Floyd Ch 5 PPT

Chapter 5

Electronics 2

EE 317

Electrical Engineering

Majmaah University

1st Semester 1433/1434 H

2012 Pearson Education. Upper Saddle River, NJ, 07458.

All rights reserved.Electronic Devices, 9th edition

Thomas L. Floyd1

Transistor Bias Circuits

Chapter Outline

51 The DC Operating Point

52 Voltage-Divider Bias

53 Other Bias Methods



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Chapter Objectives

Discuss and determine the dc operating point of a linear

amplifier.

Analyze a voltage-divider biased circuit.

Analyze an emitter bias circuit, a base bias circuit, an

emitter-feedback bias circuit, and a collector-feedback bias



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emitter-feedback bias circuit, and a collector-feedback bias

circuit.

Introduction

As you learned in Chapter 4, a transistor must be properly biased in

order to operate as an amplifier.

DC biasing is used to establish fixed dc values for the transistor currents

and voltages called the dc operating point or quiescent point (Q-point).



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In this chapter, several types of bias circuits are discussed.

This material lays the groundwork for the study of amplifiers, and other

circuits that require proper biasing.

The DC Operating Point



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1.2 3.4 5.6

The DC Operating Point



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Linear Operation

Bias establishes the operating point (Q-point) of a transistor

amplifier; the ac signal

moves above and below

IC (mA)



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moves above and below

this point.

For this example, the dc

base current is 300 A. When the input causes the

base current to vary between

200 A and 400 A, the collector current varies

between 20 mA and 40 mA.

0VCE (V)

400 A

300 A = IBQ

200 A

A

B

Q

1.2 3.4 5.6VCEQ

ICQ

Vce

IbIc

20

30

40

Load line

Waveform Distortion

A signal that swings

outside the active

region will be clipped.

I BQ

IC

Input

signal



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region will be clipped.

For example, the bias

has established a low Q-

point.

As a result, the signal

will be clipped because

it is too close to cutoff.

VCCVCE

Cutoff

QICQ

Cutoff 0

Vce

VCEQ

Waveform Distortion



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High Q- point. The signal will be

clipped because it is too close to

saturation.

Input signal too large. The signal

will be clipped from both sides.

QuizQuizQuiz

Q. A signal that swings outside the active area will be

a. clamped

b. clipped



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c. unstable

d. all of the above

Voltage-Divider Bias

A practical way to establish a Q-point is to form a voltage-

divider from VCC.

R1 and R2 are selected to establish VB. If the

divider is stiff, I is small compared to I . Then,

+VCC



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divider is stiff, IB is small compared to I2. Then,

RCR1

RER2

2B CC

1 2

RV V

R R

+

I2

IB

QuizQuizQuiz

Q. A stiff voltage divider is one in which

a. there is no load current

b. divider current is small compared to load current



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c. the load is connected directly to the source voltage

d. loading effects can be ignored


+VCC+15 V



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RCR1

RER2

DC = 200

27 k

12 k 680

1.2 k

Determine the base voltage for the circuit.

( )

2B CC

1 2

12 k15 V

27 k 12 k

RV V

R R

= +

= + = + 4.62 V


+VCC+15 V

What is the emitter voltage, VE, and current, IE?



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RCR1

RER2

DC = 200

27 k

12 k 680

1.2 k

4.62 V

VE is one diode drop less than VB:

VE = 4.62 V 0.7 V = 3.92 V

3.92 VApplying Ohms law:

EE

E

3.92 V

680

VI

R= = =

5.76 mA

QuizQuizQuiz

Q. Assuming a stiff voltage-divider for the circuit shown,

the emitter voltage is

a. 4.3 V+VCC+15 V



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b. 5.7 V

c. 6.8 V

d. 9.3 V

RCR1

RER2

DC = 200

20 k

10 k 1.2 k

1.8 k

VB = 10 / (10 + 20) * 15 = 5 V

VE = 5 0.7 = 4.3 V

QuizQuizQuiz

Q. For the circuit shown, the dc load line will intersect the

y-axis at (neglecting IB)

a. 5.0 mA+VCC+15 V



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b. 10.0 mA

c. 15.0 mA

d. none of the above

RCR1

RER2

DC = 200

20 k

10 k 1.2 k

1.8 k

@ VCE=0:

IC = (15 0) / (1.8k + 1.2k) = 15 / 3k = 5 mA


The unloaded voltage divider approximation for VB gives

reasonable results.

