Electronics The Twelfth and Thirteenth Lectures

Eleventh weekهـ 1437/ 1/ 22- 91

السلمي / سمر أ

Outline for todayChapter 3: Bipolar junction transistor

Calculation gain coefficient β and α Calculation Emitter Efficiency coefficient γ The calculation of gains of voltage, current & capability to three contact methods

for BJT’s circuits I – V Characteristic of BJT in emitter common configuration and load line I – V Characteristic of BJT and his job as Switch

Finding minority carrier distributions & terminal currents in BJTFirst: the solution of diffusion equation in the base region in BJTSecond: Evaluation Values of Terminal Currents in BJT minority carrier distributions for modes of operation for BJT for npn

Chapter 4: Field Effect TransistorFET partsTypes FET

JFET junction field-effect transistor StructureWhat happens inside JFET

Wednesday from 2 to 3

Time of Periodic Exams The Second periodic exam in / 2 / 1437 - 1110هـ , Please everyone attend in her

group

The third homework

I put the third homework in my website in the university homework Due Tuesday

28 / 1/ 1437 H in my mailbox in Faculty of Physics Department , I will not accept

any homework after that , but if you could not come to university you should sent

it to me by email in the same day

Office Hours

Calculation gain coefficient β and α Gain coefficient α when base common configuration & gain coefficient β when common emitter configuration .

We start with gain coefficient β when common emitter configuration , so when we

discussed about amplification in this configuration, we mentioned how base current

effect on emitter current. thus, gain coefficient β is ratio between collector current

&base current

When we assume that few recombination in base duo to its small thickness so IE ≈ IC

Here emitter current proportional with emitter doping base current proportional with

base doping. therefore, gain coefficient β equal to is ratio between collector doping &

base doping

Here also efficiency transistor depend on gain coefficient , so when gain coefficient is big , the efficiency transistor in increases

Calculation gain coefficient β and α Gain coefficient α when base common configuration & gain coefficient β when common emitter configuration.

However, the Gain coefficient α when base common configuration equal to is ratio

between collector current & emitter current

We can find relation between gain coefficient β & gain coefficient α

Or

Calculation gain coefficient βIn the diode, we derive diffusion current density for excess of minority - carriers of electrons and holes

Since transistor is as two diodes, diode’s calculation will be the same for transistor’s calculation . current density equal to current per area (the area is the same for 3 part) therefore, for npn

And when assume base thickness is small (WB ≈ Ln ) also from previously relation

By substitute in gain coefficient

this relation is approximately

=

Calculation gain coefficient β

Final approximately relation for gain coefficient β in npn is

Where is time of minority – carriers life in base

And is time of transit electrons to base

=

Calculation Emitter Efficiency coefficient γ

It know as ratio between electronic current (npn) which injection from emitter

and total current

Since total current is , therefore:

In transistor of n+p n type , so IEp < IEn . therefore

=

The calculation of gains of voltage, current & capability to three contact methods for BJT’s circuits

Previously, we discussed about gain coefficient α and β to two of common base

configuration & common emitter configuration , respectively. In two cases the

equation was

Here, we will calculate voltage, current and power Av , Ai & Ap respectively for the three

common configurations in addition to the characteristics of their circuits.

In the three circuits, we will focus at active mode . Aslo, we must remember the equation

IE = IB + IC

=

The calculation of gains of voltage, current & capability to three contact methods for BJT’s circuits

emitter common configuration This circuit is the most important and used in amplifiers for transistor input current is base current & output current is collector current input voltage is Vin & output voltage is Vout . the output signal opposite to input signal ,mean Out of phase with 180o , so it is Inverting Amplifier circuit input resistance Rin is lower duo to forward bias & output resistance RL is higher duo to reverse bias Current gain is

Voltage gain is The gain in this circuit less than base common configuration power gain is

The calculation of gains of voltage, current & capability to three contact methods for BJT’s circuits

base common configuration input current is emitter current & output current is collector current input voltage is Vin & output voltage is Vout . the output signal the same to input signal ,mean In of phase , so it is non inverting Amplifier circuit the ratio between output resistance RL &input resistance Rin is higherCurrent gain is

