Microelectronic Circuits
Feedback
Ching-Yuan Yang
National Chung-Hsing University
Department of Electrical Engineering
Outline
The General Feedback Structure
Some Properties of Negative Feedback
The Four Basic Feedback Topologies
The Series-Shunt Feedback Amplifier
The Series-Series Feedback Amplifier
The Shunt-Shunt and Shunt-Series Feedback Amplifier
Determining the Loop Gain
The Stability Problem
Effect of Feedback on the Amplifier Poles
Stability study Using Bode Plots
Frequency Compensation
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Ching-Yuan Yang / EE, NCHU
The General Feedback Structure
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8-2
General structure of the feedback amplifier
xo = Axi , xf = βxo , xi = xs − xf
Af ≡
xo
A
=
x s 1 + Aβ
β : feedback factor
Aβ : loop gain
1 + Aβ : amount of feedback
If Aβ >> 1, then Af ≈ 1/β.
The gain of the feedback amplifier is almost entirely determined by the
feedback network.
Aβ
Feedback signal: x f =
xs
1 + Aβ
For Aβ >> 1 we see that xf ≈ xs , which implies that xi Š 0. (xi : error signal)
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Some Properties of Negative Feedback
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Some Properties of Negative Feedback
Desensitize the gain
Extend the bandwidth of the amplifier
Reduce the effect of noise
Reduce nonlinear distortion
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Gain Desensitivity
Assume that β is constant.
Af ≡
dA f
xo
A
=
x s 1 + Aβ
=
1
(1 + Aβ )2
dA f =
dA
(1 + Aβ )2
dA
dA f
Af
=
1
dA
(1 + Aβ ) A
The percentage change in Af (due to variations in some circuit parameter) is
smaller than the percentage change in A by amount of feedback.
The amount of feedback, 1 + Aβ, is also known as the desensitivity factor.
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Bandwidth Extension
Consider an amplifier whose high-frequency response is characterized by a
single pole.
AM
AM : the midband gain
A(s ) =
s
1+
ωH : the upper 3-dB frequency
ωH
Apply the negative feedback with a frequency-independent factor β :
AM
A(s )
1 + AM β
Close - loop gain A f (s ) =
=
s
1 + βA(s ) 1 +
ω H (1 + AM β )
The feedback amplifier: (high-frequency response)
AM
Midband gain AMf =
1 + AM β
Upper 3-dB frequency ωHf = ωH (1 + AM β )
The amplifier band width is increased by the same factor, (1 + AM β ), by which
its midband gain is decreased, maintain the gain-bandwidth product constant.
Similarly, the low-frequency response of the feedback amplifier has a lower 3-dB
ωL
frequency
ω Lf =
1 + AM β
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Noise Reduction
Vs : input signal
Vo : output signal
Vn : noise or interference
A1 : gain
The amplifier suffers from noise and the noise can be
assumed to be introduced at the input of the amplifier.
Singal - to - noise ratio
S Vs
=
N Vn
Negative feedback is applied to improve SNR:
(clean amp.)
Vo = Vs
A1A2
A1
+ Vn
1 + A1A2 β
1 + A1A2 β
∴
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S Vs
=
A2
N Vn
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Reduction in Nonlinear Distortion
Curve (a) shows the amplifier transfer characteristic without feedback.
Curve (b) shows the characteristic with negative feedback (β = 0.01)
applied.
The open-loop voltage transfer characteristic of the amplifier is change from
1000 to 100 and then to 0. [curve (a)]
Apply negative feedback with β = 0.01 to the amplifier [curve (b)] :
the slope of the steepest segment:
Af 1 =
the slope of the next segment:
1000
= 90.9
1 + 1000 × 0.01
Af 2 =
100
= 50
1 + 100 × 0.01
The order-of-magnitude change in slop is considerably reduced.
The price paid is a reduction in voltage gain.
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The Four Basic Feedback Topologies
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Four basic feedback topologies
Base on the quantity to be amplified (voltage or current) and on the desired
form of the output (voltage or current), amplifiers can be classified into four
categories.
Voltage amplifiers
Current amplifiers
Transconductance amplifiers
Transresistance amplifiers
Four basic feedback topologies:
Voltage-sampling series-mixing (series-shunt) topology
Current-sampling shunt-mixing (shunt-series) topology
Current-sampling series-mixing (series-series) topology
Voltage-sampling shunt-mixing (shunt-shunt) topology
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Voltage Amplifiers
Characteristics:
Input signal: voltage
output signal: voltage
Voltage-controlled voltage source
High input impedance
Low output impedance
Feedback topology
Voltage-sampling series-mixing (series-shunt) topology
The feedback network samples the output voltage, and the feedback signal xf
is a voltage that can be mixed with the source voltage in series.
“Series” refers to the connection at the input and “shunt” refers to the
connection at the output.
Characteristics:
⌦ Stabilize the voltage gain
⌦ Higher input resistance
(series connection at the input)
⌦ Lower output resistance
(parallel connection at the output)
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Voltage-sampling series-mixing (series-shunt) topology
Example
v i = V+ − V−
Vs u
vi u
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Vo u
V− u
vi v enegative feedback
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Current Amplifiers
Characteristics:
Input signal: current
output signal: current
Current-controlled current source
Low input impedance
High output impedance
Feedback topology
Current-sampling shunt-mixing (shunt-series) topology
The feedback network samples the output current, and the feedback signal
xf is a current that can be mixed in shunt with the source current.
