EE105 – Fall 2014
Microelectronic Devices and Circuits
Prof. Ming C. Wu
wu@eecs.berkeley.edu
511 Sutardja Dai Hall (SDH)
1
Amplifier Transfer Functions
Av (s) =
Vo (s)
Vs (s)
s = σ + jω
Av(s) = Frequency-dependent voltage gain
€ V (s) = Laplace Transforms of input and output
Vo(s) and
s
voltages of amplifier,
Av (s) = K
(s+ z1)(s+ z2 )...(s+ zm )
(s+ p1)(s+ p2 )...(s+ p3 )
(In factorized form)
(-z1, -z2,…-zm) = zeros (frequencies for which transfer function is zero)
€
(-p1, -p2,…-pm) = poles (frequencies for which transfer function is
infinite)
(In polar form)
Av ( jω) = A ( jω)∠Av ( jω )
v
Bode plots display magnitude of the transfer function in dB and the
€ degrees (or radians) on a logarithmic frequency scale.
phase in
2
1
Complex Transfer Functions
Amplifier has 2 frequency ranges with
constant gain. The mid-band region is
always defined as region of highest
gain and cutoff frequencies are
defined in terms of midband gain.
Av ( jω L ) = Av ( jω H ) =
Av (s) = #
%
$
#
%
$
Ks s+ω
&
(
2'
#
%
$
s+ω s+ω 3 &('#%$ s+ω 4 &('#%$ s+ω 5 &('
#
%
2
For widely spaced poles as in the
€
figure,
&#
(%
1 '$
ωH ≅ ω4 and ωL ≅ ω3,
Amid s#%$ s+ω 2 &('
Av (s) =
Amid
&#
&
s ((%% s ((
+1
+1
%% ω
((%% ω
((
$ 4
'$ 5
'
BW= f 4 − f3 =
s+ω1 &('#%$ s+ω 3 &('%
€
ω4 −ω3
2π
3
Low-Pass Amplifiers
•  Amplifies signals below a cut-off frequency,
including dc.
•  Most operational amplifiers are designed as low
pass amplifiers.
•  Simplest (single-pole) low-pass amplifier is
described by
Av ( s ) = Ao
ωH
Ao
=
s + ω H 1+ s
ωH
–  Ao = low-frequency gain or mid-band gain
–  ωH = upper cutoff frequency or upper half-power point of
amplifier.
4
2
Low-pass Amplifier
Magnitude Response
Av ( jω ) =
Aoω H
Aω
Ao
= o H =
2
2
jω +ω H
ω2
ω +ω H
+1
2
ωH
For ω <<ωH , Av ( jω ) = Ao
dB
For ω >>ωH, Av ( jω ) = Aoω H
ω
A
o
For ω =ωH , Av ( jω ) =
2
⇒ ω H : 3-dB Bandwidth in Rad /s
Low-pass
filter
symbol
⇒ fH =
ω H : 3-dB Bandwidth in Hz
2π
•  Gain is unity (0 dB) at ω = AoωH = ωT called gain-bandwidth product
•  Bandwidth (frequency range with constant amplification ) = ωH (rad/s)
5
Arc Tangent Phase Response
If Ao positive: phase angle = 00 at dc
(If Ao negative: phase angle = 1800)
Response of tan-1(ω/ωC)
At ωC: phase = 450
One decade below ωC: phase = 5.70
One decade above ωC: phase = 84.30
Two decades below ωC: phase = 00
Two decades above ωC: phase = 900
LPF Response:
%
(
Ao
−1' ω *
∠Av ( jω) =∠
=∠A
−tan
o
'ω *
ω
& H)
1+ j
ωH
6
€
3
RC Low-pass Filter
R2
sC
Vo = Vs
R2 +
R1 +
R2 +
1
Impedance of capacitor =
sC
ωH =
1
sC
R2
sC
1
sC
!
#
!
$
Vo
R2 # 1
=#
&
Vs " R1 + R2 %# 1+ s
#
" ωH
1
( R1 R2 ) C
$
&
&
&
&
%
7
High-pass Amplifiers
•  Combines a single pole with a zero at the origin.
•  Simplest high-pass amplifier is described by
Av (s) =
Ao s
s+ω L
•  ωL = lower cutoff
frequency or lower half-power
€
point of amplifier.
8
4
High-pass Amplifier
Magnitude and Phase Response
Av ( jω ) =
AO jω
Ao ω
Ao
=
=
2
2
jω +ω L
ω +ω L 1+ω L2
2
ω
For ω >>ωL, Av ( jω ) = Ao
For ω <<ωL, Av ( jω ) =
Aoω
ωL
For ω =ωL, Av ( jω ) = Ao
2
High-pass
filter
symbol
•  Bandwidth (frequency range with constant amplification ) is infinite
•  Phase response is given by
∠Av ( jω ) =∠
Ao jω
jω + jω L
%
(
=∠Ao + 900 − tan−1''ωω **
&
L)
9
€
RC High-pass Filter
Vo =Vs
R2
1
R1 + +R2
sC
"
R2 %'"$ s %'
∴ Vo = $$
'$
'
Vs $# R1 +R2 '&$# s+ω L '&
1
&
% R + R (C
$ 1
2'
ωL = #
€
10
5
Band-pass Amplifiers
•  A band-pass characteristic is obtained by combining
high-pass and low-pass characteristics.
