Lecture 10
Frequency response

1

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topics
• Bode diagram
• BJT’s Frequency response
• MOSFET Frequency
response

2

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vi (t ) = Vi sin ω t

Amplifier

Vo
Magnitude:

Vi
Vo ( s )
T ( s) =
Vi ( s )

Magnitude:

Vi ( jω )

Phase:

φ

s = σ + jω ⇒ s = jω

Vo (ω )
T (ω ) =
Vi (ω )

Vo ( jω )

v o (t ) = Vo sin( ω t + φ )

Steady-state response

∠Vo ( jω )
Phase: ∠V ( jω )
i
3


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dB (decibel)= Decimal + Bell

dB ≡ 10 log10

p1
p2

2
v
Q p = i2z =
z

⇒ dB = 20 log10

v1
i
= 20 log10 1
v2
i2

Human hearing frequency zone : 10Hz~24kHz
Most Sensitive frequency zone : 2kHz~5kHz

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Logarithmic coordinate
Decade : dec = log10

ω2
ω1

Octave : oct = log 2 ω 2
ω1

dB

ω
ω0
1

2

3 4

10

20


Log scale

100

5

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Example(low pass)
+
Vi
−

+

R

C

Vo
−

1

1

sC V ⇒ T (ω ) =
i
1 + ( jω ) RC
R+ 1
sC
1
1
1
let ω 0 =
⇒ T (ω ) =
=
RC
1 + ( jω ) RC 1 + ( jω )

Vo =

ω0

1
2
⇒ ∠T (ω ) = 0 − tan −1 1 = −450

if

ω = ω 0 ⇒ T (ω ) =

6

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T (ω ) =

1
1+ (

Magnitude:
ω
ω
(1 + j ) −1 = −20 log 1 + ( ) 2
ω0 dB
ω0
= −10 log[1 + (

ω 2
) ]
ω0

ω
ω << ω 0 ⇒
≈0
ω0
⇒ dB = −10 log 1 = 0
Phase:

s


ω0

=
)

1
jω
1+ ( )

ω0

ω ω
ω >> ω 0 ⇒ 1 + j
≈
ω0 ω0
ω
⇒ dB ≈ −20 log
ω0
dB = −[20 log ω − 20 log ω 0 ]

ω = ω 0 ⇒ 1 + j1 ⇒ dB = −10 log 2 = −3.01

ω
−1 ω
0
∠(1 + j ) = 0 − tan
ω0
ω0

ω

ω << ω 0 ⇒
≈ 0 ⇒ ∠T (ω ) ≈ tan −1 0 = 0o
ω0
ω
ω >> ω 0 ⇒
≈ ∞ ⇒ ∠T (ω ) ≈ − tan −1 ∞ = −90o
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7


T (ω ) ( dB )

0dB

T (ω ) =

0 .1

1

ω0
ω0 + s

ω
ω0


10

− 20dB / decade
∠ T (ω )

180 0

ω = ω 0 ⇒ − 45 0

90 0

ω
ω0

− 90 0
− 180 0

8

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9

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Example(high pass)
+
Vi
−
Vo =

+
Vo
−

C
R
R
R+ 1

Vi ⇒ T (ω ) =
sC

jωRC
1 + ( jω ) RC
(

jω

)


ω0
jωRC
1
let ω 0 =
⇒ T (ω ) =
=
RC
1 + ( jω ) RC 1 + ( jω )

1
2
⇒ ∠T (ω ) = 90 − tan −1 1 = 450

if

ω0

ω = ω 0 ⇒ T (ω ) =

10

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s
T ( s) =
s + ω0


jω
T (ω ) =
jω + ω 0

Magnitude:
jω
= 20 log ω − 20 log ω 2 + ω 02
jω + ω 0 dB

ω
ω0
dB = 20 log ω − 20 log ω 0
ω = 0.1ω 0 ⇒ dB = 20 log 0.1 = −20

= 20 log ω − 10 log[ω 2 + ω 02 ]
ω >> ω0 ⇒ 20 log

ω << ω 0 ⇒ dB ≈ 20 log

1
1 + ωjω0

⇒ −20 log(1 + ωjω0 )
⇒ dB = −20 log 1 = 0

Phase:
jω
ω
∠

= 90 − tan −1
jω + ω 0
ωo
ω >> ω 0 ⇒
ω << ω 0 ⇒

ω = ω0 ⇒

1
⇒ dB = −20 log 2 = −3.01
1 + j1

ω
≈ ∞ ⇒ ∠T (ω ) ≈ 90 − tan −1 ∞ = 0o
ω0

ω
≈ 0 ⇒ ∠T (ω ) ≈ 90 − tan −1 0 = 90o
ω0

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T (ω ) ( dB )


s
T ( s) =
s + ω0

+ 20dB / decade

0dB

0 .1

∠ T (ω )

180

0

1

10

ω
ω0

ω = ω 0 ⇒ 45 0

90 0

ω
ω0


− 90 0

− 180 0
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dB ≡ 20 log T (ω )
The bandwidth represents the distance between the two
points in the frequency domain where the signal is 1 2
of the maximum signal strength.

