Frequency Response

ELC331T Amplifier Frequency Response

1

Electronics III Topic: Amplifier Frequency Response Compiled by: J. Pretorius January 2005 Introduction In Electronics I and Electronics II component- and circuit behavior were investigated involving conditions of only DC voltages and/or small signals with a constant frequency. Capacitors approached open circuit behavior at low frequencies and DC, and approached short circuit behavior at high frequencies. Which frequencies can be considered to be low and which as high? How will capacitors behave at frequencies that are neither low nor high, i.e. in the intermediate frequency range? How will other components behave at over a range of frequencies? These are all questions to be answered in Electronics III. To be able to investigate frequency response of components and circuits, some mathematical concepts need to be understood first. Frequency response is normally expressed graphically, with the frequency range indicated on the horizontal axis and the magnitude of the transfer function (gain, power or phase) on the vertical axis. Analysis will be done over a very wide range of frequencies. To be able to indicate a complete frequency range on one singular graph, the frequencies have to be expressed on a LOGARITHMIC scale. The transfer characteristic can also vary from very small values to very high values. Here the DECIBEL concept is used to express the values on the vertical axis. This graphical presentation is known as the BODE representation and referred to as the Bode Plot (gain and phase). The transfer function is used to describe the frequency-domain relationship between input and output signals appearing in different parts of a circuit. The phase plot or phase response shows the relationship between the output signal to input signal. Logarithmic scale (Frequency representation) The relationship between the variables of a logarithmic function: xba= and ax blog= for instance if b = 10 and x = 2, 100)10( 2 === xba and 2100loglog 10 === ax b Example 1.1 Find the value of x so that 100 000 = 10x. Solution

5)100000(log10 ==x Table 1.1 indicates how the logarithm of a number increases only as the exponent of the


2

number.

)10(log 010 0

)10(log 110 1

)10(log 210 2

)10(log 310 3

)10(log 410 4

)10(log 510 5

)10(log 610 6

)10(log 710 7

and so on Table 1.1 Frequency can range from DC (0Hz) up to several Giga Hertz (GHz). Table 1.2 express frequencies in logarithms. 1Hz )1(log10 0 2Hz )2(log10 0.301 3Hz )3(log10 0.477 5Hz )5(log10 0.699 7Hz )7(log10 0.845 9Hz )9(log10 0.954 10Hz )10(log10 1 50Hz )50(log10 1.699 100Hz )100(log10 2 500Hz )500(log10 2.699 1000Hz (1kHz) )1000(log10 3 1.5kHz )1500(log10 3.699 10kHz )10(log 4

10 4 100kHz )10(log 5

10 5 1000kHz or 1MHz )10(log 6

10 6 and so on

Table 1.2 Graph 1.1 is a semi-logarithmic graph showing only one cycle. The graph should be viewed sideways for the log-sale to be on horizontal axis. Graph 1.2 is a semi-logarithmic graph showing four cycles.


3

1

2

3

4

5

6

7

8

9

10

Graph 1.1. One cycle semi-logarithmic


4

1

2

3

4

5

67891

2

3

4

5

67891

2

3

4

5

67891

2

3

4

5

678910

Graph 1.2. Semi-logarithmic four cycle


5

Figure 1.1 shows the magnitude and phase plots for the transfer function

)2(1)(+

=s

sG

Figure 1.1. Frequency Response. Example of a magnitude and phase plot Decibel The magnitude of a frequency response graph (Bode plot) is indicated on the vertical axis. This magnitude can be in the form of relationship between input and output power or input and output voltage (signal). The range of this magnitude can spread over a wide range of values and is there for expressed in decibels.

Decibel to relate power levels is calculated with )(log101

210 PPGdB = dB and the

decibel to relate voltage levels is calculated with )(log201

210 VVGdB = dB

Table 1.3 below shows voltage gain in

out

VV expressed in dB.

