ele222-21604-hw7-cozumlu
TRANSCRIPT
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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!