© 2013 the mcgraw-hill companies, inc. all rights reserved. mcgraw-hill 7-1 electronics principles...

© 2013 The McGraw-Hill Companies, Inc. All rights reserved.McGraw-Hill

7-1

ElectronicsElectronics

Principles & ApplicationsPrinciples & ApplicationsEighth EditionEighth Edition

Chapter 7More About

Small-Signal Amplifiers

Charles A. Schuler


7-2

• Amplifier Coupling• Voltage Gain• FET Amplifiers• Negative Feedback• Frequency Response• Positive Feedback

INTRODUCTION


7-3

VCC

These two points are at different dc voltages.

Capacitive coupling is convenient in cascade ac amplifiers.


7-4

VCC

Direct coupling is required for dc gain.


7-5

VCC

The Darlington is a popular dc arrangement.


7-6

VCC

P S

10:1

10 ZRATIO = TRATIO

2

= 102 = 100

ZCOLLECTOR = 100 x 10 = 1000

Transformer coupling offers the advantage of impedance matching.


7-7

Transformer couplingis used in 70.7 volt

sound systems.


7-8

VCC

Transformer coupling can beused in bandpass amplifiers

to achieve selectivity.

fR

Gain


7-9

Amplifier coupling quiz

Capacitive coupling is not useful for_________ amplifiers. dc

Dc frequency response requires ________coupling. direct

Transformer coupling offers the advantageof _________ matching. impedance

Tuned transformer coupling providesfrequency _____________. selectivity

A Darlington amplifier is an example of_________ coupling. direct


7-10

RB1

EB

C

RL

VCC

RB2 RE

= 12 V

2.7 k

22 k = 2.2 k

More about solving the practical circuit for its ac conditions:

= 220

Zin = ?


7-11

RB1

EB

C

RL

VCC

RB2 RE

= 12 V

2.7 k

22 k = 2.2 k

Zin is a combination of RB1, RB2, and rin of the transistor.

= 220

rin = (RE + rE)

rin = (220 + 9.03 )

rin = 34.4 k

Note: rin = rE

when RE is bypassed.

Determine rin first:


7-12

RB1

EB

C

RL

VCC

RB2 RE

= 12 V

2.7 k

22 k = 2.2 k

= 220

Zin =1

RB2

1rin

1+

RB1

1+

++Zin =

1

2.7 k1

34.4 k1

22 k1

Zin = 2.25 k

RB1, RB2, and rin act in parallelto load the input signal.


7-13

RB1

VCC

RB2 RE

= 12 V

2.7 k

22 k RL= 2.2 k

= 220

Load = 2.2 k

What happens when an amplifier is loaded?

RL and the Load act in parallel.

RP = 1.1 k


7-14

RB1

RB2 RE

VCC = 12 V

2.7 k

22 k RL= 2.2 k

= 220

Load = 2.2 k

There are two saturation currents for a loaded amplifier.

RP = 1.1 k

ISAT(DC) = VCC

RL + RE

= 4.96 mA

ISAT(AC) = VCC

RP + RE

= 9.09 mA


7-15

0 2 4 6 8 10 12 14 16 18

2468

101214

VCE in Volts

IC in mA

20 A

0 A

100 A

80 A

60 A

40 A

There are two load lines for a loaded amplifier.

DC

TEMPORARY AC

The DC load line connects VCC and ISAT(DC).

A temporary AC load line connects VCC and ISAT(AC).


7-16

0 2 4 6 8 10 12 14 16 18

2468

101214

VCE in Volts

IC in mA

20 A

0 A

100 A

80 A

60 A

40 A

5.3 V

DC

AC

TEMP. AC

The quiescent VCE is projected to the DC load line to establish the Q-point. The AC load line is drawn through

the Q-point, parallel to the temporary AC load line.


7-17

0 2 4 6 8 10 12 14 16 18

2468

101214

VCE in Volts

IC in mA

20 A

0 A

100 A

80 A

60 A

40 A

5.3 V

AC

The AC load line shows the limits for VCE and if the Q-point is properly located.

With loaded amplifiers, the Q-point is often closer to saturation.


7-18

RB1

RB2 RE

VCC = 12 V

2.7 k

22 k RL= 2.2 k

= 220

Load = 2.2 k

What about voltage gain for a loaded amplifier?

RP = 1.1 k

AV =RP

RE + rE

AV =1.1 k

220 9.03= 4.8


7-19

VCC

Zin of the 2nd stage loads the 1st stage.

When analyzing cascade amplifiers, remember:

2nd1st


7-20

Amplifier ac conditions quiz

Emitter bypassing _________ an amplifier’sinput impedance. decreases

Loading at the output of an amplifier________ its voltage gain. decreases

A loaded amplifier has two load lines: dcand ___________. ac

The clipping points of a loaded amplifier areset by its _______ load line. ac

In a cascade amplifier, the Zin of a stage_______ the prior stage. loads


7-21

Drain

Source

Gate

VDD = 20 V

VGS = 1.5 V

RGCC

RL = 5 k

Inputsignal

Common-source JFET amplifier.

