The Miller Approximation
■
The “exact” analysis is not particularly helpful for gaining insight into the
frequency response ... consider the effect of Cµ on the input only
It
Cµ
+
Vt
+
−
gmVt
R'out = ro roc RL Vout
−
neglect the feedforward current Iµ in comparison with gm Vπ ... a good
approximation
It = (Vt - Vo) / Zµ
Vo = - gmVt RL / (RL + Rout’) = AvCµ Vt
where AvCµ is the low frequency voltage gain across Cµ
It = Vt (1- Av) / Zµ
Zeff = Vt / It = Zµ / (1 - Av)
1
1
1
1
Z eff = --------------  --------------------- = --------------------------------------------- = --------------jωC µ  1 – A vC 
jω ( C µ ( 1 – A vC ) )
jωC M
µ
µ
C M = ( 1 – A vC )C µ is the Miller capacitor
µ
EE 105 Spring 1997
Lecture 23
Generalized Miller Approximation
■
An impedance Z connected across an amplifier with voltage gain AvZ can
be replaced by an impedance to ground ... multiplied by (1-AvZ)
Z
+
Vi
−
Avo
+
Vi
+
Vo
−
−
Avo
Zeff
+
Vo
−
Zeff = Z /(1 − Avo)
■
Common-emitter and common-source:
AvZ = large and negative for Cµ or Cgd --> capacitance at the input is magnified
■
Common-collector and common-drain:
A vZ ≈ 1 --> capacitance at the input due to Cπ or Cgs is greatly reduced
EE 105 Spring 1997
Lecture 23
Voltage Gain vs. Frequency for CE Amplifier
Using the Miller Approximation
■
The Miller capacitance is lumped together with Cπ, which results in a singlepole low-pass RC filter at the input
RS
+
+
Vs
+
−
rπ
Vπ
Cπ
CM
gmVπ
R'out
Vout
−
−
CM = Cµ(1 + gm R'out)
Transfer function has one pole and no zero after Miller approximation:
–1
ω 3dB = ( r π R S ) ( C π + C M )
–1
ω 3dB = ( r π R S ) [ C π + ( 1 + g m r o r oc R L )C µ ]
–1
–1
ω 3dB ≈ ω 1 from the exact analysis (final term R out ′C µ is missing)
EE 105 Spring 1997
Lecture 23
Multistage Amplifier Frequency Response
■
Summary of frequency response of single-stages:
CE/CS: suffers from Miller effect
CC/CD: “wideband” -- see Section 10.5
CB/CG: “wideband” -- see Section 10.6
(wideband means that the stage operates to near the frequency
limit of the device ... fT)
■
How to find the Bode plot for a general multistage amplifier?
can’t handle n poles and m zeroes analytically --> SPICE
develop analytical tool for an important special case:
* no zeroes
* exactly one “dominant” pole (ω1 << ω2, ω3, ... , ωn)
Ao
V out
----------- = -------------------------------------------------------------------------------------------------------------------------V in
( 1 + j ( ω ⁄ ω1 ) ) ( 1 + j ( ω ⁄ ω2 ) ) ( … ) ( 1 + j ( ω ⁄ ωn ) )
(the example shows a voltage gain ... it could be Iout/Vin or Vout/Iin )
EE 105 Spring 1997
Lecture 23
Finding the Dominant Pole
■
Multiplying out the denominator:
V out
Ao
----------- = ---------------------------------------------------------------------------------------2
n
V in
1 + b 1 jω + b 2 ( jω ) + … + b n ( jω )
The coefficient b1 originates from the sum of jω/ωi factors --
1
1
1
b 1 = ------ + ------ + … + ------ =
ω1 ω2
ωn
n
∑ ω-----i ≈ ω------1
1
1
i
Therefore, if we can estimate the linear coefficient b1 in the demoninator
polynomial, we can estimate of the dominant pole
Procedure: see P. R. Gray and R. G. Meyer, Analysis and Design of Analog
Integrated Circuits, 3rd ed., Wiley, 1994, pp. 502-504.
1. Find circuit equations with current sources driving each capacitor
2. Denominator polynomial is determinant of the matrix of coefficients
3. b1 term comes from a sum of terms, each of which has the form:
RTj Cj
where Cj is the jth capacitor and RTj is the Thévenin resistance
across the jth capacitor terminals (with all capacitors open-circuited)
EE 105 Spring 1997
Lecture 23
Open-Circuit Time Constants
■
The dominant pole of the system can be estimated by:

