18 - 322

Fall 2003

Lecture 21

Bipolar Junction Transistor (BJT) • • • • • • •

NPN Cross-section and Masks BJT Notation Hand Analysis Models NPN Modes of Operation Ebers - Moll Model BJT Inverter TTL Chapter 2.4 Slide 1

NPN Transistor Cross-section

Slide 2

NPN Transistor Layout p n n p p

Slide 3

BipolarJunctionTransistor (BJT) Notation C

C

C n

Collector Base

B

Emitter

IC

nn

p

B

p

B

IB IE

n n E

E

E

n-p-n transistor Slide 4

NPN Regions of Operation VBC

Reverse Active

Cut-off

Saturation

Forward Active

VBE

Slide 5

NPN Forward Active Polarization VBE Forward-biased & VBC Reverse-biased E

C Electron Flow B

Base current: IB = IC / βF

(Current in opposite direction)

βF = IC / IB

βF - current gain (~100!) Slide 6

Common-Emitter Characteristics I C [mA]

14 13

I

12

C

11 10

B

= 120µA

I

B

= 100µA

I

B

= 80µA

I

B

= 60µA

I

B

= 40µA

I

B

= 20µA

9

8

B

VCE

7 6 5 4

VBE

3 2

1

E

0

1.0

2.0

3.0

4.0

5.0 VCE [V]

Slide 7

Saturation Region I C [mA]

14 13

• VBE Forward-biased • VBC Reverse-biased • No Current Gain relationship.

I

12 11 10

B

= 120µA

I

B

= 100µA

I

B

= 80µA

I

B

= 60µA

I

B

= 40µA

I

B

= 20µA

9

8 7 6 5 4 3 2

1 0

1.0

2.0

3.0

4.0

5.0 VCE [V]

Slide 8

Hand Analysis Model IB

IC IB = IS (eVbe/Vt -1)

IC (active) = βF IB

VBE VBE(on) = 0.7 V

VCE VCE(sat) = 0.1 V

Slide 9

Hand Analysis Example 5V RC RB=20kΩ 2.7V

IC

IB

IB = IC = VCE = Operation Mode:

βF=100 + -

• RC = 300Ω

IE

• RC = 1k Ω IB = IC = VCE = Operation Mode: Slide 10

BJT Parasitics • Junction Capacitances: QR

QF

CCS Cbc

Cbe

S

– Base-Emmitter – Base-Collector • Excess Base Charge:

– QR when – VBC forward-biased – Must be removed to switch modes Slide 11

NPN Transistor Doping Levels Depletion layers n

N

D

p

N 20 -3 10 cm E

2

n

A

17 10 B

2 10

15 -3 10 cm C

N

D

5

-3 2 10 cm

3

Slide 12

np Junctions Revisted • Forward-biased:

– Dominant current: diffusion of majority carriers • Reverse-biased:

– Drift of minority carriers

Slide 13

NPN Forward Active Polarization VBE > 0 & VBC < 0 forwardreversebiased biased Concentration of minority carriers

E

-

n

VBE

p

C

+

n B

+ -

VBC Slide 14

NPN Forward Active Polarization Current determined by concentration of majority carriers

Current determined by concentration of minority carriers

VBE > 0 & VBC < 0 forwardreversebiased biased

E

C

-

n

VBE

p

+

n B

+ -

VBC Slide 15

NPN Forward Active Polarization Concentration proportional to

VBE > 0 & VBC < 0

VBE e VT

forwardbiased

E

C

-

n

VBE

p

+

n B

+ -

VBC Slide 16

NPN Forward Active Polarization VBE > 0 & VBC < 0

Concentration proportional to

reversebiased

VBC e VT

E

C

-

n

VBE

p

+

n B

+ -

VBC Slide 17

NPN Forward Active Polarization VBE > 0 & VBC < 0 n E

p B

n C

Collector current: IC = IS [exp(VBE/VT) -1]

Collector current determined by the slope of the concentration of minority carriers (electrons) in the base.

IS - saturation current Slide 18

NPN Transistor Reverse Active VBE < 0 & VBC > 0

n

E

reversebiased

p

B

βF > βR

forwardbiased

n

C Slide 19

NPN Transistor Saturation VBE > 0 & VBC > 0 forwardbiased n

E

forwardbiased p

B

n

C Slide 20

NPN_Cut-off VBE < 0 & VBC < 0 n E

p B

n C

Slide 21

nn

Ebers-Moll Model C

C

α F I DE

IDC

B

B

I DE

α R IDC

E E Slide 22

Ebers - Moll Model Equations: IDE = IES [exp(VBE/VT) -1] IDC = ICS [exp(VBC/VT) -1] Typical values: αF = .99 IES = 10-15 A αR = .66 ICS = 10-15 A

Slide 23

BJT Inverter & Fan-Out Analysis • BJT Inverter • Voltage Transfer Characteristics • Logic Level Description • Fan-Out Analysis

Slide 24

Base diffusion VCC

BJT Inverter Vout

VCC = 5 V RC 1 kΩ

100 Ω/ 10 = 1kΩ

Vin

RB

Vout

10 k Ω

Vin = 5 V Vin = 0 V

Slide 25

NPN BJT Parameters VBE(on) = 0.7 V VBE(sat) = 0.8 V VCE(sat) = 0.1 V Forward active mode: • current gain βF = 70

Slide 26

VOH and VIL VEB = 0 V IC = 0 IB = 0 VCB = +5 V

VCC = 5 V

I RB

Vin = 0 V I

B

10 k Ω

RC 1 kΩ C

Vout = 5 V Q0

cut-off

n E

p B

n C

VEB = 0 - 0.7 V Cut-off

Slide 27

VOH and VIL Cut-off

Vout [V] VOH 5

BP1

4 3 2 1 Vin [V] 0 0

VIL1

0.7 V

2

3

4

5 Slide 28

VOL βF = 70 in Forward Active

VEB = 0.7 V I B = (5-0.7)/10kΩ = 0.43 mA I C = IB βF = 30 mA Vout = 5 - IC RC = -25 V ?

