Freq response of CE and CC discrete circuits

8. Frequency Response
Reading: Sedra & Smith: Sec. 1.6, Sec. 3.6 and Sec. 9
(MOS portions),
(S&S 5th Ed: Sec. 1.6, Sec. 3.7 (capacitive effects), Sec.
4.8, Sec. 4.9, ,Sec. 6. (Frequency response sections,
i.e., 6.4, 6.6, …), Sec. 7.6

Typical Frequency response of an Amplifier
F. Najmabadi, ECE102, Fall 2012 (2/59)
 Up to now we have “ignored” capacitors in circuits & computed mid-band
properties. We have to solve the circuit in the frequency domain in order
to see the impact of capacitors (a typical response is shown below):
o Lower cut-off frequency: fL
o Upper cut-off frequency: fH
o Band-width: B = fH − fL

Observation on the
frequency response of an Amplifier
Observations:
 Analytical solution of amplifiers in frequency domain is complicated!
 Response (e.g., gain) of an ideal linear amplifier should be independent
of frequency (otherwise signal “shape” would be distorted by the
amplifier). Thus:
o A practical amplifier acts as an ideal linear amplifier only for a range of
frequencies, called the “mid-band”.
o The lower and the upper cut-off frequencies (fL and fH) identify the
frequency range over which the amplifier acts linearly.
o Amplifier response at high frequencies (near the upper cut-off frequency ,
fH) is important for stability considerations (gain and phase margins).
 Thus, we are mainly interested in mid-band properties (where
capacitors can be ignored) and in poles and zeros of the amplifier
response (due to capacitors).

What do we mean by “capacitors can be ignored?”
 Capacitor impedance depends on the frequency: |Z| = 1/(jωC).
o At high frequency |Z| → 0: capacitor acts as a short circuit.
o At low frequency |Z| → ∞: capacitor acts as an open circuit.
 For the above two limits, circuit becomes a “resistive” circuit and
we do NOT need to solve the circuit in the frequency domain.
 Thus, ignoring capacitors means that we operate at either a high
enough or at a low enough frequency such that capacitors become
either open or short circuits, leading to a “resistive” circuit.
o Note that the circuit is modified by the presence of the capacitors
(e.g., elements may be shorted out).
 But “high” and/or “low” frequency compare to what?

Capacitor behavior depends
on the frequency of interest.
)/1(|||| CRZ ω=
Capacitor approximates an open circuit at low frequencies:
)/1()/1(
)/1(||||)/1(
RCCR
RCRZCR
<<→<<
≈=→<<
ωω
ωω
Capacitor approximates a short circuit at high frequencies:
)/1()/1(
0)/1()/1(||||)/1(
RCCR
CCRZCR
>>→>>
→≈=→>>
ωω
ωωω
We cannot ignore the capacitor when
This defines the reference frequency for high-f and low-f
)/1(~)/1(~ RCCR ωω →
Note: The above circuit is like a low-pass filter with a cut-off frequency of 1/RC
Example:

Finding Frequency response of amplifiers
 Capacitors typically divide into two groups: low-f capacitors
(setting fL) and high-f capacitors (setting fH).
o We need to identify low-f and high-f caps. We will use absolute
limits of f = 0 (ALL capacitors open) and f = ∞ (ALL capacitors short)
for this purpose.
 For bias (f = 0) all caps are open circuit!
 For “mid-band” properties (fL << f << fH)
o Low-f capacitors will be short circuit (because fL << f ).
o High-f capacitors will be open circuit (because f << fH).
o The resulting “resistive” circuit gives mid-band properties.
 We will use time-constant method to find fL and fH (separately)
o To find fL all high-f capacitors will be open circuit (because fL << fH)
o To find fH all low-f capacitors will be short circuit (because fH >> fL)

Impact of various capacitors depend on
the frequency of interest
f → ∞
All Caps are short.
This limit is used
to find high-
frequency Caps.
f → 0
All Caps are open.
This limit is used
to find low-
frequency Caps
Mid-band:
High-f caps are open
Low-f caps are short.
Computing fH :
High-f caps are included.
Low-f caps are short
Computing fL:
High-f caps are open.
Low-f caps included.
Impendence of capacitors (1/ωC)

