Shri Sant Gajanan Maharaj College of Engineering Department

  • Slides: 94
Download presentation
Shri Sant Gajanan Maharaj College of Engineering Department of Electronics & Telecommunications Engineering Electronic

Shri Sant Gajanan Maharaj College of Engineering Department of Electronics & Telecommunications Engineering Electronic Devices & Components Unit – VI Bipolar Junction Transistor

Unit VI Sub-Topics Theory of PNP and NPN Transistor, Transistor Configurations, Their Characteristics and

Unit VI Sub-Topics Theory of PNP and NPN Transistor, Transistor Configurations, Their Characteristics and current components. Transistor as an amplifier, Testing of Transistor using ohmmeter and CRO

Bipolar Junction Transistor

Bipolar Junction Transistor

Bipolar Junction Transistor

Bipolar Junction Transistor

Bipolar Junction Transistor

Bipolar Junction Transistor

Bipolar Junction Transistor

Bipolar Junction Transistor

Bipolar Junction Transistor After rectifier diode, LED is most widely used diode

Bipolar Junction Transistor After rectifier diode, LED is most widely used diode

Bipolar Junction Transistor After rectifier diode, LED is most widely used diode

Bipolar Junction Transistor After rectifier diode, LED is most widely used diode

Bipolar Junction Transistor After rectifier diode, LED is most widely used diode

Bipolar Junction Transistor After rectifier diode, LED is most widely used diode

Bipolar Junction Transistor After rectifier diode, LED is most widely used diode

Bipolar Junction Transistor After rectifier diode, LED is most widely used diode

Bipolar Junction Transistor After rectifier diode, LED is most widely used diode

Bipolar Junction Transistor After rectifier diode, LED is most widely used diode

Bipolar Junction Transistor After rectifier diode, LED is most widely used diode

Bipolar Junction Transistor After rectifier diode, LED is most widely used diode

Bipolar Junction Transistor After rectifier diode, LED is most widely used diode

Bipolar Junction Transistor After rectifier diode, LED is most widely used diode

Bipolar Junction Transistor After rectifier diode, LED is most widely used diode

Bipolar Junction Transistor After rectifier diode, LED is most widely used diode

What is a Transistor? • Semiconductors: ability to change from conductor to insulator •

What is a Transistor? • Semiconductors: ability to change from conductor to insulator • Can either allow current or prohibit current to flow • Useful as a switch, but also as an amplifier • Essential part of many technological advances

Bipolar Junction Transistor (BJT) • 3 adjacent regions of doped Si (each connected to

Bipolar Junction Transistor (BJT) • 3 adjacent regions of doped Si (each connected to a lead): – Base. (thin layer, less doped). – Collector. – Emitter. • 2 types of BJT: npn bipolar junction transistor – npn. – pnp. • Most common: npn (focus on it). Developed by Shockley (1949) pnp bipolar junction transistor

TWO PN JUNCTION DEVICE EMITTER P TYPE BASE N TYPE COLLECTOR P TYPE

TWO PN JUNCTION DEVICE EMITTER P TYPE BASE N TYPE COLLECTOR P TYPE

ACTUAL SCALED DIMENSIONS EMITTER BASE COLLECTOR N P P

ACTUAL SCALED DIMENSIONS EMITTER BASE COLLECTOR N P P

 • Construction – two polarities: npn and pnp

• Construction – two polarities: npn and pnp

BJT A BJT (Bipolar Junction Transistor) transistor has inside two similar semi conductive materials,

BJT A BJT (Bipolar Junction Transistor) transistor has inside two similar semi conductive materials, and between them there is a third semi conductive material of different type if the two similar materials are P and the middle one is N, then we have a P-N-P or PNP transistor. if the two materials are N and the middle one is P, then we have a N-P-N material or NPN Each transistor has 3 leads which we call base, collector and emitter, and we use the symbols b, c and e respectively The symbol of the transistor has an arrow on the emitter. If the transistor is a PNP, then the arrow points to the base of the transistor, otherwise it points to the output.

Types Of BJT Two basic types of bipolar junction transistor construction, 1. PNP 2.

