Lecture 22
Frequency Response of Amplifiers (II)
VOLTAGE AMPLIFIERS
Outline
1.
2.
3.
Full Analysis
Miller Approximation
Open Circuit Time Constant
Reading Assignment:
Howe and Sodini, Chapter 10, Sections 10.1-10.4
6.012 Spring 2007
Lecture 22
1
Common Emitter Amplifier
V+
iSUP
iOUT
RS
+
+
V
− s
RL
vOUT
−
+
V
− BIAS
V−
•
Operating Point Analysis
– vs=0, RS = 0, ro → ∞, roc → ∞, RL → ∞
– Find VBIAS such that IC=ISUP with the BJT in the
forward active region
6.012 Spring 2007
Lecture 22
2
Frequency Response Analysis of the
Common Emitter Amplifier
Cµ
RS
+
+
Vs
+
−
r
V
gmV
C
−
ro ⎢⎢roc
RL
Vout
−
(a)
•
Frequency Response
– Set VBIAS = 0.
– Substitute BJT small signal model (with capacitors)
including RS, RL, ro, roc
– Perform impedance analysis
6.012 Spring 2007
Lecture 22
3
1. Full Analysis of CE Voltage Amplifier
Cµ
RS
+
+
Vs
+
−
r
V
gmV
C
ro ⎢⎢roc
RL
Vout
−
−
(a)
Replace voltage
source and resistance
with current source
and resistance using
Norton Equivalent
Cµ
+
+
Is
R'in
V
gmV
C
R'out
−
−
R'in = RS ⎢⎢r
Node 1:
Vout
R'out = ro ⎢⎢roc ⎢⎢RL
(b)
Vπ
Is =
+ jωC π Vπ + jωCµ (Vπ − Vout )
R′in
Node 2:
Vout
g m Vπ +
= jωCµ (Vπ − Vout )
R′out
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Lecture 22
4
Full Frequency Response Analysis (contd.)
• Re-arrange 2 and obtain an expression for Vπ
• Substituting it into 1 and with some manipulation, we
can obtain an expression for Vout / Is:
(
)
− Rin
′ Rout
′ gm − jωCµ
Vout
=
Is
1 + jω Rout
′ Cµ + Rin
′ Cµ + Rin
′ Cπ + g m Rout
′ Rin
′ Cµ − ω 2 Rout
′ Rin
′ Cµ Cπ
(
)
Changing input current source back to a voltage source:
⎛ rπ ⎞⎛
Cµ ⎞
′ ⎜
−gm Rout
⎟⎜1 − jω
⎟
R
+
r
g
Vout
⎝ S π ⎠⎝
m ⎠
=
Vs 1 + jω (Rout
′ C µ + Rin′ C µ (1 + gm Rout
′ ) + Rin′ C π )− ω 2 Rout
′ Rin′ C µ C π
where R ′in = R S || rπ and R ′out = ro || roc || R L
We can ignore zero at gm/Cµ because it is higher than ωT.
The gain can be expressed as:
Vout
Avo
Avo
=
=
Vs
(1 + jωτ1 )(1 + jωτ2 ) 1 − jω(τ1 + τ2 ) − ω2 τ1τ2
where Avo is the gain at low frequency and τ1 and τ2 are
the two time constants associated with the capacitors
6.012 Spring 2007
Lecture 22
5
Denominator of the System Transfer Function
Vout
Vs
dB
− 20 decade
Avo
− 40
1/τ1
1/τ2
dB
decade
ω
log scale
τ1 + τ2 = Rout
′ Cµ (1+ gm Rout
′ ) + Rin
′ C µ + Rin
′ Cπ
τ1 • τ2 = Rout
′ Rin
′ Cµ C π
We could solve for τ1 and τ2 but is algebraically complex.
•However, if we assume that τ1 >> τ2 ,⇒ τ1 + τ2 ≈ τ1.
•This is a conservative estimate since the true τ1 is actually
smaller and hence the true bandwidth is actually larger
than:
[
]
τ1 ≈ Rin
′ Cπ + C µ (1 + g m Rout
′ ) + Rout
′ Cµ
Then:
ω3dB =
6.012 Spring 2007
1
τ1
=
1
Rin
′ Cπ + Cµ (1 + g mRout
′ ) + Rout
′ Cµ
[
]
Lecture 22
6
2. The Miller Approximation
Effect of Cµ on the Input Impedance:
It
Cµ
+
Vt
+
−
gmVt
R'out = ro ⎢⎢roc ⎢⎢RL Vout
−
The input impedance Zi is determined by applying a test
voltage Vt to the input and measuring It:
Vout = −gm Vt Rout
′ + I t Rout
′
The Miller Approximation assumes that current through
Cµ is small compared to the transconductance generator
I t << g m Vt
Vout ≈ −gm Vt Rout
′
We can relate Vt and Vout by
It
Vt − Vout =
jωC µ
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The Miller Approximation (contd.)