Voltage gain dB level 0.5 -6 0.707 -3 1 0 2 6 10 20 100 40 1000 60 10000 80 1000000 120 and so on


6

Another advantage working in dB is the logarithmic relationship of adding (summing) logs.

nT AAAAA ......321 ∗∗∗= then )(....)()()()( 321 dBAdBAdBAdBAdBA nT +++= This allows cascaded system’s gains to be added when working in dB. Figure 1.1 shows an example with the magnitude-expressed dB on the vertical axis. Plotting a frequency response (Bode Plot) The log-magnitude and phase frequency response curves as function of log ω are called Bode plots or Bode diagrams. The transfer function of a circuit in the frequency domain can take on the following general form:

))....()(())....()((

)(21

21

nm

k

pspspsszszszsK

sG++++++

=

with K a constant, z1,z2, to zk zeros, p1, p2, to pn poles and s the frequency. Simplify by working with the logarithm of the magnitude:

ω

ω

jsnm

k

pspspss

zszszsKjG

→⋅⋅+−−+−+−−

+++++++=

)(log20....)(log20)(log20log20

)(log20....)(log20)(log20log20)(log20

21

21

Magnitude plot Thus if the response of each term is known, the algebraic sum would yield the total response in dB. Further, if an approximation of each term could be made, which will only consists of straight lines, graphic addition of terms would be greatly simplified. To plot the magnitude plot a simple set of rules can be applied: Zeros have 0dB value from start freq. and then break upwards along an asymptote with 20dB/decade slope from its break point. Poles have 0dB value from start freq. and then break downwards along an asymptote with 20dB/decade slope from its break point. A zero at jω = 0 has a continues slope of +20dB/dec and a value of 0dB at ω = 1 rad/sec. A pole at jω = 0 has a continues slope of –20dB/dec and a value of 0dB at ω = 1 rad/sec. These graphs continue to infinity frequency. One decade is the start frequency multiplied with 10, and two decades is the start frequency multiplied by 100, and so on. A break point is a specific zero or pole value. Figure 1.2 below shows examples of magnitude plots.


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Figure1.2. Example of magnitude and phase plots. Example 1.2 Draw the magnitude Bode plot for the system transfer function

)1021)(

25001)(

251(

10)(5×

+++= ssssH

Solution

Let s = jω: )

1021)(

25001)(

251(

10)(5×

+++= ωωωω jjjjH

The system has three poles and no zeros. Poles are at: Pole Frequency p1 25 rad/sec p2 2500 rad/sec P3 2 x 105 rad/sec The graph can be subdivided into four magnitude graphs: Straight line gain at dBM 20)10log(20 ==


8

A straight line at 0dB until p1 and then breaking downwards with a slope of -20dB/decade. A straight line at 0dB until p2 and then breaking downwards with a slope of -20dB/decade. A straight line at 0dB until p3 and then breaking downwards with a slope of -20dB/decade.


9

The complete magnitude graph is the sum of above four graphs:


10

Phase Plot The relation between the phase of the output signal to input signal can also be plotted on a log frequency graph with the phase relation indicated in degrees on the vertical scale. To plot the magnitude plot a simple set of rules can be applied: A zero at ω = 0 has a phase of 900 at all frequencies.

• A pole at ω = 0 has a phase of -900 at all frequencies. • A constant gain term has a 00 phase at all frequencies. • The phase plot for a zero is 00 until 1 decade before the zero

break point. It then breaks upwards with 450 per decade slope until 1 decade after the zero break point. The phase stays constant for the rest of the frequencies.

• The phase plot for a pole is 00 until 1 decade before the pole break point. It then breaks downwards with 450 per decade slope until 1 decade after the pole break point. The phase stays constant for the rest of the frequencies.

See Figure 1.2 for examples of phase plots. Example 1.3 Draw the phase Bode plot for the system transfer function in Example 1.2. Solution: The solution is shown in example 1.2. Example 1.4 Draw the Bode magnitude and phase plots for the following function:

)50)(7)(1(

)20()(+++

+=

sssssG


11

Solution:

First normalize the function: )

501)(

71)(1(

)20

1()50)(7(

20

)( sss

s

sG+++

+=

• Zero break point at 20 • Pole break points at 1,7 and 50 • Constant gain factor of - 24.9dB

1.1. Problems 1.1.1. For each of the following functions plot the magnitude and phase Bode

plots.