Types Of BJT Two basic types of bipolar junction transistor construction, 1. PNP 2. NPN, which basically describes the physical arrangement of the P-type and N-type semiconductor materials n-type semiconductors the impurities result in an excess of electrons, or negative charges p-type semiconductors the material lead to a deficiency of electrons and therefore an excess of positive charge carriers or “holes. ” current regulating devices that control the amount of current flowing through them

NPN and PNP Transistor

NPN and PNP Transistor

BJT Basic construction The BJT is a three terminal device that produce two PN

BJT Basic construction The BJT is a three terminal device that produce two PN junctions Emitter ( E ) Base ( B ) Collector ( C ) Principle of operation of the two transistor types PNP and NPN Collector Biasing Base polarity of the power supply for each type Emitter

EMITTER • It is highly doped • To inject a large number of charge

EMITTER • It is highly doped • To inject a large number of charge carriers • Main function is to supply the majority carriers to the base. • It is always forward biased with respect to base

BASE • It is a middle section of a transistor • It is lightly

BASE • It is a middle section of a transistor • It is lightly doped • So that most of the charge carriers pass to the collector. • It controls flow of charges • It forms two PN junctions with Emitter and Collector

COLLECTOR • • • IT IS situated opposite to the emitter It is always

COLLECTOR • • • IT IS situated opposite to the emitter It is always reversed biased So that it can collect the majority carriers Size of collector is larger than emitter Its doping level is in the middle of base and emitter

NPN TRANSISTOR • It is constructed by two N type and One P type

NPN TRANSISTOR • It is constructed by two N type and One P type material. • Emitter and collector are of N type material • Base is of P type material • It consist of two PN junctions Sandwiching a P-type layer between two ntype layers.

OPERATION OF NPN TRANSISTOR • The base-emitter diode is forward biased • The base-collector

OPERATION OF NPN TRANSISTOR • The base-emitter diode is forward biased • The base-collector diode is reverse biased • VBE injects the electron to the Emitter. • Emitter is highly doped • Base is lightly doped • Collector creates electrostatic field which Attracts the electrons • 95 to 99% electrons diffuses in collector region C backward VCB B E Forward VBE

Transistor operation With no power applied to the transistor areas There are two depletion

Transistor operation With no power applied to the transistor areas There are two depletion zones between the two P-N contacts. Power source Connected b/w base and collector in reverse-bias With the positive of the source connected to the collector and the negative to the base. The depletion zone of the P-N contact between the base and the collector will be widened. A slight current will flow within this contact (due to impurities). This current is the reverse contact current symbol ICBO

Transistor operation Voltage supply between the emitter and the base in forward bias With

Transistor operation Voltage supply between the emitter and the base in forward bias With the positive of the source connected to the base and the negative connected to the emitter. The depletion zone between the emitter and the base will be shortened current (electrons) will flow when the voltage exceeds a specific level. This level depends on the material that the transistor is made of. Some of the electrons that go through the e-b depletion zone, will re-connect with holes in the base. This is the base current IB symbol for reference. In real life, this current is at the scale of micro-amperes (μA ):

Cont’d Most of the electrons will flow through the base (due to spilling) and

Cont’d Most of the electrons will flow through the base (due to spilling) and will be directed to the collector. When these electrons reach the depletion area between the base and the collector, they will experience a force from the electric field which exists in this zone, The electrons will pass through the depletion zone. The electrons will then re-connect with holes in the collector. The re-connected holes will be replaced with holes coming from the base-collector power supply (VCC). The movement of these holes equals to a movement of electrons in the opposite direction, from the collector to the supply. In other words, the current that flows to the emitter will be divided into the small base current and the larger collector current: IE = IB + IC

Cont’d Generally, the number of electrons that arrive at the collector is the 99%

Cont’d Generally, the number of electrons that arrive at the collector is the 99% of the total electrons, and the rest 1% causes the base current. At the collector, except the electrons that come from the emitter, there is also the reverse current from the base-collector contact Both currents flow at the same direction, so they are added IC' = IC + ICBO