How to find which capacitors contribute
to the lower cut-off frequency
Cc1 open:
vi = 0 → vo = 0
Contributes to fL
Example:
 Consider each capacitor individually. Let f = 0 (capacitor is open circuit):
o If vo (or AM) does not change, capacitor does NOT contribute to fL (i.e.,
it is a high-f cap)
o If vo (or AM) → 0 or is reduced substantially, capacitor contributes to fL
(i.e., it is a low-f cap)
CL open:
No change in vo
Does NOT contribute to fL

How to find which capacitors contribute
to the upper cut-off frequency
Cc1 short:
No change in vo
Does NOT contribute to fH
CL short:
vo = 0
Contributes to fH
 Consider each capacitor individually. Let f → ∞ (capacitor is short circuit):
o If vo (or AM) does not change, capacitor does NOT contribute to fH
(i.e., it is a low-f cap)
o If vo (or AM) → 0 or reduced substantially, capacitor contributes to fH
(i.e., it is a high-f cap)
Example:

Constructing appropriate circuits
Example:
Cc1 : Low-f capacitor
CL : High-f capacitor
Mid-band:
High-f caps are open
Low-f caps are short.
Computing fH :
High-f caps are included.
Low-f caps are short
Computing fL:
High-f caps are open.
Low-f caps included.

Low-Frequency Response

Low-frequency response of an amplifier
 Each capacitors gives a pole.
 All poles contribute to fL (exact value of fL from computation or simulation)
 A good approximation for design & hand calculations:
fL ≈ fp1 + fp2 + fp3 + …
 If one pole is at least a factor of 4 higher than others (e.g., fp2 in the above
figure), fL is approximately equal to that pole (e.g., fL ≈ fp2 in above within 20%)
321 ppp
M
sig
o
s
s
s
s
s
s
A
V
V
ωωω +
×
+
×
+
×=
Example: an amplifier with three poles
(Set s = jω to find Bode Plots)

Low-frequency response of a CS amplifier
(from detailed frequency response analysis)
Cc1 open:
vi = 0 → vo = 0
Cc2 open:
vo = 0
Cs open:
Gain is reduced substantially
(from CS amp. to CS amp. With RS)
)||||(
321
LDom
sigG
G
M
ppp
M
sig
o
RRrg
RR
R
A
s
s
s
s
s
s
A
V
V
+
−=
+
×
+
×
+
×=
ωωω
Lengthy calculations: See S&S pp689-692 for detailed calculations (S&S
assumes ro → ∞ and RS → ∞ )
,
)]1/()||[(||[
1
)||(
1
,
)(
1
2
2
3
1
1
omLDoSs
p
LoDc
p
sigGc
p
rgRRrRC
RrRCRRC
++
≈
+
=
+
=
ω
ωω
All capacitors contribute to fL (as vo is reduced when f → 0 or caps open circuit)

Finding poles by inspection
1. Set vsig = 0*
2. Consider each capacitor separately, e.g., Cn (assume all others are
short circuit!)
3. Find the total resistance seen between the terminals of the
capacitor, e.g., Rn (treat ground as a regular “node”).
4. The pole associated with that capacitor is
5. Lower-cut-off frequency can be found from
fL ≈ fp1 + fp2 + fp3 + …
* Although we are calculating frequency response in frequency domain, we
will use time-domain notation instead of phasor form (i.e., vsig instead of Vsig)
to avoid confusion with the bias values.
nn
pn
CR
f
π2
1
=

Example: Low-frequency response of a CS amplifier
(from pole inspection)
 Examination of circuit shows that ALL capacitors are low-f capacitors.
 In the following slides with compute poles introduced by each capacitor.
(Compare with the detailed calculations of slide 13!)
fL ≈ fp1 + fp2 + fp3

Low-frequency response of a CS amplifier (fp1)
1. Consider Cc1 :
∞
)(2
1
1
1
sigGc
p
RRC
f
+
=
π
2. Find resistance between
Capacitor terminals
Terminals of Cc1

1. Consider CS :
Terminals of CS
)]1/()||[(||[2
1
2
omLDoSS
p
rgRRrRC
f
++
=
π
Capacitor terminals
om
LDo
rg
RRr
+
+
1
||
om
LDo
rg
RRr
+
+
1
||
om
LDo
rg
RRr
+
+
1
||

1. Consider Cc2 : Terminals of Cc2
)||(2
1
2
3
oDLc
p
rRRC
f
+
=
π
Capacitor terminals

High-Frequency Response
o Amplifier gain falls off due to the internal
capacitive effects of transistors as well as
possible capacitors in the circuit.

