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

2. Outline for today Chapter 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 BJT First: the solution of diffusion equation in the base region in BJT Second: Evaluation Values of Terminal Currents in BJT minority carrier distributions for modes of operation for BJT for npn Chapter 4: Field Effect Transistor FET parts Types FET JFET junction field-effect transistor Structure What happens inside JFET

3. Wednesday from 2 to 3 Time of Periodic Exams The Second periodic exam in / 2 / 1437 هـ - 1110, 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

4.  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 I E ≈ I C 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

5.  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

6.  Calculation gain coefficient β In 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 (W B ≈ L n ) also from previously relation By substitute in gain coefficient this relation is approximately =

7.  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 =

8.  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 I Ep < I En. therefore =

9.  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 A v, A i & A p 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 I E = I B + I C =

10.  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 V in & output voltage is V out. the output signal opposite to input signal,mean Out of phase with 180 o, so it is Inverting Amplifier circuit input resistance R in is lower duo to forward bias & output resistance R L 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

11.  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 V in & output voltage is V out. the output signal the same to input signal,mean In of phase, so it is non inverting Amplifier circuit the ratio between output resistance R L &input resistance R in is higher Current gain is In most case approximate value is one and I B is small, thus I E ≈ I C Voltage gain is power gain is

12.  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 V in & output voltage is V out. the output signal the same to input signal,mean In of phase, so it is non inverting Amplifier circuit input resistance R in is higher & output resistance R L 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

13.  The calculation of gains of voltage, current & capability to three contact methods for BJT’s circuits Characteristic Common Base Common Emitter Common Collector Input ImpedanceLowMediumHigh Output ImpedanceVery HighHighLow Phase Angle0o0o 180 o 0o0o Voltage GainHighMediumLow Current GainLowMediumHigh Power GainLowVery HighMedium

14.  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

15.  I – V Characteristic of BJT in emitter common configuration and load line  we will deal with emitter common configuration to see relation between I C & V CE 1- we notice that I C increases rapidly by change V CE 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 I B, I C will also change duo to input voltage (means that I C control by I B ) 3- the third region called Cut Off Region which is region under current value I B =0

16.  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 I B =0 & I C =0 thus V CE = V CC Saturation point when V CE =0 thus The mean operation point is )

17.  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 V CC  V BE > 0.7V  the maxim value of collector current  The two junction at forward bias  ideal Saturation at V CE =0

18.  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  V BE < 0.7V  The two junction at reverse bias  There no current flow to collector I C = 0

19.  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 W B 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 V EB V CB WBWB xpxp 0 nn p ΔnEΔnE ΔnCΔnC

20.  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 V EB V CB WBWB xpxp 0 nn p ΔnEΔnE ΔnCΔnC

21.  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 L n the diffusion length of minority carrier electron. We must remember that thickness of base is small which means that W B ≤ L n to move all electronic current from emitter to collector. A V EB V CB WBWB xpxp 0 nn p ΔnEΔnE ΔnCΔnC

22.  First: the solution of diffusion equation in the base region in BJT Finding 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 C 1 by substituting in the first equation to obtain C 2

23.  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

24.  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 δnδn ΔnEΔnE M1ΔnEM1ΔnE M2ΔnEM2ΔnE

25.  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 x p This driven at borders of depletion region from emitter Therefore, electronic emitter current by substituting in constants since the So emitter current

26.  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

27.  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

28.  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

29.  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

30.  minority carrier distributions for modes of operation for BJT for npn Cut - off mode the figure shows the distribution of minority carriers in the base also in the emitter and collector in cut off mode

31.  minority carrier distributions for modes of operation for BJT for npn Inverted mode : the figure shows the distribution of minority carriers in the base also in the emitter and collector in inverted mode

32.  Field Effect Transistor The 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

33.  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

34.  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+P+

35.  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 V DS, the reason to current exit which come out from positive battery to negative battery in n-channel (which here is drain current direction I D ). therefore, the direction of electronic current (majority carrier in n- channel) invert I D direction and move from source to drain. However, applied voltage V GS 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.

36.  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 V D. 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.

37.  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) I D = 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 V p Channel resistivity is constant duo to impurities constant. it controls with channel resistivity by change the cross-section area of it.