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
 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
F. Najmabadi, ECE102, Fall 2012 (2/59)
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).
F. Najmabadi, ECE102, Fall 2012 (3/59)
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?
F. Najmabadi, ECE102, Fall 2012 (4/59)
Capacitor behavior depends
on the frequency of interest.
Capacitor approximates an open circuit at low frequencies:
Example:
| Z | = R || (1 / ω C )
R << (1 / ω C ) → | Z | = R || (1 / ω C ) ≈ R
R << (1 / ω C ) → ω << (1 / RC )
Capacitor approximates a short circuit at high frequencies:
R >> (1 / ω C ) → | Z | = R || (1 / ω C ) ≈ (1 / ω C ) → 0
R >> (1 / ω C ) → ω >> (1 / RC )
We cannot ignore the capacitor when
R ~ (1 / ω C ) → ω ~ (1 / RC )
This defines the reference frequency for high-f and low-f
Note: The above circuit is like a low-pass filter with a cut-off frequency of 1/RC
F. Najmabadi, ECE102, Fall 2012 (5/59)
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)
F. Najmabadi, ECE102, Fall 2012 (6/59)
Impact of various capacitors depend on
the frequency of interest
Impendence of capacitors (1/ωC)
f→∞
All Caps are short.
This limit is used
to find highfrequency Caps.
f→0
All Caps are open.
This limit is used
to find lowfrequency Caps
Computing fL:
High-f caps are open.
Low-f caps included.
F. Najmabadi, ECE102, Fall 2012 (7/59)
Mid-band:
High-f caps are open
Low-f caps are short.
Computing fH :
High-f caps are included.
Low-f caps are short
How to find which capacitors contribute
to the lower cut-off frequency
 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)
Example:
Cc1 open:
v i = 0 → vo = 0
Contributes to fL
F. Najmabadi, ECE102, Fall 2012 (8/59)
CL open:
No change in vo
Does NOT contribute to fL
How to find which capacitors contribute
to the upper cut-off frequency
 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:
Cc1 short:
No change in vo
Does NOT contribute to fH
F. Najmabadi, ECE102, Fall 2012 (9/59)
CL short:
vo = 0
Contributes to fH
Constructing appropriate circuits
Example:
Cc1 : Low-f capacitor
CL : High-f capacitor
Computing fL:
High-f caps are open.
Low-f caps included.
F. Najmabadi, ECE102, Fall 2012 (10/59)
Mid-band:
High-f caps are open
Low-f caps are short.
Computing fH :
High-f caps are included.
Low-f caps are short
Low-Frequency Response
F. Najmabadi, ECE102, Fall 2012 (11/59)
Low-frequency response of an amplifier
Example: an amplifier with three poles
Vo
s
s
s
= AM ×
×
×
Vsig
s + ω p1 s + ω p 2 s + ω p 3
(Set s = jω to find Bode Plots)
 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%)
F. Najmabadi, ECE102, Fall 2012 (12/59)
Low-frequency response of a CS amplifier
(from detailed frequency response analysis)
All capacitors contribute to fL (as vo is reduced when f → 0 or caps open circuit)
Cc1 open:
vi = 0 → vo = 0
Cc2 open:
vo = 0
Cs open:
Gain is reduced substantially
(from CS amp. to CS amp. With RS)
Lengthy calculations: See S&S pp689-692 for detailed calculations (S&S
assumes ro → ∞ and RS → ∞ )
Vo
s
s
s
= AM ×
×
×
Vsig
s + ω p1 s + ω p 2 s + ω p 3
AM = −
RG
g m (ro || RD || RL )
RG + Rsig
F. Najmabadi, ECE102, Fall 2012 (13/59)
ω p1 =
1
1
, ω p3 =
Cc1 ( RG + Rsig )
Cc 2 ( RD || ro + RL )
ω p2 ≈
1
,
Cs [ RS || [(ro + RD || RL ) /(1 + g m ro )]
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
f pn =
1
2πRn Cn
5. Lower-cut-off frequency can be found from
fL ≈ fp1 + fp2 + fp3 + …
F. Najmabadi, ECE102, Fall 2012 (14/59)
* 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.