 • Relationship between the collector current and the base current in a bipolar

• Relationship between the collector current and the base current in a bipolar transistor – characteristic is approximately linear – magnitude of collector current is generally many times that of the base current – the device provides current gain

Comparison of CB CE and CC configuration S. No Characteristics CB CE CC 1

Comparison of CB CE and CC configuration S. No Characteristics CB CE CC 1 Input impedance Low Medium High 2 Output impedance High medium low 3 Current gain Low High 4 Voltage gain High Unity 5 Power gain Medium High Low 6 Phase reversal No Yes No 7 application AF amplifiers Voltage & power amplifiers Impedance matching Department of ECE / EEE 34

BJT in Active Region Common Emitter(CE) Connection • Called CE because emitter is common

BJT in Active Region Common Emitter(CE) Connection • Called CE because emitter is common to both VBB and VCC 35

BJT in Active Region (2) • Base Emitter junction is forward biased • Base

BJT in Active Region (2) • Base Emitter junction is forward biased • Base Collector junction is reverse biased • For a particular i. B, i. C is independent of RCC transistor is acting as current controlled current source (i. C is controlled by i. B, and i. C = i. B) • Since the base emitter junction is forward biased, from Shockley equation 36

BJT in Active Region (3) • Normally the above equation is never used to

BJT in Active Region (3) • Normally the above equation is never used to calculate i. C, i. B Since for all small signal transistors v. BE 0. 7. It is only useful for deriving the small signal characteristics of the BJT. • For example, for the CE connection, i. B can be simply calculated as, or by drawing load line on the base –emitter side 37

Deriving BJT Operating points in Active Region –An Example In the CE Transistor circuit

Deriving BJT Operating points in Active Region –An Example In the CE Transistor circuit shown earlier VBB= 5 V, RBB= 107. 5 k , RCC = 1 k , VCC = 10 V. Find IB, IC, VCE, and the transistor power dissipation using the characteristics as shown below By Applying KVL to the base emitter circuit i. B 100 A By using this equation along with the i. B / v. BE characteristics of the base emitter junction, IB = 40 A 0 5 V v. BE 38

Deriving BJT Operating points in Active Region –An Example (2) By Applying KVL to

Deriving BJT Operating points in Active Region –An Example (2) By Applying KVL to the collector emitter circuit i. C 10 m. A 100 A 80 A 60 A 40 A 20 A By using this equation along with the i. C / v. CE characteristics of the base collector junction, i. C = 4 m. A, VCE = 6 V 0 20 V v. CE Transistor power dissipation = VCEIC = 24 m. W We can also solve the problem without using the characteristics if and VBE values are known 39

BJT in Cutoff Region • Under this condition i. B= 0 • As a

BJT in Cutoff Region • Under this condition i. B= 0 • As a result i. C becomes negligibly small • Both base-emitter as well base-collector junctions may be reverse biased • Under this condition the BJT can be treated as an off switch 40

BJT in Saturation Region • Under this condition i. C / i. B in

BJT in Saturation Region • Under this condition i. C / i. B in active region • Both base emitter as well as base collector junctions are forward biased • VCE 0. 2 V • Under this condition the BJT can be treated as an on switch 41

BJT in Saturation Region (2) • A BJT can enter saturation in the following

BJT in Saturation Region (2) • A BJT can enter saturation in the following ways (refer to the CE circuit) • For a particular value of i. B, if we keep on increasing RCC • For a particular value of RCC, if we keep on increasing i. B • For a particular value of i. B, if we replace the transistor with one with higher 42

BJT in Saturation Region – Example 1 In the CE Transistor circuit shown earlier

BJT in Saturation Region – Example 1 In the CE Transistor circuit shown earlier VBB= 5 V, RBB= 107. 5 k , RCC = 10 k , VCC = 10 V. Find IB, IC, VCE, and the transistor power dissipation using the characteristics as shown below Here even though IB is still 40 A; from the output characteristics, IC can be found to be only about 1 m. A and VCE 0. 2 V( VBC 0. 5 V or base collector junction is forward biased (how? )) i. C 10 m. A 100 A 80 A 60 A 40 A 20 A = IC / IB = 1 m. A/40 A = 25 100 0 20 V v. CE 43