Capacitive Effects in pn Junction
Forward Bias Reverse Bias
02 jj CC ⋅≈
T
DT
d
V
I
C
⋅
=
τ
0=dC
m
R
j
j
VV
C
C
)/1( 0
0
+
=
rD
Cj + Cd
 Charge stored in the pn junction, leading to a capacitance
 For majority carriers, stored charge is a function of applied voltage
leading to a “small-signal” junction capacitance, Cj
 For minority carriers, stored charge depends on the time for these
carriers to diffuse across the junction and recombine, leading to a
diffusion capacitance, Cd
 Both Cj and Cd depend on bias current and/or voltage.
 Junction capacitances are small and are given in femto-Farad (fF)
1 fF = 10−15 F
*See S&S pp154-156 for detailed derivations
High-f small signal
model of diode

Capacitive Effects in MOS*
1. Capacitance between
Gate and channel
(Parallel-plate capacitor)
appears as 2 capacitors:
between gate/source &
between gate/drain
3. Junction capacitance
between Source and Body
(Reverse-bias junction)
4. Junction capacitance
between Drain and Body
(Reverse-bias junction)
2. Capacitance between
Gate & Source and Gate &
Drain due to the overlap of
gate electrode
(Parallel-plate capacitor)
MOS High-frequency
small signal model
*See S&S pp154-156 for
detailed derivations

MOS high-frequency small signal model
For source connected to body
(used by S&S)
Accurate Model
(we use this model here)
Generally, transistor internal capacitances are
shown outside the transistor so that we can use
results from the mid-band calculations.

High-frequency response of a CG amplifier
gddbLL CCCC ++=′
sbgsin CCC +=
Cgs between
source & ground
“Input Pole” “Output Pole”
Low-pass
filter
Low-pass
filter
Mid-band
Amp
Cgd between
drain & ground

High-f response of a CG amplifier – Exact Solution (1)
gddbLL CCCC ++=′
sbgsin CCC +=
0)(
/1
:Node
0)(
/1
:Node
=
−
+−+
′
+
′
=
−
+−−+
−
o
io
im
L
o
L
o
o
o
oi
im
in
i
sig
sigi
i
r
vv
vg
Cs
v
R
v
v
r
vv
vg
sC
v
R
vv
v
Can be solved to find vo/vsig

)/1||(1
1
/1
/1
)1/(1
1
1
1
1
1
0)(:Node
msiginsigm
m
sig
i
sigmsiginsigmsiginsigmsig
i
isigminsigii
gRsCRg
g
v
v
RgRsCRgRsCRgv
v
vRgsCvvv
+
×
+
=
++
×
+
=
++
=
=++−
LL
Lm
i
o
imoL
L
o
o
RCs
Rg
v
v
vgvCs
R
v
v
′′+
′
=⇒=−′+
′ 1
0)(:Node
Voltage divider
(Ri= 1/gm and Rsig) “Input Pole”
“Output Pole”
Mid-band Gain
LLmsigin
Lm
sigi
i
sig
o
RCsgRsC
Rg
RR
R
v
v
′′+
×
+
×′×
+
=
1
1
)/1||(1
1
)(
Compact solution can be found by ignoring ro (i.e., ro → ∞)

Open-Circuit Time-Constants Method
b1 can be found by the open-circuit time-constants method.
1. Set vsig = 0
2. Consider each capacitor separately, e.g., Cj (assume others are open circuit!)
3. Find the total resistance seen between the terminals of the capacitor, e.g., Rj
(treat ground as a regular “node”).
4.
A good approximation to fH is:
12
1
b
fH
π
=
...
11
).../1)(/1(
...1
...1
...1
)(
21
1
21
2
21
2
21
2
21
++=
++
+++
=
+++
+++
=
pp
pp
b
ss
sasa
sbsb
sasa
sH
ωω
ωω
jj
n
j CRb 11 =Σ=

