Bipolar Junction Transistors (BJT)

NPN PNP

BJT Cross-Sections

NPN PNP

Emitter

Collector

Common-Emitter NPN Transistor

Forward bias the BEJ

Reverse bias the CBJ

Input Characteristics

• Plot IB as f(VBE, VCE)

• As VCE increases, more VBE required to turn the BE on so that IB>0.

• Looks like a pn junction volt-ampere characteristic.

Output Characteristics

• Plot IC as f(VCE, IB)• Cutoff region (off)

– both BE and BC reverse biased

• Active region– BE Forward biased– BC Reverse biased

• Saturation region (on)– both BE and BC

forward biased

Transfer Characteristics

Large-Signal Model of a BJT

KCL >> IE = IC + IB

βF = hFE = IC/IB

IC = βFIB + ICEO

IE = IB(1 + βF) + ICEO

IE = IB(1 + βF)

IE = IC(1 + 1/βF)

IE = IC(βF + 1)/βF

(1 ) ( 1)

111

1 1

E B C

CF FE

B

C F B CEO

E B F CEO B F

FE C C

F F

C F E

F FF F

F F

I I I

Ih

I

I I I

I I I I

I I I

I I

Transistor Operating Point

B BEB

B

CE CCC

C C

CE CC C C

V VI

R

V VI

R R

V V I R

DC Load Line

VCC

VCC/RC

BJT Transistor Switch

B BEB

B

CE CC C C

CE CB BE

CB CE BE

V VI

R

V V I R

V V V

V V V

BJT Transistor Switch (continued)

CC CE CC BECM

C C

CMBM

F

V V V VI

R R

II

BJT in Saturation

( )CC CE satCS

C

CSBS

F

B

BS

CSforced

B

V VI

R

II

IODF

I

I

I

Model with Current Gain

Miller Effect

vbe vce

iout

Miller Effect (continued)

( ) ( )

[1 ] [1 ]

[1 ]

out cb be ce cb be be

out cb be cb be

cb cb

d di C v v C v Av

dt dtd d

i C A v C A vdt dt

C C A

Miller Effect (continued)

• Miller Capacitance, CMiller = Ccb(1 – A)

– since A is usually negative (phase inversion), the Miller capacitance can be much greater than the capacitance Ccb

• This capacitance must charge up to the base-emitter forward bias voltage, causing a delay time before any collector current flows.

Saturating a BJT

• Normally apply more base current than needed to saturate the transistor

• This results in charges being stored in the base region

• To calculate the extra charge (saturating charge), determine the emitter current

1cse B BS BS BS

II I ODF I I I ODF

The Saturating Charge

• The saturating charge, Qs

( 1)s s e s BSQ I I ODF

storage time constant of the transistor

Transistor Switching Times

Switching Times – turn on

• Input voltage rises from 0 to V1

• Base current rises to IB1

• Collector current begins to rise after the delay time, td

• Collector current rises to steady-state value ICS

• This “rise time”, tr allows the Miller capacitance to charge to V1

• turn on time, ton = td + tr

Switching Times – turn off

• Input voltage changes from V1 to –V2

• Base current changes to –IB2

• Base current remains at –IB2 until the Miller capacitance discharges to zero, storage time, ts

• Base current falls to zero as Miller capacitance charges to –V2, fall time, tf

• turn off time, toff = ts + tf

Charge Storage in Saturated BJTs

Charge storage in the Base Charge Profile during turn-off

Example 4.2

Waveforms for the Transistor Switch

VCC = 250 V

VBE(sat) = 3 V

IB = 8 A

VCS(sat) = 2 V

ICS = 100 A

td = 0.5 µs

tr = 1 µs

ts = 5 µs

tf = 3 µs

fs = 10 kHz

duty cycle k = 50 %

ICEO = 3 mA

Power Loss due to IC for ton = td + tr

• During the delay time, 0 ≤t ≤td

• Instantaneous Power Loss

• Average Power Loss

0 0

1( )

(250 )(3 )(10 )(0.5 ) 3.75

d dt t

CC CEOd c CC CEO s d

d

V IP P t dt dt V I f t

T T

P V mA kHz s mW

( )

( ) (250 )(3 ) 0.75c CE C CC CEO

c

P t v i V I

P t V mA W

During the rise time, 0 ≤t ≤tr

( )

( )( )

max

( )

( )

( ) ( )

( )( )

( ) @

2[ ]

c CE c

CSc CC ce sat CC

r r

ce sat CCc CS CSCC ce sat CC

r r r r

c m

r CCm

CC ce sat

P t v i

ItP t V V V t

t t

V VdP t I Itt V V V

dt t t t t

P t P t t

t Vt

V V

2

max( )

2

max

(1 )(250 )0.504

2[250 2 ]

4[ ]

(250 ) (100 )6300

4[250 2 ]

m

CC CS

CC CE sat

s Vt s

V V

V IP

V V

V AP W

V V

Average Power during rise time

( )

0

1( )

2 3

(250 ) (2 250 )(10 )(100 )(1 )

2 3

42.33

rtCE sat CCCC

r c s CS r

r

r

V VVP P t dt f I t

T

V V VP kHz A s

P W

Total Power Loss during turn-on

0.00375 42.33 42.33375

42.33

on d r

on

on

P P P

P W

P W

Power Loss during the Conduction Period

( )

( ) ( )

0 0

0

( ) 100

( ) 2

( ) (100 )(2 ) 200

1( )

(2 )(100 )(10 )(48.5 ) 97

n n

n

c CS

CE CE sat

c c CE

t t

n c CE sat CS s CE sat CS s n

n

t t

i t I A

v t V V

P t i v A V W

P P t dt V I f dt V I f tT

P V A kHz s W

Power Loss during turn offStorage time

( )

( )

( ) ( )

0 0

0

( ) 100

( ) 2

( ) (2 )(100 )

( ) 200

1( )

(2 )(100 )(10 )(5 ) 10

s s

s

c CS

CE CE sat

c CE c CE sat CS

c

t t

s c CE sat CS s CE sat CS s s

s

t t

i t I A

v t V V

P t v i V I V A

P t W

P P t dt V I f dt V I f tT

P V A kHz s W

Power Loss during Fall time0

( ) 1 , 0

( ) , 0

( ) 1

( ) 11 0

3( ) @ 1.5

2 2(250 )(100 )

4 4

f

c CS CEOf

CCCE CEO

f

c CE c CC CSf f

c CC CS

f f f

fc m

CC CSm

t t

ti t I I

t

Vv t t I

t

t tP t v i V I

t t

dP t V I tt

dt t t t

t sP t P t s

V I V AP

6250W

Power Loss during Fall time (continued)

0

( )

1( )

6

(250 )(100 )(3 )(10 )125

6

6

10 125 135

ftCC CS f s

f c

f

CC foff s f CS s s CE sat

off

V I t fP P t dt

T

V A s kHzP W

V tP P P I f t V

P W

Power Loss during the off time

0

0

( )

( )

( ) (250 )(3 ) 0.75

1

(250 )(3 )(10 )((50 5 3) )

0.315

o

o

CE CC

c CEO

c CE C CC CEO

t

o CC CEO CC CEO s o

o

o

t t

v t V

i t I

P t v i V I V mA W

P V I dt V I f tT

P V mA kHz s

P W

The total average power losses

42.33 97 135 0.315

274.65

T on n off o

T

T

P P P P P

P

P W

Instantaneous Power for Example 4.2

BJT Switch with an Inductive Load

Load Lines