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
F. Najmabadi, ECE102, Fall 2012 (15/59)
Low-frequency response of a CS amplifier (fp1)
f p1 =
1. Consider Cc1 :
1
2π Cc1 ( RG + Rsig )
Terminals of Cc1
∞
F. Najmabadi, ECE102, Fall 2012 (16/59)
2. Find resistance between
Capacitor terminals
Low-frequency response of a CS amplifier (fp2)
f p2 =
1
2π CS [ RS || [(ro + RD || RL ) /(1 + g m ro )]
ro + RD || RL
1 + g m ro
1. Consider CS :
ro + RD || RL
1 + g m ro
F. Najmabadi, ECE102, Fall 2012 (17/59)
ro + RD || RL
1 + g m ro
Terminals of CS
2. Find resistance between
Capacitor terminals
Low-frequency response of a CS amplifier (fp3)
f p3 =
1. Consider Cc2 :
F. Najmabadi, ECE102, Fall 2012 (18/59)
1
2π Cc 2 ( RL + RD || ro )
Terminals of Cc2
2. Find resistance between
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.
F. Najmabadi, ECE102, Fall 2012 (19/59)
Capacitive Effects in pn Junction
 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
High-f small signal
model of diode
Forward Bias
Cj + Cd
C j ≈ 2 ⋅ C j0
Cd =
rD
F. Najmabadi, ECE102, Fall 2012 (20/59)
τT ⋅ ID
VT
Reverse Bias
Cj =
C j0
(1 + VR / V0 ) m
Cd = 0
*See S&S pp154-156 for detailed derivations
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)
2. Capacitance between
Gate & Source and Gate &
Drain due to the overlap of
gate electrode
(Parallel-plate capacitor)
4. Junction capacitance
between Drain and Body
(Reverse-bias junction)
MOS High-frequency
small signal model
F. Najmabadi, ECE102, Fall 2012 (21/59)
*See S&S pp154-156 for
detailed derivations
MOS high-frequency small signal model
Accurate Model
(we use this model here)
For source connected to body
(used by S&S)
Generally, transistor internal capacitances are
shown outside the transistor so that we can use
results from the mid-band calculations.
F. Najmabadi, ECE102, Fall 2012 (22/59)
High-frequency response of a CG amplifier
Cgs between
source & ground
Low-pass
filter
“Input Pole”
F. Najmabadi, ECE102, Fall 2012 (23/59)
Mid-band Low-pass
Amp
filter
“Output Pole”
Cgd between
drain & ground
C L′ = C L + Cdb + C gd
Cin = C gs + Csb
High-f response of a CG amplifier – Exact Solution (1)
C L′ = C L + Cdb + C gd
Cin = C gs + Csb
Can be solved to find vo/vsig
Node vi :
Node vo :
F. Najmabadi, ECE102, Fall 2012 (24/59)
vi − vsig
Rsig
+
vi
v − vo
− g m ( −vi ) + i
=0
1 / sCin
ro
vo
vo
v −v
+
+ g m (−vi ) + o i = 0
RL′ 1 / sC L′
ro
High-f response of a CG amplifier – Exact Solution (2)
Compact solution can be found by ignoring ro (i.e., ro → ∞)
Node vi : vi − vsig + ( sCin + g m ) Rsig vi = 0
vi
1
1
1
=
=
×
vsig 1 + g m Rsig + sCin Rsig 1 + g m Rsig 1 + sCin Rsig /(1 + g m Rsig )
1/ gm
vi
1
=
×
vsig 1 / g m + Rsig 1 + sCin ( Rsig || 1 / g m )
Voltage divider
(Ri= 1/gm and Rsig)
“Input Pole”
Mid-band Gain
Node vo :
vo
+ sC L′ vo − g m (vi ) = 0 ⇒
RL′
vo
g m RL′
=
vi 1 + sC L′ RL′
“Output Pole”
vo
Ri
1
1
=
× ( g m RL′ ) ×
×
vsig Ri + Rsig
1 + sCin ( Rsig || 1 / g m ) 1 + sC L′ RL′
F. Najmabadi, ECE102, Fall 2012 (25/59)
Open-Circuit Time-Constants Method
1 + a1s + a2 s 2 + ...