“Shunt” refers to the connection at the input and “series” refers to the
connection at the output.
Characteristics:
⌦ Stabilize the current gain
⌦ Lower input resistance
(shunt connection at the input)
⌦ Higher output resistance
(series connection at the output)
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Current-sampling shunt-mixing (shunt-series) topology
Example
Is u
ii (Ib1) u
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Ic1 u
Vc1 v
Ic2 (Io) v
Ie2 (Io /α) v
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If u
ii (Ib1) v enegative feedback
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Transconductance Amplifiers
Characteristics:
Input signal: voltage
output signal: current
Voltage-controlled current source
High input impedance
High output impedance
Feedback topology
Current-sampling series-mixing (series-series) topology
The feedback network samples the output current, and the feedback signal
xf is a voltage that can be mixed with the source voltage in series.
“Series” refers to the connection at the input and “series” refers to the
connection at the output.
Characteristics:
⌦ Stabilize the transconductance gain
⌦ Higher input resistance
(series connection at the input)
⌦ Higher output resistance
(series connection at the output)
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Current-sampling series-mixing (series-series) topology
Example
Vs u
vi (vbe) u
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Ic1 u
vc1 v
Ic2 (Io) v
vc2 u
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Io u
Io /α u
Vf u
vi v enegative feedback
Ching-Yuan Yang / EE, NCHU
Transresistance Amplifiers
Characteristics:
Input signal: current
output signal: voltage
Current-controlled voltage source
Low input impedance
Low output impedance
Feedback topology
Voltage-sampling shunt-mixing (shunt-shunt) topology
The feedback network samples the output voltage, and the feedback signal
xf is a current that can be mixed in with shunt the source current.
“Shunt” refers to the connection at the input and “shunt” refers to the
connection at the output.
Characteristics:
⌦ Stabilize the transresistance gain
⌦ Lower input resistance
(shunt connection at the input)
⌦ Lower output resistance
(shunt connection at the output)
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Voltage-sampling shunt-mixing (shunt-shunt) topology
Example
Norton’s
theorem
Is u
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ii u
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Vo v
If u
ii v enegative feedback
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Two-Port Network Parameters
y parameters
Equivalent circuit:
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z parameters
Equivalent circuit:
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h parameters
Equivalent circuit:
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g parameters
Equivalent circuit:
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The Series-Shunt Feedback Amplifier
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The Ideal Situation of the Series-Shunt Feedback Amplifier
Ideal structure
A circuit: a unilateral open-loop amplifier
Ri : input resistance A : voltage gain Ro : output resistance
β circuit: an ideal voltage-sampling series-mixing feedback network
The source and load resistances are included inside the A circuit.
The β circuit does not load the A circuit. Do not change A ≈ Vo /Vi.
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Equivalent circuit
The close-loop gain A f ≡
Vo
A
=
(A and β have reciprocal units)
Vs 1 + Aβ
Input resistance
Rif ≡
V + βVo
V + βAVi
Vs
Vs
V
=
= R i s = Ri i
= Ri i
= Ri (1 + Aβ )
Vi
Vi
Vi
I i Vi /Ri
The series-mixing increases the input resistance by (1 + Aβ).
General form Zif (s) = Zif (s)[1 + A(s)β(s)]
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Output resistance (Vs = 0)
Output resistance
Rof ≡
Vt − AVi
Ro
V + AβVt
I = t
Ro
where I =
Rof ≡
Vt
I
and
Vi = −V f = − βVo = − βVt (for Vs = 0)
Vt
Ro
=
I 1 + Aβ
The voltage-sampling feedback reduces the output resistance by (1 + Aβ).
General form Z of (s ) =
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Z o (s )
1 + A(s )β (s )
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The Practical Situation
Derivation of the A circuit and β circuit for the series-shunt feedback amplifier.
Block diagram of a practical series-shunt feedback amplifier
Rif and Rof are the input and output resistances, respectively, of the feedback
amplifier, including Rs and RL.
The actual input and output resistances of the feedback amplifier usually
exclude Rs and RL :
Rin = Rif − Rs
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The circuit in
and
Rout =
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1
1
1
−
Rof R L
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with the feedback network represented by its h parameters.
(h parameters)
The source and load resistances should be lumped with the basic amplifier.
In addition, the resistances in the feedback should be also lumped with the
basic amplifier.
h-parameter two-port feedback network:
Neglect h21, the forward transmission effect of the feedback network, and
thus omit the controlled source h21I1.
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The circuit in
after neglecting h21.
h21I1 is neglected.
Determine β :
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β = h12 ≡
V1
V2
I1 = 0
(port 1 open-circuited)
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Summary of the Series-Shunt Feedback Amplifier
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Example 8.1 An op amp connected in the noninverting configuration
The op amp has an open-loop gain μ, a differential input resistance Rid, and an
output resistance ro. Find expressions for A, β, the close-loop gain Vo /Vs, the
input resistance Rin, and the output resistance Rout. (μ = 104, Rid = 100kΩ, ro =
1kΩ, RL = 2kΩ, R1 = 1kΩ, R2 = 1MΩ, and Rs = 10kΩ.)
The feedback network consists
R2 and R1. This network
samples the output voltage Vo
and and provides a voltage
signal (across R1) that mixed in
series with the input source Vs.
