BJT

Biasing

Single stage amplifier

Session 4: Analog Circuits

1

1. Lab

2. Power

3. Ser/Parl

4. Small sig.

5. Applic

Outline

BJT Amplifier

2

1. Lab

2. Power

3. Ser/Parl

4. Small sig.

5. Applic

BJT: Bipolar Junction Transistor

3

D

A C

iD

VD

p n

F : Forward

R : ReverseF

R

p n p n p n

21E C

B

21E C

B

Qn

E

C

B

B

E

C

Qp

VBE

VBC

1 : F

2 : R

1 : F

2 : F

1 : R

2 : F1 : R

2 : R

SaturationActive

ReversedCut-Off

1. Lab

2. Power

3. Ser/Parl

4. Small sig.

5. Applic

BJT: Bipolar Junction Transistor

4

D

A C

iD

VD

p n

B Qn

E

C+

-

∝ /

ΔΔ 2 2 !"/#

∝ $%/ ΔΔ 2 %& & 2 !"/#

1. Lab

2. Power

3. Ser/Parl

4. Small sig.

5. Applic

BJT Configurations

5

BQn

E

C

+

-

&

&+

-

+

-B

Qn

E C

&

&+

-B

Qn

E

C

+

-

&+

-

CE: Common Emitter CB: Common Base CC: Common Collector

Output Characteristic

Input Characteristic

Transfer Characteristic

vs. for different '&

& vs. & for different ' or

& vs. or for different '&

1. Lab

2. Power

3. Ser/Parl

4. Small sig.

5. Applic

CE: Common Emitter

6

BQn

E

C

+

-

&

&+

-

Output Characteristic

Input Characteristic

&

&

( 010+,20+,30+,

&

&

' . 0.5'5602'6002'6202'

'3

'3

Early voltage

Saturation

Active

Cut-Off0.2'

& 4 (5$% 1 6 &'3 "

Active:

'& 4 0'& 7 0

very little

dependence to

output voltage

'& ↑→ : ↑→ ; ↑→ (& ↑

1. Lab

2. Power

3. Ser/Parl

4. Small sig.

5. Applic

CE: Transistor Model

7

BQn

E

C

+

-

&

&+

-

Output Characteristic

&

& <S

atu

ratio

n

Cut-Off

&

&

0.2'

(&=>

'&=>

?=>

SOA: Safe Operating Area

(

1. Lab

2. Power

3. Ser/Parl

4. Small sig.

5. Applic

CE: Output Characteristic

8

BQn

E

C

+

-

&

&+

-

&

& <S

atu

ratio

n

Cut-Off

(

4 (′5$%

Ideally linear:

& (5$% 1 6 &'3 "

4 (5; $%

; 4 & 4(&( 4

A

BC 4 ;D 4 &EF$#SPICE

BGH 4 ;& 4 &EIFIJ4 ;D1 6 &'3 "

B,H 4 K&KEIFIJ4 A

1. Lab

2. Power

3. Ser/Parl

4. Small sig.

5. Applic

Beta : Current Gain

9

1. Lab

2. Power

3. Ser/Parl

4. Small sig.

5. Applic

CB: Common Base

10

Output Characteristic

Input Characteristic

&

&

( 012,22,32,

0.4'

Saturation

Active

Cut-Off

& 4 MActive:

'& 4 0'& 7 0

very little

dependence to

output voltage

+

-B

Qn

E C

&

&+

-

M 4 ;; 6 1

ΔΔ 2 %

7

1. Lab

2. Power

3. Ser/Parl

4. Small sig.

5. Applic

;(′

Large Signal Model

11

BQn

E

C

+

-

&

&+

- &

& <S

atu

ratio

n

(

2

1

E

C

Bideal

ideal

E

C

B

0.7

0.4

(′

&B

E

C

(5 ;(

Active

E C

B

M(

Saturation

B

E

C

(5 '&N=0.2~0.3'

1. Lab

2. Power

3. Ser/Parl

4. Small sig.

5. Applic

Voltage Amplifier

12

B

Qn

E

C

+

-

&

&+

-+

-I

&P&

Q+

-

'&& Q 4 '&& P&&& 4 ; 4 (N

R 1"

