8. frequency response - tut8. frequency response tlt-8016 basic analog circuits 2005/2006 13 8.3 the...

8. Frequency Response TLT-8016 Basic Analog Circuits 2005/2006 1

8. Frequency Response


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 ; ⇒⇒


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)


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

−=

−=


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)


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.


( ) ( )

( )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.


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.


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.


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+

×+

=


( ) ( ) ( )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.


( )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.


8.3 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.


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


8.4 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.


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)


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.


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


8.5 Common - Emitter Amplifiers at High Frequencies

Figure 8.33 Common-emitter amplifier.


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.


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.


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.


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.


8.8 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)


( )( )

( )( )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.)


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.


Bypass Capacitors

Figure 8.53 Discrete common-emitter amplifier.Figure 8.54 Gain magnitude versus frequency.

8. frequency response - tut8. frequency response tlt-8016 basic analog circuits 2005/2006 13 8.3 the...

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