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A circuit:
A≡
[R L (R1 + R2 )]
V 'o
Rid
=μ
V 'i
[R L (R1 + R2 )] + ro Rid + Rs + (R1 R2 )
A ≈ 6000 V/V
β circuit:
β ≡
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V 'f
V 'o
=
R1
≈ 10 −3 V/V
R1 + R2
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6000
Vo
A
=
=
= 857 V/V
7
Vs 1 + Aβ
Input resistance Rif = Ri(1 + Aβ ) where Ri is the input resistance of the A circuit.
Voltage gain A f ≡
Ri = Rs + Rid + (R1 ||R2) ≈ 111kΩ
Rif = 111×7 = 777kΩ
Rin = Rif − Rs = 739kΩ
Ro
where Ro is the output resistance of the A circuit.
1 + Aβ
Ro = ro || RL || (R1 +R2) ≈ 667 Ω
Output resistance Rof =
667
= 95.3Ω
7
∵ Rof = Rout || RL
Rof =
Rout ≈ 100 Ω
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The Series-Series Feedback Amplifier
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The Ideal Situation of the Series-Series Feedback Amplifier
Ideal structure
A circuit: a unilateral open-loop amplifier
Ri : input resistance A : transconductance gain Ro : output resistance
β circuit: an ideal current-sampling series-mixing feedback network
The source and load resistances are included inside the A circuit.
The β circuit does not load the A circuit. Do not change A ≈ Io /Vi.
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Equivalent circuit
The close-loop gain A f ≡
Io
A
=
(A and β have reciprocal units)
V s 1 + Aβ
Input resistance
Rif ≡
Vs
Vs
V
V + βVo
V + βAVi
=
= R i s = Ri i
= Ri i
= Ri (1 + Aβ )
I i Vi /Ri
Vi
Vi
Vi
The series-mixing increases the input resistance by (1 + Aβ ).
General form Zif (s) = Zi (s)[1 + A(s)β(s)]
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Output resistance (Vs = 0)
Output resistance
Rof ≡
V
It
In this case, Vi = −V f = − βI o = − βI t (for Vs = 0)
V = (I t − AVi )Ro = (I t + AβI t )Ro
Rof ≡
V
= Ro (1 + Aβ )
It
The voltage-sampling feedback increases the output resistance by (1 + Aβ ).
General form Z of (s ) = Z o (s )[1 + A(s )β (s )]
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The Practical Situation
Derivation of the A circuit and β circuit for the series-series feedback amplifier.
Block diagram of a practical series-shunt feedback amplifier
Rif and Rof are the input and output resistances, respectively, of the feedback
amplifier, including Rs and RL.
The actual input and output resistances of the feedback amplifier usually
exclude Rs and RL :
Rin = Rif − Rs
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and
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Rout = Rof − R L
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The circuit in
with the feedback network represented by its z parameters.
(z parameters)
The source and load resistances should be lumped with the basic amplifier.
In addition, the resistances in the feedback should be also lumped with the
basic amplifier.
z-parameter two-port feedback network:
Neglect z21, the forward transmission effect of the feedback network, and
thus omit the controlled source z21I1.
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The circuit in
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after neglecting z21.
z21I1 is neglected.
Determine β :
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β = z12 ≡
V1
I2
I1 = 0
(port 1 open-circuited)
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Summary of the Series-Series Feedback Amplifier
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Example 8.2 A feedback triple is composed of three gain stages with
series-series feedback
Assume that the bias circuit causes IC1 = 0.6mA, IC2 = 1mA, and IC3 = 4mA. Using
these values and assuming hfe = 100 and ro = ∝, find the open-loop gain A, the
feedback factor β, the closed loop gain Af = Io /Vs, the voltage gain Vo /Vs, the input
resistance Rin = Rif, and the output resistance Rof (between nodes Y and Y’). Now, if
ro of Q3 is 25kΩ, estimate an approximate value of the output resistance Rout.
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A circuit:
Determine A:
− α (RC1 rπ 2 )
Vc1
=
= −14.92 V/V
'
Vi
re1 + [RE1 (RF + RE 2 )]
Vc 2
= −gm 2 RC 2 (h fe + 1)[re 3 + (R E 2 (RF + RE1 ))] = −131.2 V/V
Vc1
I o'
I
1
= e3 =
= 10.6 mA/V
Vc 2 Vb 3 re 3 + (RE 2 (RF + RE1 )
{
⇒
A≡
}
I o' Vc1 Vc 2 I o'
=
×
×
= 20.7 A/V
Vi ' Vi ' Vc1 Vi '
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β circuit:
Determine β :
β=
V f'
I o'
=
RE 2
× RE1 = 11.9 Ω
RE 2 + RF + RE1
20.7
Io
A
=
=
= 83.7 mA/V
Vs 1 + Aβ 1 + 20.7 × 11.9
Vo − I c RC 3
Voltage gain:
=
≈ − A f RC 3 = −83.7 × 10 −3 × 600 = −50.2 V/V
Vs
Vs
Input resistance of the A circuit: Ri = (h fe + 1)[re1 + (RE1 (RF + RE 2 ))] = 13.65 kΩ
Closed-loop gain: A f ≡
Input resistance of the feedback amplifier: Rif = Ri (1 + Aβ ) = 3.34 MΩ
Output resistance Ro of the A circuit: The resistance looking between Y and Y’.