A more exact solution is to Thevenize the input circuit.

Looking from the base to the left:+VCC+15 V

+VCC+15 V



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RCR1

RER2

DC = 200

27 k

12 k

+15 V

680

1.2 k

VTH = VB(no load)

= 4.62 V

RTH = R1||R2 =

= 8.31 k

The Thevenin input

circuit can be drawn

R C

R TH

R E

+V TH +

IB

+

+

I E

VBE

8.31 k

680

1.2 k

4.62 V

+15 V

DC = 200

Thevenin Resistance, RTH

To find RTH, replace voltage sources by a short circuit.

Which means VCC will be grounded.

RTH = R1||R2

+VCC+15 V

1 2 3



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RCR1

RER2

DC = 200

27 k

12 k 680

1.2 k

QuizQuizQuiz

Q. If you Thevenize the input voltage divider, the Thevenin

resistance is

a. 5.0 k

+VCC+15 V



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b. 6.67 k

c. 10 k

d. 30 k

RCR1

RER2

DC = 200

20 k

10 k 1.2 k

1.8 k

RTH = R1 R2 / (R1 + R2)

= 20 * 10 / (20 + 10) = 200 / 30 = 6.67 k


Now write KVL around the base emitter circuit and solve

for IE.

+VCC+15 V

TH B TH BE E EV I R V I R= + +

IE = IB + IC = IB + DC IB = (DC + 1) IB DC IBVTH VBE = IE RE + IB RTH IE RE + IE RTH / DC= IE ( RE + RTH / DC )



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R C

R TH

R E

+V TH +

IB

+

+

I E

VBE

8.31 k

680

1.2 k

4.62 V

+15 V

DC = 200

TH BEE

THE

DC

V VI

RR

=

+

Substituting and solving,

E

4.62 V 0.7 V

8.31 k680 +200

I

= =

5.43 mA

and VE = IERE = (5.43 mA)(0.68 k)

= 3.69 V


A pnp transistor can be biased from either a positive or negative supply.

Notice that (b) and (c) are the same circuit; both with a positive supply.

+ V V

EE

EE



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+

+

V

V

EE

EE

R

RR

2

221

11

R

RR

R

RR

C

CCR

RR

E

EE

(a) (b) (c)


Determine IE for the pnp circuit. Assume a stiff

voltage divider (no loading effect).

+VEE



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+VEE

R2

1RRC1.2 k

R E680

27 k

12 k

+15 V

( )

1B EE

1 2

27 k15.0 V 10.4 V

27 k 12 k

RV V

R R

= +

= + = +

E B BE 10.4 V 0.7 V = 11.1 VV V V= + = +

EE EE

E

15.0 V 11.1 V

680

V VI

R

= = =

5.74 mA

10.4 V 11.1 V

QuizQuizQuiz

Q. For the circuit shown, the emitter voltage is

a. less than the base voltage

b. less than the collector voltage

+VEE

R R

+15 V



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c. both of the above

d. none of the above

R2

1RRC1.2 k

R E680

27 k

12 k

pnp circuit

(higher than the base voltage by 0.7 V)

Emitter Bias

Emitter bias has excellent stability but requires both a

positive and a negative source.

VCC+15 V

For troubleshooting analysis, assume that VEfor an npn transistor is about 1 V.



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Assuming that VE is 1 V, what is IE?

VEE

RC

RE

RB68 k

+15 V

15 V

7.5 k

3.9 k

1 VEEE

E

1 V 15 V ( 1 V)

7.5 k

VI

R

= = =

1.87 mA

The minus sign means the

current is directed downwards

(opposite to initial assumption).

Emitter Bias

The approximation that VE 1 V and the neglect of DC may not be accurate

enough for design work or detailed analysis.

In this case, KVL can be applied to develop a more detailed formula for IE.

KVL applied around the base-emitter circuit in Figure 517(a), which has

been redrawn in part (b) for analysis, gives the following equation:



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Emitter Bias

VRB+ VBE + VRE

VEE = 0

IB RB + VBE + IE RE VEE = 0



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IE ( RB / DC) + IE RE = VEE VBE

IE (RE + RB / DC) = VEE VBE

QuizQuizQuiz

Q. Emitter bias

a. is not good for linear circuits

b. uses a voltage-divider on the input



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c. requires dual power supplies

d. all of the above

QuizQuizQuiz

Q. With the emitter bias shown, a reasonable assumption

for troubleshooting work is that the

a. base voltage = +1 V VCC+15 V



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b. emitter voltage = +5 V

c. emitter voltage = 1 V

d. collector voltage = VCC

VEE

RC

RE

RB68 k

15 V

7.5 k

3.9 k

Base Bias

Base bias is used in switching circuits because of its

simplicity, but not widely used in linear applications

because the Q-point is dependent.



