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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).
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.
F. Najmabadi, ECE102, Fall 2012 (5/59)
) /1(|| || CRZ ω=
Capacitor approximates an open circuit at low frequencies:
)/1( ) /1() /1(|| || ) /1(
RCCRRCRZCR
<<→<<≈=→<<
ωωωω
Capacitor approximates a short circuit at high frequencies:
)/1( ) /1(0) /1() /1(|| || ) /1(
RCCRCCRZCR
>>→>>→≈=→>>
ωωωωω
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)
F. Najmabadi, ECE102, Fall 2012 (6/59)
Impact of various capacitors depend on the frequency of interest
F. Najmabadi, ECE102, Fall 2012 (7/59)
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
F. Najmabadi, ECE102, Fall 2012 (8/59)
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
F. Najmabadi, ECE102, Fall 2012 (9/59)
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
F. Najmabadi, ECE102, Fall 2012 (10/59)
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.
F. Najmabadi, ECE102, Fall 2012 (11/59)
Low-Frequency Response
Low-frequency response of an amplifier
F. Najmabadi, ECE102, Fall 2012 (12/59)
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
Msig
o
ss
ss
ssA
VV
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)
F. Najmabadi, ECE102, Fall 2012 (13/59)
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
LDomsigG
GM
pppM
sig
o
RRrgRR
RA
ss
ss
ssA
VV
+−=
+×
+×
+×=
ωωω
Lengthy calculations: See S&S pp689-692 for detailed calculations (S&S assumes ro → ∞ and RS → ∞ )
,)]1/()||[(||[
1
)||(
1 ,)(
1
2
23
11
omLDoSsp
LoDcp
sigGcp
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 + …
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.
nnpn CR
fπ2
1=
Example: Low-frequency response of a CS amplifier (from pole inspection)
F. Najmabadi, ECE102, Fall 2012 (15/59)
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 :
∞
)( 21
11
sigGcp RRC
f+
=π
2. Find resistance between Capacitor terminals
Terminals of Cc1
F. Najmabadi, ECE102, Fall 2012 (16/59)
Low-frequency response of a CS amplifier (fp2)
1. Consider CS :
Terminals of CS
)]1/()||[(||[ 21
2omLDoSS
p rgRRrRCf
++=
π
2. Find resistance between Capacitor terminals
om
LDo
rgRRr
++1
||
om
LDo
rgRRr
++1
||
om
LDo
rgRRr
++1
||
F. Najmabadi, ECE102, Fall 2012 (17/59)
Low-frequency response of a CS amplifier (fp3)
1. Consider Cc2 : Terminals of Cc2
)||( 21
23
oDLcp rRRC
f+
=π
2. Find resistance between Capacitor terminals
F. Najmabadi, ECE102, Fall 2012 (18/59)
F. Najmabadi, ECE102, Fall 2012 (19/59)
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
F. Najmabadi, ECE102, Fall 2012 (20/59)
Forward Bias Reverse Bias
02 jj CC ⋅≈
T
DTd V
IC ⋅=τ
0=dC
mR
jj VV
CC
)/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*
F. Najmabadi, ECE102, Fall 2012 (21/59)
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
F. Najmabadi, ECE102, Fall 2012 (22/59)
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
F. Najmabadi, ECE102, Fall 2012 (23/59)
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)
F. Najmabadi, ECE102, Fall 2012 (24/59)
gddbLL CCCC ++=′
sbgsin CCC +=
0)(/1
: Node
0)(/1
: Node
=−
+−+′
+′
=−
+−−+−
o
ioim
L
o
L
oo
o
oiim
in
i
sig
sigii
rvvvg
Csv
Rvv
rvvvg
sCv
Rvv
vCan be solved to find vo/vsig
High-f response of a CG amplifier – Exact Solution (2)
F. Najmabadi, ECE102, Fall 2012 (25/59)
High-f response of a CG amplifier – Exact Solution (2)
)/1||(11
/1/1
)1/(11
11
11
0)( : Node
msiginsigm
m
sig
i
sigmsiginsigmsiginsigmsig
i
isigminsigii
gRsCRgg
vv
RgRsCRgRsCRgvv
vRgsCvvv
+×
+=
++×
+=
++=
=++−
LL
Lm
i
oimoL
L
oo RCs
RgvvvgvCs
Rvv
′′+′
=⇒=−′+′ 1
0)( : Node
Voltage divider (Ri= 1/gm and Rsig) “Input Pole”
“Output Pole”
Mid-band Gain
LLmsiginLm
sigi
i
sig
o
RCsgRsCRg
RRR
vv
′′+×
+×′×
+=
11
)/1||(11)(
Compact solution can be found by ignoring ro (i.e., ro → ∞)