1 + a1s + a2 s 2 + ...
H ( s) =
=
2
1 + b1s + b2 s + ... (1 + s / ω p1 )(1 + s / ω p 2 )...
b1 =
1
ω p1
+
1
ω p2
+ ...
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. b1 = Σ nj =1 R j C j
A good approximation to fH is: f H =
F. Najmabadi, ECE102, Fall 2012 (26/59)
1
2π b1
High-f response of a CG amplifier –
time-constant method (input pole)
C L′ = C L + Cdb + C gd
Cin = C gs + Csb
τ 1 = Cin [ Rsig || (ro +RL′ ) /(1 + g m ro )]
Terminals of Cin
=
ro + RL′
1 + g m ro
1. Consider Cin :
=
ro + RL′
1 + g m ro
F. Najmabadi, ECE102, Fall 2012 (27/59)
ro + RL′
1 + g m ro
2. Find resistance between
Capacitor terminals
High-f response of a CG amplifier –
time-constant method (output pole)
C L′ = C L + Cdb + C gd
Cin = C gs + Csb
τ 2 = C L′ [ RL′ || ro (1 + g m Rsig )]
1. Consider C’L :
≈ ro (1 +g m Rsig )
ro (1 +g m Rsig )
2. Find resistance between
Capacitor terminals
F. Najmabadi, ECE102, Fall 2012 (28/59)
High-f response of a CG amplifier –
time-constant method
C L′ = C L + Cdb + C gd
Cin = C gs + Csb
AM = +
Ri
g m (ro || RL′ )
Ri + Rsig
τ 1 = Cin [ Rsig || (ro +RL′ ) /(1 + g m ro )]
τ 2 = C L′ [ RL′ || ro (1 + g m Rsig )]
fH =
Ri = (ro +RL′ ) / g m ro
1
2π b1
=
1
2π (τ 1 + τ 2 )
Comparison of time-constant method with the exact solution (ro → ∞)
vo
1/ gm
1
1
=
× ( g m RL′ ) ×
×
vsig 1 / g m + Rsig
1 + sCin ( Rsig || 1 / g m ) 1 + sC L′ RL′
AM
F. Najmabadi, ECE102, Fall 2012 (29/59)
Input pole :
ω p1 = 1 / τ 1
Output pole :
ω p2 = 1/τ 2
High-frequency response of a CS amplifier
C L + Cdb
Cgd is between
output and input!
Two methods to find fH
1) Miller’s theorem
2) Direct calculation of
resistance between
terminals of Cgd (see
Problem Set 8, Exercise 4)
F. Najmabadi, ECE102, Fall 2012 (30/59)
“Input Pole?”
Csb is shorted out.
“Output Pole?”
Miller’s Theorem
 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.
V2 = A ⋅ V1
F. Najmabadi, ECE102, Fall 2012 (31/59)
I1 =
V1 − V2 (1 − A) ⋅ V1
=
Z
Z
I2 =
V2 − V1 ( A − 1) V1 ( A − 1) V2
=
=
Z
Z
ZA
I1 =
V1
V
= 1,
Z /(1 − A) Z1
I2 =
V2
V
Z
= 2 , Z2 =
ZA /( A − 1) Z 2
1−1/ A
Z1 =
Z
(1 − A)
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.
