ElectronicsElectronics

Principles & ApplicationsPrinciples & ApplicationsFifth EditionFifth Edition

Chapter 6Introduction to

Small-Signal Amplifiers

©1999 Glencoe/McGraw-Hill

Charles A. Schuler

• Gain• Common-Emitter Amplifier• Stabilizing the Amplifier• Other Configurations

INTRODUCTION

Amplifier Out

InGain =

In

Out= 3.33

1.5 V 5 V

1.5 V

5 VThe units cancel

Gain can be expressed in decibels (dB).

The dB is a logarithmic unit.

Common logarithms are exponents of the number 10.

102 = 100103 = 100010-2 = 0.01100 = 1103.6 = 3981

The log of 100 is 2

The log of 1000 is 3

The log of 0.01 is -2

The log of 1 is 0

The log of 3981 is 3.6

The dB unit is based on a power ratio.

dB = 10 x log POUT

PIN

50 W

1 W501.7017

The dB unit can be adapted to a voltage ratio.

dB = 20 x log VOUT

VIN

This equation assumes VOUT and VIN

are measured across equal impedances.

+10 dB -6 dB +30 dB -8 dB +20 dB

dB units are convenient for evaluating systems.

+10 dB -6 dB+30 dB -8 dB+20 dB

Total system gain = +46 dB

Gain quiz

Amplifier output is equal to the input________ by the gain. multiplied

exponents

Doubling a log is the same as _________the number it represents. squaring

System performance is found by ________dB stage gains and losses. adding

Logs of numbers smaller than one are____________. negative

Common logarithms are ________ of thenumber 10.

A small-signal amplifier can also be called a voltage amplifier.

Common-emitter amplifiers are one type.

C

BE

Start with an NPN bipolar junction transistor

VCC

Add a power supply

RL

Next, a load resistor

RB

Then a base bias resistor

CC

A coupling capacitor is often requiredConnect a signal sourceThe emitter terminal is grounded

and common to the input andoutput signal circuits.

RB RL

VCC

CC

C

BE

The outputis phase inverted.

RB

VCC

CC E

When the input signal goes positive:

B

The base current increases.

C

The collector current increases times.

RL

So, RL drops more voltage and VCE must decrease.

The collector terminal is now less positive.

RB

VCC

CC E

When the input signal goes negative:

B

The base current decreases.

C

The collector current decreases times.

RL

So, RL drops less voltage and VCE must increase.

The collector terminal is now more positive.

350 k

CC EB

C

1 k14 V

The maximum value of VCE for this circuit is 14 V.

The maximum value of IC is 14 mA.

IC(MAX) =14 V

1 k

These are the limits for this circuit.

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

The load line connects the limits.

SAT.

This end is called saturation.

CUTOFFThis end is called cutoff.

LINEAR

The linear region is between the limits.

350 k

CC EB

C

1 k14 V

IB =14 V

350 k

Use Ohm’s Law to determine the base current:

= 40 A

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

An amplifier can be operated at any point along the load line.

The base current in this case is 40 A.

Q

Q = the quiescent point

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

The input signal varies the base current above and below the Q point.

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

Overdriving the amplifier causes clipping.

The output is non-linear.

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

What’s wrong with this Q point?

How about this one?

350 k

CC EB

C

1 k14 V

IB =14 V

350 k

= 150

IC = x IB

= 40 A

= 150 x 40 A = 6 mA

VRL = IC x RL = 6 mA x 1 k = 6 V

This is a good Q point for linear amplification.VCE = VCC - VRL = 14 V - 6 V = 8 V

350 k

CC EB

C

1 k14 V

IB =14 V

350 k

= 350

IC = x IB

= 40 A (IB is not affected)

= 350 x 40 A = 14 mA (IC is higher)

VRL = IC x RL = 14 mA x 1 k = 14 V (VRL is higher)

This is not a good Q point for linear amplification.VCE = VCC - VRL = 14 V - 14 V = 0 V (VCE is lower)

is higher

RB

CC EB

C

RL

VCC

It’s dependent!

This common-emitter amplifier is not practical.

It’s also temperature dependent.

Basic C-E amplifier quiz

The input and output signals in C-E arephase ______________. inverted

The limits of an amplifier’s load line aresaturation and _________. cutoff

Linear amplifiers are normally operated nearthe _________ of the load line. center

The operating point of an amplifier is alsocalled the ________ point. quiescent

Single resistor base bias is not practical sinceit’s _________ dependent.

