8. Frequency Response TLT-8016 Basic Analog Circuits 2005/2006 1
8. Frequency Response
8. Frequency Response TLT-8016 Basic Analog Circuits 2005/2006 2
8.1 Bode Plots
Use of the Laplace Transform VariableMotivation
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.
Bode plots: simplified plots of the gain of the amplifier vs. frequency and the phase shift of the amplifier vs. frequency.
ωjs =• 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:
sCCsLL 1 ; ⇒⇒
8. Frequency Response TLT-8016 Basic Analog Circuits 2005/2006 3
Poles and ZeroesAn 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:
RCs 1
−=
Break FrequenciesFigure 8.1 Low-pass RC filter.
In the expression for Av: s = jω = j2πf
( )RCfj
fAv π211
+= (8.3)
By applying of voltage divider principle
( ) ( )sCR
sCsvvsA
in
ov 1
1+
== ( ) o45707.011
1−∠=
+=
jfA bv(8.1) (8.4)
( )1
1+
=RCs
sAv ( ) ( )bin
ov f/fjV
VfA+
==1
1 (8.5)(8.2)
RCfb π2
1= (8.6)
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Gain Magnitude Expressed in Decibels Logarithmic Frequency Scales
( )( )21
1
b
vf/f
fA+
= (8.7)
Figure 8.2 Logarithmic frequency scale.(8.8)( ) ( )2/1log20 bdBv fffA +−=
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
( ) ( )[ ]2/1log10 bdBv fffA +−= (8.9)
For f << fb |Av(f)|dB = 0.
For f >> fb
( ) ( )[ ]( )b
bdBv
ff
fffA
log20
log10 2
−=
−=
8. Frequency Response TLT-8016 Basic Analog Circuits 2005/2006 5
The Magnitude Bode PlotThe Phase Plot
( )bff /arctan−=θ
Figure 8.3 Bode plot for the low-pass RC filter.
( ) ( )[ ]2/log10 bdBv fffA −≅ (8.10) Figure 8.4 Bode plot for phase of the low-pass RC filter.
( ) ( )bdBv fffA /log20−≅ (8.11)
( ) dBfAdBbv 3−≅ (8.12)
8. Frequency Response TLT-8016 Basic Analog Circuits 2005/2006 6
Example 8.1 Bode Plot for a RC Circuit with One Pole and One Zero
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.)
Solution
CRfz
221
π= (8.15)
( )CRRf p
2121+
=π
(8.16)
( ) ( )( )p
zv ffj
ffjfA++
=11
(8.17)
( ) ( ) ( ) 11
21
2
+++
==CRRs
CsRsVVsA
in
ov ( ) ( )
( )22
1
1
z
zv
ff
fffA
+
+=
(8.14)(8.18)
( ) ( )
( )22
1log20
1log20
p
zdBv
ff
fffA
+−
+=(8.19)
( ) ( )pz ffff arctanarctan −=θ (8.20)
Figure 8.5 Circuit for Example 8.1.
8. Frequency Response TLT-8016 Basic Analog Circuits 2005/2006 7
( ) ( )
( )22
1log20
1log20
p
zdBv
ff
fffA
+−
+=(8.19)
Figure 8.6 Bode plots of the terms on the right-hand side of Equation (8.19).
Figure 8.7 Bode plot of the magnitude of Av for the circuit of Figure 8.5.
8. Frequency Response TLT-8016 Basic Analog Circuits 2005/2006 8
Figure 8.8 Approximate plots of the terms of Equation (8.20).
( ) ( )pz ffff arctanarctan −=θ (8.20)
Figure 8.9 Bode phase plot of the voltage-transfer function for the circuit of Figure 8.5.
8. Frequency Response TLT-8016 Basic Analog Circuits 2005/2006 9
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).
At very high frequency (f →∞):•Very small impedance of C (ZC → 0);•The voltage gain is defined from the voltage divider R1 – R2:
Figure 8.5 Circuit for Example 8.1.