RC 2
= 143.9 Ω
Ro = [RE 2 (RF + RE1 )] + re 3 +
h fe + 1
Output resistance of the feedback amplifier:
Rof = Ro (1 + Aβ ) = 35.6 kΩ
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Find the output resistance from the equivalent circuit.
Rout = ro 3 + (1 + gm 3ro 3 )(Rof rπ 3 )
≈ ro 3 [1 + gm 3 (Rof rπ 3 )]
= 25[1 + 100(35.6 1)
= 2.5 MΩ
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Homework #1
Problems 6, 14, 24, 25, 27, 37
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The Shunt-Shunt Feedback Amplifier
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The Ideal Situation of the Shunt-Shunt Feedback Amplifier
Ideal structure
A circuit: a unilateral open-loop amplifier
Ri : input resistance A : transresistance gain Ro : output resistance
β circuit (transconductance): an ideal voltage-sampling shunt-mixing
feedback network
The source and load resistances are included inside the A circuit.
The β circuit does not load the A circuit. Do not change A ≈ Vo /Ii.
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Equivalent circuit
The close-loop gain A f ≡
Vo
A
=
(A and β have reciprocal units)
I s 1 + Aβ
Input resistance
Rif ≡
Vi I i Ri
I
Ii
Ii
Ii
Ri
=
= Ri i = Ri
= Ri
= Ri
=
Is
Is
Is
Ii + I f
I i + βVo
I i + βAI i 1 + Aβ
The shunt-mixing reduces the input resistance by (1 + Aβ ).
General form
Z if (s ) =
Z i (s )
1 + A(s )β (s )
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Output resistance (Vs = 0)
Output resistance
Rof ≡
Vt − AI i
Ro
V + AβVt
I = t
Ro
where I =
Rof ≡
Vt
I
and
I i = − I f = − βVo = − βVt (for I s = 0)
Vt
Ro
=
I 1 + Aβ
The voltage-sampling feedback reduces the output resistance by (1 + Aβ).
General form Z of (s ) =
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Z o (s )
1 + A(s )β (s )
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The Practical Situation
Derivation of the A circuit and β circuit for the series-shunt feedback amplifier.
Block diagram of a practical shunt-shunt feedback amplifier
Rif and Rof are the input and output resistances, respectively, of the feedback
amplifier, including Rs and RL.
The actual input and output resistances of the feedback amplifier usually
exclude Rs and RL :
Rin =
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The circuit in
1
1
1
−
Rif Rs
and
Rout =
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1
1
1
−
Rof R L
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with the feedback network represented by its y parameters.
(y parameters)
The source and load resistances should be lumped with the basic amplifier.
In addition, the resistances in the feedback should be also lumped with the
basic amplifier.
y-parameter two-port feedback network:
Neglect y21, the forward transmission effect of the feedback network, and
thus omit the controlled source y21I1.
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The circuit in
after neglecting y21.
y21I1 is neglected.
Determine β :
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β = y12 ≡
I1
V2
V1 = 0
(port 1 short-circuited)
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Summary of the Shunt-Shunt Feedback Amplifier
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Example 8.3 A feedback circuit of the shunt-shunt type
Determine the small-signal voltage gain Vo /Vs, the input resistance Rin , and
the output resistance Rout = Rof . The transistor has β = 100.
dc analysis:
⎧⎪VC = 0.7 + (I B + 0.07)47 = 3.99 + 47I B
⎨ 12 − VC = ( β + 1)I + 0.07
B
⎪⎩ 4.7
I B ≈ 0.015mA, I C = 1.5mA, and VC = 4.5V
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Small-signal analysis
A circuit:
Transfer gain:
(
Vπ = I 'i Rs R f rπ
)
V 'o = −gmVπ (R f RC )
A=
Input resistance of the A circuit
Ri = Rs || Rf || rπ = 1.4 kΩ
Output resistance of the A circuit
Ro = RC || Rf = 4.27 kΩ
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V 'o
I 'i
(
= −gm (R f RC ) Rs R f rπ
= −358.7 kΩ
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)
Determine β :
β ≡
I 'f
V 'o
=−
1
1
=−
Rf
47kΩ
Determine the closed-loop characteristics:
Closed - loop gain A f ≡
Vo
A
− 358.7
= −41.6 kΩ
=
=
I s 1 + Aβ 1 + 358.7 / 47
Since Vs = I s Rs , thus the voltage gain
Vo
V
− 41.6
= o =
≈ −4.16 V/V
Vs I s R s
10
Ri
1.4
=
= 162.2 Ω
1 + Aβ 8.63
4.27
Ro
Output resistance with feedback Rof =
=
= 495 Ω
1 + Aβ 8.63
Input resistance with feedback Rif =
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The Shunt-Series Feedback Amplifier
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The Ideal Situation of the Shunt-Series Feedback Amplifier
Ideal structure
A circuit: a unilateral open-loop amplifier
Ri : input resistance A : current gain Ro : output resistance
β circuit : an ideal current-sampling shunt-mixing feedback network
The source and load resistances are included inside the A circuit.
The β circuit does not load the A circuit. Do not change A ≈ Io /Ii.
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Equivalent circuit
The close-loop gain A f ≡
Io
A
=
(A and β have reciprocal units)
I s 1 + Aβ
Input resistance
Rif ≡
Vi I i Ri
I
Ii
Ii
Ii
Ri
=
= Ri i = Ri
= Ri
= Ri
=
Is
Is
Is
Ii + I f
I i + βI o
I i + βAI i 1 + Aβ
The series-mixing reduces the input resistance by (1 + Aβ ).