1.1.1.1. )4)(2(

1)(++

=sss

sG

1.1.1.2. )4)(2(

)5()(++

+=

ssssG

1.1.1.3. )4)(2()5)(3()(

++++

=ssssssG

1.1.2. For each of the following functions plot the magnitude and phase Bode plots.

1.1.2.1. )

101)(

50001(

)90

1)((900)(

5ωω

ωωω jj

jjjH

++

+=


12

1.1.2.2. )

1021)(

25001)(

251(

10)(5×

+++= ωωωω jjjjH

1.1.2.3. 2

2

)5000

1(

)90

1(200)( ω

ω

ω j

j

jH+

+=

1.2. Sources of Capacitance and Inductance in electronic circuits

The most common origins of capacitance include the discrete capacitors used in single element design, the stray capacitances contributed by interconnections such as wires or printed circuit board paths, and the internal capacitances that originate within electronic devices (ex. transistors).

1.2.1. Internal capacitance in Bipolar Junction Transistors

Figure 1.3. Small signal model of BJT including internal junction capacitors. On data sheets of discrete BJT’s , Cµ is also referred to as Cob. The lateral base resistance rx, which interacts with Cπ and Cµ at high frequencies is also shown in the device model in Fig 1.3. With Cπ and Cµ included, the dependent current source can still be considered a function of the total base current Ib if β0 is represented as a function of frequency β(ω).

Figure 1.4. Small signal BJT model with frequency dependent β(ω)


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µωπ CgCT

m −= With ωT the unity gain frequency.

µππππ

βωCC

gCCr

mT +

=+

=)(

0

On manufacturers’ data sheets, β(ω) is sometimes labeled with the symbol hfe which is short hand notation for hfe(ω).

1.2.2. Internal capacitance in Field Effect Transistors.

Figure 1.5.

1.3. Frequency response of electronic circuits and devices

Figure 1.6. RC circuit with capacitor as shunt element.

RCjCjRCj

VV

in

out

ωω

ω+

=+

=1

11

1


14

Figure 1.7. Bode plots for RC circuit in Fig. 1.6.

Figure 1.8. RC circuit with C as series element.

RCjRCj

CjRR

VV

in

out

ωω

ω +=

+=

11


15

Figure 1.9. Bode plots for RC circuit in Fig. 1.8. 1.3.1. Bode Plot of systems with complex transfer functions.

The task of constructing the Bode plot of any circuit, no matter how complex, is greatly simplified if its system function is expressed in the

general form )....1)(1()....1)(1()(

31

42

ωωωωωωωωω

ωjjjjjAjH

++++

=

with the numbered frequencies ω1….ωn the break points of the system function and A constant. The solitary factor jω is not present for all circuits. The super position of poles approximation can be applied to any system function that can be put in the form a midband-gain multiplied by separate low-frequency and high-frequency system functions.

+++

×

+++

=

⋅⋅=

)1)...(1()1(1

)1()(

...)1()(

)1()/)(

)(

210

0

nm

m

b

b

a

a

HL

jjjjj

jj

jj

A

HHAjH

ωωωωωωωωωω

ωωωω

ωωωω

ω

The breakpoints ωa….ωm of HL jointly define the low-frequency limit of the midband, and the breakpoints ω1….ωn of HH jointly define the high-frequency limit of the midband. The high-frequency –3dB point ωH of the system function which constitutes the upper limit of the midband region, is given by


16

nn

H ωωωωωω

ω ....1....1121

1

21

=

+++=

−

This is known as the superposition-of-poles approximation at the high-frequency end of the midband. The high-frequency poles with the lowest frequency will make the most contribution to ωH. If one pole is significantly lower than the others, it will dominate ωH. Poles located near each other will make nearly equal contributions to ωH. Poles well above ωH will make little contribution to the value of ωH. The low-frequency –3dB point ωL of the system function which constitutes the lower limit of the midband region, is given by

mbaL ωωωω +++= .... This is known as the superposition-of-poles approximation at the low-frequency end of the midband. The low-frequency poles with the highest frequency will make the most contribution to ωL. If one pole is significantly higher than the others, it will dominate ωL. Poles located near each other will make nearly equal contributions to ωL. Poles well below ωL will make little contribution to the value of ωL.