Other parts of the circuit
V2 = A ⋅ V1
V2 = A ⋅ V1
Z
Z1 =
1− A
F. Najmabadi, ECE102, Fall 2012 (32/59)
Z2 =
Z
1−
1
A
Example of Miller’s Theorem: Inverting amplifier
OpAmp : vo = A0 ⋅ (v p − vn ) = − A0 ⋅ vn
Recall from ECE 100, if A0 is large
Rf
vo
=−
vi
R1
Solution using Miller’s theorem:
Rf 1
vn
=
vi R1 + R f 1
vo
vn − A0 R f 1
= − A0 =
vi
vi R1 + R f 1
Z
Z1 =
1− A
Rf1 =
Rf
1 + A0
≈
Z
Z2 =
1−1/ A
Rf
A0
F. Najmabadi, ECE102, Fall 2012 (33/59)
Rf 2 =
Rf
1 + 1 / A0
≈ Rf
=
− A0 ( R f / A0 )
R1 + ( R f / A0 )
vo − R f
≈
vi
R1
=
− Rf
R1 + ( R f / A0 )
Applying Miller’s Theorem to Capacitors
V2 = A ⋅ V1
Z
Z1 =
1− A
Z=
Large capacitor at
the input for A >> 1
F. Najmabadi, ECE102, Fall 2012 (34/59)
Z2 =
Z
1−
1
A
1
jω C
Z
Z1 =
⇒ C1 = (1 − A) C
1− A
Z
Z2 =
⇒ C2 = (1 − 1 / A) C
1−1/ A
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.
A=
vd
= − g m (ro || RL′ )
vg
C gd ,i = C gd (1 − A) = C gd [1 + g m (ro || RL′ )]
C gd ,o = C gd (1 − 1 / A) = C gd [1 + 1 / g m (ro || RL′ )]
≈ C gd *
Cin = C gs + C gd ,i
C L′ = Cdb + C gd ,o + C L
Note: Cgd appears in the input (Cgd,i)
as a “much larger” capacitor.
* Assuming gmR’L >> 1
F. Najmabadi, ECE102, Fall 2012 (35/59)
High-f response of a CS amplifier –
Miller’s Theorem and time-constant method
Cin = C gs + C gd [1 + g m (ro || RL′ )]
C L′ = Cdb + C gd + C L
Input Pole (Cin ):
Output Pole (C’L ):
ro
∞
τ 1 = Cin Rsig
1
2π f H
F. Najmabadi, ECE102, Fall 2012 (36/59)
τ 2 = C L′ (ro || RL′ )
= b1 = Cin Rsig + C L′ (ro || RL′ )
High-f response of a CS amplifier – Exact solution
Solving the circuit (node voltage method):
− g m (ro || RL′ ) × (1 − sC gd / g m )
vo
=
1 + b1s + b2 s 2
vsig
b1 = Cin Rsig + C L′ (ro || RL′ )
b2 = [(C L + Cdb )(C gs + C gd ) + C gs C gd ] ×
Rsig (ro || RL′ )
Cin = C gs + C gd [1 + g m (ro || RL′ )]
C L′ = Cdb + C gd + C L
F. Najmabadi, ECE102, Fall 2012 (37/59)
Miller & time-constant method:
1. Same b1 and same fH as the exact solution
1
= b1 = Cin Rsig + C L′ (ro || RL′ )
2π f H
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!
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 midband 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).
F. Najmabadi, ECE102, Fall 2012 (38/59)
Finding the “zero” of the CS amplifier
i = g m v gs
1) Definition of Zero: vo(s = sz) = 0
i=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:
v gs − 0 = Z i =
sz =
F. Najmabadi, ECE102, Fall 2012 (39/59)
gm
,
C gd
g m v gs
s z C gd
fz =
gm
2π C gd
Zero of CS amplifier can play an important
role in the stability of feedback amplifiers
Substantial change in
gain and phase margins!
f p1
Case of
fz
f p1
f z >> f p 2 > f p1
Case of
f p2
fz
f p2
f p 2 > f z > f p1
Note: Since the input pole is at
1/(2πτ 1 ) = 1 /( 2π Cin Rsig )
F. Najmabadi, ECE102, Fall 2012 (40/59)
Small Rsig can push fp2 to very large values!
