Floyd Ch 5 PPT

Chapter 5

Electronics 2

EE 317

Electrical Engineering

Majmaah University

1

_{Semester 1433/1434 H}

Chapter Outline

5–1 The DC Operating Point

5–2 Voltage-Divider Bias

5–3 Other Bias Methods

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

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

).

• 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.

Linear Operation

Bias establishes the operating point (

Q-point

) of a transistor

amplifier; the ac signal

moves above and below

IC (mA)

moves above and below

this point.

For this example, the dc

base current is 300 µA.

When the input causes the

base current to vary between

200 µA and 400 µA, the

collector current varies

between 20 mA and 40 mA.

0 VCE (V) 400 A 300 A = I_BQ 200 A A B Q 1.2 3.4 5.6 V_CEQ I_CQ Vce I_b I_c 20 30 40 µ µ µ

Load line

Waveform Distortion

A signal that swings

outside the active

region will be

clipped

.

I

I

Input

signal

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

.

V

V

Cutoff

Q

I

Cutoff

0

V

V

Waveform Distortion

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.

Quiz

Quiz

Quiz

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

a. clamped

b. clipped

c. unstable

Voltage-Divider Bias

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

voltage-divider from V

_CC

.

R

₁

and R

₂

are selected to establish V

_B

. If the

divider is stiff,

I is small compared to I

. Then,

+

V

divider is stiff,

I

_B

is small compared to I

₂

. Then,

R

R

R

R

2

B

CC

1

2

R

V

V

R

R





≈ 

₊







I

₂

I

_B

Quiz

Quiz

Quiz

Q. A

stiff

voltage divider is one in which

a. there is no load current

b. divider current is small compared to load current

c. the load is connected directly to the source voltage

d. loading effects can be ignored

Voltage-Divider Bias

+

V

+15 V

R

R

R

R

β

= 200

27 kΩ

12 kΩ

680 Ω

1.2 kΩ

Determine the

b

ase voltage for the circuit.

(

)

2

B

CC

1

2

12 k

15 V

27 k

12 k

R

V

V

R

R





= 

₊







Ω





=

_

_

+

=

Ω +

Ω





4.62 V

Voltage-Divider Bias

+

V

+15 V

What is the emitter voltage, V

_E

, and current, I

_E

?

R

R

R

R

β

= 200

27 kΩ

12 kΩ

680 Ω

1.2 kΩ

4.62 V

V

_E

is one diode drop less than V

_B

:

V

_E

= 4.62 V – 0.7 V =

3.92 V

3.92 V

Applying Ohm’s law:

E

E

E

3.92 V

680

V

I

R

=

=

=

Ω

5.76 mA

Quiz

Quiz

Quiz

Q. Assuming a

stiff

voltage-divider for the circuit shown,

the

e

mitter voltage is

a. 4.3 V

+

V

+15 V

b. 5.7 V

c. 6.8 V

d. 9.3 V

R

R

R

R

β

= 200

20 kΩ

10 kΩ

1.2 kΩ

1.8 kΩ

V

_B

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

V

= 5 – 0.7 =

4.3 V

Quiz

Quiz

Quiz

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

y-axis

at (neglecting I

_B

)

a. 5.0 mA

+

V

+15 V

b. 10.0 mA

c. 15.0 mA

d. none of the above

R

R

R

R

β

= 200

20 kΩ

10 kΩ

1.2 kΩ

1.8 kΩ

@

V

_CE

=0

:

I

_C

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

Voltage-Divider Bias

The

unloaded

voltage divider

approximation

for V

_B

gives

reasonable

results.

A

more exact

solution is to Thevenize the input circuit.

Looking from the base to the left:

+

V

+15 V

+ V

_CC

+15 V

R

R

R

R

β

= 200

27 kΩ

12 kΩ

+15 V

680 Ω

1.2 kΩ

V

_TH

= V

_{B(no load)}

=

4.62 V

R

_TH

= R

₁

||R

₂

=

=

8.31 kΩ

The Thevenin input

circuit can be drawn

R

_C

R

_TH

R

_E

+V

_TH

+

–

I

B

+

+

–

–

I

_E

V

8.31 kΩ

680 Ω

1.2 kΩ

4.62 V

+15 V

β

= 200

Thevenin Resistance, R

_TH

To find R

_TH

, replace

voltage sources

by a

short

circuit.

Which means

V

_CC

will be

grounded

.

