8. Frequency Response
8. Frequency Response
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8.1 Bode Plots
Use of the Laplace Transform Variable
Motivation
Frequency range of the signals is different
depending on application:
• audio signals: 20Hz till 20kHz;
• electrocardiograms: 0.05Hz to 100Hz;
• video signals: dc to 4.5MHz.
When amplify these signals it is necessary to
know how is changed the magnitude and the
phase of the frequency components.
Investigation of the feedback circuit:
• How the feedback affects the magnitude and the
phase shift of the amplifier?
• Stability investigation.
s = jω
• Simplifies the analysis of the circuit, it avoids
dealing with the complex numbers during the
analysis.
• From the results after the analysis in s-domain
can be derived time domain properties of the
circuit (by applying inverse Laplace transform)
as well frequency domain properties (s = jω).
• Poles and zeros: useful tool in the circuit
description.
s-domain impedances:
Bode plots: simplified plots of the gain of the
amplifier vs. frequency and the phase shift of the
amplifier vs. frequency.
8. Frequency Response
TLT-8016 Basic Analog Circuits
L ⇒ sL;
C⇒
1
sC
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Poles and Zeroes
An Example: The Lowpass RC Circuit
Poles: the roots of the denominator.
Zeros: roots of numerator.
For the circuit from the example: no zeros, one
pole at:
1
s=−
RC
Break Frequencies
Figure 8.1 Low-pass RC filter.
In the expression for Av: s = jω = j2πf
1
Av ( f ) =
1 + j 2πRCf
By applying of voltage divider principle
Av (s ) =
vo
(s ) = 1 sC
vin
R + 1 sC
(8.1)
1
RCs + 1
(8.2)
Av (s ) =
Av ( f b ) =
1
= 0.707 ∠ − 45o
1 + j1
(8.4)
Vo
1
=
Vin 1 + j ( f / f b )
(8.5)
1
2π RC
(8.6)
Av ( f ) =
fb =
8. Frequency Response
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(8.3)
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Gain Magnitude Expressed in Decibels
Av ( f ) =
1
(8.7)
1 + ( f / fb )
2
Av ( f ) dB = −20 log 1 + ( f / f b )
[
Av ( f ) dB = −10 log 1 + ( f / f b )
2
]
(8.9)
For f << fb |Av(f)|dB = 0.
[
Av ( f ) dB = −10 log ( f f b )
= −20 log( f f b )
8. Frequency Response
Figure 8.2 Logarithmic frequency scale.
(8.8)
2
For f >> fb
Logarithmic Frequency Scales
2
]
decade: change the frequency of 10 times.
octave: change the frequency 2 times.
dB/decade: increasing or decreasing the gain in dB
when frequency increases 10 times.
dB/octave: increasing or decreasing the gain in dB
when frequency increases 2 times.
Example:
20 dB/decade = 6 dB/octave
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The Magnitude Bode Plot
The Phase Plot
θ = − arctan( f / f b )
Figure 8.3 Bode plot for the low-pass RC filter.
[
Av ( f ) dB ≅ −10 log ( f / f b )
2
]
(8.10)
Av ( f ) dB ≅ −20 log ( f / f b )
(8.11)
Av ( f b ) dB ≅ −3 dB
(8.12)
8. Frequency Response
Figure 8.4 Bode plot for phase of the low-pass RC filter.
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Example 8.1 Bode Plot for a RC Circuit with
One Pole and One Zero
1
2π R2C
(8.15)
1
2π (R1 + R2 )C
(8.16)
fz =
Prepare Bode Plots of magnitude and phase of the
voltage transfer function Av(f)=Vo / Vin for the
circuit shown in Figure 8.5. (The component
values have been selected to result in convenient
break frequency.)
fp =
Av ( f ) =
Solution
Av (s ) =
Vo
(s ) = sR2C + 1
Vin
s (R1 + R2 )C + 1
(8.14)
Av ( f ) =
1 + j( f f z )
1 + j( f f p )
(8.17)
1+ ( f fz )
2
(8.18)
1+ ( f fz )
2
Av ( f ) dB = 20 log 1 + ( f f z )
2
− 20 log 1 + ( f f p )
2
θ = arctan ( f f z ) − arctan ( f f p )
(8.19)
(8.20)
Figure 8.5 Circuit for Example 8.1.