General form
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Z if (s ) =
Z i (s )
1 + A(s )β (s )
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Output resistance (Vs = 0)
Output resistance
In this case,
Rof ≡
V
It
I i = − I f = − βI o = − βI t (for I s = 0)
V = (I t − AI i )Ro = (I t + AβI t )Ro
Rof ≡
V
= Ro (1 + Aβ )
It
The current-sampling feedback increases the output resistance by (1 + Aβ ).
General form Z of (s ) = Z o (s )[1 + A(s )β (s )]
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Ching-Yuan Yang / EE, NCHU
The Practical Situation
Derivation of the A circuit and β circuit for the series-shunt feedback amplifier.
Block diagram of a practical shunt-series feedback amplifier
Rif and Rof are the input and output resistances, respectively, of the feedback
amplifier, including Rs and RL.
The actual input and output resistances of the feedback amplifier usually
exclude Rs and RL :
Rin =
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1
1
1
−
Rif Rs
and
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Rout = Rof − R L
Ching-Yuan Yang / EE, NCHU
The circuit in
with the feedback network represented by its g parameters.
(g parameters)
The source and load resistances should be lumped with the basic amplifier.
In addition, the resistances in the feedback should be also lumped with the
basic amplifier.
g-parameter two-port feedback network:
Neglect g21, the forward transmission effect of the feedback network, and
thus omit the controlled source g21V1.
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The circuit in
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Ching-Yuan Yang / EE, NCHU
after neglecting g21.
g21V1 is neglected.
Determine β :
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β = g12 ≡
I1
I2
V1 = 0
(port 1 short-circuited)
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Summary of the Shunt-Series Feedback Amplifier
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Ching-Yuan Yang / EE, NCHU
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Example 8.4 A feedback circuit of the shunt-series type
Find the current gain Iout /Iin, the input resistance Rin , and the output resistance
Rout. The transistor has β = 100 and VA = 75V.
dc analysis: (neglect the effect of finite transistor β and VA )
VC1 ≈ 12 − 10 × 1 = 2 V
15
= 1.57 V
100 + 15
≈ 1.57 − 0.7 = 0.87 V
VB1 ≈ 12
VE 1
I E1
VE 2 ≈ 2 − 0.7 = 1.3 V
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1.3
= 0.4 mA
3.4
≈ 12 − 0.4 × 8 = 8.8 V
I E1 ≈
0.87
≈
= 1 mA
0.87
VC 2
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Ching-Yuan Yang / EE, NCHU
Small-signal equivalent circuit
A circuit:
Vπ 1 = I 'i [Rs (RE 2 + R f ) RB r π 1 ]
{
}
Vb 2 = −gmVπ 1 ro1 RC1 [rπ 2 + ( β + 1)(RE 2 R f )
Vb 2
I'
I 'o ≈
⇒ A ≡ o ≈ −201.45 A/A
re 2 + (RE 2 R f )
I 'i
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Ching-Yuan Yang / EE, NCHU
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Determine β :
β ≡
I 'f
I 'o
=−
RE 2
= −0.254
RE 2 + R f
Determine the closed-loop characteristics:
Find input/output impedance:
Ri = Rs (RE 2 + R f ) RB rπ 1 = 1.535 kΩ
RC1 ro1
= 2.69 kΩ
β +1
Ri
= 29.5 Ω
1 + Aβ
⇒ Rof = Ro (1 + Aβ ) ≈ 140.1 kΩ
1
≈ 29.5 Ω
1/Rif − 1/Rs
⇒ Rout = ro 2[1 + gm (rπ 2 Rof ) ≈ 18.1 MΩ
⇒ Rif =
⇒ Rin =
Ro = (RE 2 R f ) + re 2 +
Find the required current gain:
RC 2 I c
RC 2 I o
I out I out
≈
=
≈
= −3.44 A/A
I in
Is
R L + RC 2 I s R L + RC 2 I s
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Summary of Relationships for the Four Feedback_Amplifier Topologies
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Determining The Loop Gain
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Determining the Loop Gain
Loop gain Aβ is a very important quantity that characterizes a feedback loop.
Find the loop gain:
Let the external source xs = 0 .
Open the feedback loop by breaking the connection of xo to the feedback
network and apply a test signal xt .
Determine the loop gain:
xf = βxt
xi = −xf = −βxt
xo = Axi = −Aβxt
Loop gain Aβ = −xo /xt
The loop gain Aβ is given by the negative of the ratio of the returned signal to
the applied test signal.
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A conceptual feedback loop is broken at XX’ and a test voltage xt is applied. The
impedance Zt is equal to that previously seen looking to the left of XX’. The loop gain
Aβ = −Vr /Vt , where Vr is the returned voltage.
Aβ = −
Vr
Vt
As an alternative, Aβ can be determined by finding the open-circuit transfer function Toc
and the short-circuit transfer function Tsc and combining them as indicated.