( )dB20 1.0101103
10133
3
21
2 −=×+×
×=
+=
RRRAv
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 TLT-8016 Basic Analog Circuits 2005/2006 10
Example 8.2 Bode Plot for a High Pass RC Filter
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:
From the voltage divider principle:
2
2
RZRAv +
=
sCRZ 1
1 +=where
( ) ( ) 121
2
++=
CRRsCsRsAv (8.21)
( )CRRf p
2121+
=π (8.22)
( )p
pv ffj
ffjRR
RfA++
=121
2Figure 8.10 Circuit of Example 8.2. (8.23)
( )( )221
2
1 p
pv
f/f
f/fRR
RfA+
×+
=
8. Frequency Response TLT-8016 Basic Analog Circuits 2005/2006 11
( ) ( ) ( )221
2 /1log20/log20log20 ppdBv ffffRR
RfA +−++
= (8.24)
( ) ( ) ( )298.3/1log2098.3/log2012 fffAdBv +−+−= (8.25)
Figure 8.11 Plots of the terms on the right-hand side of Equation (8.25).
Figure 8.12 Magnitude Bode plot for the circuit of Figure 8.10.
8. Frequency Response TLT-8016 Basic Analog Circuits 2005/2006 12
( )98.3/arctan90 f−= oθ (8.26)
Figure 8.13 Plots of the terms on the right-hand side of Equation (8.26).
Figure 8.14 Bode phase plot for the voltage-transfer function of the circuit shown in Figure 8.10.
8. Frequency Response TLT-8016 Basic Analog Circuits 2005/2006 13
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.
oif VVV −=
ivo A VV =
( )vif A−= 1VV
( )f
vi
f
ff Z
AZ
−==
1VVI
v
f
AZ
Z−
=1Millerin,
1Millerout, −=
v
vf
AAZ
ZFigure 8.24 A feedback impedance can be replaced by
impedances in parallel with the input and output terminals.Av < 0, otherwise Zin, Miller is negative.
8. Frequency Response TLT-8016 Basic Analog Circuits 2005/2006 14
Miller Effect Applied to Feedback Capacitance
Cf
Amplifier Cin, MillerAmplifier Cout, Miller
Amplifier, having parallel feedback capacitance Cf, and its Miller equivalent circuit.
ff Cj
Zω
1=
( )vf ACC −= 1Millerin,
( )vf ACjZ
−=
11
Millerin, ω
( )1Millerout, −=
v
vf
AAC
C
( )1Millerout, −=
vf
v
ACjAZ
ω
8. Frequency Response TLT-8016 Basic Analog Circuits 2005/2006 15
8.4 The Hybrid - π Model for the BJT
The Hybrid - π ModelThe rπ - β 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 TLT-8016 Basic Analog Circuits 2005/2006 16
Figure 8.29 Hybrid- π equivalent circuit..
gm – transconductance and accounts for amplifying properties. (10..600mS)
rx – base spreading resistance. It is the ohmic resistance of the base region. (~10Ω)
rπ – dynamic resistance of base-emitter junction (1..2..3kΩ)
Cπ – capacitance (diffusion + depletion) of base-emitter junction (~102pF)
rµ – reflects base-width modulation. Few MΩ and usually is neglected.
Cµ – depletion capacitance of collector-base junction. (Few pF)
T
CQm V
Ir
g ==π
β(8.47)
ro – reflects the upward slope of output characteristics of the BJT. (10..100kΩ)
CQ
Ao I
Vr ≅ (8.43)
8. Frequency Response TLT-8016 Basic Analog Circuits 2005/2006 17
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.
8. Frequency Response TLT-8016 Basic Analog Circuits 2005/2006 18
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 - πequivalent circuit for a typical 2N2222A transistor at a Q - point of ICQ=10 mA and VCEQ=10 V. Assume that VT=26mV.
Results:Sgm 385.0= Figure 8.31 Hybrid-π model for the 2N2222A at ICQ =10 mA
and VCEQ = 10V. For these values, β ≈ 225.225=βΩ= 585πr
Ω= M5.1µr
Ω19=xr
pFC 8=µ
pFC 196≅π
8. Frequency Response TLT-8016 Basic Analog Circuits 2005/2006 19
8.5 Common - Emitter Amplifiers at High Frequencies
Figure 8.33 Common-emitter amplifier.
8. Frequency Response TLT-8016 Basic Analog Circuits 2005/2006 20
Figure 8.33 Common-emitter amplifier.
oCL'L r||R||RR = (8.49)
( )[ ]sBx's R||Rr||rR += π (8.50)
(8.51)21 R||RRB = 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.