1.3.2. High- and Low-frequency Capacitors

Capacitors only influence frequency response below or above the midband region. Conversely, the midband represents the frequency range over which circuit behavior is unaffected by circuit capacitors. Certain capacitors influence the region below the midband range, called low-frequency capacitors and others the region above the midband range, and called high-frequency capacitors. High-frequency capacitors degrade the gain above the midband and low-frequency capacitors degrade the gain below the midband. Low-frequency capacitors behave like short circuit in midband and the high-frequency capacitors like open circuit in the midband. General rules:

• Capacitors will function as low-frequency capacitors if they are in series with a circuit’s input or output terminal.

• Capacitors will function as high-frequency capacitors if they shunt input or output terminals to small-signal ground.

• All internal device and stray lead capacitors are high-frequency capacitors.

• All external series capacitors act as low-frequency capacitors. • A bypass capacitor that shunts a device terminal to ground acts

as a low-frequency capacitor. (Example CE)


17

+

-gmVpi CL

+

-

Vs1

Cu

Cpi

Ce

CcCs

RLRc

Re

RpiRb

Rs

Cpi

Figure 1.10. Small-signal model showing all capacitors.

In Fig1.10 above all capacitors are shown. This circuit can be split into a low-frequency model showing only the low-frequency capacitors and a high-frequency model showing only the high frequency capacitors. Low-frequency capacitors: Cs, Cc, and CE High-frequency capacitors: CL, Cπ, and Cµ

Each capacitor is important to only one end of the frequency spectrum.

• At frequencies well below high-frequency end of midband: o All high-frequency capacitors can be treated as open

circuits. • At frequencies well above low-frequency end of midband:

o All low-frequency capacitors can be treated as short circuits.

• A model where low-frequency capacitors are short circuits and high-frequency capacitors are open circuits is called the midband model.

Example 1.5

Identify the high- and low-frequency capacitors in the following circuit and compute the midband gain and separately examine the high- and low-frequency ends of the Bode plot. Assume the following conditions for the BJT: VT = 25mV Cµ = 0 rx =10Ω fT = 450MHz βF = βo = 100 Vf = 0.7V ro = ∞


18

+ Vcc12V

+

-

Vs110V

CL14pF

Cc10uF

Cs10uF RL

1M

Rc5.1k

RB1M

Rs1k

Q1

Figure1.11. Common emitter amplifier Solution

• Identify the low- and high-frequency capacitors. Low-frequency caps: CS and CC High-frequency caps: Cπ and CL

• Compute bias current through Q1 so that gm, rπ and Cπ can be determined.

B

fCCB R

VVI

−= and mAII BFC 13.1≈= β

VAmVI

gT

Cm 45≈=

η and Ω≈= k

gr

m

o 2.2β

π

pFCfg

CT

m 162

≈−= µπ π

• Evaluate the low-frequency model of the circuit

VoutVb

+

-gmVpi

+

-

Vs1

CcCs

RLRcRpi

Rb

Rs

Figure 1.12. Low-frequency small-signal model With voltage division Vb can be found:

