Phan Nhu Quan, PhD Dec 02, 2017

Transcription

1 Phan Nhu Quan, Ph ec 0, 07

2 Course learnng outcomes (COs) CO: Provde the background of the electronc components and ther applcaton crcuts CO: Explan and demonstrate the operaton prncple of the electronc crcuts based on the characterstc of the electronc components and the bass of the mathematcal CO3: Analyze and solve the problems related to electronc crcuts CO4: Analyze and smulate the electronc crcuts by the specalzed software

3 Outlne Chapter Chapter Chapter 3 Chapter 4 Chapter 5 Chapter 6 Chapter 7 Materal structure Semconductor and ode Bpolar juncton transstor H parameter & equvalent crcut Feld effect transstor C coupled amplfer Opamp & applcaton crcuts

4 Chapter. Materals structure.. Materals structure... Materal structure An atom s composed of a nucleus, whch contans postvely charged protons and neutral neutrons, and negatvely charged electrons that, n the classcal sense, orbt the nucleus. The electrons are dstrbuted n varous "shells" at dfferent dstances from the nucleus Nucleus +8 Electrons Fg... Atom structure

5 Chapter. Materals structure... Atom structure +8 share one electron +8 eceve one electron Postve on Negatve on Fg... Atom structure

6 Chapter. Materals structure... Atom structure Order of orbtals (fllng) n mult-electron atom For example: Z Gemanum = 3: correspondng: s s p 6 3s 3p 6 4s 3d 0 4p Ge Fg..3. Electrons dstrbuton

7 Chapter. Materals structure..3. Materals..3. Conductor The substances that the outermost shell have only one or two electrons. For example slver, gold, copper and alumnum..3. nsulator The substances that the outermost shell have 8 electrons For example glass, porcelan, rubber and paper..3.3 Semconductor The substances that n the outermost shell have 4 electrons The semconductor resstance s greater than the conductor resstance but smaller than the nsulator resstance For example slcon, germanum, etc.

8 Chapter. Materals structure.. Semconductor materals Most electronc devces are fabrcated by usng semconductor materals along wth conductors and nsulators... Negatve semconductor S S Free- electron S S S S P S S S Fg..4. Pure semconductor Fg..5. Negatve semconductor

9 Chapter. Materals structure... Postve semconductor S Holes S n S S Fg..6. ndum atom dstrbuton Note: n normally, both the N and the P-type semconductor s neutralzaton atom. t just shows the ablty to share or receve the free-electron

10 Outlne Chapter Chapter Chapter 3 Chapter 4 Chapter 5 Chapter 6 Chapter 7 Materal structure Semconductor and ode Bpolar juncton transstor H parameter & equvalent crcut Feld effect transstor C coupled amplfer Opamp & applcaton crcuts

11 Chapter. Semconductor ode.. Semconductor ode... Symbol and shape... Structure OE PN juncton U =0.3V or 0.7V (potental barrer) P - + N Anode or P-type Holes Cathode or N-type Electrons Holes epleton regon Electrons Fg... Semconductor dode

12 Chapter. Semconductor ode..3. Forward Bas ode allow the current through t P N Where, qu kt s e s e qu kt P U C - U C N Fg..3. Forward bas voltage s reverse saturaton current (leakage current, typcally 0 - A ) e Euler s constant (.78888) q charge of electron ( coulombs or J/V) U voltage appled across dode (V) K Boltzmann s constant ( J/K) T=tC+73 juncton temperature (Kelvn)

13 Chapter. Semconductor ode..4. everse Bas When reverse bas, ode does not allow the current through t, therefore the current flow through the depleton regon very small P N s e qu kt s P S N - + U C Fg... everse bas voltage U C

14 Chapter. Semconductor ode..5. Volt-Ampere characterstcs of ode Fg..4. V-A characterstc curve

15 Chapter. Semconductor ode.. rect current load lne (C) The crcut has only C source V C V-A of ode + V s / V S (t) V - When v s not change n tme, the capacty C opened Q 0 Q C V S V V V. V. V VS V S V Q V s Fg..5. ntersecton between V-A characterstc curve and C V