Aβ = −
open-circuit transfer function Toc
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1
1
1
+
Toc Tsc
short-circuit transfer function Tsc
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Ching-Yuan Yang / EE, NCHU
Rx
Closed-loop circuit:
+
Vx
AxVx
−
X
Y
Rt
X’
Y’
Break XX’ to determine the loop gain Aβ :
Rx
+
+
Vt
Vx
AxVx
Rt
Vr
−
−
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∵ Vr = AxVt
Aβ = −
Rt
R x + Rt
Vr
Rt
= − Ax
Vt
R x + Rt
Ching-Yuan Yang / EE, NCHU
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Break YY’ to determine Toc and Tsc :
Open-circuit transfer function Toc
Rx
Vt
Vx
Rt
∵ Voc = AxVt
+
+
AxVx
−
Voc
∴ Toc ≡
−
Short-circuit transfer function Tsc
∵ V x = I t Rt
Rx
and I sc =
+
It
Rt
Vx
A x Vx
−
Thus,
−
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1
1
1
+
Toc Tsc
Voc
= Ax
Vt
=−
1
1
Rx
+
Ax Ax Rt
Isc
∴ Tsc ≡
AxV x Ax I t Rt
=
Rx
Rx
I sc
R
= Ax t
It
Rx
= Aβ
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Ching-Yuan Yang / EE, NCHU
Equivalence of Circuits from a Feedback-Loop Point of View
Break XX’ to determine the loop gain Aβ :
Vr = − μV1
R1 (Rid + R )
Rid
{ R L [R2 + R1 (Rid + R )]} + ro [R1 (Rid + R )] + R2 Rid + R
L = Aβ = −
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{ R L [R2 + R1 (Rid + R )]}
{ R L [R2 + R1 (Rid + R )]}
R1 (Rid + R )
Vr
V
Rid
=− r =μ
Vt
V1
{ R L [R 2 + R1 (Rid + R )]} + ro [R1 (Rid + R )] + R2 Rid + R
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The Stability Problem
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Stability Problem of a Feedback Amplifier
The closed-loop transfer function of a feedback amplifier
A f (s ) =
A(s )
1 + A(s )β (s )
where A(s) : the open-loop transfer function
β (s) : the feedback transfer function
For physical frequencies s = jω ,
A f ( jω ) =
A( jω )
1 + A( jω )β ( jω )
L( jω) ≡ A( jω)β ( jω) = |A( jω)β ( jω)|e−jφ(ω )
If for s = jω0, L(jω0) = −1, then the closed-loop gain approaches infinity at ω0.
Under this condition, the circuit amplifies its own noise components at ω0 indefinely.
Barkhausen criteria: If a negative-feedback circuit has a loop gain that satisfies two
conditions:
|L(jω0)| ≥ 1 and ∠L(jω0) = 180o,
then the circuit may oscillate at ω0.
In order to ensure oscillation in the presence of temperature and process
variations, we typically choose the loop gain to be at least twice or three times the
required value.
Oscillations could occur in a negative-feedback amplifier, we wish to find methods
to prevent their occurrence.
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Nyquist Plot
Nyquist criterion:
If the intersection occurs to the left of (−1, 0), the magnitude of loop gain at
this frequency is greater than unity and the amplifier will be unstable.
If the intersection occurs to the right of (−1, 0), the amplifier will be stable.
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Effect Of Feedback On The Amplifier Poles
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Stability and Pole Location
Consider an amplifier with a pole pair at s = σ0 ± jωn .
v(t) = eσ 0t [e +jωnt + e −jωnt ] = 2 eσ 0t cos (ωnt )
Relationship between pole location and transient response:
The existence of any
right-half-plane poles
results in instability.
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Poles of the Feedback Amplifier
The closed-loop transfer function of a feedback amplifier
A f (s ) =
A(s )
1 + A(s )β (s )
where A(s) : the open-loop transfer function
β (s) : the feedback transfer function
Find poles: The poles of the feedback amplifier are the zeros of 1 + A(s)β (s).
1 + A(s)β (s) = 0 (the characteristic equation of the feedback loop.)
It should therefore be apparent that applying feedback to an amplifier
changes its poles.
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Ching-Yuan Yang / EE, NCHU
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Amplifier with Single-Pole Response
An amplifier whose open-loop transfer function is characterized by a single pole:
A0
A(s ) =
1 + s /ω p
A0 /(1 + A0 β )
The closed-loop transfer function A f (s ) =
1 + s /ω p (1 + A0 β )
open - loop
Midband Gain
A0
→
Pole
ωP
→
closed - loop
A0
Aof =
1 + Ao β
ωPf = ωP (1 + Ao β )
Effect of feedback on the pole location and the frequency response:
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A0ω P
≈ A(s )
s
At such high frequencies the loop gain is much smaller than unity and the
feedback is ineffective.
Effect of feedback on the frequency response:
For frequencies ω >> ωP(1 + A0β ),
A f (s ) ≈
Applying negative feedback to an amplifier results in extending its
bandwidth at the expense of a reduction in gain.
Since the pole of the closed-loop amplifier never enters the right half of
the s plane, the single-pole amplifier is stable for any value of β.
Unconditionally stable
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Ching-Yuan Yang / EE, NCHU
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Amplifier with Two-Pole Response
An amplifier whose open-loop transfer function is characterized by two poles:
Ao
A(s ) =
(1 + s /ω P 1 )(1 + s /ωP 2 )
Close-loop poles
1 + A(s )β = 0
⇒ s 2 + s(ω P 1 + ω P 2 ) + (1 + A0 β )ω P 1ω P 2 = 0
⇒ s=−
Root-locus diagram
ωP1 + ωP 2
2
±
1
(ωP 1 + ωP 2 )2 − 4(1 + A0 β )ω P 1ωP 2
2
The feedback amplifier also is unconditionally stable.