R

_TH

= R

₁

||R

₂

+

V

+15 V

1

2

3

R

R

R

R

β

= 200

27 kΩ

12 kΩ

680 Ω

1.2 kΩ

Quiz

Quiz

Quiz

Q. If you Thevenize the input voltage divider, the

Thevenin

resistance

is

a. 5.0 kΩ

Ω

+

V

+15 V

b. 6.67 kΩ

c. 10 kΩ

d. 30 kΩ

R

R

R

R

β

= 200

20 kΩ

10 kΩ

1.2 kΩ

1.8 kΩ

R

_TH

= R

₁

R

₂

/ (R

₁

+ R

₂

)

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

Voltage-Divider Bias

Now write

KVL

around the

b

ase

e

mitter circuit and solve

for I

_E

.

+ V

_CC

+15 V

TH

B

TH

BE

E

E

V

=

I R

+

V

+

I R

I

_E

= I

_B

+ I

_C

= I

_B

+ β

_DC

I

_B

= (β

_DC

+ 1) I

_B

≈ β

_DC

I

_B

V

_TH

– V

_BE

= I

_E

R

_E

+ I

_B

R

_TH

≈ I

_E

R

_E

+ I

_E

R

_TH

/ β

_DC

= I

_E

( R

_E

+ R

_TH

/ β

_DC

)

R

_C

R

_TH

R

_E

+V

_TH

+

–

I

B

+

+

–

–

I

_E

V

8.31 kΩ

680 Ω

1.2 kΩ

4.62 V

+15 V

β

= 200

TH

BE

E

TH

E

DC

β

V

V

I

R

R

−

=

+

Substituting and solving,

E

4.62 V

0.7 V

8.31 k

680 +

200

I

=

−

=

Ω

Ω

5.43 mA

and V

_E

= I

_E

R

_E

= (5.43 mA)(0.68 kΩ)

=

3.69 V

Voltage-Divider Bias

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

−

+

+

V

V

EE

EE

R

R

R

2

2

2

1

1

1

R

R

R

R

R

R

C

C

C

R

R

R

E

E

E

Voltage-Divider Bias

Determine I

_E

for the

pnp

circuit. Assume a

stiff

voltage divider (

no loading effect

).

+V

_EE

+V

_EE

R

₂

1

R

R

_C

1.2 kΩ

R

_E

680 Ω

27 kΩ

12 kΩ

+15 V

(

)

1

B

EE

1

2

27 k

15.0 V

10.4 V

27 k

12 k

R

V

V

R

R





= 

₊







Ω





=

_

_

+

=

Ω +

Ω





E

B

BE

10.4 V 0.7 V = 11.1 V

V

=

V

+

V

=

+

EE

E

E

E

15.0 V 11.1 V

680

V

V

I

R

−

−

=

=

=

Ω

5.74 mA

10.4 V

11.1 V

Quiz

Quiz

Quiz

Q. For the circuit shown, the

e

mitter voltage is

a. less than the

b

ase voltage

b. less than the

c

ollector voltage

+V

_EE

R

R

+15 V

c. both of the above

d. none of the above

R

₂

1

R

R

_C

1.2 kΩ

R

_E

680 Ω

27 kΩ

12 kΩ

pnp

circuit

Emitter Bias

Emitter bias has excellent stability but requires both a

positive

and a

negative

source.

V

_CC

+15 V

For troubleshooting analysis, assume that V

_E

for an

npn

transistor is about

−1 V

.

Assuming that V

_E

is

−1 V

, what is I

_E

?

V

_EE

R

_C

R

R

_B

68 kΩ

+15 V

−15 V

7.5 kΩ

3.9 kΩ

−1 V

EE

E

E

1 V

15 V ( 1 V)

7.5 k

V

I

R

−

−

−

− −

=

=

=

Ω

−1.87 mA

The

minus

sign means the

current is directed downwards

(opposite to initial assumption).

Emitter Bias

The approximation that V

_E

≈ −1 V and the neglect of β

_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 I

_E

.

KVL applied around the

b

ase-

e

mitter circuit in Figure 5–17(a), which has

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

Emitter Bias

V

_RB

+ V

_BE

+ V

_RE

− V

_EE

= 0

I

_B

R

_B

+ V

_BE

+ I

_E

R

_E

− V

_EE

= 0

−

I

_E

( R

_B

/ β

_DC

) + I

_E

R

_E

= V

_EE

− V

_BE

I

_E

(R

_E

+ R

_B

/ β

_DC

) = V

_EE

− V

_BE

Quiz

Quiz

Quiz

Q. Emitter bias

a. is not good for linear circuits

b. uses a voltage-divider on the input

c. requires dual power supplies

Quiz

Quiz

Quiz

Q. With the

emitter bias

shown, a reasonable assumption

for troubleshooting work is that the

a.

b

ase voltage = +1 V

V

CC

+15 V

b.

e

mitter voltage = +5 V

c.

e

mitter voltage = −1 V

d.

c

ollector voltage = V

_CC

V

_EE

R

_C

R

R

_B

68 kΩ

−15 V

7.5 kΩ

3.9 kΩ

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

β dependent

.