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RC

R B

+VCC

Base current is derived from the collector supply

through a large base resistor.

What is IB?

CCB

B

0.7 V 15 V 0.7 V

560 k

VI

R

= = =

25.5 A

RC

R B

+VCC

560 k

+15 V

1.8 k

IB

+

VBE

Base Bias

Compare VCE for the case where = 100 and = 300.

( )( )C B 100 25.5 A 2.55 mAI I= = =For = 100:V V I R=



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RC

R B

+VCC

560 k

+15 V

1.8 k

10.4 V( )( )CE CC C C

15 V 2.55 mA 1.8 k

V V I R=

= =

For = 300: ( ) ( )C B 300 25.5 A 7.65 mAI I= = =

( )( )CE CC C C

15 V 7.65 mA 1.8 k

V V I R=

= = 1.23 V

IC

+

VCE

The Q-point is dependent.

QuizQuizQuiz

Q. The circuit shown is an example of

a. base bias

b. collector-feedback bias+VCC



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c. emitter bias

d. voltage-divider bias

RC

R B

Emitter-Feedback Bias

An emitter resistor (RE) changes base bias into emitter-

feedback bias, which is more predictable.

The emitter resistor (RE) is a form of negative feedback.

If IC tries to increase, VE increases, causing an



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R

R

C

E

R B

+VCCIf IC tries to increase, VE increases, causing an

increase in VB because VB = VE + VBE.

This increase in VB reduces the voltage across RB(VRB

= VCC VB), thus reducing IB and keeping ICfrom increasing.

A similar action occurs if IC tries to decrease.

Emitter-Feedback Bias

The equation for emitter current (IE) is found by writing KVL

around the base circuit.

IE RE + IB RB = VCC VBE



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R

R

C

E

R B

+VCC

IB

IE+

+

+

IE RE + IB RB = VCC VBE

IE RE + (IE / DC) RB = VCC VBE

IE (RE + RB / DC) = VCC VBE

Collector-Feedback Bias

Collector feedback bias uses another form of negative

feedback to increase stability. (more on next slide)

Instead of returning the base resistor (RB) to VCC, it is

returned to the collector.



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The equation for collector current (IC) is found

by writing KVL around the base circuit.

(details on page 247 in the book)

The result isCC BE

CB

CDC

V VI

RR

=

+

+VCC

RCRB


Collector feedback bias uses another form of negative

feedback to increase stability. Instead of returning the base

resistor to VCC, it is returned to the collector.

The negative feedback creates an offsetting effect

that tends to keep the Q-point stable.



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+VCC

RCRB

If IC tries to increase, it drops more voltage across

RC, thereby causing VC to decrease.

When VC decreases, there is a decrease in voltage

across RB, which decreases IB.

The decrease in IB produces less IC which, in turn,

drops less voltage across RC and thus offsets the

decrease in VC.


Compare IC for the case when = 100 with the case when = 300.



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When = 100,

CC BEC

BC

DC

15 V 0.7 V

330 k1.8 k100

V VI

RR

= = =

++

+VCC

RCR B

330 k

1.8 k

+ 15 V

2.80 mA

When = 300,

CC BEC

BC

DC

15 V 0.7 V

330 k1.8 k300

V VI

RR

= = =

++4.93 mA

QuizQuizQuiz

Q. The circuit shown is an example of

a. base bias

b. collector-feedback bias

+VCC

RC



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c. emitter bias

d. voltage-divider bias

RCRB

Key TermsKey TermsKey Terms

Q-point

DC load line

Linear region

The dc operating (bias) point of an amplifier

specified by voltage and current values.

A straight line plot of IC and VCE for a

transistor circuit.

The region of operation along the load line



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Linear region

Stiff voltage

divider

Feedback

The region of operation along the load line

between saturation and cutoff.

A voltage divider for which loading effects

can be ignored.

The process of returning a portion of a

circuits output back to the input in such a way

as to oppose or aid a change in the output.

Floyd Ch 5 PPT

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