High-f response of a CG amplifier –
time-constant method (input pole)
1. Consider Cin :
)]1/()(||[1 omLosigin rgRrRC +′+=τ
Capacitor terminals
Terminals of Cin
om
Lo
rg
Rr
+
′+
1
gddbLL CCCC ++=′
sbgsin CCC +=
om
Lo
rg
Rr
+
′+
=
1
om
Lo
rg
Rr
+
′+
=
1

time-constant method (output pole)
1. Consider C’L :
)]1(||[2 sigmoLL RgrRC +′′=τ
Capacitor terminals
)1( sigmo Rgr +≈
)1( sigmo Rgr +
gddbLL CCCC ++=′
sbgsin CCC +=

time-constant method
omLoi
Lom
sigi
i
M
rgRrR
Rrg
RR
R
A
/)(
)||(
′+=
′
+
+=
)(2
1
2
1
)]1(||[
)]1/()(||[
211
2
1
ττππ
τ
τ
+
==
+′′=
+′+=
b
f
RgrRC
rgRrRC
H
sigmoLL
omLosigin
gddbLL CCCC ++=′
sbgsin CCC +=
LLmsigin
Lm
sigm
m
sig
o
RCsgRsC
Rg
Rg
g
v
v
′′+
×
+
×′×
+
=
1
1
)/1||(1
1
)(
/1
/1
Comparison of time-constant method with the exact solution (ro → ∞)
AM 11 /1
:poleInput
τω =p 22 /1
:poleOutput
τω =p

High-frequency response of a CS amplifier
dbL CC +
Csb is shorted out.
Cgd is between
output and input!
“Input Pole?” “Output Pole?”
Two methods to find fH
1) Miller’s theorem
2) Direct calculation of
resistance between
terminals of Cgd (see
Problem Set 8, Exercise 4)

Miller’s Theorem
12 VAV ⋅=
Z
VA
Z
VV
I 121
1
)1( ⋅−
=
−
=
)1(
,
)1/(
1
1
11
1
A
Z
Z
Z
V
AZ
V
I
−
==
−
=
A
Z
Z
Z
V
AZA
V
I
/11
,
)1/(
2
2
22
2
−
==
−
=
AZ
VA
Z
VA
Z
VV
I
)1()1( 2112
2
−
=
−
=
−
=
 Consider an amplifier with a gain A with an impedance Z attached
between input and output
 V1 and V2 “feel” the presence of Z only through I1 and I2
 We can replace Z with any circuit as long as a current I1 flows out of
V1 and a current I2 flows out of V2.

Miller’s Theorem – Statement
 If an impedance Z is attached between input and output an
amplifier with a gain A , Z can be replaced with two impedances
between input & ground and output & ground.
12 VAV ⋅=
Other parts of the circuit
A
Z
Z
1
1
2
−
=
A
Z
Z
−
=
1
1
12 VAV ⋅=

Example of Miller’s Theorem: Inverting amplifier
1
0101
00
11
10
0
)/()/(
)/(
R
R
v
v
ARR
R
ARR
ARA
RR
RA
v
v
A
v
v
f
i
o
f
f
f
f
f
f
i
n
i
o
−
≈
+
−
=
+
−
=
+
−
=−=
nnpo vAvvAv ⋅−=−⋅= 00 )(:OpAmp
1R
R
v
v f
i
o
−=
Recall from ECE 100, if A0 is large
Solution using Miller’s theorem:
f
f
f R
A
R
R ≈
+
=
0
2
/1100
1
1 A
R
A
R
R
ff
f ≈
+
=
11
1
f
f
i
n
RR
R
v
v
+
=
A
Z
Z
/11
2
−
=
A
Z
Z
−
=
1
1

Applying Miller’s Theorem to Capacitors
A
Z
Z
1
1
2
−
=
A
Z
Z
−
=
1
1
12 VAV ⋅=
)/11(
/11
)1(
1
1
22
11
CAC
A
Z
Z
CAC
A
Z
Z
Cj
Z
−=⇒
−
=
−=⇒
−
=
=
ω
Large capacitor at
the input for A >> 1

High-frequency response of a CS amplifier –
Using Miller’s Theorem
Use Miller’s Theorem to replace capacitor
between input & output (Cgd ) with two
capacitors at the input and output.
)]||(1[)1(, Lomgdgdigd RrgCACC ′+=−=
*
)]||(/11[)/11(,
gd
Lomgdgdogd
C
RrgCACC
≈
′+=−=
)||( Lom
g
d
Rrg
v
v
A ′−==
* Assuming gmR’L >> 1
igdgsin CCC ,+= LogddbL CCCC ++=′ ,
Note: Cgd appears in the input (Cgd,i)
as a “much larger” capacitor.