After some Algebra:
Vt
1
1
= Z eff =
=
It
′ ) jω C 1 − A
jω Cµ (1+ gm Rout
µ
vC
(
µ
)
The effect of Cµ at input is that Cµ is “Miller multiplied”
by (1-AvCµ)
Generalized “Miller Effect”
Z
+
Vi
−
Avo
+
Vi
+
Vo
−
−
Avo
Zeff
+
Vo
−
Zeff = Z /(1 − Avo)
• An impedance connected across an amplifier with
voltage gain Avo can be replaced by an an impedance
to ground … divided by (1-Avo)
• Avo is large and negative for common-emitter and
common-source amplifiers
• Capacitance at input is magnified.
Zeff =
6.012 Spring 2007
Z
(1− Avo )
Lecture 22
8
Frequency Response of the CE Voltage
Amplifier Using Miller Approximation
RS
+
+
Vs
+
−
r
V
CM
C
g mV
R'out
Vout
−
−
CM = Cµ(1 + gm R'out)
• The Miller capacitance is lumped together with Cπ,
which results in a single pole low pass filter at the
input
⎛ rπ ⎞
Vout
⎟⎟ R out
= − g m ⎜⎜
′
Vs
r
+
R
⎝ π
S⎠
⎡
⎤
1
⎥
⎢
ω
C
+
C
R
||
r
1
+
j
( π M )( S π )⎦⎥
⎢⎣
• At low frequency (DC) the small signal voltage gain is
⎛ r
⎞
Vout
π
⎟⎟ Rout
′
= −gm ⎜⎜
Vs
⎝ rπ + RS ⎠
• The frequency at which the magnitude of the voltage
gain is reduced by 1/√2 is
⎤
⎤⎡
⎡
1
1
1
⎥
⎥⎢
ω3dB =
=⎢
′ )Cµ ⎥⎦
(Rs || rπ )(Cπ + CM ) ⎢⎣(Rs || rπ )⎥⎦⎢⎣ Cπ + (1 + gm Rout
6.012 Spring 2007
Lecture 22
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3. Open Circuit Time Constant Analysis
Assumptions:
• No zeros
• One “dominant” pole (1/τ1 << 1/τ2 , 1/τ3 … 1/τn )
• N capacitors
Vout
Avo
=
Vs
(1 + jωτ1 )(1 + jωτ2 )(1 + jωτn )
The example shows a voltage gain; however, it could be
Iout/Vs or Vout/Is.
Multiplying out the denominator:
Vout
Avo
=
Vs
1 + b1( jω) + b2 ( jω)2 + ... + bn ( jω)n
where b1 = τ1 + τ2 + τ3 +….+ τn
It can be shown that the coefficient b1 can be found
exactly [see Gray & Meyer, 3rd Edition, pp. 502-506]
⎛N
⎞ ⎛N
⎞
b1 = ⎜⎜ ∑ RTi Ci ⎟⎟ = ⎜⎜ ∑ τ Cio ⎟⎟
⎝ i=1
⎠ ⎝ i
⎠
• τCio is the open-circuit time constant for capacitor Ci
• Ci is the ith capacitor and RTi is the Thevenin
resistance across the ith capacitor terminals (with all
capacitors open-circuited)
6.012 Spring 2007
Lecture 22
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Open Circuit Time Constant Analysis
Estimating the Dominant Pole
The dominant pole of the system can be estimated by:
b1 = τ1 + τ2 + τ3 +….+ τn
⎛N
⎞
b1 = ⎜⎜ ∑ RTi Ci ⎟⎟
⎝ i=1
⎠
≈
τ1 =
1
ω1
RTiCi is the open-circuit time constant for capacitor Ci
Power of the Technique:
• Estimates the contribution of each capacitor to the
dominant pole frequency separately
• Enables the designer to understand what part of a
complicated circuit is responsible for limiting the
bandwidth of amplifier
• The approximate magnitude of the Bode Plot is
6.012 Spring 2007
Lecture 22
11
Common Emitter Amplifier Analysis
Using OCT
Cµ
RS
+
+
Vs
+
−
r
V
gmV
C
ro ⎢⎢roc
RL
−
Vout
−
(a)
From the Full Analysis
⎛ r
⎞⎛
Cµ ⎞
π
⎟⎟ ⎜⎜1− jω ⎟⎟
′ ⎜⎜
−gmRout
gm ⎠
Vout
⎝ RS + rπ ⎠ ⎝
=
Vs 1+ jω Rout
′ Cµ + Rin
′ Cµ(1+ gm Rout
′ )+ Rin
′ Cπ − ω2 Rout
′ Rin
′ CµCπ
(
)
where R ′in = RS || rπ and R ′out = ro || roc || RL
b1 = Rout
′ ) + Rin
′ Cπ
′ Cµ + Rin
′ C µ (1 + g m Rout
ω3dB ≈
6.012 Spring 2007
1
1
=
′ Cµ (1+ +g m Rout
′ ) + Rin
′ Cπ
b1 Rout
′ C µ + Rin
Lecture 22
12
Common Emitter Amplifier Analysis
Using OCT—Procedure
1.
