BJT & FET Frequency Responsespringzabdesk.szabist.edu.pk/CoursePortFolioFiles...3/29/2012 2 Chapter...

3/29/2012

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Chapter 4

BJT & FET

Frequency Response

Spring 2012

4th Semester Mechatronics

SZABIST, Karachi

CH 4

Frequency Response

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Course Support

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Office: 100 Campus (404)

Official: ZABdesk

Subsidiary:

https://sites.google.com/site/zabistmechatronics/home/spring-2012/ecd

ebooks:

https://sites.google.com/site/zabistmechatronics/home/ebooks

2

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Frequency Response

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Chapter Contents

• BJT & JFET Frequency Response− Introduction

• Logarithms and Decibels

• General Frequency Considerations

• Bode plot − Low Frequency Analysis

• Low Frequency Response − BJT Amplifier

• Low Frequency Response − FET Amplifier

• High Frequency Response − BJT Amplifier

• High Frequency Response − FET Amplifier

• Multistage Frequency Effects*

3Frequency Response

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4

Introduction

Frequency Response

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Frequency Response:

Phase and amplitude plots and equations of an amplifier

Frequency Response Prerequisites:

1. Logarithms

2. Semi-log plots

3. Decibels

4. Normalization

Introduction 5

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Logarithms

6Frequency Response

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Logarithms:

The logarithm of a number is the exponent by which another fixed value, the base,

has to be raised to produce that number.

Common logarithms:

Natural logarithms:

Relationship of CL and NL:

Benefits:

• Plotting of a variable between wide limits

• Compression of large data

Logarithms 7

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Frequency Response

,

2.3

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Logarithms:

Logarithms 8

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Frequency Response

A nautilus displaying a

logarithmic spiral

Broccoli, which grows

in a logarithmic spiral

A low pressure area over

Iceland shows an

approximately logarithmic

spiral pattern

The whirlpool Galaxy

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Example 9-1:

Using the calculator, determine the logarithm of the following numbers to the base

indicated:

a. log10 106

b. loge e3

c. log10 10−2

d. loge e−1

Example 9-2:

Using the calculator, determine the logarithm of the following numbers:

a. log10 64

b. loge 64

c. log10 1600

d. log10 8000

Logarithms 9

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Example 9-3:

Using calculator, determine the antilogarithm of the following expressions:

a. 1.6 = log10 a

b. 0.04 = loge a

Example 9-4:

Using calculator, determine the logarithm of the following numbers:

a. log10 0.5

b. log10 (4000/250)

c. log10 (0.6 x 30)

Logarithms1

0

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Frequency Response

1 0

1

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Semi−−−−logPlots

11Frequency Response

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Semilog Plots 12

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Frequency Response

≅ 30%log102=0.3010

≅ 48%Log103 = 0.4771 log104 = 0.6021

(≅ 60%)

log10 9 = 0.9543

log10 8 = 0.9031

log10 7 = 0.8451

log10 6 = 0.7781

log10 5 = 0.6999

Linear

1

2

Semilog graph paper

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Semilog Plots 13

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Frequency Response

Identifying the numerical values of the tic marks on a log scale

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Example 9-5:

Determine the value of the point appearing on the logarithmic plot in Fig. 9-4 using

the measures made by a ruler (linear).

Semilog Plots 14

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Frequency Response

10 10 ⁄

10 10

d1

d2

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Decibels


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Decibels 16

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Frequency Response

! "#"$%

!& 10 "#"$'(%

!&) 10 **+,$-.)/% $'(0%

!1& 20 22 $'(%

|415 | =|41| ∙ |41| ∙ |417 |⋯⋯ |419|

!&5 = !&+ !&+ . . . . . + !&9

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Example 9-6:

Find the magnitude gain corresponding to a voltage gain of 100 dB.

