Basic Transistor Amplifier-Basic Electronics-Lecture 05-Electronic and Information Engineering, Lecture notes of Basic Electronics

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EIE209 Basic Electronics

Basic Transistor Amplifiers

Contents

• Biasing
• Amplification principles
• Small-signal model development for BJT

Aim of this chapter

To show how transistors can

be used to amplify a signal.

amplifier

Biasing the transistor

To set the transistor to a certain DC level = To set V CE

and I C

R

R L

B

V

BE

V

CE

I

C

I

B

Transistor:

b = 100

V

CC

=10V

Suppose we want the following biasing condition:

I

C

= 10 mA and V CE

= 5 V

Find R B

and R L

Start with V BE

≈ 0.7 V.

Then, I B

= (10 – V

BE

) / R

B

= (10 – 0.7) / R

B

I

C

= bI B

= 100 (10 – 0.7) / R

B

= 10 mA

So, R

B

= 94kΩ

Also, V CE

= 10 – R

L

I

C

Hence, 5 = 10 – 10 R L

So, R

L

= 0.5kΩ

b dependent biasing — bad biasing

R

R L

B

V

BE

V

CE

I

C

I

B

Transistor:

b = 100

V

CC

=10V

This is a bad biasing circuit!

because it relies on the accuracy of b,

but b can be ±50% different from what is

given in the databook.

Now, let’s go to the lab and try using R B

= 94kΩ and R L

= 0.5kΩ, and see if we get what we want.

…totally wrong! We don’t get I C

= 10mA and V CE

= 5V

A much better biasing method —

emitter degeneration

R

R L

B

V

CE

I

C

I

B

V

CC

=10V

Again, our objective is to find the resistors such that

I

C

= 10mA and V CE

= 5V.

R

B R E

Set V E

= 2V, say. Then, R E

= 2V/10mA = 0.2kΩ.

Surely, R L

= 0.5kΩ in order to get V CE

= 5V.

Finally, we have V B

= V

E

• 0.6. Therefore, if I B

is

small compared to I RB

and I RB

, we have

V

E

R

B 1

R

B 2

Hence, R B

= 740Ω and R B

NOTE: b is never used in calculation!!

Stable (good) biasing

R

R L

B

V

CE

I

C

I

B

V

CC

=10V

R

B R E

V

E

Summary of biasing with emitter degeneration:

Choose V E

, I C

and V CE

.

R E

R L

Use V BE

≈ 0.6 to get V B

.

Then use

to choose R B

and R B

such that I B

is much smaller the

current flowing in R B

and R B

.

V

B

R B 1

R B 2

=

10 - V B

V B

Alternative view of biasing

R

L

V

BE

V

CE

I

C

V

CE

V

R

V

CC

V

CC

I

C

V

CE

I

C

V

CC

R

L (^) Load line

Slope=–1/ R L

operating point

What controls the operating point?

R

L

V

BE

V

CE

I

C

V

CC

V

CE

I

C

Load line

Slope=–1/ R L

V

CC

operating point

a bigger V BE

a smaller R L

V BE

or I

B

controls the OP

R L

also controls the OP

CONCLUSION:

Animation to show amplifier action

Derivation of voltage gain

Question: what is?

D V

o

D V

in

D V

CE

D V

BE

R

L

V

BE

V

CE

I

C

V

CC

Then, what relates DI C

and DV BE

Clearly, Ohm’s law says that

V

CE

= V

CC

• I C

R

L

fi D V

CE

= - R

L

.D I

C

Last lecture: transconductance

g m

D I

C

D V

BE

Hence, (^) D V CE

D V BE

= - g m

R L

How do we inject signal into the amplifier?

R

L

V

BE

v CE

I

C

V

CC

R

B

R

B

v

in

±20mV

= V

CE

• v CE

~

~

Dv

CE

or

Dv

in

Note on symbols

v CE

= V

CE

• v CE

~

total signal

(large signal)

operating point

or

DC value

or

quiescent point

small signal

or

ac signal

Total signal

a

DC point

A

Small signal

a or Da

The wonderful voltage source: capacitor

R

L

V

BE

V

CE

I

C

V

CC

R

B

R

V B

C

The capacitor voltage is

exactly equal to V BE

because DC current

must be zero

0A

Solution — insert coupling capacitor

R

L

v BE

v CE

I

C

V

CC

R

B

R

B

v

in

±20mV

DC voltage equal

to exactly the

same biasing V BE

This is called a

coupling capacitor

~