I

& Q

I'&&

cut-off

act

ive saturation

0.1v

10v

1. Lab

2. Power

3. Ser/Parl

4. Small sig.

5. Applic

Biasing: VBE= cte

13

+

-I

&P&

Q+

-

'&&

I

& Q

I'&&

cut-off

act

ive

saturation

0.1v

10v

5'

1. Lab

2. Power

3. Ser/Parl

4. Small sig.

5. Applic

Biasing: IB= cte

14

+

-

I&P&

Q+

-

'&&( 4 (#(& 4 ;(

5

(#∞

+

-

I&P&

Q+

-

'&&

∞

5

P ( 4 '&& 'P

(& 4 12, ; 4 100(# 4 10+,

→ P& ~5UΩ

'&& 4 10'

For max swing: '&~5'→ P& ~9.7UΩFor max gain: '&~0.3'

→ P4 930UΩ'&& 4 10'(& 4 12,

What is the problem?

Replace it with transistor with ; 4 250(& 4 2.52, '& 4 10 5 Y 2.5 4 2.5 Z '&N=

1. Lab

2. Power

3. Ser/Parl

4. Small sig.

5. Applic

Biasing: IB= cte

15

+

-

I&P&

Q+

-

'&&

∞

5

P

P 4 '& '( 4 ; '& '(&

(& 4 12, ; 4 100'&& 4 10'For max swing: '&~5'

→ P4 430UΩ

; ↑ → (& ↑→ '& ↓

→ ( ↓→ (& ↓

Negative feedback!

1. Lab

2. Power

3. Ser/Parl

4. Small sig.

5. Applic

Biasing: IE= cte

16

+

-I

&P&

Q+

-

'&&

Q

I

'&& 10v

5'

P+

-I

&P&

Q+

-

'&&

5 (

( 4 ' 'P

' ≫ 'independent of '(

'~2.7'

12,P&

10'

P2.7' 2'

4 2UΩ

6'4UΩ 4

I&

P&

0.7' 6.7'

Q+

-

+

-

10'5UΩ

Where is the

trade-off?

1. Lab

2. Power

3. Ser/Parl

4. Small sig.

5. Applic

Biasing: IE= cte

17

Q

'&& ( 4 (&;Assume:

12,

P&10'

P

2.7'2'

4 2UΩ

6'4 4UΩ

P

P∞

I (

(( ≫ ((&

P

P(

('&& 10'

'' 4 PP 6 P '&&

P

'&&P&10'

''

' 4 ' 'Q'&

(

( 4 'P (& 4 M( ((& '& 4 '&& (&P&

Q

'&&

12,

P&10'

P

2.7'2'4 2UΩ

6'4 4UΩ

P

P∞

I 10+,

0.12, 4 73UΩ

4 27UΩThis is for design

how about analysis

?

1. Lab

2. Power

3. Ser/Parl

4. Small sig.

5. Applic

Biasing: IE= cte

18

' 4 2'Assume: ( ≫ (

P

P(

('&&

'' 4 2.7'Q

'&&

12,

P&10'

P

2.7'2'

2UΩ

6'4UΩ

P

P∞

I 10+,

0.12, 73UΩ

27UΩ( 4 12, ( 4 0.012,

( 4 0.12,

What if ; was 10!

P

P(

('&&

''&& PP 6 P

P ∥ P Q

'&&

(&P&10'

P 1M (&

(&/;KVL

PP 6 P '&& 4 (&P ∥ P; 6 '^ 6 1M P(& → (& 4 ⋯

For the above numbers: (& 4 21.01 Y 2U 6 0.01 Y 19.7U 4 0.922,

1. Lab

2. Power

3. Ser/Parl

4. Small sig.

5. Applic

Biasing: IE= cte

19

(P 6 ' 4 '^ 6 (PQ

'&&

(

P&

PP

P

similarD(

( '&&P 6 P ( 4 PP (

'&&P&

PP

P

(

( 4 '&& '^P 6 P

(P 4 (P'&&

P&P( < <

( 4 '&& '^P( ( 4 (5 $%/ 1

+

- ( 4 (5 $%/ 1

(( 4(5(5 4

Only in Integrated Circuits!