16 Chapter. Semconductor ode.3. Alternatng current load lne (AC) The crcut has only AC source V C V s /( // ) AC V-A of ode + V s / V S (t) V - Q Q C When v S capacty C shorted: v alternates n tme, the ( t) ( // ). ( t) v ( t) v // s ( t) v ( t) ( t). t // v t v s S t // 0 V Q Q ( // U V s Fg..6. ntersecton between V-A characterstc curve, C and AC V

17 Chapter. Semconductor ode.4. Parameters of ode.4.. Off resstance or C resstance r f U Ur r f f (forward resstance, few ohms to tens of ohms) (reverse resstance, several hundred k).4.. On resstance or AC resstance r d du d qu kt kt s e U q du f kt kt rd d q q f s ln S s

18 Chapter. Semconductor ode kt q 5mV 5mV rd ( ) ( ma) The more r d decreases, the more forward current ncreases. r d du d r When reverse bas, the current r kt q s about zeros.4.3. Several of parameters Maxmum peak reverse voltage U rmax Maxmum average forward rectfed current max Maxmum power P max Maxmum frequency f max

19 Chapter. Semconductor ode.5 Applcaton crcuts of ode.5.. Half-wave rectfer crcut N : N U U Waveform U T 0 t(s) U 0ut 0 t(s) Fg..7. Half-ware rectfer and ts waveform

20 Chapter. Semconductor ode.5.. Half-wave rectfer crcut wth flter capactor N : N U U C Waveform U T 0 t(s) U 0ut 0 t(s) Fg..8. Half-ware rectfer wth flter capactor and ts waveform

21 Chapter. Semconductor ode.5.3. Full-wave rectfer crcut N ; N U U U 3 T U Waveform 0 t(s) U 0ut 0 Fg..9. Full-ware rectfer t(s)

22 Chapter. Semconductor ode.5.4. Full-wave rectfer crcut wth flter capactor N ; N U U U 3 C T Waveform U 0 t(s) U 0ut 0 t(s) Fg..0. Full-ware rectfer wth flter capactor

23 Chapter. Semconductor ode.5.5. Full wave brdge rectfer crcut U N ; N U - + Waveform T U 0 t(s) U 0ut 0 t(s) Fg... Full-ware brdge rectfer

24 Chapter. Semconductor ode.5.6. Full wave brdge rectfer crcut wth flter capactor N ; N U U - + C Waveform T U 0 t(s) U 0ut 0 t(s) Fg... Full-ware brdge rectfer wth flter capactor

25 Chapter. Semconductor ode.5.7. Voltage multpler crcut U U m N : N C C 0 -U m U C t(s) U T U U m 0 -U m t(s) U C U m 0 t(s) -U m Fg..3. Voltage multpler

26 Chapter. Semconductor ode.5.8. Postve clpper crcut N : N U U E U 0ut T Waveform U U m E U U 0ut 0 t(s) U m Fg..4. Postve clpper crcut

27 Chapter. Semconductor ode.5.9. Postve and negatve clpper crcut N : N U U U0ut T E E Waveform U U m U E U 0ut 0 t(s) U m -E Fg..5. Postve & negatve clpper crcut

28 Chapter. Semconductor ode.5.0. Clampng crcut N : N C Waveform U U U 0ut E T U U m E 0 t(s) U 0ut U m +E U 0utC E 0 t(s) Fg..6. Clampng crcut

29 Chapter. Semconductor ode.6 Problems Problem : ode wth threshold voltage U = 0.V, r d s nsgnfcant. oad resstor = 9, source nternal resstor r =.. Power supply U wth sne waveform, ampltude V. Plot the waveform and determne the load voltage U (t). Calculate U (t=/) Soluton At the tme that U > U, dode on: r = U =9 U U 9 ( t) ( U U ) U r r 9 At the tme that U < U, dode off: t U t 0. U t 9 t 0 0 U

30 Chapter. Semconductor ode When U (t) small, need consder the U. ode only on at the tme that U > U U t t ( U r U ) sn t At the tme / ( 0.)9 t U t V U U 0 U t 0 t t t 3 t 4 t (s) T