sss

ss

sb

SSSxB

xBb

VRrCjRrCj

Rrr

VV

VRCjrrR

rrRV

)(1)(

)1()()(

+++

+==

+++

+=

π

π

π

ππ

π

π

ωω

ω

with (rπ + rx) ≈ rπ and RB (rπ + rx) ≈ rπ


19

π

π

ωω

ω

VRRCjRRCj

RRRR

gV

RVgCjRR

RV

RIV

CLC

CLC

LC

LCmout

LmCLC

Cout

Loutout

)(1)(

)(1

+++

+−=

−++

=

⋅=

Substituting the equation for Vπ into the Vout equation will yield

)(1)(

)(1)(

)(1)(

)(1)(

CLC

CLC

Ss

SS

LC

L

SCm

S

out

SCLC

CLC

Ss

SS

LC

LC

SCmout

RRCjRRCj

RrCjRrCj

RRR

Rrr

RgVV

VRRCjRRCj

RrCjRrCj

RRRR

Rrr

RgV

+++

+++

++−=

+++

+++

++−=

ωω

ωω

ωω

ωω

π

π

π

π

π

π

π

π

The above expression is in the form H(jω)=A0HL.

Hzf

radRRC

Hzf

radRrC

CLC

SS

016.02

sec1.0)(

1

52

sec3.31)(

1

22

2

11

1

==

=+

=

==

=+

=

πω

ω

πω

ωπ

• Evaluate the high-frequency model of the circuit

rx

+

-gmVpi CL

+

-

Vs1 CpiRLRc

RpiRb

Rs

Figure 1.13. High-frequency small-signal model

VoutVpi

gmVpiCL

+

-

VTh CpiRLRc

Rp

gmVpi

Figure 1.14. Simplified high-frequency small-signal model


20

SS

xSBSB

BSTh

xSBp

Rrr

VrRRr

rRR

RVV

rrRRR

+=

+++=

+=

π

π

π

π

π

)(

)(

SPS

ThP

VRCjRr

rV

RCjCj

Vππ

π

π

ππ ωω

ω++

=+

=1

111

[ ]

)(1)(

)()1(

LCL

LCmout

mLLCout

RRCjRR

VgV

VgCjRRV

ω

ω

π

π

+−=

−=

Substituting Vπ into Vout equation:

MHzf

RC

MHzf

RRC

RRCjRCjRrr

RRgVV

P

LCL

LCLPSLCm

S

out

5.142

1

24.22

)(1

)(11

11)(

44

4

33

3

==

=

==

=

+++−=

πω

ω

πω

ω

ωω

π

ππ

π

• Evaluate the midband-frequency modal for the circuit.

(Determination of midband gain) The midband gain can be determined from the low-frequency model equation by letting the frequency be much higher than ω1 or the high-frequency model equation by letting the frequency be much lower than ω3.

dBRR

RRrr

RgaLC

L

SCmv 44157≡−=

++−=

π

π

with CmRg− the basic midband gain

and SRr

r+π

π the input-loading factor

and LC

L

RRR+

the output-loading factor.


21

Students have to be able to do above example with a FET transistor instead of a BJT. This is assumed to be self-study.

1.4. Problems (Frequency response of Transistor amplifiers) 1.4.1. Find the low- and high-frequency –3dB endpoints of the circuit. The

MOSFET has internal capacitance Cgs = 8pF, and negligible Cgd and Cds. The output resistance is ro = 100kΩ and K = 1mA/V2 and VTR = 4V so that ID = 1mA and gm = 2(KID)½ = 2mA/V.

C210uF

1kHz

V1-1/1V

+ VDD10V

CL1pF

C110uF

RL10k

RD5k

R21M

R11M

Rs10k Q1

Figure 1.15 Circuit for problem 1.8.1

1.4.2. Find the –3dB midband of the endpoints of the BJT amplifier using the superposition-of-poles technique. Assume the BJT to be at room temperature with parameters Cµ = 0, rx = 10Ω, fT = 450MHz, βF = βo = 100, Vf = 0.7V, η = 1, and ro = ∞.