The maximum phase shift of A(s) is 180o (90o per
pole), but this value is reached at ω = ∝.
The open-loop amplifier might have a dominant pole,
but this is not necessarily the case for the closedloop amplifier.
As is the case with second-order responses
generally, the closed-loop response can show a peak.
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The characteristic equation of a second-order network:
s2 + s
ω0
Q
+ ω02 = 0
ω0 : pole frequency Q : pole Q factor
Definition of ω0 and Q of a pair
of complex conjugate poles:
Determine Q factor:
(1 + A0 β )ω P 1ωP 2
ωP1 + ωP 2
Q=
Normalized magnitude response:
Q > 0.7, peaking. Q ≤ 0.7, no peaking.
Q = 0.7 (poles at 45o), maximally flat.
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Ching-Yuan Yang / EE, NCHU
8-86
Example 8.5 Positive-feedback circuit
Open-loop circuit:
Find the loop transmission L(s) and the characteristic
equation. Sketch a root-locus diagram for varying K, and
find the value of K that results in a maximally flat response,
and the value of K that makes the circuit oscillate.
(Assume that the amplifier has infinite input impedance
and output impedance.)
Solution
The loop transmission:
V
L (s ) ≡ A(s )β (s ) = − r = −KT (s )
Vt
where T(s) is the transfer function of the two-port RC
network.
V
s(1/CR )
T (s ) ≡ r = 2
V1 s + s(3 /CR ) + (1/CR )2
− s(K /CR )
s + s(3 /CR ) + (1/CR )2
The characteristic equation: 1 + L(s) = 0
L (s ) =
Thus
2
2
s2 + s
3 ⎛ 1 ⎞
K
+⎜
=0
⎟ −s
CR
CR ⎝ CR ⎠
∴ Pole frequency: ω0 =
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1
CR
2
s2 + s
3−K ⎛ 1 ⎞
+⎜
⎟ =0
CR
⎝ CR ⎠
Q factor: Q =
1
3−K
Ching-Yuan Yang / EE, NCHU
Root-locus diagram:
ω0 =
Q=
1
CR
1
3−K
The maximally flat response:
Q = 0.707
K = 1.586
The poles are at 45o.
K ≥ 3, the circuit is unstable.
The feedback is in fact positive,
and the circuit will oscillate at the
frequency for which the phase of
T (jω) is zero.
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Amplifier with Three or More Poles
A value of A0β exits at which this pair of complex-conjugate poles enters the right
half of the s plane, thus causing the amplifier to become unstable.
An amplifier with three poles:
Root-locus diagram
Since the amplifier with three poles
has a phase shift that reach −270o as
ω reaches ∝, there exists a finite
frequency ω180o, at which the loop gain
has 180o phase shift.
Frequency compensation
There exists a maximum value for β above which
the feedback amplifier becomes unstable.
Alternatively, there exists a minimum value value for
the closed-loop gain Af 0 below which the amplifier
becomes unstable.
To obtain lower values of the closed-loop gain one needs to alter the loop
transfer function L(s). This is the process of frequency compensation.
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Stability Study Using Bode Plots
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Gain and Phase Margins
Gain margin:
The difference between the value of
|Aβ | at ω180o and unity (in dB).
Phase margin:
The difference between the phase
angle at the 0-dB frequency and
180o.
The gain margin represents the amount
by which the loop gain can be increased
while stability is maintained.
If at the frequency of unity loop-gain
magnitude, the phase lag is in excess of
180o, the amplifier will be unstable.
Feedback amplifiers are usually designed
to have sufficient gain margin and phase
margin to allow for the inevitable changes
in loop gain with temperature, time, and
so on.
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Effect of Phase Margin on Closed-Loop Response
A( jω1 )β = 1 × e − jθ
where θ = 180o − phase margin
Close-loop gain (at ω1):
A f ( jω1 ) =
⇒
A( jω1 )
(1/ β )e − jθ
=
1 + A( jω1 )β
1 + e − jθ
A f ( jω1 ) =
1/ β
1 + e − jθ
⌦ For a phase margin of 45o, θ = 135o,
1
Gain peaking
A f ( jω1 ) = 1.3
β
⌦ The peaking increase as the phase margin
is reduced, eventually reaching ∞ when
the phase margin is zero.
⌦ The closed-loop gain at low frequencies is
approximately 1/β .
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Ching-Yuan Yang / EE, NCHU
An Alternative Approach for Stability Analysis Using Bode Plot of |A|
Assume that β is independent of
frequency, we can plot 20log(1/β )
as a horizontal straight line on the
same plane used for 20log|A|.
1
20 log A( jω ) − 20 log = 20 log Aβ
β
(expressed in dB)
Study stability by examining the
difference between the two plots.
If we wish to evaluate stability for a
different feedback factor we simply
draw another horizontal straight line
at the level 20log(1/β ).