31 Chapter. Semconductor ode Problem : The crcut s gven as follows. Use the Thevenn transform to solve the crcut. Plot the waveform of the current and the voltage of the oad, gve the saturaton current of ode s = 0-6 ma A K A K A K =k U =5V r =.5k U =0snt U =.4k U u r U U Th Th U U Soluton U only, u = 0V: u only, U = 0V: Both of them: U U U AK AK AK U u U Th r r r U r r u r.5 0snt 5 3 4snt.5.5 V

32 Chapter. Semconductor ode Thevenn equvalent resstance At the tme r.5 Th / / r.4 k r.5 t 0; ; ;...; 3 U Th = 3V; 6.46V; 7V;.. et U to be a forward voltage on ode, to be a current of ode, and s the oad. U Th U Th UTh 3 4snt U U Th Th ma The relatonshp between and U s the lne, ntersecton wth the vertcal axs at U th / th, ntersecton wth the horzontal axs at U Th, wth the slope s / th, called the C.

33 Chapter. Semconductor ode At the tme t=0; /3; /, we have the lnes AB, C, and EF. (ma) 6 4 E C Q Q A Q 0 B F U Th U t V Th =3+4snt

34 Chapter. Semconductor ode Except for ths relatonshp, and U characterstc of ode stll mutually dependent accordng to the The ntersecton between C and the characterstc curves of ode gve us the nstantaneous values of, U. s e qu kt (ma) 6 4 E C Q Q A Q 0 U e B F U Th U t V Th =3+4snt

35 Chapter. Semconductor ode Problem 3: ode wth reverse saturaton current S =30A. etermne the On resstor of ode r at the temperature of 5C wth forwardng bas voltage 0.V n both the forward and reverse drectons. Soluton: For ON and OFF states: wth S = 30A, S e qu KT qu KT,38.0, U U e 9U

36 Chapter. Semconductor ode n forward drecton, U 0.V e e 9 U r d du e U Smen r n reverse drecton, U 0.V e r r d du 0 0Smen

37 Chapter. Semconductor ode Problem 4: A rectfer power supply conssts of a transformer, a dode wth r = 5, and a load = 400. The waveform of the AC source at two ends of the secondary wndngs s sne wth an effectve value of U = 300V. etermne: The average current through the oad. The effectve current through the oad. The shape factor of the current. The wave factor of the current. T r max (A) U 0 t

38 Chapter. Semconductor ode Soluton: Suppose that, the transformer s deal, the load s provded by generator wth e=300 t V, and the nternal resstor r=r =5 m m m avg m sn t. dt cost 0 avg 0 m Em 300 0,38A r 3,

39 The shape factor 3,4,57 m m avg F The wave factor wave avg avg wave avg F,,57 F F wth and 4 4 sn 4 cos. sn m m m m m t t t d t t d t A m m 0, Chapter. Semconductor ode

40 Chapter. Semconductor ode Problem 5: The crcut s gven as follows. The V-A characterstc of ode s an exponental functon Plot the C. u 0,, U U 0 0 Plot the AC 00 C Calculate the On resstor of ode Calculate the voltage on oad E=0V u=0,coswt(v) U

41 Chapter. Semconductor ode Soluton C source only: E = 0V, u(t) = 0, the capactor C opened E U E 0 U U ma U 5mA (C) 0,4 0,4 0,4 The ntersecton between C and V-A curves s the stable operatng pont Q wth coordnate Q(8mA, 3V) U E AC source only: E=0V, u(t)=0.cost(v), the capactor C shorted td / / , 5k t u t 0, t u 5 u t t u t 0,5 0,5 (AC) ths curves s lmted by two lnes

42 Chapter. Semconductor ode t t u 0,5 U 0,5 U 0,5 max t u 0,5 mn t (ma) Q The On resstor of ode U(V) 3 g 4u. Smen u d du The value of the voltage on the oad Q u 0 3V r 83, 33 g u r t u t 0,cos t 0,045 cosv ,33