8. Frequency Response
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Av ( f ) dB = 20 log 1 + ( f f z )
2
− 20 log 1 + ( f f p )
2
(8.19)
Figure 8.6 Bode plots of the terms on the right-hand side of
Equation (8.19).
8. Frequency Response
Figure 8.7 Bode plot of the magnitude of Av for the circuit
of Figure 8.5.
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θ = arctan ( f f z ) − arctan ( f f p )
(8.20)
Figure 8.8 Approximate plots of the terms of Equation (8.20).
8. Frequency Response
Figure 8.9 Bode phase plot of the voltage-transfer function
for the circuit of Figure 8.5.
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Simple Check of the Bode Plot
At very low frequencies (f → 0):
•Very high impedance of C (ZC → ∞);
•No current flows;
•No voltage drop over R1;
•Vo = Vin and Av = 1 (0dB).
Figure 8.5 Circuit for Example 8.1.
At very high frequency (f → ∞):
•Very small impedance of C (ZC → 0);
•The voltage gain is defined from the voltage
divider R1 – R2:
R2
1× 103
Av =
=
= 0.1 (− 20dB)
3
3
R1 + R2 3 ×10 + 1×10
The corners in the Bode plot are determined from
the pole and the zero (the corner frequencies).
Figure 8.7 Bode plot of the magnitude of Av for the circuit
of Figure 8.5.
8. Frequency Response
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Example 8.2 Bode Plot for a High Pass RC
Filter
From the voltage divider principle:
Prepare Bode plots of the magnitude and phase of
the voltage transfer function Av=Vo / Vin (f) for the
circuit illustrated in Figure 8.10.
Solution:
Av =
R2
Z + R2
Z = R1 +
where
Av (s ) =
fp =
1
sC
sR2C
s (R1 + R2 )C + 1
1
2π (R1 + R2 )C
j f fp
R2
Av ( f ) =
R1 + R2 1 + j f f p
Figure 8.10 Circuit of Example 8.2.
(8.21)
(8.22)
(8.23)
f / fp
R2
Av ( f ) =
×
2
R1 + R2
1+ ( f / f p )
8. Frequency Response
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Av ( f ) dB = 20 log
R2
2
+ 20 log( f / f p ) − 20 log 1 + ( f / f p )
R1 + R2
(8.24)
Av ( f ) dB = −12 + 20 log( f / 3.98) − 20 log 1 + ( f / 3.98)
2
Figure 8.11 Plots of the terms on the right-hand side of
Equation (8.25).
8. Frequency Response
(8.25)
Figure 8.12 Magnitude Bode plot for the circuit of Figure 8.10.
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θ = 90o − arctan( f / 3.98)
Figure 8.13 Plots of the terms on the right-hand side of
Equation (8.26).
8. Frequency Response
(8.26)
Figure 8.14 Bode phase plot for the voltage-transfer
function of the circuit shown in Figure 8.10.
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8.3 The Miller Effect
Miller effect: an impedance, connected between
input and output of an amplifier (parallel feedback
impedance) can be replaced equivalently by two
impedances in parallel to the input and output.
V f = Vi − Vo
Vo = Av Vi
V f = Vi (1 − Av )
If =
Vf
Zf
=
Vi (1 − Av )
Zf
Z in, Miller =
Zf
1 − Av
Z out, Miller =
Figure 8.24 A feedback impedance can be replaced by
impedances in parallel with the input and output terminals.
8. Frequency Response
Z f Av
Av − 1
Av < 0, otherwise Zin, Miller is negative.
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Miller Effect Applied to Feedback Capacitance
Cf
Amplifier
Cin, Miller
Amplifier
Cout, Miller
Amplifier, having parallel feedback capacitance Cf, and its Miller equivalent circuit.