BJT in Saturation Region – Example 2 In the CE Transistor circuit shown earlier

BJT in Saturation Region – Example 2 In the CE Transistor circuit shown earlier VBB= 5 V, RBB= 43 k , RCC = 1 k , VCC = 10 V. Find IB, IC, VCE, and the transistor power dissipation using the characteristics as shown below Here IB is 100 A from the input characteristics; IC can be found to be only about 9. 5 m. A from the output characteristics and VCE 0. 5 V( VBC 0. 2 V or base collector junction is forward biased (how? )) = IC / IB = 9. 5 m. A/100 A = 95 100 Transistor power dissipation = VCEIC 4. 7 m. W Note: In this case the BJT is not in very hard saturation 44

BJT in Saturation Region – Example 2 (2) i. C i. B 10 m.

BJT in Saturation Region – Example 2 (2) i. C i. B 10 m. A 100 A 80 A 60 A 40 A 20 A 0 0 20 V v. CE 5 V v. BE Output Characteristics Input Characteristics 45

BJT in Saturation Region – Example 3 In the CE Transistor circuit shown earlier

BJT in Saturation Region – Example 3 In the CE Transistor circuit shown earlier VBB= 5 V, VBE = 0. 7 V RBB= 107. 5 k , RCC = 1 k , VCC = 10 V, = 400. Find IB, IC, VCE, and the transistor power dissipation using the characteristics as shown below By Applying KVL to the base emitter circuit Then IC = IB= 400*40 A = 16000 A and VCE = VCC-RCC* IC =10 - 0. 016*1000 = -6 V(? ) But VCE cannot become negative (since current can flow only from collector to emitter). Hence the transistor is in saturation 46

BJT in Saturation Region – Example 3(2) Hence V 0. 2 V CE IC

BJT in Saturation Region – Example 3(2) Hence V 0. 2 V CE IC = (10 – 0. 2) /1 = 9. 8 m. A Hence the operating = 9. 8 m. A / 40 A = 245 47

Biasing Effect as an amplifier

Biasing Effect as an amplifier

Biasing Effect as an amplifier Low value Vbe medium value Vbe High value Vbe

Biasing Effect as an amplifier Low value Vbe medium value Vbe High value Vbe

Inverting and Non-Inverting Output Input Non-Inverting Output

Inverting and Non-Inverting Output Input Non-Inverting Output

Transistor Biasing Collector N P IC Base N IE = I B + I

Transistor Biasing Collector N P IC Base N IE = I B + I C Emitter IB Forward bias IE Reverse bias

28 -5: Checking a Transistor with an Ohmmeter § To check the base-emitter junction

28 -5: Checking a Transistor with an Ohmmeter § To check the base-emitter junction of an npn transistor, first connect the ohmmeter as shown in Fig. 28 -9 (a) and then reverse the ohmmeter leads as shown in (b). § For a good p-n junction made of silicon, the ratio RR/RF should be equal to or greater than 1000: 1. Fig. 28 -9 Copyright © The Mc. Graw-Hill Companies, Inc. Permission required for reproduction or display.

28 -5: Checking a Transistor with an Ohmmeter § To check the collector-base junction,

28 -5: Checking a Transistor with an Ohmmeter § To check the collector-base junction, first connect the ohmmeter as shown in Fig. 28 -10 (a) and then reverse the ohmmeter leads as shown in (b). § For a good p-n junction made of silicon, the ratio RR/RF should be equal to or greater than 1000: 1. § Although not shown, the resistance measured between the collector and emitter should read high or infinite for both connections of the meter leads. Fig. 28 -10 Copyright © The Mc. Graw-Hill Companies, Inc. Permission required for reproduction or display.