,`,`area

1. Lab

2. Power

3. Ser/Parl

4. Small sig.

5. Applic

Biasing: Example 01

20

(′3.9(′ 6 0.7 4 5.60.9 (′; "

Q1

Q2

4.7kΩ

5.6kΩ 3.9kΩ

56kΩ 3.3kΩ

150kΩ

15V ; 4 ; 4 100 'ab 4 0.7< <

(& [mA]

'& [V]4'

0.072,

3.3'12,

9.4' 10.1'1.262,

1.262,5.9'

0.012,1 1.266.1 4.2

Q1

41kΩ

150||564.08

3.3kΩ

(

(/;

KVL

5.6kΩ

15V

( 4 4.083.3 6 40.8/101 4 0.912,

3'

Q2

3.9kΩ

15V

4.7kΩ

(′/;0.9

(′ 4 1.132,

9.89'

KVL

10.59'5.31'

1. Lab

2. Power

3. Ser/Parl

4. Small sig.

5. Applic

Biasing: Example 02

21

; 4 ; 4 100 'ab 4 0.7< <

(& [mA]

'& [V]

1.492,

1.05 1.491.4 9.3Q2

8.2kΩ

Q1

470Ω

Vcc=10V

0.7'

1.4' (& 4 1.492,

1.052,

( 4 1.052, 0.01√

0.015 √

1. Lab

2. Power

3. Ser/Parl

4. Small sig.

5. Applic

Biasing: Example 03

22

; 4 ; 4 100'ab 4 0.7 < <

(& [mA]

'& [V]

2 1.72.4 3.56

Vcc=9V

47kΩ

Q2

2.2 kΩ3.3 kΩ

Q1

1 kΩ

0.7'0.7'

0.72,

0.72,7.46'1.4'

2.32,

0.007 √

0.023 ××××

1.7'1.72,

1.72,5.26'2.4'

22,

8.3 330 4 0.747

47kΩ

Q2

2.2 kΩ3.3 kΩ

Q1

1 kΩ

9V

0.747U

6 0.7

8.3 3.3U

⇒ 4 1.65

1. Lab

2. Power

3. Ser/Parl

4. Small sig.

5. Applic

Biasing: Example 04

23

; 4 ; 4 100 'ab 4 0.7< <

(& [mA]

'& [V]

1.1 1.11.5 3.11

Vcc=10V

Q2

3.9 kΩ

18 kΩ

Q1

1 kΩ

1.8'0.12, 15 kΩ

68 kΩ

P3.3'

1.1'1.12,

2.6'5.71'

P 4 8.2UΩFind bias points if

Q2

8.2 kΩ

P

2.6'3.3'

' 4 10'

0.2'

'& 4 10 1.1 Y 8.2 2.64 1.622.8' 0.882, < <

(& [mA]

'& [V]

1.1 0.881.5 0.2

1. Lab

2. Power

3. Ser/Parl

4. Small sig.

5. Applic

Linear BJT Amplifier

24

Vin

R

VCC

Vo

Qn

& 4 (5$% 1 6 &'3 " (5

$%

' 4 262' '3 4 200'

4 ' 6 e sinij

Q 4 '&& P& 4 '&& P(5$ ek lmn o (&

4 '&& P(& 1 6 e' sinij 6e2' sin

ij 6 ⋯ '&& P(& P(& e' sinij'Q ^

, 4 ^ 4P&(&' 4 pP& p 4 (&'

1. Lab

2. Power

3. Ser/Parl

4. Small sig.

5. Applic

BJT Small Signal Model (h-π)

25

Trance-Conductance:

BQn

E

C

+

-

&&+

-

A qr pr q#r+

-

Input resistance:

qr ≡ KKOutput resistance:

4 KK 4 ; K&K

& 4 (5 %$ 1 1 6 &'3 "

≅ (5%$ 1 6 &'3 "(&

4 ; (5' %$

4 ;'(&4

;p 4 ;q

q ≡ K&K& 4 K&K& 4 (&'3

4 '3(&

p ≡ K&K 4 (5' %$ 4 (&' 4

1q

1. Lab

2. Power

3. Ser/Parl

4. Small sig.

5. Applic

N

Example 01 -CE

26

(&P&

Q+

-

10'