Zf =
Z in, Miller =
Z out, Miller
8. Frequency Response
1
jω C f
Cin, Miller = C f (1 − Av )
1
jω C f (1 − Av )
Cout, Miller =
Av
=
jω C f ( Av − 1)
TLT-8016 Basic Analog Circuits
C f Av
( Av − 1)
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8.4 The Hybrid - π Model for the BJT
The rπ - β Model
The Hybrid - π Model
Figure 8.29 Hybrid- π equivalent circuit. It is an expansion of the
rπ– β model of the BJT.
Figure 8.27 The rπ– β model for the BJT.
8. Frequency Response
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Figure 8.29 Hybrid- π equivalent circuit..
rx – base spreading resistance. It is the ohmic
resistance of the base region. (~10Ω)
gm – transconductance and accounts for
amplifying properties. (10..600mS)
rπ – dynamic resistance of base-emitter junction
(1..2..3kΩ)
Cπ – capacitance (diffusion + depletion) of baseemitter junction (~102pF)
rµ – reflects base-width modulation. Few MΩ and
usually is neglected.
gm =
rπ
=
I CQ
(8.47)
VT
ro – reflects the upward slope of output
characteristics of the BJT. (10..100kΩ)
Cµ – depletion capacitance of collector-base
junction. (Few pF)
8. Frequency Response
β
TLT-8016 Basic Analog Circuits
VA
ro ≅
I CQ
(8.43)
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Figure 8.30 Hybrid-π model with rx = 0, rµ = ∞, ro= ∞, and
the capacitors replaced by open circuits. This approximate
low-frequency model is equivalent to the rπ– β model of
Figure 8.27.
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Example 8.7 Determining the
Hybrid - π Parameter by Using the
Data Sheet
Use the data sheet in Appendix B to
determine values for the hybrid - π
equivalent circuit for a typical
2N2222A transistor at a Q - point of
ICQ=10 mA and VCEQ=10 V. Assume
that VT=26mV.
Results:
g m = 0.385 S
β = 225
Figure 8.31 Hybrid-π model for the 2N2222A at ICQ =10 mA
and VCEQ = 10V. For these values, β ≈ 225.
rπ = 585Ω
rµ = 1.5MΩ
rx = 19 Ω
C µ = 8 pF
Cπ ≅ 196 pF
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8.5 Common - Emitter Amplifiers at High Frequencies
Figure 8.33 Common-emitter amplifier.
8. Frequency Response
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Figure 8.33 Common-emitter amplifier.
RL' = RL || RC || ro
(8.49)
Rs' = rπ || [rx + (RB || Rs )]
(8.50)
RB = R1 || R2
8. Frequency Response
(8.51)
Figure 8.34 Equivalent circuit of Figure 8.33b after
removing rµ, replacing ro, RC, and RL by their parallel
equivalent, and replacing the circuit to the left-hand side of
b' by its Thévenin equivalent.
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Figure 8.35 Simplified equivalent
circuit for the common-emitter
amplifier.
Vo = − g m rπ RL'
Avb' =
Vo
= − g m RL'
Vπ
(
CT = Cπ + C µ 1 + g m R 'L
fH =
8. Frequency Response
(8.52)
1
2π Rs' CT
(8.53)
)
(8.54)
(8.55)
Figure 8.36 High-frequency behavior of the commonemitter amplifier.
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Example 8.8 High Frequency Response of the
Common - Emitter Amplifier
Consider the common - emitter amplifier shown in
Figure 8.37. Initially, we assume that RE1=0, and
then this circuit has the same small signal
equivalent circuit as the circuit illustrated in Figure
8.33. It can be demonstrated that the Q-point is at
approximately ICQ=10 mA and VCEQ=10 V. The
values of the hybrid - π parameters for the
transistor at this Q-point are shown in Figure 8.31.
Use the result derived in this section to find the
upper half-power frequency and the formulas of
Chapter 4 to find the midband value for Avs.
8. Frequency Response
TLT-8016 Basic Analog Circuits
Figure 8.37 Circuit for Example 8.8.
