Electronic Circuits for Mechatronics ELCT 609 (Spring 2019 ...eee.guc.edu.eg/Courses/Electronics/ELCT609 Elecronic Circuits...BJT I-V characteristics 3. NPN Transistor in Reverse Active

Electronic Circuits for

Mechatronics ELCT 609

(Spring 2019)

Lecture 4: Bi-polar Junction

Transistor Physical StructureDr. Eman Azab

Assistant Professor

Office: C3.315

E-mail: [email protected]

Dr. Eman Azab

Electronics Dept., Faculty of IET

The German University in Cairo

1

Introduction Why we call it Transistor?

The name came as an abbreviation of the device job!

Dr. Eman Azab



2

Input Voltage

Scaled with a predefined Trans-conductance gain

Output Current

Transistor Function

Bipolar Junction

Transistor (BJT)Physical Structure and I-V Characteristics

Dr. Eman Azab



3

BJT Physical Structure Two back to back PN Junctions

NPN or PNP Transistor

Three terminal device, for NPN: Base (P-Type), Emitter and

Collector (N-Type)

Emitter-Base Junction (EBJ)

Collector-Base Junction (CBJ)

Emitter doping is higher than Collector doping

Dr. Eman Azab



4

NPN BJT Transistor PNP BJT Transistor

BJT I-V characteristics1. BJT NPN Transistor in Cutoff Mode

Emitter-Base is Reverse biased

Collector-Base Junction is Reverse biased

Dr. Eman Azab



5

EBJ Reverse

CBJ Reverse

Cutoff Mode C B EI I I 0

Q

BJT I-V characteristics2. BJT NPN Transistor in Active Mode

EBJ is Forward biased and electrons pass through the Base

to the collector due to the Base small area

Electrons from the Emitter are collected at the Collector

side

The transistor’s Collector current can be modeled by a

current dependent current source

Is depends on doping and width of the Base

Dr. Eman Azab



6

C F BI I

BEC s

T

VI I exp

V

Q

BJT I-V characteristics3. NPN Transistor in Reverse Active mode

Collector-Base Junction is Forward biased

Emitter-Base Junction is reverse biased

Emitter and collector reverse their roles

However, BJT has an asymmetrical physical structure

The current gain from Base to Emitter is very small

Dr. Eman Azab



7

E R BI I R F

EBJ Reverse

CBJ Forward

Reverse Active Mode

Q

𝑉𝐵𝐶 ≅ 0.5𝑉

BJT I-V characteristics4. NPN Transistor in Saturation Mode

Both junctions are Forward biased

The total current is the EBJ diffusion current subtracted from

CBJ diffusion current

BJT could be used as a closed switch in Saturation mode

Dr. Eman Azab



8

BEJ Forward

BCJ Forward

Saturation Mode

𝑉𝐵𝐸 ≅ 0.7𝑉

𝑉𝐵𝐶 ≅ 0.5𝑉C F BI I

Q

𝑉𝐶𝐸 ≅ 0.2𝑉

BJT NPN I-V Characteristics IC versus VBE and VCE

Dr. Eman Azab



9

IC = Is expVBEVT

For Active

ONLY

Q

BJT NPN I-V Characteristics IC versus VCE

The Early effect

Dr. Eman Azab



10

CEBEC S

T A

vVi I exp( ) 1

V V

C Ao

CE C

i Vr

v I

Q

BJT Large Signal Model in

Active Mode

Dr. Eman Azab



11

Q

BJT PNP Physical Structure BJT PNP Transistor

PNP is the NPN Complementary

structure

Two back to back PN Junctions

Three terminal device: Base (N-

type), Emitter and Collector (P-type)

Emitter-Base Junction

Collector-Base Junction

Emitter doping is higher than

Collector doping

Same I-V Characteristics as NPN

Transistor

Dr. Eman Azab



12

Q

PNP BJT Transistor

BJT PNP Physical Structure BJT PNP Transistor in Active Mode

Dr. Eman Azab



13

EBJ Forward

CBJ Reverse

Active Mode

Q

BJT Modes of OperationElectrical Equations of BJT

Dr. Eman Azab



14

BJT NPN Modes of Operation

Mode BEJ BCJ Equations Condition

Cutoff Reverse Reverse 𝐼𝐶 = 𝐼𝐸 = 𝐼𝐵=0𝑉𝐵𝐸 < 0.7𝑉𝐵𝐶 < 0.5

Active

(Forward)