∞5

P

P 4 10 0.70.012,

(& 4 12,; 4 100Assume

Design for

r 4 N qr ∥ Pqr ∥ P 6 PN

PN1U

'3~∞and maximum swing

Find ,, P, P^v

AC:

Qwindow

for

10'cutoff

saturation0.2'~0? 'Q 4 5 P& 4 5UΩ4 930UΩDC:

PN1UP

^P&qr p

r

r+-

~N qrqr 6 PN# 4 prP&

, 4 #N 4 pP&qrqr 6 PN 4

;P&qr 6 PN4P&q 6 PN ;⁄ 4

Collector resistance

Emitter’s circuit resistance

if , 4 pP&PN → 0:

1. Lab

2. Power

3. Ser/Parl

4. Small sig.

5. Applic

Q

Example 02 -CE

27

Q12,

P&10'

P

2.7'2'2UΩ

6'4UΩ

P

P∞10+,

0.12, 73UΩ

27UΩ

y 6 Py rqr 6 r 6 Prqr 1 6 ; 4 0

PN ∞

Find ,, P, P^v

PNP P P

P&

P ∥ P

y Py

yPy

P

P&qr+

-

r pr

r qrz

r 1qr 6 p

'|

4 rqr 1 6 ;

Q~20UΩ

1UΩ

y 4 P ∥ PP ∥ P 6 PN Py 4 P ∥ P ∥ PN'|:

r 4 y qrPy 6 qr 6 P 1 6 ;

^ 4 prP&

,~ 4 ^y 4;P&Py 6 qr 6 P 1 6 ;

4 P&Py 6 qr; 6 P 1 6 ;;, 4 ^N 4

yN ∙^y 4

2021 ∙

43.5100 6 2 Y 101100

4 1.8

; 4 100Assume '3~∞

AC

circuit

1. Lab

2. Power

3. Ser/Parl

4. Small sig.

5. Applic

Example 02 -CE

28

QP&

10'

P2UΩ

4UΩ

P

P 73UΩ

27UΩ

PN1UΩ

, 4 1.8

How we can increase gain?

Q

QP&

10'

P2UΩ

4UΩ

P

P 73UΩ

27UΩ

PN1UΩ Bypass capacitance

H

1. Lab

2. Power

3. Ser/Parl

4. Small sig.

5. Applic

Example 03 -CC

29

Q

10'

P

P

6 P5 rqr 6 r 6 Prqr 6 pr 4 0

PN

P5

P

qr+

-

r pr

r qrz

r 1qr 6 p

'|'|:

r 4 P5qr 6 1 6 P p 6 1qr

, 4 ^

Design a buffer window for

9.3'saturation

0'? ' 4 5

P 4 5UΩ

~10

cutoff

(& 4 12,

P 4 10 5.70.012 4 430UΩ

PNP P

AC circuit

^^ 4 r 1

qr 6 p P

41qr 6 p P

P5qr 6 1 6 P p 6 1qr4 PP5 6 qr1 6 ; 6 P

~1

^

1. Lab

2. Power

3. Ser/Parl

4. Small sig.

5. Applic

Example 03 -CC

30

4 qr 6 P 1 6 ;

P5

P

qr+

-

r pr

r qrz

r 1qr 6 p

P 4r 6 P rqr 6 prrqr

PN

P

^

^ 4PP5 6 qr1 6 ; 6 P^

P PN

P ^P

P 4 qr 6 P 1 6 ;

PN

P ^P^v

P^v 4 P ∥ q

q

4 P ∥ P5 6 qr1 6 ;

q

1. Lab

2. Power

3. Ser/Parl

4. Small sig.

5. Applic

Example 04 –Multi-stage Amplifier

31

P&!10'

2UΩ

73UΩ

27UΩ

1UΩP& Q

Design an amplifier:

(& 4 12,, 1000; 4 100' 4 10'P5 4 1UΩ

930UΩ

P&! 4 10 312 4 7UΩP& 4 10 512 4 5UΩ

AC circuit

1UΩ

27 ∥ 73

20UΩ

< <

<<

7UΩ 930UΩ 5UΩP! ^ , 4 ^ 4

⋅ ⋅

^P

420 ∥ qr!

