ElectronicsElectronics

Principles & ApplicationsPrinciples & ApplicationsSixth EditionSixth Edition

Chapter 7More About

Small-Signal Amplifiers(student version)

©2003 Glencoe/McGraw-Hill

Charles A. Schuler

• Amplifier Coupling

• Voltage Gain

• FET Amplifier

• Negative Feedback

• Frequency Response

INTRODUCTION

Dear Student:

This presentation is arranged in segments. Each segment is preceded by a Concept Preview slide and is followed by a Concept Review slide. When you reach a Concept Review slide, you can return to the beginning of that segment by clicking on the Repeat Segment button. This will allow youto view that segment again, if you want to.

Concept Preview• Cascade amplifiers can use capacitive coupling.

• When dc gain is required, direct coupling is required.

• The Darlington configuration is an example of direct coupling.

• Transformer coupling offers the advantage of impedance matching.

• The impedance ratio is equal to the square of the turns ratio.

• Tuned transformers provide selectivity.

VCC

These two points are at different dc voltages.

Capacitive coupling is convenient in cascade ac amplifiers.

VCC

Direct coupling is required for dc gain.

VCC

The darlington is a popular dc arrangement.

VCC

P S

10:1

10 ZRATIO = TRATIO

2

= 102 = 100

ZCOLLECTOR = 100 x 10 = 1000

Transformer coupling offers the advantage of impedance matching.

VCC

Transformer coupling can beused in bandpass amplifiers

to achieve selectivity.

fR

Gain

Amplifier Coupling Quiz

Capacitive coupling is not useful for_________ amplifiers. dc

Dc frequency response requires ________ coupling. direct

Transformer coupling offers the advantage of _________ matching. impedance

Tuned transformer coupling provides frequency _____________. selectivity

A darlington amplifier is an example of _________ coupling. direct

Concept Review• Cascade amplifiers can use capacitive coupling.

• When dc gain is required, direct coupling is required.

• The Darlington configuration is an example of direct coupling.

• Transformer coupling offers the advantage of impedance matching.

• The impedance ratio is equal to the square of the turns ratio.

• Tuned transformers provide selectivity.

Repeat Segment

Concept Preview

• The input impedance of a C-E amplifier is equal to the equivalent parallel resistance of the base divider and rin of the transistor.

• rin is times the sum of the emitter resistances when the emitter resistor is not bypassed.

• Loading the output circuit changes the clipping points and decreases the voltage gain.

• The clipping points are shown by the ac load line.

• The ac load line passes through the same Q-point as the dc load line.

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 = ?

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:

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.

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

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

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).

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.

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.

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

VCC

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

When analyzing cascade amplifiers, remember:

2nd1st

Amplifier ac Conditions Quiz

Emitter bypassing _________ an amplifier’s input impedance.

decreases

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

A loaded amplifier has two load lines: dc and ___________. ac

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

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

Concept Review• The input impedance of a C-E amplifier is

equal to the equivalent parallel resistance of the base divider and rin of the transistor.

• rin is times the sum of the emitter resistances when the emitter resistor is not bypassed.

• Loading the output circuit changes the clipping points and decreases the voltage gain.

• The clipping points are shown by the ac load line.

• The ac load line passes through the same Q-point as the dc load line.

Repeat Segment

Concept Preview• A common-source JFET amplifier uses the gate

as the input and the drain as the output.

• The forward transfer admittance (Yfs) can be determined from the drain family of curves.

• Voltage gain is equal to Yfs times RL.

• Source bias produces negative feedback and decreases the voltage gain.

• The gain with feedback is determined by the feedback ratio and the open-loop gain.

• The feedback can be eliminated with a source bypass capacitor.

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

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

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

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.

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.

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

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

AB +1AVin Vout

Use the simplified model:

A(WITH NEG. FEEDBACK) =8

(8)(0.16) + 1= 3.51

CS

DG

VDD

RGCC

RL

RS

The source bypass capacitor will eliminate the ac negative feedback

and restore the voltage gain.

JFET Amplifier Quiz

In a common-source amplifier, the input signal 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 equal to Yfs x _________. load resistance

Source bias is produced by current flow through the _______ resistor. source

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

Concept Review• A common-source JFET amplifier uses the gate

as the input and the drain as the output.• The forward transfer admittance (Yfs) can be

determined from the drain family of curves.• Voltage gain is equal to Yfs times RL.• Source bias produces negative feedback and

decreases the voltage gain.• The gain with feedback is determined by the

feedback ratio and the open-loop gain.• The feedback can be eliminated with a source

bypass capacitor.

Repeat Segment

Concept Preview

• Dc negative feedback stabilizes the Q-point.

• Ac negative feedback decreases gain.

• Ac negative feedback increases bandwidth.

• Ac negative feedback reduces distortion.

• Amplifier gain is maximum at mid-band.

• The break frequencies are where the gain drops by 3 dB.

• Amplifier bandwidth is found by subtracting the lower break frequency from the upper break frequency.

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

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

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.

Gai

n in

dB

0

50

Frequency10 Hz 100 MHz

BW1

BW2

Gain and bandwidth with and without the emitter bypass

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.

Concept Review• Dc negative feedback stabilizes the Q-point.

• Ac negative feedback decreases gain.

• Ac negative feedback increases bandwidth.

• Ac negative feedback reduces distortion.

• Amplifier gain is maximum at mid-band.

• The break frequencies are where the gain drops by 3 dB.

• Amplifier bandwidth is found by subtracting the lower break frequency from the upper break frequency.

Repeat Segment

REVIEW

• Amplifier Coupling

• Voltage Gain

• FET Amplifier

• Negative Feedback

• Frequency Response