•  Transfer function of a band-pass amplifier is given by
Av (s) =
Ao sω H
(s+ω L )(s+ω H )
= Ao
s
1
#
(s+ωL ) % s &(
+1(
%ω
$ H
'
•  An ac-coupled amplifier has a band-pass characteristic:
€ –  Capacitors added to circuit cause low frequency roll-off
–  Inherent frequency limitations of solid-state devices cause
high-frequency roll-off.
11
Band-pass Amplifier
Magnitude and Phase Response
•  The mid-band range of frequencies is given by
ωL ≤ ω ≤ ωH
Av ( jω ) ≅ Ao
•  Transfer characteristic is
€
Av ( jω) =
€
€
Ao ( jω )ω H
( jω +ω L )( jω +ω H )
=
Aoωω H
2
(ω +ω L2 )(ω 2 +ω H2 )
12
6
Band-pass Amplifier
Magnitude and Phase Response (cont.)
At both ωH and ωL, assuming ωL<<ωH
Av ( jω L ) = Av ( jω H ) = Ao
2
Bandwidth = ωH - ωL
The phase response is given by
∠Av ( jω ) =∠
Ao jω
jω + jω L
%
(
%
=∠Ao + 900 − tan−1''ωω ** − tan−1''ωω
&
L)
&
(
*
*
H)
€
13
Narrow-band or High-Q Band-pass
Amplifiers
•  Gain maximum at center frequency ωo and
decreases rapidly by 3 dB at ωH and ωL.
•  Bandwidth defined as ωH - ωL, is a small
fraction of wo with width determined by:
Q=
ωo
ω H −ω L
=
fo
=
fH − fL
fo
BW
•  For high Q, poles will be complex and
€
s
Av (s) = Ao
s +s
2
ωo
Q
ωo
+ωo2
Band-pass
Q
filter
symbol
•  Phase response is given by:
€
%
(
*
∠Av ( jω ) =∠Ao + 900 − tan−1'' 1 ωω
' Q ω 2 − ω 2 **
&
)
o
O
14
€
7
Band-Rejection Amplifier or Notch
Filter
•  Gain maximum at frequencies far from
ωο and exhibits a sharp null at ωo.
•  To achieve sharp null, the transfer
function has a pair of zeros on the jω
axis at notch frequency ωo , and the
poles are complex.
Av (s) = Ao
Bandreject
filter
symbol
s2 +ωo2
ω
s2 + s o +ωo2
Q
•  Phase response is given by:
€
%
(
&
)
%
1 (*%' ωωo (*
*
' Q *''ω 2 − ω 2 **
)
& )& o
∠Av ( jω ) =∠Ao +∠'ω o2 −ω 2 * − tan−1''
€
15
All-pass Function
•  Uniform magnitude response at all frequencies.
•  Can be used to tailor phase characteristics of a
signal
•  Transfer function is given by:
s−ω
Av (s) = Ao s+ωo
o
•  For positive Ao,
Av ( jω ) = Ao
€
%
ω (*
∠Av ( jω ) = −2tan−1''ω
*
&
o)
16
€
8
Op Amp Building Blocks:
Generalized Inverting Amplifier
17
Op Amp Building Blocks:
Single-pole Low-Pass Filter
R2
sC
AV =
vo
=−
vi
ωH =
1
R2C
"
1
$
sC = − R2 "$ 1 %' = − R2 $ 1
R1
R1 # 1+ sR2C &
R1 $ 1+ s
$
# ωH
R2 +
%
'
'
'
'
&
18
9
Op Amp Building Blocks:
Integrator
Since iC = ii
1
∫ dv = ∫ − RC v (τ ) dτ
o
i
1
RC
vo ( 0 ) = VC ( 0 )
∴ vo (t ) = −
•  Feedback resistor R2 in the
inverting amplifier is replaced by
capacitor C.
∫ v (τ ) dτ + v (0)
i
o
•  Output voltage is
proportional to the
integral of the input
•  The circuit uses frequency
-dependent feedback.
ii =
vi
R
iC = −C
dvO
dt
19
Op Amp Building Blocks:
Differentiator
iR = −
vO
R
Since iR = ii :
•  Input resistor R1 in the inverting
amplifier is replaced by capacitor C.
ii = C
dvi
dt
vO = RC
dvi
dt
Output is proportional to
the derivative of input
voltage.
•  Derivative operation emphasizes
high-frequency components of
input signal, hence is less often
used than the integrator.
20
10