High-f response of a CS amplifier –
Miller’s Theorem and time-constant method
Output Pole (C’L ):
)||(2 LoL RrC ′′=τ1 sigin RC=τ
Input Pole (Cin ):
)||(
2
1
1 LoLsigin
H
RrCRCb
f
′′+==
π
or
∞
LgddbL
Lomgdgsin
CCCC
RrgCCC
++=′
′++= )]||(1[

High-f response of a CS amplifier – Exact solution
Miller & time-constant method:
1. Same b1 and same fH as the exact solution
2. Although, we get the same fH, there is a
substantial error in individual input and
output poles.
3. Miller approximation did not find the
zero!
Solving the circuit (node voltage method):
)||(
]))([(
)||(
1
)/1()||(
2
1
2
21
Losig
gdgsgdgsdbL
LoLsigin
mgdLom
sig
o
RrR
CCCCCCb
RrCRCb
sbsb
gsCRrg
v
v
′
×+++=
′′+=
++
−×′−
=
LgddbL
Lomgdgsin
CCCC
RrgCCC
++=′
′++= )]||(1[
)||(
2
1
1 LoLsigin
H
RrCRCb
f
′′+==
π

Miller’s Theorem vs Miller’s Approximation
 For Miller Theorem to work, ratio of V2/V1 (amplifier gain) should be
calculated in the presence of impedance Z.
 In our analysis, we used mid-band gain of the amplifier and ignored changes
in the gain due to the feedback capacitor, Cgd. This is called “Miller’s
Approximation.”
o In the OpAmp example the gain of the chip, A0 , remains constant when Rf is
attached (as the output resistance of the chip is small).
 Because the amplifier gain in the presence of Cgd is smaller than the mid-
band gain (we are on the high-f portion of the Bode gain plot), Miller’s
approximation overestimates Cgd,i and underestimates Cgd,o
o There is a substantial error in individual input and output poles. However, b1
and fH are estimated well.
 More importantly, Miller’s Approximation “misses” the zero introduced by
the feedback capacitor (This is important for stability of feedback amplifiers
as it affects gain and phase margins).
o Fortunately, we can calculate the zero of the transfer function easily (next slide).

Finding the “zero” of the CS amplifier
1) Definition of Zero: vo(s = sz) = 0
2) Because vo = 0, zero current will flow in ro, CL+Cdb and R’L
3) By KCL, a current of gmvgs will flow in Cgd.
4) Ohm’s law for Cgd gives:
gdz
gsm
gs
Cs
vg
iZv ==− 0
gsmvgi =
0=i
gd
m
z
gd
m
z
C
g
f
C
g
s
2
,
π
==

Zero of CS amplifier can play an important
role in the stability of feedback amplifiers
zf2pf1pf zf 2pf1pf
12ofCase ppz fff >>> 12ofCase pzp fff >>
Substantial change in
gain and phase margins!
Note: Since the input pole is at
Small Rsig can push fp2 to very large values!
)2/(1)1/(2 1 sigin RCππτ =

Comparison of CS and CG amplifiers
 Both CS and CG amplifiers have a high gain of
 CS amplifier has an infinite input resistance while CG amplifier
suffers from a low input resistance.
 CG amplifier has a much better high-f response:
o CS amplifier has a large capacitor at the input due to the
Miller’s effect: compared to that
of a CG amplifier
o In addition a CS amplifier has a zero.
 The Cascode amplifier combines the desirable properties of a
high input resistance with a reasonable high-f response. (It has
a much better high-f response than a two-stage CS amplifier)
)||( Lom Rrg ′
)]||(1[ Lomgdgsin RrgCCC ′++=
sbgsin CCC +=

