC 

 Rbe

 R

i

o

 g r k 

 R

 g 

r k 

 R

vv

v

vv

 EC 

 EC 

. This was expected,

because this is a common-collector circuit. Finally, 91,0=⋅==o

i

i

o

 s

ov

v

v

v

v

v

v A .

(c)

 BB

k beb

k be

 BB

 xb

 x

 x

 xi

 Rk 

vvi

vv

 Rk 

vi

v

i

v R

+

++

+=

++

==

1010

2

2  

with

+⋅=+= Πmbebembk    g r vk v g ik v

1

2)(22 .

1818

20

2)1(1

2)1(k 

k 

k r  g r 

k r  g r 

m

m =++

+

++

ΠΠ

ΠΠ Ri  =

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GOOD LUCK!