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Solution:
rx = 19Ω
rπ = 585Ω
Cπ = 196pF
ro = 22.5kΩ
g m = 0.385S
rµ = 1.5MΩ
Rin = RB || [rπ + (β + 1)RE1 ]
= (10kΩ ) || [585 + 226 × 0] = 553Ω
C µ = 8pF
Avs =
RL' = RL || RC || ro = 510 || 510 || 22.5kΩ = 252Ω
Rin
vo
= Av
Rs + Rin
vs
= −96.9
Rs' = rπ || [rx + (RB || Rs )] = 585 || [19 + (10kΩ || 50 )]
= 61.5Ω
(
CT = Cπ + Cµ 1 + g m R 'L
553
= −88.9
50 + 553
)
= 196 + 8(1 + 0.358 × 252) = 196 + 784 = 980pF
fH =
1
1
=
= 2.64MHz
'
−12
2π Rs CT 2π × 61.5 × 980 × 10
β ≅ rπ g m = 585 × 0.385 = 225
vo
− β RL'
− 225 × 252
Av =
=
=
vin rπ + (β + 1)RE1 585 + (225 + 1)× 0
= −96.9
8. Frequency Response
Figure 8.38 Gain magnitude versus frequency for the
common-emitter amplifier of Example 8.8.
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8.8 Low Frequency Response of RC - Coupled Amplifiers
Vo
RL
j( f / f 2 )
=
×
V y Ro + RL 1 + j ( f / f 2 )
Coupling Capacitors
f2 =
1
2π (Ro + RL )C2
Avo =
Figure 8.47 Amplifier with coupling capacitors.
Vo Vx V y Vo
Avs =
= × ×
Vs Vs Vx V y
(8.71)
Vx
Rin
j ( f / f1 )
=
×
Vs Rs + Rin 1 + j ( f / f1 )
(8.72)
f1 =
1
2π (Rs + Rin )C1
8. Frequency Response
Avs =
Vy
(8.74)
(8.75)
(8.76)
Vx
Rin
j ( f / f1 )
RL
j( f / f 2 )
×
× Avo ×
×
Rs + Rin 1 + j ( f / f1 )
Ro + RL 1 + j ( f / f 2 )
Avsmid =
Rin
RL
× Avo ×
Ro + RL
Rs + Rin
Avs = Avsmid ×
j ( f / f1 )
j( f / f2 )
×
1 + j ( f / f1 ) 1 + j ( f / f 2 )
(8.77)
(8.78)
(8.73)
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Avs = Avsmid ×
j ( f / f1 )
j( f / f 2 )
×
1 + j ( f / f1 ) 1 + j ( f / f 2 )
(8.78)
Two break frequencies in the terms
j ( f / f1 )
1 + j ( f / f1 )
and
j( f / f 2 )
1 + j( f / f 2 )
Figure 8.48 Magnitude Bode plot for the amplifier of Figure
8.47. (We have assumed that f1 > f2.)
8. Frequency Response
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Exercise 8.17
Find the break frequency for the amplifier of
Figure 8.52. If we want to reduce the lower half power frequency of the amplifier, which
capacitor is most critical? Should we increase or
reduce its value?
Solution:
The amplifier has three coupling capacitors,
determining three break frequencies:
1
f1 =
2π (Rs + Ri1 )C1
=
1
= 0.796Hz
3
3
−6
2π 100 × 10 + 100 × 10 × 1× 10
(
)
f2 =
1
2π (Ro1 + Ri 2 )C2
1
= 53.1Hz
3
3
−6
2π 1× 10 + 2 × 10 × 1× 10
1
f3 =
2π (Ro 2 + RL )C3
=
=
(
)
1
= 15.9kHz
2π (2 + 8)× 1× 10 −6
Since f3 >> f1 and f3 >> f2 , the break frequency is
f3 = 15.9kHz.
C3 must be changed to >100µF.
Figure 8.52 Amplifier for Exercise 8.17.
8. Frequency Response
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Bypass Capacitors
Figure 8.53 Discrete common-emitter amplifier.
8. Frequency Response
Figure 8.54 Gain magnitude versus frequency.
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