22

CM4pF

RE3.3k

+ Vee6V

+ Vcc6V

Vout

+

C2100uF

+

C12.2uF

1kHz

V1-10m/10mV

Q1

Rs1k RM

1M

Rc5k

R210k

R120k

Figure 1.16. Circuit for problem 1.8.2

1.5. Effect of Transverse Capacitance on amplifier response

Up to now the transverse capacitance Cµ was assumed to be negligible. This capacitance is almost always present in a real transistor. Two methods can be used to determine the influence of this capacitance on the frequency response of a circuit. One method uses the Thevenin-resistance to find the contribution of Cµ to the high-frequency midband point, and the other method is called the Miller’s Theorem. The later will be discussed and used. 1.5.1. Miller’s Theorem and Miller Multiplication

Miller’s Theorem is valid for all resistive, capacitive, inductive and circuits containing general linear complex impedance elements. A transverse impedance like Zµ can be modeled by the equivalent parallel impedance ZA.

Input

Va

Output

Vb

Input

Va

CA

Zu

Figure 1.17. Miller’s Theorem )1( abA VVCC −= µ Example1.6.


23

Find the –3dB high-frequency endpoint using Miller’s theorem for Figure 1.16. Omit the load elements RM an CM and assume the BJT to be biased at IC = 1mA with parameters Cµ = 2pF, fT = 450MHz, ro = ∞, rx = 10Ω, βo = 100, gm = 40mA/V, and rπ = 2.5kΩ. Solution

gmVpi+

-

Vs1

Cu

Cpi

rx

RcRpi

Rb

Rs

↓

CA

gmVpi+

-

Vs1Cpi

rx

RcRpi

Rb

Rs

CA

Figure 1.18. Replacing Cµ with CA

with

1.5.2. High-Frequency poles with feedback resister

When a device is connected in the follower configuration, or when a resistor is shared between the input and output loops of an inverter, Miller’s Theorem must be used with care.

[ ] Ω=+=

=+

=

=−=−=

+=

650)(

591)(2

1402)1(

)(

21 RRRrrr

kHzCCr

f

pFVVCCRVgVCCCC

Sxth

AthA

outA

Cmout

AA

ππ

ππ

πµ

π

ππ

π


24

Vout

gmVpi+

-

Vs1

Cu

Cpi

rx

Rc

Re

RpiRb

Rs

Figure 1.19. High-frequency model of BJT follower. Using Miller’s theorem will result in CA being in parallel with Cπ. CA appears between the base-emitter terminals and not between the base and ground. This will result in a tedious calculation to find the Miller’s capacitance CA. The high-frequency response of an amplifier with feedback resistor is best derived using the Thevenin-resistance method and the superposition of poles. Example 1.7 (Feedback resistor and Thevenin-resistance method) Find the upper –3dB endpoint of the circuit for which Cµ = 2pF, fT = 400MHz and rx = 10Ω. Suppose βo = 100 and IC = 2.5mA so that gm = 100mV/A and rπ = 1kΩ at room temperature. Remember CS behaves like a short circuit at the frequencies of interest. Note that the poles determined using Thevenin-resistance method apply whether the amplifier is used as a follower or an inverter.

RE1k

+

C110uF

1kHz

V1-10m/10mV

Q1

Rg1k

Rc5k

R26.8k

R127k

1kHz

V1-10m/10mV

Figure 1.20 The small-signal model of Fig1.19 can represent this amplifier.


25

Let

pFCg

C

VRRR

RRV

RRR

v

RRRRRR

T

m

gS

Sb

bin

gbSs

38

845

121

21

21

≈−=

+=

+=

Ω===

µπ ω

The high-frequency –3dB point can be found by evaluating the small-signal Thevenin-resistance seen by Cπ and Cµ.