Forward

Reverse

𝑉𝐵𝐸 = 0.7𝐼𝐸 = 𝐼𝐶 + 𝐼𝐵

𝐼𝐶 = 𝛽𝐹𝐼𝐵 = 𝛼𝐹𝐼𝐸

𝛼𝐹 =𝛽𝐹

1 + 𝛽𝐹

𝑉𝐵𝐶 < 0.5Or

𝑉𝐶𝐸 > 0.2

Saturation

Forward

Forward

𝑉𝐵𝐸 = 0.7𝑉𝐵𝐶 = 0.5𝑉𝐶𝐸 = 0.2𝐼𝐸 = 𝐼𝐶 + 𝐼𝐵

𝐼𝐶 < 𝛽𝐹𝐼𝐵

Reverse Active

Reverse

Forward

𝑉𝐵𝐶 = 0.5𝐼𝐶 = 𝐼𝐸 + 𝐼𝐵

𝐼𝐸 = 𝛽𝑅𝐼𝐵 = 𝛼𝑅𝐼𝐶

𝛼𝑅 =𝛽𝑅

1 + 𝛽𝑅

𝑉𝐵𝐸 < 0.7

Dr. Eman Azab



15

Q

Q

BJT PNP Modes of Operation

Mode EBJ CBJ Equations Condition

Cutoff Reverse Reverse 𝐼𝐶 = 𝐼𝐸 = 𝐼𝐵=0𝑉𝐸𝐵 < 0.7𝑉𝐶𝐵 < 0.5

Active

(Forward)

Forward

Reverse

𝑉𝐸𝐵 = 0.7𝐼𝐸 = 𝐼𝐶 + 𝐼𝐵

𝐼𝐶 = 𝛽𝐹𝐼𝐵 = 𝛼𝐹𝐼𝐸

𝛼𝐹 =𝛽𝐹

1 + 𝛽𝐹

𝑉𝐶𝐵 < 0.5Or

𝑉𝐸𝐶 > 0.2

Saturation

Forward

Forward

𝑉𝐸𝐵 = 0.7𝑉𝐶𝐵 = 0.5𝑉𝐸𝐶 = 0.2𝐼𝐸 = 𝐼𝐶 + 𝐼𝐵

𝐼𝐶 < 𝛽𝐹𝐼𝐵

Reverse Active

Reverse

Forward

𝑉𝐶𝐵 = 0.5𝐼𝐶 = 𝐼𝐸 + 𝐼𝐵

𝐼𝐸 = 𝛽𝑅𝐼𝐵 = 𝛼𝑅𝐼𝐶

𝛼𝑅 =𝛽𝑅

1 + 𝛽𝑅

𝑉𝐸𝐵 < 0.7

Dr. Eman Azab



16

Q

Q

Calculating DC operating

pointSolved Exercise

Dr. Eman Azab



17

Solved Example Find the DC Operating point of the Transistors?

Given: VBE=0.7V ,β=10

(Ans.: IB=0.023mA, IC=0.23mA, IE=0.253mA, VCE=9.54V, Active )

Dr. Eman Azab



18

Solution Steps:1. Identify the BJT Type

2. Place the terminals name on the circuit

3. Write a KVL in the INPUT Loop

Input loop for BJT is any loop containing VBE

4. Assume the BJT mode (most of the time active)

5. Calculate the currents and voltages

6. Write KVL in the OUTPUT loop

Output loop for BJT is any loop containing VCE

7. Verify your assumption!

Dr. Eman Azab



19

Example Find the DC Operating point of the Transistors?

Given: VEB=0.7V ,β=10

(Ans.: IB=93µA, IC=0.93mA, IE=1.023mA, VEC=8.14V, Active )

Dr. Eman Azab



20

Documents

Electronic Circuits for Mechatronics ELCT 609 (Spring 2019 ...eee.guc.edu.eg/Courses/Electronics/ELCT609 Elecronic Circuits...BJT I-V characteristics 3. NPN Transistor in Reverse Active