high-frequency response of a cg...
TRANSCRIPT
High-frequency response of a CG amplifier
gddbLL CCCC ++=′
sbgsin CCC +=
Cgs between source & ground
Cgd between drain & ground
“Input Pole” “Output Pole”
Low-pass filter
Low-pass filter
Mid-band Amp
High-f response of a CG amplifier – Exact Solution (1)
gddbLL CCCC ++=′
sbgsi 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)
)/1||(11
/1/1
)1/(11
11
11
0)( : Node
msigisigm
m
sig
i
sigmsigisigmsigisigmsig
i
isigmisigii
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
LLmsigiLm
sigi
i
sig
o
RCsgRsCRg
RRR
vv
′′+×
+×′×
+=
11
)/1||(11)(
Compact solution can be found by ignoring ro (i.e., ro → ∞)
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
=
Open-Circuit Time-Constants Method
...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)
1. Consider Ci :
)]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)
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-frequency response of a CG amplifier
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
dbL CC +
Csb is shorted out. Cgd is between output and input!
“Input Pole?” “Output Pole?”
Miller’s Theorem
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
12 VAV ⋅=
Other parts of the circuit
A
ZZ 112
−=
AZZ−
=11
12 VAV ⋅=
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
−≈
+
−=
+
−=
+
−=−=
Example of Miller’s Theorem: Inverting amplifier
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
A
ZZ 112
−=
AZZ−
=11
12 VAV ⋅=
)/11( /11
)1( 1
1
21
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
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
Output Pole (C’L ):
)||( 2 LoL RrC ′′=τ 1 siginRC=τ
Input Pole (Cin ):
)||( 21
1 LoLsiginH
RrCRCbf
′′+==π
or
∞
LgddbL
Lomgdgsin
CCCCRrgCCC
++=′
′++= )]||(1[
Miller & time-constant method: 1. Same b1 and same fH as 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!
High-f response of a CS amplifier – Exact solution
Solving the circuit (node voltage method):
gdgsgdgsdbL
LoLsigin
mgdLom
sig
o
CCCCCCbRrCRCb
sbsbgsCRrg
vv
+++=
′′+=
++
−×′−=
))((
)||(
1)/1()||(
2
1
221
LgddbL
Lomgdgsin
CCCCRrgCCC
++=′
′++= )]||(1[
)||( 21
1 LoLsiginH
RrCRCbf
′′+==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 (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 gain Bode 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 resistor (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).
1) By definition,
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:
Finding the “zero” of the CS amplifier
gdz
gsmgs Cs
vgi Z v ==− 0
0 )( :Zero == zo ssv
gsmvgi =0=i
gd
mz C
gs =
Zero of CS amplifier can play an important role in stability of feedback amplifiers
zf 2pf 1pf zf 2pf 1pf
Since the input pole is at
Small Rsig can push fp2 to very large values!
) 2/(1)1/(2 1 siginRCππτ =
12 of Case ppz fff >>> 12 of Case pzp fff >>
Caution: 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= =