Open-Circuit Time-Constants Method
F. Najmabadi, ECE102, Fall 2012 (26/59)
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:
1 21
bfH π
=
...11
).../1)(/1(...1
...1...1)(
211
21
221
221
221
++=
+++++
=++++++
=
pp
pp
b
sssasa
sbsbsasasH
ωω
ωω
jjnj CRb 11 =Σ=
High-f response of a CG amplifier – time-constant method (input pole)
F. Najmabadi, ECE102, Fall 2012 (27/59)
1. Consider Cin :
)]1/()(||[1 omLosigin rgRrRC +′+=τ
2. Find resistance between Capacitor terminals
Terminals of Cin
om
Lo
rgRr
+′+
1
gddbLL CCCC ++=′
sbgsin CCC +=
om
Lo
rgRr
+′+
=1
om
Lo
rgRr
+′+
=1
High-f response of a CG amplifier – time-constant method (output pole)
F. Najmabadi, ECE102, Fall 2012 (28/59)
1. Consider C’L :
)]1(||[2 sigmoLL RgrRC +′′=τ
2. Find resistance between Capacitor terminals
)1( sigmo Rgr +≈
)1( sigmo Rgr +
gddbLL CCCC ++=′
sbgsin CCC +=
High-f response of a CG amplifier – time-constant method
F. Najmabadi, ECE102, Fall 2012 (29/59)
omLoi
Lomsigi
iM
rgRrR
RrgRR
RA
/)(
)||(
′+=
′+
+=
)( 21
21
)]1(||[
)]1/()(||[
211
2
1
ττππ
τ
τ
+==
+′′=
+′+=
bf
RgrRCrgRrRC
H
sigmoLL
omLosigin
gddbLL CCCC ++=′
sbgsin CCC +=
LLmsiginLm
sigm
m
sig
o
RCsgRsCRg
Rgg
vv
′′+×
+×′×
+=
11
)/1||(11)(
/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
F. Najmabadi, ECE102, Fall 2012 (30/59)
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
F. Najmabadi, ECE102, Fall 2012 (31/59)
12 VAV ⋅=
ZVA
ZVVI 121
1)1( ⋅−
=−
=
)1( ,
)1/( 11
111 A
ZZZV
AZVI
−==
−=
AZZ
ZV
AZAVI
/11 ,
)1/( 22
222 −
==−
=
AZVA
ZVA
ZVVI
)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.
F. Najmabadi, ECE102, Fall 2012 (32/59)
12 VAV ⋅=
Other parts of the circuit
A
ZZ 112
−=
AZZ−
=11
12 VAV ⋅=
Example of Miller’s Theorem: Inverting amplifier
F. Najmabadi, ECE102, Fall 2012 (33/59)
1
0101
00
11
100
)/()/()/(
RR
vv
ARRR
ARRARA
RRRA
vvA
vv
f
i
o
f
f
f
f
f
f
i
n
i
o
−≈
+
−=
+
−=
+
−=−=
nnpo vAvvAv ⋅−=−⋅= 00 )( :OpAmp
1RR
vv f
i
o −=
Recall from ECE 100, if A0 is large
Solution using Miller’s theorem:
ff
f RA
RR ≈
+=
02 /1100
1 1 AR
AR
R fff ≈
+=
11
1
f
f
i
n
RRR
vv
+=
AZZ
/112 −=
AZZ−
=11
Applying Miller’s Theorem to Capacitors
F. Najmabadi, ECE102, Fall 2012 (34/59)
A
ZZ 112
−=
AZZ−
=11
12 VAV ⋅=
)/11( /11
)1( 1
1
22
11
CACA
ZZ
CACA
ZZ
CjZ
−=⇒−
=
−=⇒−
=
=ω
Large capacitor at the input for A >> 1
High-frequency response of a CS amplifier – Using Miller’s Theorem
F. Najmabadi, ECE102, Fall 2012 (35/59)
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
CRrgCACC
≈
′+=−=
)||( Lomg
d RrgvvA ′−==
* 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
F. Najmabadi, ECE102, Fall 2012 (36/59)
Output Pole (C’L ):
)||( 2 LoL RrC ′′=τ 1 siginRC=τ
Input Pole (Cin ):
)||( 21
1 LoLsiginH
RrCRCbf
′′+==π
or
∞
LgddbL
Lomgdgsin
CCCCRrgCCC
++=′
′++= )]||(1[
High-f response of a CS amplifier – Exact solution
F. Najmabadi, ECE102, Fall 2012 (37/59)
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
221
Losig
gdgsgdgsdbL
LoLsigin
mgdLom
sig
o
RrRCCCCCCb
RrCRCbsbsb
gsCRrgvv
′
×+++=
′′+=
++
−×′−=
LgddbL
Lomgdgsin
CCCCRrgCCC
++=′
′++= )]||(1[
)||( 21
1 LoLsiginH
RrCRCbf
′′+==π
Miller’s Theorem vs Miller’s Approximation
F. Najmabadi, ECE102, Fall 2012 (38/59)
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
F. Najmabadi, ECE102, Fall 2012 (39/59)
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
gsmgs Cs
vgi Z v ==− 0
gsmvgi =0=i
gd
mz
gd
mz C
gfCgs
2 ,
π==
Zero of CS amplifier can play an important role in the stability of feedback amplifiers
F. Najmabadi, ECE102, Fall 2012 (40/59)
zf 2pf 1pf zf 2pf 1pf
12 of Case ppz fff >>> 12 of Case 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 siginRCππτ =
Comparison of CS and CG amplifiers
F. Najmabadi, ECE102, Fall 2012 (41/59)
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
F. Najmabadi, ECE102, Fall 2012 (42/59)
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!)
