ElectronicsElectronicsPrinciples & ApplicationsPrinciples & Applications

Sixth EditionSixth Edition

Solving DC Analysis and Plotting Load Line with Q Point

•Load Line

•Q Point

•Common-Emitter Amplifier

•Voltage Divider Bias

INTRODUCTION

Amplifier Out

InGain =

In

Out= 3.33

1.5 V 5 V

1.5 V5 V

The units cancel

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

E

The outputis phase inverted.

B

350 kΩ

CC E

C

1 kΩ14 V

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

The maximum value of IC is 14 mA.

IC(MAX) =14 V1 kΩ

B

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 μA

0 μA

100 μA80 μA60 μA

40 μA

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 kΩ

CC E

C

1 kΩ

B

14 V

IB =14 V

350 kΩ

Use Ohm’s Law to determine the base current:

= 40 μA

0 2 4 6 8 10 12 14 16 18

2468

101214

VCE in Volts

IC in mA

20 μA

0 μA

100 μA80 μA60 μA

40 μA

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

The base current in this case is 40 μA.

Q

Q = the quiescent point

0 2 4 6 8 10 12 14 16 18

2468

101214

VCE in Volts

IC in mA

20 μA

0 μA

100 μA80 μA60 μA

40 μA

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 μA

0 μA

100 μA80 μA60 μA

40 μA

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 μA

0 μA

100 μA80 μA60 μA

40 μA

What’s wrong with this Q point?

How about this one?

350 kΩ

CC E

C

1 kΩ

B

14 V

IB =14 V

350 kΩ

β = 150

IC = β x IB = 150 x 40 μAVRL

= IC x RL = 6 mA x 1 kΩ

= 40 μA

= 6 mA= 6 V

This is a good Q point for linear amplification.VCE = VCC - VRL

= 14 V - 6 V = 8 V

350 kΩ

CC E

C

1 kΩ

B

14 V

IB =14 V

350 kΩ

β = 350

IC = β x IB = 350 x 40 μAVRL

= IC x RL = 14 mA x 1 kΩ

= 40 μA (IB is not affected)

= 14 mA (IC is higher)= 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)

β is higher

0 2 4 6 8 10 12 14 16 18

2468

101214

VCE in Volts

IC in mA

20 μA

0 μA

100 μA80 μA60 μA

40 μA

The output is non-linear.

The higher β causessaturation.

RB

CC E

C

RL

B

VCC

It’s β dependent!This common-emitter amplifier is not practical.

It’s also temperature dependent.

Basic C-E Amplifier QuizThe input and output signals in C-E are phase ______________. inverted

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

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

The operating point of an amplifier is also called the ________ point. quiescent

Single resistor base bias is not practical since it’s _________ dependent. β

RB1

CC

E

C

RL

B

VCC

RB2 RE

This common-emitter amplifier is practical.

It uses voltage divider bias andemitter feedback to reduce β 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 + RB2VCC

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

RB1

EB

C

RL

VCC

RB2 RE = 220 Ω

= 12 V

2.7 kΩ

22 kΩ = 2.2 kΩ

Solving the practical circuit for its dc conditions:

VB = RB2

RB1 + RB2x VCC

VB = 2.7 kΩ

22 kΩ2.7 kΩ +x 12 V

VB = 1.31 V

RB1

EB

C

RL

VCC

RB2 RE = 220 Ω

= 12 V

2.7 kΩ

22 kΩ = 2.2 kΩ

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 Ω

= 12 V

2.7 kΩ

22 kΩ = 2.2 kΩ

Solving the practical circuit for its dc conditions:

IE = RE

VE

IE = 0.61 V220 Ω

= 2.77 mA

IC ≅ IE

RB1

EB

C

RL

VCC

RB2 RE = 220 Ω

= 12 V

2.7 kΩ

22 kΩ = 2.2 kΩ

Solving the practical circuit for its dc conditions:

VRL= IC x RL

VRL= 2.77 mA x 2.2 kΩ

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 Ω

= 12 V

2.7 kΩ

22 kΩ = 2.2 kΩ

Solving the practical circuit for its ac conditions:

The ac emitter resistance is rE:

rE = 25 mVIE

rE = 25 mV2.77 mA = 9.03 Ω

RB1

EB

C

RL

VCC

RB2 RE = 220 Ω

= 12 V

2.7 kΩ

22 kΩ = 2.2 kΩ

Solving the practical circuit for its ac conditions:

The voltage gain from base to collector:

AV =RL

RE + rE

AV =2.2 kΩ

220 Ω + 9.03 Ω= 9.61

RB1

EB

C

RL

VCC

RB2 RE

= 12 V

2.7 kΩ

22 kΩ = 2.2 kΩ

Solving the practical circuit for its ac conditions:

AV =RL

rE

AV =2.2 kΩ

9.03 Ω= 244

An emitter bypass capacitorcan be used to increase AV:

CE

Practical C-E Amplifier Quiz

β-dependency is reduced with emitterfeedback and voltage _________ bias. divider

To find the emitter voltage, VBE is subtracted from ____________. VB

To find VCE, VRL and VE are subtracted from _________. VCC

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

Voltage gain can be increased by ________ the emitter resistor. bypassing