BJT BIASING

in active region aF

transistor utilization as amplifier (CE)

in active region (aF), the transistor operates around a dc operating point (OP)

Necessity for dc transistor biasing

VPS – dc supply

VI – sets the OP: (VO, IO)

vi – input voltage

(to be amplified)

vo – output voltage

(amplified voltage)

• superposition of the

variable signal over the

dc regime

DC biasing – setting the OP

Operation of the transistor as amplifier:

• the transistor biased as close as possible to the middle of the aF

• the instantaneous (mobile) operating point - in the active region

• the input variable signal kept small (linear region around OP)

OP:

stabile and predictibile

independent of the transistor parameters

BJT biasing, unipolar (single) supply usual approach in discrete circuits

oppositely to MOSFET, for BJT appears:

- base current IB different from zero

- through collector and emitter do not flow exactly same current

CBBCEIIIII

1)1(

ECII One can approximate

BCII

• precise calculation: make use of IB

• approximate calculation: neglects IB

compared to the current through the

resistive voltage divider in the base of

the transistor

),(Q CCE IV

Approximate calculation

IB much smaller than the current flowing through the base divider (through RB1 ,RB2)

PS

BB

BBB V

RR

RV

21

2

E

BEBBEC

R

VVII

)( ECCPS

EECCPSCE

RRIV

RIRIVV

• RE is very important for setting

and stabilizing the OP, through a negative feedback mechanism

IC↑; IE↑; VRE↑; VBE↓; IC↓

Use Thevenin theorem on the full circuit, between the base terminal

and ground to determine VBB, RB

IC=IE+IB ≈ IE

( )

CE PS C C E E

CE PS C C E

V V I R I R

V V I R R

)1/(

BE

BEBBE

RR

VVI

IE=(β+1)IB

EEBEBBBB IRVIRV

Precise calculation

Precise calculation

Stability analysis

• IE insensitive to β variations

)1(

B

E

RR

B

E

RR 10

•RB1, RB2 small values for the

independence of OP on β

•RB1, RB2 high values for the high

input resistance of the future

amplifier

• IE insensitive to temperature variations due to VBE

V1.0BBV

a ΔVBE variation of 0.1V can be neglected

vis a vis a VBB=3…5V

)1/(

BE

BEBBE

RR

VVI

VPS=15 V; RB1=10 kΩ; RB2=4.7 kΩ;

RE =1.5 kΩ; RC =1.8 kΩ; β =150

Approximate calculation

Exact calculation

IC = ?

VCE =?

VC = ?

VE = ?

IC = 2.73 mA

VCE = 6 V

VC = 10.1V

VE = 4.1V

IC = ? IC = 2.7 mA

Problem 1

Example 2

Size the resistances so

that T is biased in

aF @ IC=2mA ?

VPS =12V, β =100

Usually choose: V4123

1

3

1 PSBB VV

k65.12

7.012)3/1(

E

BEBBE

I

VVR

PSPS

BB

BBB VV

RR

RV

3

1

21

2

21 2 BB RR

mA84.1)1100/(7.1465.1

7.04

)1/(

BE

BEBBE

RR

VVI

Verification (precise calculation):

RB2=22 kΩ; RB1=44 kΩ

B

E

RR 10 E

BB

BB RRR

RR10

21

21

k75.242BR

Adjust k5.1ER mA2EI

BJT biasing, differential supply

Voltage across the current source:

VE

VB

VC

C

BECE

IIIIII ;;

CCPSC IRVV

BBBBB IRIRV 0

V7.0 BBEBE VVVV

ECCE VVV

PSEPSE VVVV )(

EECCPSCE

BE

BEPS

BE

PSBEE

EBPSEEBEBB

IRIRVV

RR

VV

RR

VVI

IIVIRVIR

2

)1/()1/(

)(0

)1/(;0

Problem 3

±VPS = ± 9 V; RB = 180 kΩ;

RC = 3.3 kΩ; β =100; I = 2 mA

IC = ?

VCE =?

VC = ? VE = ?VB = ?

Voltage across the current source = ?