VCC = 5 V

I RB

Vin = 5 V I

B

10 k Ω

RC 1 kΩ C

NOT Forward Active !

Vout = 0.1V

I C < IB βF = 30 mA βF < 70

Q0

Saturation n E

Vout = 0.1 V

p B

n C

Slide 29

VOL Cut-off

Vout [V] VOH 5

BP1

4 3 2

Saturation

1

0.1 V

Vin [V]

VOL 0 0

VIL 1

2

3

4

5

0.7 V Slide 30

VIH Edge of saturation: VCC = 5 V

I RB

Vin = ? I

B

10 k Ω

RC 1 kΩ

βF = 70 V = 0.8 V EBs

V

out

= 0.1V

C

Q0

Vout = 0.1V

I B = ((5-0.1)/1kΩ)/70 = 70 µA V = VEBs + I B R B = 0.8 + 0.7 Vin = 1.5 V Slide 31

VIH Vout [V] VOH 5

Cut-off BP1

4 3 2

Saturation

1

0.1 V

Vin [V]

BP2

VOL 0 0

VIL1 VIH 2

0.7 V 1.5 V

3

4

5 Slide 32

Transition Region VCC = 5 V RC 1 kΩ V in = 0.7 - 1.5V RB 10 k Ω

V out = 5.0 - 0.1 V Q0

Forward Active !

IB = (Vin - 0.7)/R B I C = IB βF Vout = VCC - IC RC Slide 33

Transition Region: Load Line Analysis I C [mA]

14 13 12 11 10

I

B

= 120µA

I

B

= 100µA

I

B

= 80µA

I

B

= 60µA

IB = (Vin - 0.7)/R B I C = IB βF Vout = VCC - IC RC

9

8 7 6

IB = (1.5 - 0.7)/ 10kΩ = 80 µΑ Vin = 0.7 - 1.5V

5 4

I

B

= 40µA

I

B

= 20µA

3 2

1 0

1.0

2.0

3.0

4.0

5.0 VCE [V] Slide 34

Voltage Transfer Characteristic Cut-off

Vout [V] VOH 5

BP1

4

Forward Active 3 2

Saturation

1

Vin [V]

BP2

VOL 0 0

VIL

1

VIH

2

3

4

5 Slide 35

Voltage Transfer Characteristic Vout [V] VOH 5

• Transition Width: TW= VIH - VIL = .8 V

BP1

4

• High Noise Margin: NMH = VOH - VIH = 3.5 V

3 2 1 VOL

Vin [V]

BP2

0 0

VIL

1

VIH

2

3

4

• Low Noise Margin: NML = VIL -VOL = .6 V

5

• Logic Swing: LS = VOH - VOL = 4.9 V

Slide 36

Inverter Fan-Out VCC = 5 V RC 1 kΩ

Vin

RB 10 k Ω

Vin = 0 RB

n

Q0

RC 1 kΩ V out RB 10 k Ω

Q1 RC 1 kΩ RB 10 k Ω

QN

Slide 37

Inverter Fan-Out • LOAD =1 VOH = 4.6 V

VCC = 5 V

RC 1 kΩ

V in= 0.1 V

RB 10 k Ω

Vout = ( V

CC

RC 1 kΩ

Vout = 4.6 V RB

Q0 cut-off 10 k Ω

- 0.8 )

( 5.0 - 0.8 )

RB R

B

10

+RC

10

+1

Q1 saturation

+ 0.8 = + 0.8 = 4.61 V Slide 38

Equivalent Circuit VCC = 5 V

VOH = ?

RC 1 kΩ RB

Vin

10 k Ω

Vout = ( V

CC

- 0.8 )

V out Q0 RB/N

RB / N R /N B

( 5.0 - 0.8 )

VBE(sat)

+RC

10 /10 10 /10 + 1

+ 0.8 =

+ 0.8 = 2.9 V Slide 39

Maximum Number of Load Gates: VOH = VIH VCC = 5 V

VOH = ?

RC 1 kΩ

VIH = ? N=?

RB

Vin

Vout = ( V

V out Q0

10 k Ω

CC

- 0.8 )

RB / N R /N B

( 5.0 - 0.8 )

10 /N 10 /N + 1

+R C

RB/N

VBE(sat)

+ 0.8

+ 0.8 = 1.5 V Slide 40

Transistor-Transistor Logic (TTL) • Disadvantages of TTL gates: • Large # of components, area --> VLSI not feasible • Large power consumption • Saturated transistors in either high or low state • Large propagation times • Advantage: • Can drive large capacitive loads since large output currents available

Slide 41

TTL Inverter V R1 4 kΩ

Q A

= 5CC V R3 1.6 k Ω

R4 130 Ω Q4

1

Q2

D

1

R2 1 kΩ

INPUT STAGE

PHASE-SPLITTER or LEVEL-SHIFT

Q

3

OUTPUT STAGE

Slide 42

TTL NAND Gate V R1 4 kΩ

= 5CC V R3 1.6 k Ω

R4 130 Ω Q4

Q A B

Q2

1A

D

1

Q

1B

R2 1 kΩ

INPUT STAGE

PHASE-SPLITTER or LEVEL-SHIFT

Q

3

OUTPUT STAGE

Slide 43

Multi-Emitter Transistor B

E1

E1 E2 B

C

C p

E2

n

B E1

C

E2 Slide 44