In most case approximate value is one and IB is small, thus IE ≈ IC

Voltage gain is

power gain is

The calculation of gains of voltage, current & capability to three contact methods for BJT’s circuits

collector common configuration input current is base current & output current is emitter current input voltage is Vin & output voltage is Vout . the output signal the same to input signal ,mean In of phase , so it is non inverting Amplifier circuit input resistance Rin is higher & output resistance RL is lower in the circuit input voltage contact direct to base while output voltage is taken from load resistance Current gain is

Voltage gain is

Always is less than one duo to power gain is

The calculation of gains of voltage, current & capability to three contact methods for BJT’s circuits

CharacteristicCommon

BaseCommonEmitter

CommonCollector

Input Impedance Low Medium High

Output Impedance Very High High Low

Phase Angle 0o 180o 0o

Voltage Gain High Medium Low

Current Gain Low Medium High

Power Gain Low Very High Medium

I – V Characteristic of BJT

In chapter two, we study I – V Characteristic of diode in lecture and Lab, so we

know about proportional relation in addition to load line. Also, in this chapter we

will study I – V Characteristic of BJT and its calculation and duo to different

contact methods, there will be different curves , however, the relation always

proportional

I – V Characteristic emitter common I – V Characteristic emitter common

I – V Characteristic of BJT in emitter common configuration and load line we will deal with emitter common configuration to see relation between IC & VCE

1- we notice that IC increases rapidly by change VCE

at beginning in region called Saturation Region. Than,

The increase becomes very slow unnoticed which

consider as constant in region called

Amplification Region or Active Region.

2- when we change IB , IC will also change

duo to input voltage (means that IC

control by IB)

3- the third region called Cut Off Region

which is region under current value IB =0

I – V Characteristic of BJT in emitter common configuration and load line

4 - load line will cut y- axis Saturation point (S –point) and x- axis cut off point

(C –point)

5- at operation point (Q- point) cut of load line with I – V Characteristic

load line equation

6- from equation and figure, we notice

cut off point when IB =0 & IC =0 thus

VCE = VCC

Saturation point when VCE =0 thus

The mean operation point is )

I – V Characteristic of BJT and his job as Switch

we can describe job’s BJT as switch

1- Saturation Region:

which represent an close switch (Fully-ON)

input voltage and base contact with VCC

VBE > 0.7V

the maxim value of collector current

The two junction at forward bias

ideal Saturation at VCE =0

I – V Characteristic of BJT and his job as Switch

we can describe job’s BJT as switch

2- Cut Off Region :

which represent an close switch (Fully-OFF)

input voltage and base contact with

grounded 0V

VBE < 0.7V

The two junction at reverse bias

There no current flow to collector

IC = 0

Finding minority carrier distributions & terminal currents in BJT

we deal with active mode and base common configuration to npn . At the

beginning and to make calculation ease, we assume the following :

1- the thickness of base WB is small, therefore, the diffusion here for electronic

current from emitter to collector . In addition, we neglects drift current at base.

2- emitter efficiency γ ≈ 1 because emitter current is only electronic current

3- reverse leakage current or reverse saturation current is neglected

4- the action part in base and two

junctions have regular section area

and electronic current move in one

direction or one dimension which is x

5- all currents and voltages are stable

A

VEB VCB

WB

xp0

n np

ΔnEΔnC

First: the solution of diffusion equation in the base region in BJT

we deal with active mode and base common configuration to npn . Previously, we

study about a excess of minority -carriers of electrons concentration in p-type in

base region.

Electronic current enter to base from emitter, also come out from base to collector.

To calculate excess of electrons to two end-sides of base at two depletion regions

from emitter side and collector side to obtain :

excess of electrons concentration from

end-side depletion regions of emitter

excess of electrons concentration from

end-side depletion regions of collector

A

VEB VCB

WB

xp0

n np

ΔnEΔnC

First: the solution of diffusion equation in the base region in BJT

If we assume that emitter & base junction has strong forward bias which means

and base & collector junction has strong reverse bias which means

, we will obtain

We mentioned before of minority carrier diffusion equations which is quadratic

equation, hence the solution are

Where Ln the diffusion length of minority

carrier electron . We must remember that

thickness of base is small which means that

WB ≤ Ln to move all electronic current from

emitter to collector.