20 log T (ω )
if

1
ω = ω 0 ⇒ T (ω ) =

2

1
if T (ω ) =
2
⇒ 20 log T (ω ) = 10 log 2 = 3dB

14

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Y ( s)
k ( s + z1 )( s + z 2 ) L
=
2
R( s ) ( s + p1 )( s + p2 )( s + as + b) L
GH (dB)

Case I : k

ω

Magnitude:
0. 1

1


10

k dB = 20 log k (dB)
∠GH

1800

Phase:

⎧ 0
∠k = ⎨ o
⎩180
o

,k f 0
,k p 0

900

ω
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Case II :


1
sp

Magnitude:
1
( jω ) p

0. 1

dB

1
( jω ) p

p=2

p =1

= −20 p log ω (dB)

ω
1

10

∠GH

Phase:
∠


GH (dB)

= (−90 o ) × p

900
− 90

0

− 1800

ω

p =1
p=2

16

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Case III : s p
p=2

GH (dB)


Magnitude:
( jω ) p

dB

ω

= 20 p log ω (dB)
0. 1

∠GH

Phase:
∠( jω ) = (90 ) × p
p

p =1

180

0

90

0

1

10


p=2
p =1

ω

o

− 900
− 1800
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a
1
−1
or
(
s
+
1
)
Case IV : ( s + a)
a

Magnitude:

(1 + j

ω
a

a =1

GH (dB)

ω

) −1
dB

= −20 log 1 + ( ) 2
a

ω

ω

= −10 log[1 + ( ) 2 ]
a
ω
ω pp a ⇒ ≈ 0 ⇒ dB = −10 log1 = 0
a
ω ω
ω
ω ff a ⇒ 1 + j ≈ ⇒ dB ≈ −20 log
∠GH

a a
a

1800

dB = −[20 log ω − 20 log a]

ω = a ⇒ 1 + j1 ⇒ dB = −10 log 2 = −3.01

Phase:
ω

∠(1 + j

a

) = 0 0 − tan

0. 1

1

10

ω = a ⇒ −450

900

−1 ω


− 900

a

− 1800

ω

ω
ω pp a ⇒ ≈ 0 ⇒ ∠GH ≈ tan −1 0 = 0o
a
ω
Circuit by
ω ff a ⇒ ≈ ∞ ⇒ ∠GH ≈ − tan −1 ∞ =Microelectric
−90o

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Case V : ( s + a) or ( 1 s + 1)
a

Magnitude:

(1 + j

ω
a

a =1

a

GH (dB)

ω

)
dB

= 20 log 1 + ( ) 2
a

ω

ω

= 10 log[1 + ( ) 2 ]
a
ω
ω pp a ⇒ ≈ 0 ⇒ dB = 10 log1 = 0
a
ω ω
ω

ω ff a ⇒ 1 + j ≈ ⇒ dB ≈ 20 log
a a
dB = 20 log ω − 20 log a

a

ω = a ⇒ 1 + j1 ⇒ dB = 10 log 2 = 3.01

Phase:
∠(1 + j
ω pp a ⇒
ω ff a ⇒

ω
a

ω
a

ω
a

) = tan

0. 1

∠GH

1800


10

ω = a ⇒ 450

900

−1 ω

− 900

a

− 1800

≈ 0 ⇒ ∠GH ≈ tan −1 0 = 0o

1

ω

19

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ω n2
T (s) = 2
s + 2ξω n s + ω n2

Case VI :
T ( jω ) =

T ( jω ) =

ω n2
(ω n − ω 2 ) + 2 jξω nω
2

1

ω 2
ω
(1 − ( ) ) + j 2ξ
ωn
ωn


∠ T ( jω ) = − tan −1

2ξωω n
2
(ω n − ω 2 )

ω
2ξ
ωn
−1
∠ T ( jω ) = − tan
ω 2
1− ( )
ωn

ω
⎧
ω
pp 1
,
0
,
pp
1
⎪
0
ωn
ω
⎧
0

n
⎪
ω
⎪
ω
⎪
0
=1
T ( jω ) = ⎨ − 20 log(2ξ ) ,
= 1 ∠T ( jω ) = ⎨ − 90 ,
ωn
⎪
⎪− 180o ω n
ω
ω
⎩
⎪
ω
−
40
log(
)
,
ff
1
,
ff 1
⎪
ω
ω

n
n
⎩
ωn

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ω = ωn

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BJT high frequency model

Splitting resistance (refinement the lumped-component circuit)
Depletion capacitance

Cπ >> Cμ


Emitter-base capacitance
= diffusion capacitance + Base-Emitter junction capacitance

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KTC9013 Technical data

How to find

Cπ by datasheet ?

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1
sC μ

h fe ≡


T (ω ) =

Ic
I
⇔β= C
Ib
IB

1+ (

Vπ
I c = g mVπ −
= ( g m − sC μ )Vπ
1
sCμ
Ib =

1
)

1
ωβ =
(Cπ + Cμ )rπ

β 0 = g m rπ

g m − sC μ

⇒ h fe ≈


ω0

3-dB frequency

Vπ
(rπ // Cπ // C μ )

Ic
h fe ≡ =
I b 1 rπ + s (Cπ + Cμ )

s

Low frequency ß

β0
g m rπ
=
1 + s (Cπ + Cμ )rπ Microelectric
1 + s (Cπ +Circuit
Cμ )rby
π

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