Transistors as Switches

Transistors as Switches

Basic Transistor Switch Circuit • Transistor “switch” circuit: (Lab 4– 9) (BC junction forward

Basic Transistor Switch Circuit • Transistor “switch” circuit: (Lab 4– 9) (BC junction forward biased) VB 0. 6 V (The Art of Electronics, Horowitz and Hill, 2 nd Ed. ) 0. 2 V VC 0 V VE – With switch open, transistor is off and lamp is off – With switch closed, IB = (10 – 0. 6) V / 1 k = 9. 4 m. A – However, IC = IB 940 m. A (assuming = 100) • • When collector current IC = 100 m. A, lamp has 10 V across it To get a higher current, collector would need to be below ground Transistor can’t do this, so it goes into saturation Collector voltage gets as close to emitter voltage as it can (about 0. 2 V higher) and IC remains constant (IC is “maxed out”)

Transistor Operation • The basic operation will be described using the pnp transistor. The

Transistor Operation • The basic operation will be described using the pnp transistor. The operation of the pnp transistor is exactly the same if the roles played by the electron and hole are interchanged. • One p-n junction of a transistor is reverse-biased, whereas the other is forward-biased. Forward-biased junction of a pnp transistor Reverse-biased junction of a pnp transistor

 • Both biasing potentials have been applied to a pnp transistor and resulting

• Both biasing potentials have been applied to a pnp transistor and resulting majority and minority carrier flows indicated. • Majority carriers (+) will diffuse across the forwardbiased p-n junction into the n-type material. • A very small number of carriers (+) will through n-type material to the base terminal. Resulting IB is typically in order of microamperes. • The large number of majority carriers will diffuse across the reverse-biased junction into the p-type material connected to the collector terminal.

 • Majority carriers can cross the reverse-biased junction because the injected majority carriers

• Majority carriers can cross the reverse-biased junction because the injected majority carriers will appear as minority carriers in the n-type material. • Applying KCL to the transistor : IE = I C + I B • The comprises of two components – the majority and minority carriers IC = ICmajority + ICOminority • ICO – IC current with emitter terminal open and is called leakage current.

Common-Base Configuration • Common-base terminology is derived from the fact that the : -

Common-Base Configuration • Common-base terminology is derived from the fact that the : - base is common to both input and output of the configuration. - base is usually the terminal closest to or at ground potential. • All current directions will refer to conventional (hole) flow and the arrows in all electronic symbols have a direction defined by this convention. • Note that the applied biasing (voltage sources) are such as to establish current in the direction indicated for each branch.

 • To describe the behavior of common-base amplifiers requires two set of characteristics:

• To describe the behavior of common-base amplifiers requires two set of characteristics: - Input or driving point characteristics. - Output or collector characteristics • The output characteristics has 3 basic regions: - Active region –defined by the biasing arrangements - Cutoff region – region where the collector current is 0 A - Saturation region- region of the characteristics to the left of V CB = 0 V

 • The curves (output characteristics) clearly indicate that a first approximation to the

• The curves (output characteristics) clearly indicate that a first approximation to the relationship between IE and IC in the active region is given by IC ≈IE • Once a transistor is in the ‘on’ state, the base-emitter voltage will be assumed to be VBE = 0. 7 V

 • In the dc mode the level of IC and IE due to

• In the dc mode the level of IC and IE due to the majority carriers are related by a quantity called alpha = IC = IE + ICBO • It can then be summarize to IC = IE (ignore ICBO due to small value) • For ac situations where the point of operation moves on the characteristics curve, an ac alpha defined by • Alpha a common base current gain factor that shows the efficiency by calculating the current percent from current flow from emitter to collector. The value of is typical from 0. 9 ~ 0. 998.

Biasing • Proper biasing CB configuration in active region by approximation IC IE (IB

Biasing • Proper biasing CB configuration in active region by approximation IC IE (IB 0 u. A)

Transistor as an amplifier

Transistor as an amplifier

Simulation of transistor as an amplifier

Simulation of transistor as an amplifier

Common-Emitter Configuration • It is called common-emitter configuration since : - emitter is common

Common-Emitter Configuration • It is called common-emitter configuration since : - emitter is common or reference to both input and output terminals. - emitter is usually the terminal closest to or at ground potential. • Almost amplifier design is using connection of CE due to the high gain for current and voltage • Two set of characteristics are necessary to describe the behavior for CE ; input (base terminal) and output (collector terminal) parameters.