• Thevenin-resistance seen by Cπ

Vin = 0

e

c

ie

i(test)xb

+

-Vtest

gmVpi

rx

Rc

Re

Rpi

Rs

Figure 1.21 Applying KCL to nodes x and e:

)(1

)1(11

)(

)()(

xsE

EmE

xstest

xs

Etesttest

testmtesttest

test

Etesttest

testmEtestmEee

xs

etesttesttest

rRRrRgR

rRrv

rR

Rirvvgv

rvi

RirvvgRiivgRiv

rRvv

rvi

++

++

++

=

+

−+++=∴

−+=−+==

++

+=

ππ

π

π

πππ

π

The small-signal Thevenin-resistance seen by Cπ is given by the expression

Ω≈++

++

++== 18

)1(11)(1

ππ

π

rRgR

rRr

rRRiv

rE

mExs

xsE

test

testth

MHzCr

fth

23321

==ππ

π π


26

• Thevenin-resistance seen by Cµ

ic

Vin = 0

e

c

ie

xb

+ -

Vtest

gmVpi

rx

Rc

Re

Rpi

Rs

+ -

Vtest

Figure 1.22 Applying KCL to nodes c and x yields

)()(

))((

)(

Cmxs

Cxstesttest

Ccxstesttest

Ccxsscxtets

testtests

mtestc

RgrrR

vRrRiv

RirRrv

iv

RirRivvvrv

iiii

vgii

++

−+++=∴

++−=∴

++=−=

−=−=

+=

ππ

π

π

π

ππ

π

Apply KVL around input loop:

MHzCr

f

kRgrrR

rRRrgrrrR

RrRiv

r

RgrrR

rRRrgrrrRi

RrRiv

rRRrgrrrRi

v

Rvgrv

vrRrv

i

Rvgrv

vrRi

th

Cmxs

xsEm

xsCxs

test

testth

Cmxs

xsEm

xstestCxstesttest

xsEm

xstest

Emxstest

Emxss

96.721

10))()()1(

)(()(

))()()1(

)(()(

)()1()(

)())((

)()(

≈=

Ω≈++

−++++

++++==∴

++

−++++

++++=∴

+++++

=

++=+−∴

++=+

µµµ

πππ

πµ

πππ

π

ππ

ππ

ππ

ππ

π

π

ππ

ππ

π Using the superposition of poles to find the upper –3dB point:

MHzMHzMHz

fffH 7.7)96.71

2331( 1 ≈+== −

µπ


27

Important notice:

The equations for rthπ and rthµ show that the Thevenin-resistance seen by Cπ and Cµ can be reduced by increasing the value of RE which will increase the bandwidth of the amplifier. This increase comes with the expense of midband gain that will decrease as RE increase.

1.6. Frequency Response with Bypass Capacitor

The simultaneous goals of amplifier bias stability and large gain are often not compatible. A large resister like RE shared by the input and output terminals produces bias stability but reduce gain. The gain can be improved by bypassing (shunting) RE down to signal ground with a large capacitor that will act as a short circuit at frequencies of interest where high gain is required.

gmVpiCL

+

-

Vs1

Cu

Cpi

Ce

CcCs rx

RLRc

Re

RpiRb

Rs

Figure 1.23 The small-signal gain without CE of the above circuit was determined in Electronics II.

E

c

EbS

c

in

outv R

RRrRR

Rvv

A−

≈+++

−==

)1( 0

0

ββ

π

With CE is present and the frequency high that CE function as a incremental short circuit to signals, the gain is given by:

π

βrRR

Rvv

AbS

c

in

outv +

−== 0

The frequency where CE undergoes the transition from open- to short-circuit is of importance. The problem can be addressed by treating RE and CE in parallel as a complex impedance ZE.

EE

E

EEE CRj

RCj

RZωω +

==1

1

Substitution of ZE in the place of RE in the equation for Av will yield:


28

++

=

=

++

+++−

=

1

1

111

)1(

0

2

1

2

1

0

0

β

ω

ω

ωωωω

ββ

π

π

rRRC

CR

jj

RrRRR

vv

SEE

EE

EbS

C

in

out

The zero ω1 represents the frequency at which CE begins to bypass the emitter node to ground and increasing the magnitude of the gain. The pole ω2 represents the frequency at which complete bypassing is achieved.