Comparison of CS and CG amplifiers
 Both CS and CG amplifiers have a high gain of g m (ro || RL′ )
 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: Cin = C gs + C gd [1 + g m (ro || RL′ )] compared to that
of a CG amplifier Cin = C gs + Csb
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)
F. Najmabadi, ECE102, Fall 2012 (41/59)
Caution: Time-constant Method &
Miller’s Approximation

The time constant method approximation to fH (see S&S page 724).
b1 = Σ nj =1 R j C j ≈

Since,
b1 =
1
ω p1
+
1
ω p2
1
ωH
+ ... ⇒
1
ωH
≈
1
ω p1
+
1
ω p2
+ ...
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)
1
ωH

=
1
ω
2
p1
+
1
ω
2
p2
+
1
ω
2
p3
+ ...
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!)
F. Najmabadi, ECE102, Fall 2012 (42/59)
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).
F. Najmabadi, ECE102, Fall 2012 (43/59)
High-f response of a Source Follower
Cgd between gate
and ground
Cgs between
output and input
F. Najmabadi, ECE102, Fall 2012 (44/59)
Cdb is shorted out.
 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
Node vi :
Node vo :
vi − vsig
Rsig
+ sC gd vi + sC gs (vi − vo ) = 0
vo
+ s (C L + C sb )vo + g m (vi − vo ) + sC gs (vo − vi ) = 0
RL′ || ro
(1 + sC gs / g m )
vo
g m (ro || RL′ )
=
×
vsig 1 + g m (ro || RL′ ) 1 + b1s + b2 s 2
Mid-band gain
F. Najmabadi, ECE102, Fall 2012 (45/59)
Zero : s z = −
gm
,
C gs
fz =
gm
2π C gs
Lengthy analysis is needed to find
b1, b2, and two poles
High-f response of a source follower –
time-constant method (1)
Cgd :
CL + Csb:
∞
1/ gm
τ 1 = C gd Rsig
F. Najmabadi, ECE102, Fall 2012 (46/59)
τ 2 = (C L + Csb )(1 / g m || RL′ )
High-frequency response of a follower –
Time-constant method (2)
3. Cgs (cannot use elementary R forms)
F. Najmabadi, ECE102, Fall 2012 (47/59)
High-frequency response of a follower –
Time-constant method (3)
3. Cgs (cannot use elementary R forms), continued
Vx = v gs
KVL Vx = I x Rsig + ( RL′ || ro )( I x − g m v gs )
Vx = I x Rsig + ( RL′ || ro ) I x − g m ( RL′ || ro )Vx
Vx [1 + g m ( RL′ || ro )] = I x [ Rsig + ( RL′ || ro )]
Vx Rsig + ( RL′ || ro )
Rgs =
=
I x 1 + g m ( RL′ || ro )
τ 3 = C gs Rgs
1
2π f H
= b1 = τ 1 + τ 2 + τ 3 = C gd Rsig
F. Najmabadi, ECE102, Fall 2012 (48/59)
Rsig + ( RL′ || ro )
+ (C L + C sb )(1 / g m || RL′ ) + C gs ×
1 + g m ( RL′ || ro )
Finding the “zero” of a source follower
i=0
i = g m v gs
i=0
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:
0 − v gs = Z i =
sz = −
F. Najmabadi, ECE102, Fall 2012 (49/59)
gm
,
C gs
g m v gs
s z C gs
fz =
gm
2π C gs
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.
F. Najmabadi, ECE102, Fall 2012 (50/59)
High-f response of a CS amplifier
with current-source/active load
Csb1 , Csb2 , & Cgs2
are shorted out.