R

_C

R

_B

+V

_CC

Base current is derived from the collector supply

through a

large base resistor

.

What is I

_B

?

CC

B

0.7 V

15 V 0.7 V

560 k

V

I

R

−

−

=

=

=

Ω

25.5 µA

R

_C

R

_B

+V

_CC

560 kΩ

+15 V

1.8 kΩ

I

_B

+

−

V

_BE

Base Bias

Compare

V

_CE

for the case where β = 100 and β = 300.

(

)(

)

C

β

B

100 25.5 µA

2.55 mA

I

=

I

=

=

For β = 100:

V

=

V

−

I R

R

_C

R

_B

+V

_CC

560 kΩ

+15 V

1.8 kΩ

10.4 V

(

)(

)

CE

CC

C

C

15 V

2.55 mA 1.8 k

V

=

V

−

I R

=

−

Ω =

For β = 300:

_I

_C

=

_β

_I

_B

=

(

_{300 25.5 µA}

)(

)

=

_{7.65 mA}

(

)(

)

CE

CC

C

C

15 V

7.65 mA 1.8 k

V

=

V

−

I R

=

−

Ω =

1.23 V

I

_C

+

V

_CE

−

Quiz

Quiz

Quiz

Q. The circuit shown is an example of

a. base bias

b. collector-feedback bias

+V

CC

c. emitter bias

d. voltage-divider bias

R

_C

R

_B

Emitter-Feedback Bias

An emitter resistor (

R

_E

) changes base bias into

emitter-feedback bias, which is more predictable.

The emitter resistor (

R

_E

) is a form of

negative feedback

.

If

I

_C

tries to

increase

,

V

_E

increases

, causing an

R

R

C

E

R

_B

+V

_CC

If

I

_C

tries to

increase

,

V

_E

increases

, causing an

increase

in

V

_B

because V

_B

= V

_E

+ V

_BE

.

This increase in V

_B

reduces

the

voltage across R

_B

(V

_RB

= V

_CC

− V

_B

), thus

reducing I

_B

and keeping I

_C

from increasing.

Emitter-Feedback Bias

The equation for

e

mitter current (

I

_E

) is found by writing

KVL

around the

b

ase circuit.

I

_E

R

_E

+ I

_B

R

_B

= V

_CC

− V

_BE

R

R

C

E

R

_B

+V

_CC

I

_B

I

_E

+

+

+

−

−

−

I

_E

R

_E

+ I

_B

R

_B

= V

_CC

− V

_BE

I

_E

R

_E

+ (I

_E

/ β

_DC

) R

_B

= V

_CC

− V

_BE

I

_E

(R

_E

+ R

_B

/ β

_DC

) = V

_CC

− V

_BE

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 (

R

_B

) to V

_CC

, it is

returned to the

c

ollector.

The equation for collector current (I

_C

) is found

by writing KVL around the base circuit.

(details on page 247 in the book)

The result is

CC

BE

C

B

C

DC

β

V

V

I

R

R

−

=

+

+V

_CC

R

_C

R

_B

Collector-Feedback Bias

Collector feedback bias uses another form of

negative

feedback

to increase stability. Instead of returning the base

resistor to V

_CC

, it is returned to the collector.

The negative feedback creates an “offsetting” effect

that tends to keep the

Q-point stable

.

+V

_CC

R

_C

R

_B

If

I

_C

tries to

increase

, it drops more voltage across

R

_C

, thereby causing

V

_C

to

decrease

.

When

V

_C

decreases

, there is a decrease in voltage

across R

_B

, which

decreases I

_B

.

The decrease in I

_B

produces

less I

_C

which, in turn,

drops

less voltage across R

_C

and thus offsets the

Collector-Feedback Bias

Compare I

_C

for the case when β = 100 with the case when β = 300.

When β = 100,

CC

BE

C

B

C

DC

15 V 0.7 V

330 k

1.8 k

100

β

V

V

I

R

R

−

−

=

=

=

Ω

Ω +

+

+V

_CC

R

_C

R

_B

330 kΩ

1.8 kΩ

+ 15 V

2.80 mA

When β = 300,

CC

BE

C

B

C

DC

15 V 0.7 V

330 k

1.8 k

300

β

V

V

I

R

R

−

−

=

=

=

Ω

Ω +

+

4.93 mA

Quiz

Quiz

Quiz

Q. The circuit shown is an example of

a. base bias

b. collector-feedback bias

+V

_CC

R

_C

c. emitter bias

d. voltage-divider bias

R

_C

R

_B

Key Terms

Key Terms

Key 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 I

_C

and V

_CE

for a

transistor circuit.

The region of operation along the load line

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

circuit’s output back to the input in such a way

as to oppose or aid a change in the output.