Example 9-7:

The input power to a device is 10,000 W at a voltage of 1000 V. The output power is

500 W and the output impedance is 20 Ω.

Example 9-8:

An amplifier rated at 40 W output is connected to a 10 Ω speaker. Calculate:

a) The input power required for full power output if the power gain is 25 dB

b) The input voltage for rated output if the amplifier voltage gain is 40 dB

Decibels 17

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Frequency Response

General Frequency

Considerations


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General Frequency Considerations:

The frequency response of an amplifier refers to the frequency range in

which the amplifier will operate with negligible effects from capacitors

and device internal capacitance.

This range of frequencies can be called the mid-range.

• At frequencies above and below the midrange, capacitance and any inductance

will affect the gain of the amplifier.

• At low frequencies the coupling and bypass capacitors lower the gain.

• At high frequencies stray capacitances associated with the active device lower the

gain.

• Also, cascading amplifiers limits the gain at high and low frequencies.

Freq. Considerations

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19



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Frequency Response

• A Bode plot indicates the

frequency response of an

Amplifier:

• The horizontal scale

indicates the frequency (in

Hz) and the vertical scale

indicates the gain (in dB)

• The mid-range frequency

range of an amplifier is

called the bandwidth of the

amplifier

• The bandwidth is defined by

the lower and upper cutoff

frequencies

• Cutoff – any frequency at

which the gain has dropped

by dB

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Normalization


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Normalization Process:

In communication, a decibel plot vs frequency is normally provided

rather than gain vs frequency

A process in which the vertical parameter is divided by a specific level or

quantity sensitive to a combination or variables of the system

The band frequencies define a level where the gain or quantity of interest

will be 70.7% or its maximum value

Normalization

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Normalization Process:

Normalization

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Normalized gain versus frequency plot

Decibels plots of the normalized gain versus frequency plot

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Example 9-9:Given the frequency response:

a) Find the cutoff frequency f1 and f2 using the measurements provided

b) Find the bandwidth of the response

c) Sketch the normalized response

Normalization

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Example 9-9:

Normalization

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dB Plot:

Normalization

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Decibel plot of the normalized gain versus frequency plot

Av/Avmid Av/Avmid|dB

1 0

0.707 -3

0.5 -6

0.35 -9

0.25 -12

Low Frequency Analysis


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Low Frequency RC Circuit Analysis:

LF Analysis−−−− Bode Plot

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Frequency Response

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Low frequency response for the R-C circuit



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Frequency Response

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41 0.707|;<=,

> 1

2?@A41

11 B >>

41$CD% 20 1

1 >>

#

41$CD% 20 >> E≪E

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Frequency Response

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31

f f1/f Av(dB)

f1 1 0

½ f1 2 -6

¼ f1 4 -12

1/10 f1 10 -20



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Frequency Response

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f f1/f Av(dB)

f1 1 0

½ f1 2 -6

¼ f1 4 -12

1/10 f1 10 -20

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Low Frequency RC Circuit Analysis:• The piecewise linear plot of the asymptotes and associated breakpoints is called a

Bode plot of the magnitude versus frequency

• A change in frequency by a factor of 2, equivalent to 1 octave, results in a

6-dB change in the ratio as noted by the change in gain from f1/2 to f1.

• For a 10:1 change in frequency, equivalent to one decade, there is a 20-dB change

in the ratio as noted by the change in gain from f1/10 to f1.


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Frequency Response

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33

41 GH 10IJ$CD%/L M NOP >

>



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Frequency Response

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Phase response for the RC circuit Example 9-9

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Example 9-10:

For the network of fig. 9-20: (R = 5 kΩ, C = 0.1 µF)

a) Determine the break frequency

b) Sketch the asymptotes and locate the −3 dB point

c) Sketch the frequency response curve

d) Find the gain at Av(dB) = − 6 dB


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Example 9-10:


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Example 9-10:


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Computer Analysis

% bode plot of Example 9-10

f = 10:10^4;

fo = 318.5;

A = 20*log(1./(1+(fo./f).^2).^(1/2));

semilogx(f,A), xlabel('f (log scale)'),

ylabel('Av(dB)')

grid

101

102

103

104

-70

-60

-50

-40

-30

-20

-10

0

f (log scale)

Av(d

B)

Low Frequency Response

BJT amplifiers


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BJT Amplifiers:

.