Miller’s
Between input &
ground
C L′ = C L + C gd 1,o + C gd 2 + Cdb1 + Cdb 2
Cin = C gs1 + C gd 1,i
F. Najmabadi, ECE102, Fall 2012 (51/59)
Between output &
ground
High-f response of a CS amplifier
with current-source/active load
ro1
∞
v
A = d 1 = − g m (ro1 || ro 2 || RL ) ≡ − g m RL′
v g1
C gd 1,i = C gd 1 (1 − A) = C gd 1 (1 + g m RL′ )
C gd 1,o = C gd (1 − 1 / A) = C gd 1 (1 + 1 / g m RL′ )
Cin = C gs1 + C gd 1,i
C L′ = C L + C gd 1,o + C gd 2 + Cdb1 + Cdb 2
Cin : R = Rsig || ∞ = Rsig
⇒ τ 1 = Cin Rsig
C L′ : R = ro1 || ro 2 || RL
⇒ τ 2 = C L′ (ro1 || ro 2 || RL )
1
2π f H
sz =
F. Najmabadi, ECE102, Fall 2012 (52/59)
= b1 = Cin Rsig + C L′ (ro1 || ro 2 || RL )
gm
,
C gd
fz =
gm
2π C gd
High-f response of Cascode amplifiers
Between output &
ground
C L′ = C L + C gd 2 + Cdb 2
C1 = C gd 1,o + Cdb1 + C gs 2 + Csb 2
Cin = C gs1 + C gd 1,i
Miller’s
Between D1
& ground
Between input &
ground
Csb1 is
shorted out.
F. Najmabadi, ECE102, Fall 2012 (53/59)
High-f response of cascode amplifiers
Av 2 = g m 2 (ro 2 || RL )
Ro2
Mid-band Gains & resistances
Ro 2 = ro1 + ro 2 + g m 2 ro1ro 2 & Ri 2 =
ro 2 +RL
= RL1
1 + g m 2 ro 2
Ri2
Av1 = − g m1 (ro1 || Ri 2 )
∞
C gd 1,i = C gd 1 (1 − Av1 )
C gd 1,o
Miller’s
= C gd (1 − 1 / Av1 ) Capacitors
ro1
C L′ = C L + C gd 2 + Cdb 2
C1 = C gd 1,o + C gs 2 + Cdb1 + Csb 2
Cin = C gs1 + C gd 1,i
Cin : R = Rsig || ∞ = Rsig ⇒ τ 1 = Cin Rsig
C1 : R = ro1 || Ri 2
C L′ : R = Ro 2 || RL
1
2π f H
= b1 = τ 1 + τ 2 + τ 3
F. Najmabadi, ECE102, Fall 2012 (54/59)
⇒ τ 2 = C1 (ro1 || Ri 2 )
⇒ τ 3 = C L′ ( Ro 2 || RL )
sz =
g m1
,
C gd 1
fz =
g m1
2π C gd 1
High-f response of differential amplifiers
 For a symmetric differential amplifier (i.e. when we can use the
half-circuit concept):
vo 2 = −vo1 → vo ,d = vo 2,d − vo1,d = −2vo1,d
Ad =
vo ,d
vd
= −2 ×
vo ,1d
vd
=
vo ,1d
− 0.5vd
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),
F. Najmabadi, ECE102, Fall 2012 (55/59)
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)
F. Najmabadi, ECE102, Fall 2012 (56/59)
Dominant Pole by adding CL
Generic Form:
Example:
Ro
1 / f p = 2π C L ( RL || Ro )
 Typically the required CL is large and CL is located outside the
chip (i.e., between output terminal and ground).
F. Najmabadi, ECE102, Fall 2012 (57/59)
Dominant Pole via Miller’s effect
Generic Form:
 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!
1 / f p = 2π | A | CM ( Rsig || Ri )
F. Najmabadi, ECE102, Fall 2012 (58/59)
Example:
Ri
Dominant Pole via Miller’s effect
Generic Form:
1 / f p = 2π | A | CM ( Rsig || Ri )
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.
F. Najmabadi, ECE102, Fall 2012 (59/59)