43 Chapter. Semconductor ode Problem 6: Usng a resstor k, a power supply V and a deal ode. esgn the crcut wth V-A curves as bellow fgures (ma) (ma) u (V) 0 Fg u (V) - Fg (ma) (ma) u (V) 0 u (V) - Fg 3 Fg 4

44 Chapter. Semconductor ode Soluton Fgure Segment : U<, wth =0 Segment : U>, wth U=+ Fgure Segment : U=-, wth U<- Segment : =0, wth U>- + U _ k V + U _ k V Fgure 3 Segment : U=+, wth <- Segment : U=0, wth >- + + Fgure 4 Segment : U=0, wth < Segment : U=-, wth > U _ k V U _ k V

45 Chapter. Semconductor ode.7. Zener ode Zener ode s a partcular type of ode, wth P-N juncton, suffer very hgh temperature, Therefore, when havng the reverse voltage t wll not break down. Base on ths advantage, ode Zener usually used as a voltage stablzer. Note : Zener ode operatng when reverse bas. Fg..7. V-A characterstc curve of Zener ode

46 Chapter. Semconductor ode.7.. Parameters of Zener ode Stablty voltage U z On resstor r d = duz/d, the smaller r d the hgher stable Off resstor t = Uz/z Stablty factor S d du z / z d z U z t when r d smaller than t then stablty hgher z / U z du Temperature coeffcent of stablty voltage z z r d T U z du dt z const ependng on the avalanche breakdown or tunnel breakdown T negatve. n general, T = (4)0-3 /C n fact, Zener dode wth U z < 4V: T <0 U z > 6V: T > 0 4 < U z < 6: T maybe negatve or postve z postve or

47 Chapter. Semconductor ode To mnmum the temperature coeffcent we connect a normal ode seral to Zener Table: Several of Zener dode Name Materal Maxmum dsspaton power (W) z U fw +Uz Stablty voltage U z (V) everse current(ma) N 70 S N 703 S N 706 S N 707A S N 708 S N 7A S N 74 S N 758 S N 55 S On resstor r d

48 Chapter. Semconductor ode.8 Problems of Zener ode Problem : The crcut s gven as follows. = 300, = 00. etermnes the voltage change range U to keep stable the voltage of load at0v =300 K Soluton U +U z A =.k U Choose the type of Zener dode N758 wth U z = 0V, P max = 0.4W Choose mn = 0mA, max = 30mA U U 0 8. ma 00 U U V 3 mn mn z z mn z 5 U V U max max U z z max z 5 Thus U = 5.5V.5V

49 Chapter. Semconductor ode Problem : The crcut s gven as follows, U = V + U, = 300, = 00. Zener ode zener N708 wth U z = 5.6V, and r d = 3.6 Calculate the stablty coeffcent and the voltage change on the load etermnes the range of change of operaton pont when U = 0% Soluton: Usng Thevenn method between A and K 00 U AK U Th U 9. 6V U+U K A z U Th // U Th Th K z A

50 Chapter. Semconductor ode Th K U AK =U U Th z Call to be a current through zener dode A U Th U Th Th U V Th Th 0.4 U U 40 ma The -U curve subject to ths equaton s the load lne MN ntersects the vertcal axs at 40mA, and ntersects the horzontal axs at 9.6V On the other hands, and U z (or U ) have the relatonshp accordng to the characterstc of zener dode The ntersecton of two curves s the operaton pont Q when U = V

51 Chapter. Semconductor ode K Th K U+U z U U Th z A A When U = U + U = 0% = 0.8V 3.V the load lne change from AB to C, the operaton pont change from Q to Q Stablty coeffcent S: S U U r r d d // // r r d d U U S 0% %

52 Chapter. Semconductor ode Problem 3: The crcut s shown as follows. U =30V, varable. U z = 5V, max = 65mA, = 00. Calculate the range of change of to keep stable the voltage on load at 5V K U z Soluton: The nput voltage small, so when small then U small, can not breakdown dode U U U U U When ncrease then U (or U z ) ncrease untl s broke down zener mn U Z 5 00 U U Z A U U Z mA 00 We consder = 0 when zener breakdown