Caution: Time-constant Method &
Miller’s Approximation
 The time constant method approximation to fH (see S&S page 724).
 Since,
This is the correct formula to find fH
 However, S&S gives a different formula in page 722 (contradicting
formulas of pp724). Ignore this formula (S&S Eq. 9.68)
 Discussions (and some conclusions re Miller’s theorem) in Examples 9.8
to 9.10 are incorrect. The discrepancy between fH from Miller’s
approximation and exact solution is due to the use of Eq. 9.68 (Not
Miller’s fault!)
...
1111
2
3
2
2
2
1
+++=
pppH ωωωω
...
111
...
11
2121
1 ++≈⇒++=
ppHpp
b
ωωωωω
H
jj
n
j CRb
ω
1
11 ≈Σ= =

Caution: Miller’s Approximation
 The main value of Miller’s Theorem is to demonstrate that a large
capacitance will appear at the input of a CS amplifier (Miller’s
capacitor).
 While Miller’s Approximation gives a reasonable approximation to
fH, it fails to provide accurate values for each pole and misses the
zero.
o Miller’s approximation should be used only as a first guess for
analysis. Simulation should be used to accurately find the
amplifier response.
o Stability analysis (gain and phase margins) should utilize
simulations unless a dominant pole far away from fH is
introduced.
 Miller’s approximation breaks down when gain is close to 1 (See
source follower, following slides).

High-f response of a Source Follower
Cdb is shorted out.
Cgs between
output and input
Cgd between gate
and ground
 Because mid-band gain is close to 1 and
positive, Miller approximation will not
work well.
o Need to apply Miller’s theorem exactly
o Lengthy calculations

High-f response of a source follower –
Exact Solution
0)()()(
||
:Node
0)(:Node
=−+−+++
′
=−++
−
iogsoimosbL
oL
o
o
oigsigd
sig
sigi
i
vvsCvvgvCCs
rR
v
v
vvsCvsC
R
vv
v
2
211
)/1(
)||(1
)||(
sbsb
gsC
Rrg
Rrg
v
v mgs
Lom
Lom
sig
o
++
+
×
′+
′
= gs
m
z
gs
m
z
C
g
f
C
g
s
2
,:Zero
π
=−=
Mid-band gain
Lengthy analysis is needed to find
b1, b2, and two poles

High-f response of a source follower –
time-constant method (1)
CL + Csb:
)||/1)((2 LmsbL RgCC ′+=τ1 siggd RC=τ
Cgd :
mg/1∞

High-frequency response of a follower –
Time-constant method (2)
3. Cgs (cannot use elementary R forms)

High-frequency response of a follower –
Time-constant method (3)
3. Cgs (cannot use elementary R forms), continued
)||(1
)||(
)]||([)]||(1[
)||()||(
))(||(KVL
oLm
oLsig
x
x
gs
oLsigxoLmx
xoLmxoLsigxx
gsmxoLsigxx
gsx
rRg
rRR
I
V
R
rRRIrRgV
VrRgIrRRIV
vgIrRRIV
vV
′+
′+
==
′+=′+
′−′+=
−′+=
=
3 gsgs RC=τ
)||(1
)||(
)||/1)((
2
1
3211
oLm
oLsig
gsLmsbLsiggd
H rRg
rRR
CRgCCRCb
f ′+
′+
×+′++=++== τττ
π

Finding the “zero” of a source follower
gsmvgi =
0=i
0=i
1) Definition of Zero: vo(s = sz) = 0
2) Because vo = 0, zero current will flow in ro, CL, Csb and R’L
3) By KCL, a current of gmvgs will flow in Cgs.
4) Ohm’s law for Cgs gives:
gsz
gsm
gs
Cs
vg
iZv ==−0
gs
m
z
gs
m
z
C
g
f
C
g
s
2
,
π
=−=

Examples of Computing high-f response
of various amplifiers
Procedure:
1. Include internal-capacitances of NMOS and simplify the circuit.
2. Use Miller’s approximation for “Miller” capacitors in
configurations with large (and negative) A.
3. Use time-constant method to find fH
4. Do not forget about zeros in CS and CD configurations.

High-f response of a CS amplifier
with current-source/active load
Between output &
ground
Miller’s
Between input &
ground
212,1 dbdbgdogdLL CCCCCC ++++=′
igdgsin CCC ,11 +=
Csb1 , Csb2 , & Cgs2
are shorted out.

