TRANSISTOR

BJT :DC BIASING

Transistor Currents

Emitter current (IE) is the sum of the collector current (IC) and the base current (IB) .

Kirchhoff’s current law;

BCE III …(Eq. 3.1)

Collector Current (IC) Collector current (IC) comprises two components;

majority carriers (electrons) from the emitter

minority carriers (holes) from reverse-biased BC junction → leakage current, ICBO

Total collector current (IC);

Since leakage current ICBO is usually so small that it can be ignored.

EC IImajority

CBOC II minority

CBOEC III …(Eq. 3.2)

Collector Current (IC) Then;

The ratio of IC to IE is called alpha (α), values typically range from 0.95 to 0.99.

E

C

I

I …(Eq. 3.3)

Base Current (IB) IB is very small compared to IC;

The ratio of IC to IB is the dc current gain of a transistor, called beta (β)

The level of beta typically ranges from about 50 to over 400

B

C

I

I …(Eq. 3.4)

Current & Voltage Analysis Consider below figure. Three dc currents and three dc voltages can be

identified

IB: dc base current

IE: dc emitter current

IC: dc collector current

VBE: dc voltage across base-emitter junction

VCB: dc voltage across collector-base junction

VCE: dc voltage from collector to emitter Transistor bias circuit.

Current & Voltage Analysis When the BE junction is forward-biased, it is like a forward-

biased diode. Thus; (Si = 0.7, Ge = 0.3)

From HVK, the voltage across RB is

By Ohm’s law;

Solving for IB

V7.0VBE

BEBBR VVVB

BBR RIVB

B

BEBBB R

VVI

…(Eq. 3.5)

…(Eq. 3.6)

Current & Voltage Analysis The voltage at the collector is;

The voltage drop across RC is

VCE can be rewritten as

The voltage across the reverse-biased CB junction is

CCR RIVC

CRCCCE VVV

CCCCCE RIVV

BECECB VVV …(Eq. 3.8)

…(Eq. 3.7)

Transistor as Amplifier Transistor is capable to amplify

AC signal : (output signal > input signal)

Eg: Audio amplifier that amplify the sound of a radio

Transistor Amplifier Circuit Analysis There are 2 analysis;

DC Analysis AC Analysis

Transistor will operate when DC voltage source is applied to the amplifier circuit

Q-point must be determined so that the transistor will operate in active region (can operate as an amplifier)

Transistor Amplifier Circuit Analysis Q-Point

Operating point of an amplifier to state the values of collector current (ICQ) and collector-emitter voltage (VCEQ).

Determined by using transistor output characteristic and DC load line

Q-Point

DC LOAD LINE DC Load Line

A straight line intersecting the vertical axis at approximately IC(sat) and the horizontal axis at VCE(off).

IC(sat) occurs when transistor operating in saturation region

VCE(off) occurs when transistor operating in cut-off region

DC Load Line

Cutoff Region

Saturation Region

Q-Point

0

CE

sat

VC

CCC R

VI

0)(

Coff ICCCCCE RIVV

DC LOAD LINE (Example)VCC = 8V

RB = 360 kΩRC = 2 kΩ

Draw DC Load Line and Find Q-point.Answers;

mAk

V

R

VI

CE

sat

VC

CCC 4

2

8

0

0)(

Coff ICCCCCE RIVV

VVV CCCE off8

)(

DC LOAD LINE (Example)

Draw DC Load Line and Find Q-point.Answers;

Q-point can be obtained by calculate the half values of maximum IC and VCE

4V

2 mA

DC Analysis of Amplifier Circuit

Amplifier Circuit Amplifier Circuit w/o capacitor

DC Analysis of Amplifier Circuit Refer to the figure, for DC analysis:

Replace capacitor with an open-circuit

R1 and R2 create a voltage-divider circuit that connect to the base

Therefore, from DC analysis, you can find: IC VCE

Amplifier Circuit w/o capacitor

DC Analysis of Amplifier Circuit

Amplifier Circuit w/o capacitor Simplified Circuit

Thevenin Theorem;

DC Analysis of Amplifier Circuit Important equation for DC Analysis

1

2

21

2121 //

RR

RRRRRTH

CCTH VRR

RV

21

2

BCETH

BETHB II

RR

VVI

;)1(

)( ECCCCCE RRIVV

1

2

From HVK;

From Thevenin Theorem;

TRANSISTOR

BJT BIASING CIRCUIT

BJT BIASING CIRCUIT Fixed Base Bias Circuit

(Litar Pincangan Tetap)

Fixed Bias with Emitter Resistor Circuit(Litar Pincangan Pemancar Terstabil)

Voltage-Divider Bias Circuit(Litar Pincangan Pembahagi Voltan)

Feedback Bias Circuit(Litar Pincangan Suap-Balik Voltan)

FIXED BASE BIAS CIRCUIT

This is common emitter (CE) configuration

Solve the circuit using HVK 1st step: Locate capacitors

and replace them with an open circuit

2nd step: Locate 2 main loops which; BE loop CE loop

FIXED BASE BIAS CIRCUIT 1st step: Locate capacitors and replace them with an open circuit

FIXED BASE BIAS CIRCUIT 2nd step: Locate 2 main loops.