2.
3.
Eliminate all independent sources [e.g. Vs → 0]
Open-circuit all capacitors
Find the Thevenin resistance by applying it and
measuring vt.
Time Constant for Cπ
Result obtained by inspection
RT π = RS || rπ
τ C = RT πC π
πo
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Lecture 22
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Common Emitter Amplifier Analysis
Using OCT—Time Constant for Cµ
Using the same procedure
Let R in
′ = RS || rπ and R ′out = ro || roc || RL
vπ
−it =
Rin
′
vt + v π
it =
+ g m vπ
Rout
′
Eliminate vπ:
vt
= RTµ = Rout
′ + Rin
′ (1 + g m Rout
′ )
it
τ Cµo = RT µ Cµ = [Rout
′ + Rin
′ (1 + g m Rout
′ )]Cµ
6.012 Spring 2007
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Common Emitter Amplifier Analysis
Using OCT—Dominant Pole
Summing individual time constants
b1 = RT πC π + RTµ Cµ
b1 = Rout
′ Cµ + Rin
′ C µ (1 + g m Rout
′ ) + Rin
′ Cπ
Assume τ1 >> τ2
b1 = τ1 + τ2 ≈ τ1
b1 = Rout
′ Cµ + Rin
′ C µ (1 + g m Rout
′ ) + Rin
′ Cπ
ω3dB ≈
1
1
=
′ Cµ (1+ gm Rout
′ ) + Rin
′ Cπ
b1 Rout
′ C µ + Rin
This result is very similar to the Miller Effect calculation
Additional term R’outCµ taken into account
6.012 Spring 2007
Lecture 22
15
Compare the Three Methods of Analyzing
the Frequency Response of CE Amplifier
Full Analysis—
ω3dB ≈
1
τ1
=
1
′ Cµ (1 + gm Rout
′ ) + Rin
′ Cπ
Rout
′ Cµ + Rin
Miller Approximation—
⎤
⎡ 1 ⎤⎡
1
⎥
ω3dB = ⎢
⎥⎢
′ ⎦ ⎢⎣ Cπ + (1 + gm Rout
′ )Cµ ⎥⎦
⎣ Rin
Open Circuit Time Constant—
ω3dB ≈
6.012 Spring 2007
1
1
=
′ Cµ (1+ gm Rout
′ ) + Rin
′ Cπ
b1 Rout
′ C µ + Rin
Lecture 22
16
What did we learn today?
Summary of Key Concepts
•
Full Analysis
– Assumes that τ1 + τ2 ≈ τ1
1
1
ω3dB ≈ =
τ1 Rout
′ Cµ (1 + gm Rout
′ ) + Rin
′ Cπ
′ Cµ + Rin
•
Miller Approximation
– Does not take into account R’out
⎤
⎡ 1 ⎤⎡
1
⎥
ω3dB = ⎢
⎥⎢
′ ⎦ ⎢⎣ Cπ + (1 + gm Rout
′ )Cµ ⎦⎥
⎣ Rin
•
Open Circuit Time Constant (OCT)
– Assumes a dominant pole as full analysis
1
1
ω3dB ≈ =
′ Cµ (1+ gm Rout
′ ) + Rin
′ Cπ
b1 Rout
′ C µ + Rin
6.012 Spring 2007
Lecture 22
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