Proper Biasing common-emitter configuration in active region

Proper Biasing common-emitter configuration in active region

 • IB is microamperes compared to miliamperes of IC. • IB will flow

• IB is microamperes compared to miliamperes of IC. • IB will flow when VBE > 0. 7 V for silicon and 0. 3 V for germanium • Before this value IB is very small and no IB. • Base-emitter junction is forward bias • Increasing VCE will reduce IB for different values. Input characteristics for a common-emitter NPN transistor

Output characteristics for a common-emitter npn transistor • • • For small VCE (VCE

Output characteristics for a common-emitter npn transistor • • • For small VCE (VCE < VCESAT, IC increase linearly with increasing of VCE > VCESAT IC not totally depends on VCE constant IC IB(u. A) is very small compare to IC (m. A). Small increase in IB cause big increase in IC IB=0 A ICEO occur. Noticing the value when IC=0 A. There is still some value of current flows.

Beta ( ) or amplification factor • The ratio of dc collector current (IC)

Beta ( ) or amplification factor • The ratio of dc collector current (IC) to the dc base current (IB) is dc beta ( dc ) which is dc current gain where IC and IB are determined at a particular operating point, Qpoint (quiescent point). • It’s define by the following equation: 30 < dc < 300 2 N 3904 • On data sheet, dc=h. FE with h is derived from ac hybrid equivalent cct. FE are derived from forward-current amplification and common-emitter configuration respectively.

 • For ac conditions an ac beta has been defined as the changes

• For ac conditions an ac beta has been defined as the changes of collector current (IC) compared to the changes of base current (IB) where IC and IB are determined at operating point. • On data sheet, ac=hfe • It can defined by the following equation:

Example From output characteristics of common emitter configuration, find ac and dc with an

Example From output characteristics of common emitter configuration, find ac and dc with an Operating point at IB=25 A and VCE =7. 5 V.

Solution:

Solution:

Relationship analysis between α and β

Relationship analysis between α and β

Common – Collector Configuration • Also called emitter-follower (EF). • It is called common-emitter

Common – Collector Configuration • Also called emitter-follower (EF). • It is called common-emitter configuration since both the signal source and the load share the collector terminal as a common connection point. • The output voltage is obtained at emitter terminal. • The input characteristic of common-collector configuration is similar with common-emitter. configuration. • Common-collector circuit configuration is provided with the load resistor connected from emitter to ground. • It is used primarily for impedance-matching purpose since it has high input impedance and low output impedance.

Notation and symbols used with the common-collector configuration: (a) pnp transistor ; (b) npn

Notation and symbols used with the common-collector configuration: (a) pnp transistor ; (b) npn transistor.

 • For the common-collector configuration, the output characteristics are a plot of IE

• For the common-collector configuration, the output characteristics are a plot of IE vs VCE for a range of values of IB.

Limits of Operation • Many BJT transistor used as an amplifier. Thus it is

Limits of Operation • Many BJT transistor used as an amplifier. Thus it is important to notice the limits of operations. • At least 3 maximum values is mentioned in data sheet. • There are: a) Maximum power dissipation at collector: PCmax or PD b) Maximum collector-emitter voltage: VCEmax sometimes named as VBR(CEO) or VCEO. c) Maximum collector current: ICmax • There are few rules that need to be followed for BJT transistor used as an amplifier. The rules are: i) transistor need to be operate in active region! ii) IC < ICmax ii) PC < PCmax

Note: VCE is at maximum and IC is at minimum (ICmax=ICEO) in the cutoff

Note: VCE is at maximum and IC is at minimum (ICmax=ICEO) in the cutoff region. IC is at maximum and VCE is at minimum (VCE max = VCEsat = VCEO) in the saturation region. The transistor operates in the active region between saturation and cutoff.