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|A|
Example: Stability Analysis
Using Bode Plot of |A|
The open-loop gain of the amplifier
20 log A( jω ) − 20 log
1
β
= 20 log Aβ
We can find
A( f ) =
105
f ⎞⎛
f ⎞⎛
f ⎞
⎛
⎜1 + j 5 ⎟⎜1 + j 6 ⎟⎜1 + j 7 ⎟
10 ⎠⎝
10 ⎠⎝
10 ⎠
⎝
3.2×106 Hz
and
⎛ f ⎞
A = 100 − 20 log 1 + ⎜ 5 ⎟
⎝ 10 ⎠
2
2
⎛ f ⎞
⎛ f ⎞
− 20 log 1 + ⎜ 6 ⎟ − 20 log 1 + ⎜ 7 ⎟
10
⎝
⎠
⎝ 10 ⎠
⎡
f
f
f
2
⎤
φ = − ⎢tan−1 ⎛⎜ 5 ⎞⎟ + tan−1 ⎛⎜ 6 ⎞⎟ + tan−1 ⎛⎜ 7 ⎞⎟⎥
⎝ 10 ⎠⎦
⎝ 10 ⎠
⎝ 10 ⎠
⎣
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Frequency Compensation
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Theory of Frequency Compensation
|A|
A’ :Introduce an additional
pole at fD .
poles: fD, fP1, fP2, fP3, ⋅⋅⋅
β = 10−2
A” :Move the original lowfrequency pole to f ’D .
(i.e., fP1 p f ’D )
poles: f ’D, fP2, fP3, ⋅⋅⋅
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Implementation of Frequency Compensation
Two-Cascaded gain stages of a multistage amplifier
Equivalent circuit for the interface
between the two stages
f P1 =
The circuit with a compensating
capacitor CC added
1
2πC x R x
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f 'D =
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1
2π (C x + CC )R x
Ching-Yuan Yang / EE, NCHU
Miller Compensation and Pole Splitting
A gain stage in a multistage amplifier before compensation
A gain stage in a multistage amplifier with a compensating capacitor in the
feedback loop
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(sC f − gm )R1R 2
Vo
=
I i 1 + s[C1R1 + C 2R2 + C f (gm R1R2 + R1 + R2 )] + s 2[C1C 2 + C f (C1 + C 2 )]R1R2
⎛ 1
⎛
1 ⎞
s ⎞⎛
s ⎞
s2
⎟⎟ = 1 + s ⎜⎜
⎟⎟ +
⎟⎟⎜⎜1 +
+
D (s ) = ⎜⎜1 +
ω 'P 1 ⎠⎝ ω 'P 2 ⎠
⎝
⎝ ω 'P 1 ω 'P 2 ⎠ ω 'P 1 ω 'P 2
≈1+
⇒ ω 'P 1 =
s
s2
+
ω 'P 1 ω 'P 1 ω 'P 2
(ω’P1 << ω’P2 )
1
1
≈
C1R1 + C 2R2 + C f (gm R1R2 + R1 + R2 ) gm R2C f R1
ω 'P 2 ≈
gmC f
C1C 2 + C f (C1 + C 2 )
(8.87)
(8.88)
jω
jω
Before Compensation
After Compensation
ρ
ρ
Pole splitting
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Example 8.6: Frequency Compensation
The open-loop transfer function of the operational amplifier:
A( f ) =
105
f ⎞⎛
f ⎞⎛
f ⎞
⎛
⎜1 + j 5 ⎟⎜1 + j 6 ⎟⎜1 + j 7 ⎟
10
10
10
⎝
⎠⎝
⎠⎝
⎠
We wish to compensate the op amp so that the closed-loop amplifier with resistive
feedback is stable for any gain (that is, for β up to unity) and PM (phase margin) ≥ 45o.
Assume that the op amp circuit includes a stage such as that of Fig.a with C1 = 100pF,
C2 = 5pF, and gm = 40mA/V, that the pole at fP1 is caused by by the input circuit of that
stage, and that the pole is connected either between the input node B and ground or
in the feedback path of the transistor.
Fig.a:
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Ching-Yuan Yang / EE, NCHU
8-100
Find the compensating capacitor CC connected across the input terminals of the
transistor stage.
(a) Determine R1 and R2 :
f P 1 = 0.1MHz =
f P 2 = 1MHz =
(b) Determine f ’D :
1st pole from fP1 to f ’D :
f 'D =
1
2πC1R1
R1 =
1
2πC 2R2
1
2π (C1 + CC )R1
R2 =
105
Ω
2π
105
π
Ω
and the 2nd pole remains unchanged.
PM ≥ 45o : The required value for f ’D is determined by drawing a −20dB/dec line from the 1-MHz
point on the 20log(1/β ) = 20log1 = 0dB line. This line will intersect the 100-dB dc gain line at 10Hz.
Thus
f 'D = 10Hz =
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1
2π (C1 + CC )R1
∴ CC = 1μF
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Find the compensating capacitor Cf connected in the feedback path of the transistor.
By Eqs. (8.58) and (8.59):
f 'P 1 ≈
1
2πgm R2C f R1
and
f 'P 2 ≈
gmC f
2π [C1C 2 + C f (C1 + C 2 )]
gm
= 60.6 MHz
2π (C1 + C 2 )
The pole f ’P2 moves to a frequency higher than fP3 (= 10MHz).
nd
The new 2 pole is at fP3. This requires that the 1st pole be located at 100Hz for PM ≥ 45o :
1
f 'P 1 = 100Hz =
∴ C f = 78.5pF
2πgm R2C f R1
Determine fP2 : Assume Cf >> C2 , then
f 'P 2 ≈
Conclusion: Using Miller compensation not only results in a much smaller compensating
capacitor but, owing to pole splitting, also enables us to place the dominant pole a
decade higher in frequency.
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Homework #2
Problems 42, 58, 61, 67, 70, 80
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