1.7. Problems (Bypass capacitor) 1.7.1. The following amplifier must provide a midband gain magnitude of at

least 120, beginning at a frequency no higher than 100Hz. Resisters R1, R2, and RE have been chosen to set the BJT bias current to about 1.4mA. Choose appropriate values for RC, CE and CS such that the gain objectives are met. The pole of CE should dominate the low-frequency response.

RE1k

+Vcc10V

Vout

+

CE

+

Cs

1kHz

V1-10m/10mV

Q1

Rs

Rc

R227k

R1100k

+Vcc10V

Figure 1.24

1.7.2. Consider the circuit in 1.7.1 with β0 = 90. Draw the low-frequency end of the Bode plot of Vout/Vs if RC = 2k2Ω, CS = 150µF and CE = 10µF.


29

1.8. Active Filter Frequency Response (First Order) 1.8.1. Introduction

Filters of some sort are essential to the operation of most electronic circuits. It is therefore in the interest of anyone involved in electronic circuit design to have the ability to develop filter circuits capable of meeting a given set of specifications. Unfortunately, many in the electronics field are uncomfortable with the subject, whether due to a lack of familiarity with it, or a reluctance to grapple with the mathematics involved in a complex filter design. In circuit theory, a filter is an electrical network that alters the amplitude and/or phase characteristics of a signal with respect to frequency. Ideally, a filter will not add new frequencies to the input signal, nor will it change the component frequencies of that signal, but it will change the relative amplitudes of the various frequency components and/or their phase relationships. Filters are often used in electronic systems to emphasize signals in certain frequency ranges and reject signals in other frequency ranges. Such a filter has a gain which is dependent on signal frequency. As an example, consider a situation where a useful signal at frequency f1 has been contaminated with an unwanted signal at f2. If the contaminated signal is passed through a circuit (Figure 1.25) that has very low gain at f2 compared to f1, the undesired signal can be removed, and the useful signal will remain. Note that in the case of this simple example, we are not concerned with the gain of the filter at any frequency other than f1 and f2. As long as f2 is sufficiently attenuated relative to f1, the performance of this filter will be satisfactory.

In general, however, a filter's gain may be specified at several different frequencies, or over a band of frequencies. Since filters are defined by their frequency-domain effects on signals, it makes sense that the most useful analytical and graphical descriptions of filters also fall into the frequency domain. Thus, curves of gain vs frequency and phase vs frequency are commonly used to illustrate filter characteristics, and the most widely-used mathematical tools are based in the frequency domain.

Figure 1.25. Using a Filter to Reduce the Effect of an Undesired Signal at Frequency f2, while Retaining Desired Signal at Frequency f1.


30

1.8.2. The Basic Filter Types There are five basic filter types (band-pass, notch, low-pass, high-pass, and all-pass).

1.8.3. Band-pass Filter

Figure 1.26. Examples of band-pass filter responses

1.8.4. Band-reject Filter (Notch)

Figure 1.27. Examples of band-reject filter responses.


31

1.8.5. Low-pass Filter

Figure 1.28. Examples of Low-pass filter responses.

VoutVin

Z2

Z1

Figure 1.29. Inverting amplifier topology with feedback. In the inverting amplifier in Fig. 1.29. let Z1 be purely resistive R1. Let Z2 be a resistor R2 and capacitor C in parallel.

CRjRR

ZZ

VV

CRjR

CjR

CjR

CjRXRZ

RZZZ

VV

in

out

C

in

out

21

2

1

2

2

2

2

2

222

11

1

2

11

11

11

ω

ωω

ωω

+−=−=

+=

+===

=

−=


32

Figure 1.30. Frequency response of an active low-pass filter.

1.8.6. High-pass Filter

Figure 1.31. Examples of high-pass filter responses.

VoutVin

Z2

Z1

Figure 1.32. Inverting amplifier. For the inverting amplifier in Fig. 1.32., let Z1 be R1 and C in series, and Z2 be R2. It is left to the reader to proof that:

CRjCRj

VV

in

out

1

2

1 ωω+

−=

and to plot the frequency response to indicate that it is a high-pass filter.

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Frequency Response