...11112
32
221
+++=pppH ωωωω
...111 ...11
21211 ++≈⇒++=
ppHpp
bωωωωω
Hjj
nj CRb
ω1
11 ≈Σ= =
Caution: Miller’s Approximation
F. Najmabadi, ECE102, Fall 2012 (43/59)
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
F. Najmabadi, ECE102, Fall 2012 (44/59)
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
F. Najmabadi, ECE102, Fall 2012 (45/59)
0)()()(||
: Node
0)( : Node
=−+−+++′
=−++−
iogsoimosbLoL
oo
oigsigdsig
sigii
vvsCvvgvCCsrR
vv
vvsCvsCR
vvv
2211
)/1()||(1
)||( sbsb
gsCRrg
Rrgvv mgs
Lom
Lom
sig
o
++
+×
′+′
= gs
mz
gs
mz C
gfCgs
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)
F. Najmabadi, ECE102, Fall 2012 (46/59)
CL + Csb:
)||/1)(( 2 LmsbL RgCC ′+=τ 1 siggd RC=τ
Cgd :
mg/1∞
High-frequency response of a follower – Time-constant method (2)
F. Najmabadi, ECE102, Fall 2012 (47/59)
3. Cgs (cannot use elementary R forms)
High-frequency response of a follower – Time-constant method (3)
F. Najmabadi, ECE102, Fall 2012 (48/59)
3. Cgs (cannot use elementary R forms), continued
)||(1)||(
)]||([)]||(1[
)||()||(
))(||( KVL
oLm
oLsig
x
xgs
oLsigxoLmx
xoLmxoLsigxx
gsmxoLsigxx
gsx
rRgrRR
IVR
rRRIrRgVVrRgIrRRIV
vgIrRRIVvV
′+
′+==
′+=′+
′−′+=
−′+=
=
3 gsgs RC=τ
)||(1)||(
)||/1)(( 21
3211oLm
oLsiggsLmsbLsiggd
H rRgrRR
CRgCCRCbf ′+
′+×+′++=++== τττ
π
Finding the “zero” of a source follower
F. Najmabadi, ECE102, Fall 2012 (49/59)
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
gsmgs Cs
vgi Zv ==− 0
gs
mz
gs
mz C
gfCgs
2 ,
π=−=
F. Najmabadi, ECE102, Fall 2012 (50/59)
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
F. Najmabadi, ECE102, Fall 2012 (51/59)
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
F. Najmabadi, ECE102, Fall 2012 (52/59)
212,1
,11
1,1
11,1
)/11()/11(
)1()1(
dbdbgdogdLL
igdgsin
Lmgdgdogd
Lmgdgdigd
CCCCCCCCC
RgCACCRgCACC
++++=′
+=
′+=−=
′+=−=
LmLoomg
d RgRrrgvvA ′−≡−== )||||( 21
1
1
∞
ro1
sigin
sigsigin
RCRRRC
=⇒
=∞=
1
|| :
τ
)||||( |||| :
212
21
LooL
LooL
RrrCRrrRC
′=⇒=′
τ
gd
mz
gd
mz C
gfCgs
2 ,
π==
)||||( 21
211 LooLsiginH
RrrCRCbf
′+==π
High-f response of Cascode amplifiers
F. Najmabadi, ECE102, Fall 2012 (53/59)
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
F. Najmabadi, ECE102, Fall 2012 (54/59)
∞ ro1
Ri2
Ro2
)||(
1
&
)||(
2111
122
22212212
222
iomv
Lom
Loioomooo
Lomv
RrgA
Rrg
RrRrrgrrR
RrgA
−=
=+
+=++=
= Mid-band Gains & resistances
igdgsin
sbdbgsogd
dbgdLL
vgdogd
vgdigd
CCCCCCCC
CCCCACC
ACC
,11
212,11
22
1,1
11,1
)/11(
)1(
+=
+++=
++=′
−=
−= Miller’s Capacitors
3211
232
2112211
1
21
)||( || : )||( || :
|| :
τττπ
ττ
τ
++==
′=⇒=′=⇒=
=⇒=∞=
bf
RRCRRRCRrCRrRC
RCRRRC
H
LoLLoL
ioio
siginsigsigin
1
1
1
1
2 ,
gd
mz
gd
mz C
gfCgs
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),
F. Najmabadi, ECE102, Fall 2012 (55/59)
d
do
d
do
d
dod
dodododooo
vv
vv
vv
A
vvvvvv
5.02
2
1,1,,
,1,1,2,12
−=×−==
−=−=→−=
Dominant Pole Compensation
F. Najmabadi, ECE102, Fall 2012 (56/59)
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
F. Najmabadi, ECE102, Fall 2012 (57/59)
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
F. Najmabadi, ECE102, Fall 2012 (58/59)
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
F. Najmabadi, ECE102, Fall 2012 (59/59)
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=