High-f response of a CS amplifier
with current-source/active load
212,1
,11
1,1
11,1
)/11()/11(
)1()1(
dbdbgdogdLL
igdgsin
Lmgdgdogd
Lmgdgdigd
CCCCCC
CCC
RgCACC
RgCACC
++++=′
+=
′+=−=
′+=−=
LmLoom
g
d
RgRrrg
v
v
A ′−≡−== )||||( 21
1
1
∞
ro1
sigin
sigsigin
RC
RRRC
=⇒
=∞=
1
||:
τ
)||||(
||||:
212
21
LooL
LooL
RrrC
RrrRC
′=⇒
=′
τ
gd
m
z
gd
m
z
C
g
f
C
g
s
2
,
π
==
)||||(
2
1
211 LooLsigin
H
RrrCRCb
f
′+==
π

High-f response of Cascode amplifiers
Between output &
ground
Between D1
& ground
Miller’s
Between input &
ground
22 dbgdLL CCCC ++=′
221,11 sbgsdbogd CCCCC +++=
igdgsin CCC ,11 +=
Csb1 is
shorted out.

High-f response of cascode amplifiers
∞
ro1
Ri2
Ro2
)||(
1
&
)||(
2111
1
22
2
2212212
222
iomv
L
om
Lo
ioomooo
Lomv
RrgA
R
rg
Rr
RrrgrrR
RrgA
−=
=
+
+
=++=
= Mid-band Gains & resistances
igdgsin
sbdbgsogd
dbgdLL
vgdogd
vgdigd
CCC
CCCCC
CCCC
ACC
ACC
,11
212,11
22
1,1
11,1
)/11(
)1(
+=
+++=
++=′
−=
−= Miller’s
Capacitors
3211
232
2112211
1
2
1
)||(||:
)||(||:
||:
τττ
π
τ
τ
τ
++==
′=⇒=′
=⇒=
=⇒=∞=
b
f
RRCRRRC
RrCRrRC
RCRRRC
H
LoLLoL
ioio
siginsigsigin
1
1
1
1
2
,
gd
m
z
gd
m
z
C
g
f
C
g
s
π
==

High-f response of differential amplifiers
 For a symmetric differential amplifier (i.e. when we can use the
half-circuit concept):
o Therefore, the poles and zero of the differential amplifier
(differential gain) are the same as those of differential half circuit.
 Detailed analysis is necessary for asymmetric differential
amplifier (e.g., IC differential amplifier with a single-ended
output of slides 21-29 of Lecture set 7B),
d
do
d
do
d
do
d
dodododooo
v
v
v
v
v
v
A
vvvvvv
5.0
2
2
1,1,,
,1,1,2,12
−
=×−==
−=−=→−=

Dominant Pole Compensation
1. Very often we need to purposely introduce an additional “pole”
in the circuit (in order to provide gain or phase margin in
feedback amplifiers). This is called the dominant pole
compensation.
2. This pole has to be the “dominant pole” (several octave below
any zero or pole).
3. As such, we can ignore transistor internal capacitances in the
analysis as poles introduced by these capacitances would be at
higher frequencies and does not impact the dominant pole.
4. Dominant pole is introduce by the addition of a capacitor either
a) Between output & ground or
b) Between input & output of one stage (using Miller Effect)

Dominant Pole by adding CL
 Typically the required CL is large and CL is located outside the
chip (i.e., between output terminal and ground).
)||(2/1 oLLp RRCf π=
Ro
Generic Form: Example:

Dominant Pole via Miller’s effect
Ri
Generic Form: Example:
 Dominant pole is produced by |A|CM at the
input (not CM in the output). Otherwise we
would have connected CM in the output and
not as a Miller capacitor!
)||(|A|2/1 isigMp RRCf π=

Dominant Pole via Miller’s effect
Generic Form:
Note: fp1 is proportional to Rsig
1. Usually not used in the first-stage as we do NOT know what Rsig is.
2. For second or latter stages, Rsig is Ro of previous stage and can be
large. Because CM appears as a large capacitor due to Miller’s effect,
this is the preferred method for including a capacitor inside the chip.
3. Note Miller’s approximation does NOT give the correct value for
poles. Simulation should be used to confirm the value of CM . Also
note that CM introduces a zero.
)||(|A|2/1 isigMp RRCf π=

Freq response of CE and CC discrete circuits

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