12

1

2

BE Loop CE Loop

FIXED BASE BIAS CIRCUIT BE Loop Analysis

1

From HVK;

IBB

BECCB

BEBBCC

R

VVI

0VRIV

A

FIXED BASE BIAS CIRCUIT CE Loop Analysis

From HVK;

As we known;

Subtituting with

BC II 2

IC CCCCCE

CECCCC

RIVV

0VRIV

B

A B

B

BECCDCC R

VVI

FIXED BASE BIAS CIRCUIT DISADVANTAGE

Unstable – because it is too dependent on β and produce width change of Q-point

For improved bias stability , add emitter resistor to dc bias.

FIXED BASE BIAS CIRCUIT Example 1 Find IC, IB, VCE, VB,

VC, VBC? (Silikon transistor);

Answers;IC = 2.35 mAIB = 47.08 μAVCE = 6.83VVB = 0.7VVC = 6.83VVBC = -6.13V

FIXED BIAS WITH EMITTER RESISTOR An emitter resistor, RE is added

to improve stability Solve the circuit using HVK 1st step: Locate capacitors and

replace them with an open circuit

2nd step: Locate 2 main loops which; BE loop CE loop

Resistor, RE added

FIXED BIAS WITH EMITTER RESISTOR 1st step: Locate capacitors and replace them with an open circuit

FIXED BIAS WITH EMITTER RESISTOR 2nd step: Locate 2 main loops.

12

2

BE Loop CE Loop

1

FIXED BIAS WITH EMITTER RESISTOR BE Loop Analysis

From HVK;

Recall;

Subtitute for IE

0 EEBEBBCC RIVRIV

EB

BECCB

EBBEBBCC

RR

VVI

RIVRIV

)1(

0)1(

BE II )1(

1

FIXED BIAS WITH EMITTER RESISTOR CE Loop Analysis

From HVK;

Assume;

Therefore;

CE II

0 EECECCCC RIVRIV

2

)( ECCCCCE RRIVV

FIXED BIAS WITH EMITTER RESISTOR Example 2 Find IC, IB, VCE, VB,

VC, VE & VBC? (Silikon transistor);

Answers;IC = 2.01 mAIB = 40.1 μAVCE = 13.97VVB = 2.71VVE = 2.01VVC = 15.98VVBC = -13.27V

VOLTAGE DIVIDER BIAS CIRCUIT Provides good Q-point stability

with a single polarity supply voltage Solve the circuit using HVK 1st step: Locate capacitors and

replace them with an open circuit 2nd step: Simplified circuit using

Thevenin Theorem 3rd step: Locate 2 main loops

which; BE loop CE loop

VOLTAGE DIVIDER BIAS CIRCUIT 1st step: Locate capacitors and replace them with an open circuit

VOLTAGE DIVIDER BIAS CIRCUIT

Simplified Circuit

Thevenin Theorem;

2nd step: : Simplified circuit using Thevenin Theorem

21

2121 //

RR

RRRRRTH

CCTH VRR

RV

21

2

From Thevenin Theorem;

VOLTAGE DIVIDER BIAS CIRCUIT 2nd step: Locate 2 main loops.

1

2

BE Loop CE Loop

1

2

VOLTAGE DIVIDER BIAS CIRCUIT BE Loop Analysis

From HVK;

Recall;

Subtitute for IE

0 EEBETHBTH RIVRIV

ERTH

BETHB

EBBETHBTH

RR

VVI

RIVRIV

)1(

0)1(

BE II )1(

1

VOLTAGE DIVIDER BIAS CIRCUIT CE Loop Analysis

From HVK;

Assume;

Therefore;

CE II

0 EECECCCC RIVRIV

)( ECCCCCE RRIVV

2

VOLTAGE DIVIDER BIAS CIRCUIT Example 3 Find RTH, VTH, IC, IB, VCE,

VB, VC, VE & VBC? (Silikon transistor);

Answers;RTH = 3.55 kΩVTH = 2VIC = 0.85 mAIB = 6.05 μAVCE = 12.22VVB = 1.978VVE = 1.275VVC = 13.5VVBC = -11.522V