Fixed bias

ISAT = 20 V

5 k= 4 mA

Phase-invertedoutput


7-22

0

2

4

1

VDS in Volts

ID in mA

5 10 15 20 25

3

-2.5

-2.0

-1.5

-1.0

-0.5

0

N-channel JFET characteristic curves

VG

S in

Vol

ts

Load line

The Q-point is set by the fixed bias.

8 VP-P

1 VP-P

AV = 8


7-23

0

2

4

1

VDS in Volts

ID in mA

5 10 15 20 25

3

-2.5

-2.0

-1.5

-1.0

-0.5

0Determining forward transfer admittance:

Yfs = ID

VGS

VG

S in

Vol

ts

VDS

1.6 mA

= 1.6 mS


7-24

D

S

G

VDD = 20 V

VGS = 1.5 V

RGCC

RL = 5 k

When the forward transfer admittance is known,the voltage gain can be determined using:

AV = Yfs x RL

= 1.6 mS x 5 k

= 8

This agrees with the graphic solution.


7-25

D

S

G

VDD

VGS = ID x RS

RGCC

RL

RS

Source bias eliminates the need for a separate VGS supply.

IS = ID

This resistor also providesac negative feedback whichdecreases the voltage gain.


7-26

JFET amplifier quiz

In a common-source amplifier, the inputsignal goes to the _______. gate

In a common-source amplifier, the inputto output phase relationship is ____. 180o

The voltage gain of a C-S amplifier is equalto Yfs x _________. load resistance

Source bias is produced by current flowthrough the _______ resistor. source

An unbypassed source resistor _______ thevoltage gain of a C-S amp. decreases


7-27

Vin - BVout

A(Vin - BVout)

BVout

A = open loop gain

Summingjunction

Vin VoutA

B Feedback

A negative feedback model

B = feedback ratio

Vout = A(Vin - BVout)Vout = AVin - ABVout

AVin

Vout

1 = - ABAVin

Vout

AB +1 =Vin

Vout

AB +1A

=Vin

Vout

AB +1

A=

AB +1AVin Vout

A simplified model


7-28

D

S

G

VDD

RGCC

RL

RS

= 5 k

= 800

The feedback ratio (B) for this circuitis easy to determine since the source and

drain currents are the same.

B = 800 5 k

= 0.16


7-29

AB +1AVin Vout

Use the simplified model:

A(WITH NEG. FEEDBACK) =8

(8)(0.16) + 1= 3.51


7-30

CS

DG

VDD

RGCC

RL

RS

The source bypass capacitor will eliminate the ac negative feedback

and restore the voltage gain.


7-31

Amplifier Negative Feedback

• DC reduces sensitivity to device parameters

• DC stabilizes operating point

• DC reduces sensitivity to temperature change

• AC reduces gain• AC increases

bandwidth• AC reduces signal

distortion and noise• AC may change

input and output impedances


7-32

0.707 Amax

A

f

The frequency response curve of an ac amplifier

Bandwidth

The gain is maximum in the midband.

Amax

Midband

The bandwidth spans the -3 dB points which are called the break frequencies.

-3dB


7-33

50

10 F

10 F

1 k

100 1k

6.8 k

The emitter bypass capacitor in this amplifier hasa significant effect on both gain and bandwidth.


7-34G

ain

in d

B

0

50

Frequency10 Hz 100 MHz

BW1

BW2

Gain and bandwidth with and without the emitter bypass


7-35

Amplifier frequency response

• The lower break frequency is partly determined by coupling capacitors.

• It is also influenced by emitter bypass capacitors.

• The upper break frequency is partly determined by transistor internal capacitance.

• Both break frequencies can be influenced by negative feedback.


7-36

Positive Feedback

• Is the opposite of negative feedback

• Increases gain and reduces bandwidth

• Can be used in some circuits to reduce the effects of noise

• The next slide shows a circuit with a noise problem.


7-37

This circuit is supposed to convert the input signal to a rectangular output signal. It works, but the output waveform shows an extra pulse caused by noise.

The trip points are equal.


7-38

UTP LTP

This circuit has positive feedback and two trip points. The hysteresis is the difference between the trip points (UTP and LTP) and that makes this circuit less sensitive to noise. The output waveform is noise free.

R5 provides positive feedback back from the output amplifier to the input amplifier.


7-39

REVIEW

• Amplifier Coupling• Voltage Gain• FET Amplifier• Negative Feedback• Frequency Response• Positive Feedback

© 2013 the mcgraw-hill companies, inc. all rights reserved. mcgraw-hill 7-1 electronics principles...

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