A

VEB VCB

WB

xp0

n np

ΔnEΔnC

First: the solution of diffusion equation in the base region in BJTFinding constants C with boundary conditions at the two end-sides of base

When multiply the first equation with , we obtain

And collected with second equation to obtain C1

by substituting in the first equation to obtain C2

First: the solution of diffusion equation in the base region in BJT

by substituting with constants C in

we mentioned that high reverse voltage to obtain

We can write equation in this from

where

First: the solution of diffusion equation in the base region in BJT

the figure shows the distribution of minority carriers in the base also in the emitter

and collector

where

δn

ΔnE

M1ΔnE

M2ΔnE

Second: Evaluation Values of Terminal Currents in BJT

We will find current by known current density to two end-sides of base

At driven excess of electron in base reign in p-type respect of xp

This driven at borders of depletion region from emitter

Therefore, electronic emitter current

by substituting in constants

since the

So emitter current

Second: Evaluation Values of Terminal Currents in BJT

This derivation at the borders of the depletion region from collector side

Therefore, collector current is

By substituting in constants

Second: Evaluation Values of Terminal Currents in BJT

Finally, we can find base current from emitter and collector currents

When taking into consideration , we obtain threes

approximate value

minority carrier distributions for modes of operation for BJT for npn

Active mode: the figure shows the distribution of minority carriers in the base

also in the emitter and collector in active mode

minority carrier distributions for modes of operation for BJT for npn

Saturation mode: the figure shows the distribution of minority carriers in the

base also in the emitter and collector in saturation mode

minority carrier distributions for modes of operation for BJT for npnCut - off mode the figure shows the distribution of minority carriers in the base also in the emitter and collector in cut off mode

minority carrier distributions for modes of operation for BJT for npnInverted mode :the figure shows the distribution of minority carriers in the base also in the emitter and collector in inverted mode

Field Effect TransistorThe field effect transistor uses electric field to control the conduction shape of

channel for type extrinsic semiconductor either n-type or p-type . Therefore, it

called n-channel or p-channel, respectively. Also, we will focus at majority

carriers of channel. because of this, it called also unipolar junction transistor

FET parts

As any transistor, it has three parts. Their name are

Source: a majority carrier enters from it which is the start point of channel.

Drain : a majority carrier comes out from it which is the end point of channel

Gates: the third part which controls of conduction of a majority carrier of channel

which is above the channel

Types FET

famous types in field-effect transistor are

1- (JFET (junction field-effect transistor

2- (MESFET) metal–semiconductor field-effect transistor

3- (MISFET) metal–Insulator–semiconductor field-effect transistor

4- (MOSFET) metal–oxide–semiconductor field-effect transistor

We will focus on the first type JFET and last type MOSFET in our study

JFET junction field-effect transistor

JFET Structure

As BJT structure, it distribute three extrinsic types of n-type & p-type in

respectively way. Here the channel is in the center

of 3 three types either n-channel or p-channel.

Source (S) and Drain (D) are metal bars at start

& end of the channel. Gates (G) is metal bar

contact with other extrinsic type (which has

more impurities than in channel),and places

above and blow the channel.

P +

JFET junction field-effect transistor

What happens inside JFET

We will focus at n-channel. At the beginning, we contact transistor with two voltage sources

. One of them contacts with circle between source (S) and drain (D), the other contacts as

voltage bias to gates (G). We must notice that source contacts with grounded.

When we look at figure below, we notice that battery voltage VDS , the reason to current exit

which come out from positive battery to negative battery in n-channel (which here is drain

current direction ID ). therefore, the direction of electronic current (majority carrier in n-

channel) invert ID direction and move from source to drain.

However, applied voltage VGS at gates p+ and

n-channel is reverse bias or reverse voltage.

Thus, length of the depletion region

between p+n and n p+ junctions is big

which controls of moving electronic current

which move from source S to drain D.

JFET junction field-effect transistor What happens inside JFET

Notice that the depletion region between p+n and n p+ junctions depend on

dimension x . the width of depletion region different in place x =0 from place x

=L

Where is wide in L duo to applied voltage change VD.

The voltage at source is

less than voltage at drain.

Thus, it controls with the

cross-section area of

conduction n-channel and

narrows from drain side.

The depletion region works

as side door or gate which

open and close the channel.

JFET junction field-effect transistor What happens inside JFET

Notice from below figure The depletion region cause closing gate from drain.

Therefore, there no electronic current get out, also drain current (real current in

circuit) ID = 0 (correct to be constant). This condition is called saturated state to

current. At the beginning of saturated, we called pinch–off or pinch–off voltage Vp

Channel resistivity is constant duo to impurities constant. it controls with channel

resistivity by change the cross-section area of it.