Problems for Amplifier Section
Lecture notes: Sec. 6
- F. Najmabadi, ECE65, Winter 2012
Problems for Amplifier Section Lecture notes: Sec. 6 F. Najmabadi, - - PowerPoint PPT Presentation
Problems for Amplifier Section Lecture notes: Sec. 6 F. Najmabadi, ECE65, Winter 2012 Exercise 1: Find the bias point and the amplifier parameters of the circuit below. (Si BJT with = 200, V A = 150 V, ignore Early effect in bias
V 95 . 4 9 k 18 k 22 k 22 k 90 . 9 k 18 || k 22 = × + = = =
BB B
V R
A 3 . 20 / mA 05 . 4 10 7 . ) 1 /( 10 9 . 9 4.95 4.95 : KVL BE
3 3
µ β β = = ≈ = + + + × = + + = −
C B C E E E E E BE B B
I I I I I I R I V R I
CE C BE
V 0.7 V 5 10 10 4 9 9 : KVL CE
3 3
= > = × × + = + = −
− D CE CE E E CE
V V V R I V V 95 . 4 k 9 . 9 = =
BB B
V R
k 28 . 1 r k . 37 10 x 4 150 r mA/V 156 10 26 10 4.05
3
3
= = = = = ≈ = × × = =
m B T
A
C m
g I V I V V I g β
π
3
− v L E
L E
L E
i
= =
π m
g
i L E
i
π
π π
B
= =
π m
g
6 3 2 2 2 6 3 1 1 1
− −
p c L
p c sig i p
2 1
p p p
V 22 . 2 15 k 9 . 5 k 34 k 9 . 5 k . 5 k 9 . 5 || k 34 = × + = = =
BB B
V R
A 2 . 14 / mA 84 . 2 510 7 . ) 1 /( 10 . 5 2.22 2.22 : KVL BE
3
µ β β = = ≈ = + + + × = + + = −
C B C E E E E E BE B B
I I I I I I R I V R I
CE C BE
V 0.7 V 5 . 10 ) 510 10 ( 10 84 . 2 15 15 : KVL CE
3 3
= > = + × × + = + + = −
− D CE CE E E CE C C
V V V R I V R I V 22 . 2 k . 5 = =
BB B
V R
k 83 . 1 r k 8 . 52 10 84 . 2 150 r mA/V 10.9 10 26 10 2.84
3
3
= = = = × = ≈ = × × = =
m B T
A
C m
g I V I V V I g β
π
k 83 . 1 r k 8 . 52 r mA/V 10.9
= =
π m
g Amplifier Parameters This is a CE amplifier with RE
i E B i
π
3 3
− − i
C
C E m L C m i
C
B E E
π
v i
i i sig
9 3 3 2 2 2 6 3 1 1 1
− −
p c L
p c sig i p
2 1
p p p
= =
π m
g
A . 5 / mA . 1 10 3 . 2 3 : KVL BE
3
µ β = = ≈ = + × = −
C B C E EB E
I I I I V I
EC C EB
V 0.7 V 4 . 1 10 10 6 . 4 6 3 10 3 . 2 10 3 . 2 3 : KVL CE
3 3 3 3
= > = × × − = − × + + × = −
− D EC EC C EC E
V V V I V I
k 26 . 5 r k 150 10 150 r mA/V 38.5 10 26 10
3
= = = = = ≈ = × = =
m B T C A T C m
g I V I V V I g β
π
3
− i
i i v i
C
i
= =
π m
g Hz 82 16 67
2 1
p p p
π π
B i
V 353 . 1 8 . 1 100 33 100 V
G
= + = ⇒ = k k k IG Assume Saturation
2 3 2
OV OV
n D
−
t OV GS
OV DS
2 3
− OV D
GS G S
D D D
S D DS
OV
D S t OV D S GS G
2 3 3
OV OV
−
2
OV OV
k 3 . 36 10 75 2 . 10 r mA/V 1.82 0.303 10 0.275 2 2
3
= × = = = × × = =
A
D m
I V V I g 45 . 3 45 . 3 ) || k 2 || k 3 . 36 ( 10 1.82 ) || || (
3
− i
i i v i
D
i
G i
G
V 7 . − = − =
E E BE
V V V
CE C BE
k 21 . 1 r k 9 . 34 10 4.3 150 r mA/V 16.5 10 26 10 4.3
3
3
= = = = × = ≈ = × × = =
m B T
A
C m
g I V I V V I g β
π
A 5 . 21 / mA 3 . 4 µ β = = ≈ =
C B C E
I I I I V 0.7 V 7 . 4 4
0 =
> = − =
D E CE
V V V
E
V
= =
π m
g 1 427 1 427 427 ) || || ( k 9 . 25 k 100 || || k 9 . 34 || || ) || || ( 1 ) || || ( ≈ + = = = = ∞ = + =
i
L E
L E
E
L E
i
π π
sig B E
i L E
i
π
= =
π m
g Cut-off frequency Hz 34 . 3 10 47 . ) 61 10 100 ( 2 1 ] [ 2 1
6 3 2
−
p c
p
2 2 6 4 4 4
− OV OV OV OV tp OV D SG D
2 OV
p D
t OV SG
OV DS
2 6
− OV D
S S SG
S SD
OV
3
= × × = = = × × = =
− − D OV D m
I V I g λ
m S
i L S
L S
v
6 3 3 2
−
p c
p
2 2 6 4 4 4
− OV OV OV OV D t OV D GS G
2 OV
n D
t OV GS
OV DS
2 6
− OV D
4 4
DS D DS D
OV
G
G
3
− sig S m
D S i v L S
v
2 1 2 2 1 1
p p p c
p c sig i p
k 250 10 0.4 01 . 1 1 r mA/V 0.80 1 10 0.4 2 2
3
= × × = = = × × = =
− − D OV D m
I V I g λ
3 2
n D
t OV GS
OV DS
OV
S
D
GS S S G GS
S D DS
m S
i L S
L S
v
6 3 2
−
p c
p
k 141 10 0.71 01 . 1 1 r mA/V 1.42 1 10 0.71 2 2
3
= × × = = = × × = =
− − D OV D m
I V I g λ Signal circuit Real circuit