RB1

CC

EB

C

RL

VCC

RB2 RE

This common-emitter amplifier is practical.

It uses voltage divider bias and

emitter feedback to reduce sensitivity.

+VCC

RL

RE

RB1

RB2

Voltage divider bias

{RB1 and RB2 form a voltage divider

+VCC

RB1

RB2

+VB

Voltage dividerbias analysis:

VB =RB2

RB1 + RB2

VCC

The base current is normallymuch smaller than the dividercurrent so it can be ignored.

RB1

EB

C

RL

VCC

RB2 RE = 220

= 12 V

2.7 k

22 k = 2.2 k

Solving the practical circuit for its dc conditions:

VB = RB2

RB1 + RB2

x VCC

VB = 2.7 k

22 k2.7 k +x 12 V

VB = 1.31 V

RB1

EB

C

RL

VCC

RB2 RE = 220

= 12 V

2.7 k

22 k = 2.2 k

Solving the practical circuit for its dc conditions:

VE = VB - VBE

VE = 1.31 V - 0.7 V = 0.61 V

RB1

EB

C

RL

VCC

RB2 RE = 220

= 12 V

2.7 k

22 k = 2.2 k

Solving the practical circuit for its dc conditions:

IE = RE

VE

IE = 0.61 V

220 = 2.77 mA

IC IE

RB1

EB

C

RL

VCC

RB2 RE = 220

= 12 V

2.7 k

22 k = 2.2 k

Solving the practical circuit for its dc conditions:

VRL = IC x RL

VRL = 2.77 mA x 2.2 k

VRL = 6.09 V

VCE = VCC - VRL - VE

VCE = 12 V - 6.09 V - 0.61 V

VCE = 5.3 V

A linear Q point!

Review of the analysis thus far:

1. Calculate the base voltage using the voltage divider equation.

2. Subtract 0.7 V to get the emitter voltage.

3. Divide by emitter resistance to get the emitter current.

4. Determine the drop across the collector resistor.

5. Calculate the collector to emitter voltage using KVL.

6. Decide if the Q-point is linear.

7. Go to ac analysis.

RB1

EB

C

RL

VCC

RB2 RE = 220

= 12 V

2.7 k

22 k = 2.2 k

Solving the practical circuit for its ac conditions:

The ac emitter resistance is rE:

rE = 25 mV

IE

rE =25 mV

2.77 mA= 9.03

RB1

EB

C

RL

VCC

RB2 RE = 220

= 12 V

2.7 k

22 k = 2.2 k

Solving the practical circuit for its ac conditions:

The voltage gain from base to collector:

AV =RL

RE + rE

AV =2.2 k

220 9.03= 9.61

RB1

EB

C

RL

VCC

RB2 RE

= 12 V

2.7 k

22 k = 2.2 k

Solving the practical circuit for its ac conditions:

AV =RL

rE

AV =2.2 k

9.03= 244

An emitter bypass capacitormay be used to increase AV:

CE

Practical C-E amplifier quiz

-dependency is reduced with emitter feedbackand voltage _________ bias. divider

To find the emitter voltage, VBE is subtractedfrom ____________. VB

To find VCE, VRL and VE are subtractedfrom _________. VCC

Voltage gain is equal to the collector resistance_______ by the emitter resistance. divided

Voltage gain can be increased by ________the emitter resistor. bypassing

RB1

EB

C

RL

VCC

RB2 RE CE

The common-emitter configuration is used most often.

It has the best power gain.

RB1

EB

C

RC

VCC

RB2 RL

The common-collector configuration is shown below.

Its input impedance and current gain are both high.

It’s often called an emitter-follower.

In-phaseoutput

RB1

EB

C

RL

VCC

RB2 RE

The common-base configuration is shown below.

Its voltage gain is high. It’s used mostat RF.

In-phaseoutput

Amplifier configuration quiz

In a C-E amplifier, the base is the input andthe __________ is the output. collector

In an emitter-follower, the base is the inputand the ______ is the output. emitter

The only configuration that phase-inverts isthe ________. C-E

The configuration with the best power gainis the ________. C-E

In the common-base configuration, the________ is the input terminal. emitter

REVIEW

• Gain• Common-Emitter Amplifier• Stabilizing the Amplifier• Other Configurations