LF Response −−−− BJT Amplifiers

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Effects of Cs on the LF response:


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Effects of CC on the LF response:


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Effects of CE on the LF response:


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Effects of Cs and CE on the LF response:


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LC

o L c

1f

2 ( )Cπ R R=

+

o C oR R ||r=

The cutoff frequency due to CC can

be calculated with

where

The cutoff frequency due to CS can be

calculated by

Ls

s i s

1f

2 (R R )Cπ=

+

i 1 2 eR R ||R ||βr=

where

Example 9-11:a) Determine the lower cutoff frequency for the network of Fig. 9.23 using the

following parameters:

CS = 10 µF, CE = 20 µF, CC = 1 µF,

RS = 1 kΩ, R1 = 40 kΩ, R2 = 10 kΩ, RE = 2 kΩ, RC = 4 kΩ, RL = 2.2 kΩ,

β = 100, ro = ∞ Ω, VCC = 20 V

a) Sketch the frequency response using a Bode plot

b) Verify the result using a Simulator.


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Example 9-11:


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Low Frequency Response

FET amplifiers


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FET Amplifiers:

LF Response −−−− FET Amplifiers

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FET Amplifiers:


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sig i G

1

2 (R R )CLG

fπ

=+

i GR R=

The cutoff frequency due to

CG can be calculated with

where

o L G

1

2 (R R )CLC

fπ

=+

O D G||R R r=


CC can be calculated with

where

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FET Amplifiers:


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ReqLS

eq S

1

2 R Cf

π=

d

eq S

m Ω

1||

gr

R R

≅ ∞

=


CS can be calculated with

where

Example 9-12:a) Determine the lower cutoff frequency for the network of Fig. 11.32 using the

following parameters:

CG = 0.01 F, CC = 0.5 F, CS = 2 F

Rsig = 10 k, RG = 1 M, RD = 4.7 k, RS = 1 k, RL = 2.2 k

IDSS = 8mA, VP= − 4 V rd = ∞ Ω , VDD = 20 V

b) Sketch the frequency response using a Bode plot.


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Example 9-12:


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High Frequency Response

FET amplifiers


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FET Amplifiers:

HF Response −−−− FET Amplifiers

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FET Amplifiers:


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FET Amplifiers:

Capacitances that affect the high-frequency response are

• Junction capacitances

Cgs, Cgd, Cds

• Wiring capacitances

Cwi, Cwo

• Coupling capacitors

CG, CC

• Bypass capacitor

CS


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FET Amplifiers:


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Hi

Thi i

1

2 Cf

πR=

i Wi gs MiC C CC= + +

Mi v gd(1 A )CC = −

Thi sig GR R ||R= Ho

Tho o

1

2 R Cf

π=

o Wo ds MoC C CC = + +

Mo gd

v

11 C

AC

= −

Tho D L dR ||R ||rR =

Figure 9-64 (a) & (b)

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Example 9-14:


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Square Wave

Testing


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Square Wave Testing:

Square Wave Testing

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Square Wave Testing:

Square Wave Testing

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Example 9-15:The application of a 1-mV, 5-kHz square wave to an amplifier resulted in the output

waveform of Fig. 9-72.

(a) Write the Fourier series expansion for the square wave through the ninth

harmonic.

(b) Determine the bandwidth of the amplifier

(c) Calculate the low cutoff frequency.

Square Wave Testing

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Reading:

1. Summary

2. Equations

3. Computer analysis

Problems:

1. Sec 8.2: (odd)

2. Sec 8.3: 17,18

3. Sec 8.4: 19,21

4. Sec 8.5:23,25

5. Sec 8.6: 27,29

6. Sec 8.7: 31

7. Sec 8.8: 33,35,37

8. Sec 8.10: 39,41

9. Sec 8.11: 43

10. Sec 8.12: 45

11. Sec 8.14: 47

12. Sec 8.15: 49

Home Task 62Frequency Response

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CH 1

References 63FET

1. Bolestad

2. Paynter

Frequency Response

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