1
ω 1 ≈ ----- = 
b1

n
∑
j

R Tj C j

–1

= 

n
∑
1

τ j

–1
,
where τj = RTj Cj is the open-circuit time constant for capacitor Cj
■
This technique is valuable because it estimates the contribution of each capacitor
to the dominant pole frequency separately ... which enables the designer to
understand what part of a complicated circuit is responsible for limiting the
bandwidth of the amplifier.
EE 105 Spring 1997
Lecture 23
Example: Revisit CE Amplifier
■
Small-signal model:
Cµ
RS
Vs
■
+
−
+
+
rπ
Vπ
Cπ
gmVπ
ro roc
−
RL
Vout
−
Apply procedure to each capacitor separately
1. Cπ’s Thévenin resistance is found by inspection as the resistance across its
terminals with all capacitors open-circuited:
R Tπ = R S r π = R in ′ --> τ C = R Tπ C π
πo
2. Cµ’s Thévenin resistance is not obvious --> must use test source and network
analysis
EE 105 Spring 1997
Lecture 23
Time Constant for Cµ
■
Circuit for finding RTµ
− vt
+
R'in = RS rπ
vπ
+
it
gmvπ
R'out = ro roc RL
−
vπ is given by:
v π = – i t ( R s r π ) = – i t R in ′
vo is given by:
v o = – i o R out ′ = ( i t – g m v π )R out ′ = i t ( g m R in ′ + 1 )R out ′
vt is given by:
v t = v o – v π = i t ( ( 1 + g m R in ′ )R out ′ + R in ′ )
solving for RTµ = vt / it
R Tµ = R in ′ + R out ′ + g m R in ′R out ′
τC
µo
= R Tµ C µ = ( R in ′ + R out ′ + g m R in ′R out ′ )C µ
EE 105 Spring 1997
Lecture 23
Estimate of Dominant Pole for CE Amplifier
■
Estimate dominant pole as inverse of sum of open-circuit time constants
–1
ω 1 = ( R Tπ C π + R Tµ C µ ) = R in ′C π + ( R in ′ + R out ′ + g m R in ′R out ′ )C µ
■
inspection --> identical to “exact” analysis (which also assumed ω 1 « ω 2 )
Advantage of open-circuit time constants: general technique
Example: include Ccs and estimate its effect on ω1
EE 105 Spring 1997
Lecture 23
Multistage Amplifier Frequency Response
■
Applying the open-circuit time constant technique to find the dominant pole
frequency -- use CS/CB cascode as an example
V+
iSUP
iOUT
+
Q2
RL
RS
M1
Vs
+
−
VBIAS
+
−
vOUT
−
V−
■
Systematic approach:
1. two-port small-signal models for each stage (not the device models!)
2. carefully add capacitances across the appropriate nodes of two-port models,
which may not correspond to the familiar device configuation for some
models
EE 105 Spring 1997
Lecture 23
Two-Port Model for Cascode
■
The base-collector capacitor Cµ2 is located between the output of the CB stage
(the collector of Q2) and small-signal ground (the base of Q2)
Cgd1
RS
Vs
+
−
I2
+
+
Vgs1
Cgs1
Iout
gm1Vgs1
ro1
1/gm2
Cπ2
−I2
βo2ro2
Cµ2
−
RL Vout
−
We have omitted Cdb1, which would be in parallel with Cπ2 at the output of the
CS stage, and Ccs2 which would be in parallel with Cµ2. In addition, the current
supply transistor will contribute additional capacitance to the output node.
■
Time constants
τC
τC
gd1o
gs1o
= R S C gs1
= ( R in ′ + R out ′ + g m1 R in ′R out ′ )C gd1
1
1
where R in ′ = R S and R out ′ = r o1  --------- ≈ --------g  g
m2
m2
Since the output resistance is only 1/gm2, the Thévenin resistance for Cgd1 is not
magnified (i.e., the Miller effect is minimal):
τC
gd1o
 g m1 

1
=  R S + --------- +  --------- R S C gd1 ≈ R S ( 1 + g m1 ⁄ g m2 )C gd1
g m2  g m2 

EE 105 Spring 1997
Lecture 23
Cascode Frequency Response (cont.)
■
The base-emitter capacitor of Q2 has a time constant of
τC
■
The base-collector capacitor of Q2 has a time constant of
τC
■
π2o
1
=  --------- C π2
g 
m2
µ2o
= ( β o2 r o2 r oc R )C µ2 ≈ R L C µ2
L
Applying the theorem, the dominant pole of the cascode is approximately
–1
ω 3db ≈ τ C
gs1o
+ τC
gd1o
+ τC
π2o
+ τC
µ2o
–1
1
ω 3db ≈ R S C gs1 + R S ( 1 + g m1 ⁄ g m2 )C gd1 +  --------- C π2 + R L C µ2
g 
m2
EE 105 Spring 1997
Lecture 23
Gain-Bandwidth Product
■
A useful metric of an amplifier’s frequency response is the product of the lowfrequency gain |Avo| and the 3 dB frequency ω3dB
For the cascode, the gain is |Avo| = |-gm1RL| and the gain-bandwidth product is
g m1 R L
A vo ω 3dB ≈ ------------------------------------------------------------------------------------------------------------------------------------------1
R S C gs1 + R S ( 1 + g m1 ⁄ g m2 )C gd1 +  --------- C π2 + R L C µ2
g 
m2
■
If the voltage source resistance is small, then
g m1 R L
A vo ω 3dB ≈ ---------------------------------------------------(C ⁄ g + R C )
π2
m2
L µ2
which has the same form as the common-base gain-bandwidth product (and
which is much greater than the Miller-degraded common-source)
EE 105 Spring 1997
Lecture 23
EE 105 Spring 1997
Lecture 23