Refer to the fig. Step 1: The maximum collector power dissipation, PD=ICmax x VCEmax

Refer to the fig. Step 1: The maximum collector power dissipation, PD=ICmax x VCEmax (1) = 18 m x 20 = 360 m. W Step 2: At any point on the characteristics the product of and must be equal to 360 m. W. Ex. 1. If choose ICmax= 5 m. A, subtitute into the (1), we get VCEmax. ICmax= 360 m. W VCEmax(5 m)=360/5=7. 2 V Ex. 2. If choose VCEmax=18 V, subtitute into (1), we get VCEmax. ICmax= 360 m. W (10) ICmax=360 m/18=20 m. A

Derating PDmax • PDmax is usually specified at 25°C. • The higher temperature goes,

Derating PDmax • PDmax is usually specified at 25°C. • The higher temperature goes, the less is PDmax • Example; – A derating factor of 2 m. W/°C indicates the power dissipation is reduced 2 m. W each degree centigrade increase of temperature.

INTRODUCTION • What is an amplifier? The function of amplifier is to provide an

INTRODUCTION • What is an amplifier? The function of amplifier is to provide an output which is greater than input. Amplifier Main types are: i) Voltage Amplifier: Intended to provide voltage gain. ii) Current Amplifier: Intended to give current gain without gain in voltage. iii) Power Amplifier: Both current and voltage can be amplified. iv) Inverting Amplifier: Gives an amplified output of phase from input.

CURRENT AMPLIFIER • The BJT(Bipolar Junction Transistor) behaves as an Current Amplifier. • Ic=hfe.

CURRENT AMPLIFIER • The BJT(Bipolar Junction Transistor) behaves as an Current Amplifier. • Ic=hfe. Ib and Ie=Ib+Ic. • The input signal is applied to the base Ib. The output can be taken from collector or emitter. Hence, acting as a Current Amplifier.

AC VOLTAGE AMPLIFIER • Any signal that is to be amplifiedmust be between 0.

AC VOLTAGE AMPLIFIER • Any signal that is to be amplifiedmust be between 0. 6 V and 0. 72 V. • Any lower than 0. 6 V, transistor will be off. • Any higher than 0. 72 V, the transistor is saturated. • So it amplifies only one half of the input signal i. e above 0. 7 V input signal. In order to amplify the whole input signal we need biasing of the transistor circuit as an amplifier.

BIASING OF THE CIRCUIT • We bias the circuit by adding something like resistor

BIASING OF THE CIRCUIT • We bias the circuit by adding something like resistor to the base to turn the transistor ON even during other half cycle. • Amplifier biased to produce maximum swing. Output is set at half the supply voltage with no input. • Value of bias or base resistance is chosen to produce base current. This adds about 1 V to the input signal. • Base current produces collector current i. e it makes the transistor to conduct. • Output is supply voltage less voltage drop across collector resistance.

POTENTIAL DIVIDER BIASING

POTENTIAL DIVIDER BIASING

POTENTIAL DIVIDER BIASING • Potential divider bias doesn’t Depend upon transistor h. • Quiescent

POTENTIAL DIVIDER BIASING • Potential divider bias doesn’t Depend upon transistor h. • Quiescent output voltage set to Vcc/2. • Voltage across Rc gives current through collector which is nearly equal to emitter current Ie and Ib=Ic/hfe. • Re is the emitter resistance that prevent thermal runaway but it reduces voltage gain. • This can be removed by adding decoupling capacitor. It blocks dc and allows ac to pass. fe

SUMMARY(IMPORTANT) • When a transistor is biased such that i) Its Base Emitter junction

SUMMARY(IMPORTANT) • When a transistor is biased such that i) Its Base Emitter junction is forward biased. This implies low value of base emitter resistance thus high value of Ie. ii) Its Base Collector junction is reversed biased which implies high base collector resistance.

SUMMARY CONT… • In the given figure we can see, the value of Ie

SUMMARY CONT… • In the given figure we can see, the value of Ie is very high. • This passes through the base to collector side giving a very small amount of current Ib. Now, the collector side has high resistance due to reverse biasing which produces a very high output voltage. Thus, pnp transistor(here) works as an amplifier.

Exercise 13. 3 A certain transistor operated with forward bias of the base-emitter junction

Exercise 13. 3 A certain transistor operated with forward bias of the base-emitter junction and reverse bias of the base-collector junction has i. C = 9. 5 m. A and i. E = 10 m. A. Find the value of i. B, and .