Electronics The Twelfth and Thirteenth Lectures Eleventh week 91- 22/ 1/ 1437 هـ أ / سمر...
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Transcript of Electronics The Twelfth and Thirteenth Lectures Eleventh week 91- 22/ 1/ 1437 هـ أ / سمر...
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.