20 ∥ qr! 6 1 4 0.69

4 p! 7 ∥ 930 ∥ qr 4 73^ 4 p5 4 200

, 4 10143

< <7UΩ 5UΩ

1UΩ^

buffer

1. Lab

2. Power

3. Ser/Parl

4. Small sig.

5. Applic

Example 05 –CE , CB, CC

32

10'

2

73

27PN

Q

AC circuit

<<

<<

7<

73

27 2

4430

512,2.723

7.38 12,

4 12,5.7 5

20PN

7 2< 4 Q

< <1.5

< 4 Q

PN ≪?

5

5430

1. Lab

2. Power

3. Ser/Parl

4. Small sig.

5. Applic

Common Base

33

PPN

NQPNP&

PQP& B

E

C

qr+

- prE

B

rC

NE

B

C

≡

≡E

B

C

qr

prr

+

-

E C

qrr+

-

≡B

prpr

≡ E C

qrr+

-

B

prq

E C

qr+

-

B

pr

q 4 q ∥ qr 4 q ;; 6 1 ≅ q

NPN

PQP&

, 4 ^N 4P&P 6 qr; 6 1 6 PN

P 4 P 6 qr; 6 1P^ 4 P

P P^

1. Lab

2. Power

3. Ser/Parl

4. Small sig.

5. Applic

Example 05 –CE , CB, CC

34

< <1.5

< 4 Q

N 4 p 1.5 ∥ P

5 P

4 1 4 6p 4 ∥ qr 6 ;5" 4 160^ 4

55 6 qr;

≅ 1

, 4 ^N 4N ⋅

⋅^ 4 160

P

P 4 qr! 4 2.5P^ 4 4 ∥ 4

6 qr; 6 1 4 63Ω

P^

1. Lab

2. Power

3. Ser/Parl

4. Small sig.

5. Applic

neglecting

base

current

CascodeAmplifier , CE-CB

35

Q(

18'1.5

AC circuit:

6.8

5.6

1

4.7<

<

( 4 186.8 6 5.6 6 4.7 4 1.13

5

11.2

4.34.3

10.511.5 Q

1.5

<

<

4 p! q 4 1

q

^ 41.5qr;

4 245

, 4 245

1. Lab

2. Power

3. Ser/Parl

4. Small sig.

5. Applic

Some Notes:

36

qr pr q#r+-

E

BC

qr pr q#r+-

E

B C

qr pr q#r+-

E

B C

qr pr q#r+-

< < q≡ ≡

1. Lab

2. Power

3. Ser/Parl

4. Small sig.

5. Applic

Summary

37

NPN

PQP&

^N 4P&P 6 qr; 6 1 6 PN

P 4 P 6 qr; 6 1P^ 4 P

P^ PN

P

^ 4PP5 6 qr1 6 ; 6 P

^P

P 4 qr 6 P 1 6 ;

P^

P^ 4 P ∥ P5 6 qr1 6 ;

QPNP

P&P

PP^

^N 4P&PN 6 qr; 6 1 6 P

P 4 qr 6 P 1 6 ;

P^ 4 P&

1. Lab

2. Power

3. Ser/Parl

4. Small sig.

5. Applic

? VA

38

PN

P

^ 4PP5 6 qr1 6 ; 6 P

^P

P 4 qr 6 P 1 6 ;

P^

P^ 4 P ∥ P5 6 qr1 6 ;

q PN

P ∥ q ^PP^

q 4 ∞^ 4

P ∥ qP5 6 qr1 6 ; 6 P ∥ qP 4 qr 6 P ∥ q 1 6 ;

P^ 4 P ∥ q ∥ P5 6 qr1 6 ;

1. Lab

2. Power

3. Ser/Parl

4. Small sig.

5. Applic

Input / Output Impedances

39

PPj

,''+

-

'+

-

'j >>

P P^

Inp

ut

po

rt

>

Ou

tpu

t p

ort

>