8. Frequency Response TLT-8016 Basic Analog Circuits 2005/2006 21
Figure 8.35 Simplified equivalent circuit for the common-emitter amplifier.
(8.52)'Lmo RrgV π−=
'Lm
o'vb Rg
VVA −==π
(8.53)
( )'1L
RgCCC mT ++= µπ (8.54)
T's
H CRf
π21
= (8.55) Figure 8.36 High-frequency behavior of the common-emitter amplifier.
8. Frequency Response TLT-8016 Basic Analog Circuits 2005/2006 22
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. Figure 8.37 Circuit for Example 8.8.
8. Frequency Response TLT-8016 Basic Analog Circuits 2005/2006 23
Solution:Ω=19xrΩ= 585πrΩ= k5.22orΩ= M5.1µr
pF196=πC
pF8=µCS385.0=mg
Ω=Ω== 252k5.22||510||510||||'oCLL rRRR
( )[ ] ( )[ ]Ω=
Ω+=+=5.61
50||k1019||585||||'sBxs RRrrR π
( )( ) pF980784196252358.018196
1 '
=+=×++=
++=L
RgCCC mT µπ
MHz64.2109805.612
12
112' =
×××== −ππ Ts
H CRf
2253850585 =×=≅ .gr mπβ
( ) ( )9.96
01225585252225
1 1
'
−=×++
×−=
++−
==E
L
in
ov Rr
RvvA
ββ
π
( )( ) [ ] Ω=×+Ω=
++=5530226585||k10
]1[|| 1EBin RrRR βπ
9.8855350
5539.96 −=+
−=
+==
ins
inv
s
ovs RR
RAvvA
Figure 8.38 Gain magnitude versus frequency for the common-emitter amplifier of Example 8.8.
8. Frequency Response TLT-8016 Basic Analog Circuits 2005/2006 24
8.8 Low Frequency Response of RC - Coupled AmplifiersCoupling Capacitors ( )
( )2
2
1 f/fjf/fj
RRR
VV
Lo
L
y
o
+×
+= (8.74)
( ) 22 2
1CRR
fLo +
=π (8.75)
x
yvo V
VA = (8.76)
( )( )
( )( )2
2
1
1
11 f/fjf/fj
RRRA
f/fjf/fj
RRRA
Lo
Lvo
ins
invs +
×+
××+
×+
=Figure 8.47 Amplifier with coupling capacitors.
y
o
x
y
s
x
s
ovs V
VVV
VV
VVA ××== (8.71)
Lo
Lvo
ins
invsmid RR
RARR
RA+
××+
= (8.77)
( )( )1
1
1 f/fjf/fj
RRR
VV
ins
in
s
x
+×
+= (8.72) ( )
( )( )( )2
2
1
1
11 f/fjf/fj
f/fjf/fjAA vsmidvs +
×+
×= (8.78)
( ) 11 2
1CRR
fins +
=π
(8.73)
8. Frequency Response TLT-8016 Basic Analog Circuits 2005/2006 25
( )( )
( )( )2
2
1
1
11 f/fjf/fj
f/fjf/fjAA vsmidvs +
×+
×= (8.78)
Two break frequencies in the terms
( )( )1
1
1 f/fjf/fj
+( )( )2
2
1 f/fjf/fj
+and
Figure 8.48 Magnitude Bode plot for the amplifier of Figure 8.47. (We have assumed that f1 > f2.)
8. Frequency Response TLT-8016 Basic Analog Circuits 2005/2006 26
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?
( )
( ) Hz1.531011021012
12
1
633
2212
=×××+×
=
+=
−π
π CRRf
io
( )
( ) kHz9.15101822
12
1
6
323
=××+
=
+=
−π
π CRRf
LoSolution:The amplifier has three coupling capacitors, determining three break frequencies:
( )
( ) Hz796.010110100101002
12
1
633
111
=×××+×
=
+=
−π
π CRRf
isSince 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 TLT-8016 Basic Analog Circuits 2005/2006 27
Bypass Capacitors
Figure 8.53 Discrete common-emitter amplifier.Figure 8.54 Gain magnitude versus frequency.