53 Chapter. Semconductor ode K U z A Contnues to ncrease then ncrease ( = const) decrease, but the current through dode mn Z max mA U 5 max. 5k 0 mn Therefore = (00.5k)

54 Chapter. Semconductor ode Problem 4: The crcut s shown as bellow. etermne the change of voltage of zener V Z V 0 V max 0 mn 470 V= 9-49(V) Z V Z = 9V P Z = W Z = 30 V 0 Soluton 49V 9V Z max 0. 08A V 9V Z mn 0. 0A V V VZ. Z max Z V 0 max VZ. Z mn Z V 0 mn VZ VZ max VZ mn V

55 Chapter. Semconductor ode Advantages of zener dode Power dsspaton capacty s very hgh Hgh accuracy Small sze ow cost.8 Applcaton crcuts of Zener ode t s normally used as voltage reference Zener dodes are used n voltage stablzers or shunt regulators. Zener dodes are used n swtchng operatons Zener dodes are used n clppng and clampng crcuts. Zener dodes are used n varous protecton crcuts

56 Chapter. Semconductor ode.8.. Seres voltage regulator Q V Opamp 6 - V 0 4 Z 3 Operaton prncple Suppose that V ncreases, the output voltage ncreases (V 0 = V S ), the amount of feedback voltage on voltage dvder, 3 large. Opamp wll compare the voltage of Zener to the feedback voltage, the output of Opamp negatve saturaton, transstor Q off, then the output voltage decreased. On contrary, f the V decreases, the output voltage decreases (V 0 = V S ), the amount of feedback voltage on voltage dvder, 3 small. Opamp wll compare the voltage of Zener to the feedback voltage, the output of Opamp postve saturaton, transstor Q on, then the output voltage ncreased Therefore the output voltage s kept stable

57 Chapter. Semconductor ode.8.. Parallel voltage regulator S V Z Opamp 6 Q V 0 3 Operaton prncple: Suppose that V ncreases, the output voltage ncreases (V 0 = V S ), the amount of feedback voltage on voltage dvder, 3 large. Opamp wll compare the voltage of Zener to the feedback voltage, the output of Opamp postve saturaton, transstor Q on, then the output voltage decreased. On contrary, f the V decreases, the output voltage decreases (V 0 = V S ), the amount of feedback voltage on voltage dvder, 3 small. Opamp wll compare the voltage of Zener to the feedback voltage, the output of Opamp negatve saturaton, transstor Q off, then the output voltage ncreased Therefore the output voltage s kept stable

58 Outlne Chapter Chapter Chapter 3 Chapter 4 Chapter 5 Chapter 6 Chapter 7 Materal structure Semconductor and ode Bpolar juncton transstor H parameter & equvalent crcut Feld effect transstor C coupled amplfer Opamp & applcaton crcuts

59 Chapter 3. Bpolar juncton transstor 3.. Structure of Transstor 3... Structure 3... Symbol 3... structure Q NPN Q PNP N P N P N P E N B P N C E P B N P C E B C E B C Fg.3.. Symbol, shape and structure of transstor

60 Chapter 3. Bpolar juncton transstor 3... Operaton prncple E N P N C N P N E C B E B C _ + _ + _ B + _ + Fg.3.. Operaton prncple of transstor

61 3..3. Transstor parameters, for example C85

62 Chapter 3. Bpolar juncton transstor 3.. Bas transstor crcuts For transstors, basng means to set the proper voltage and current of the transstor base, thus settng the operatng pont, also known as quescent pont (Q) Fxed bas Ths s the most rarely used basng method wth transstor amplfers, but t s wdely used when the transstor operates as a swtch. The base current B s controlled by the base resstor B. V CC =0V K: V V V V, h CC BE B B BE B C B fe B =80K C =00 B VCC VBE A B B C β=300 6 Vout V Cc BC V C Fg.3.3. Fxed bas

63 Chapter 3. Bpolar juncton transstor Now suppose that the temperature rses, ths wll ncrease the current gan. An ncrement by 5% s a realstc and rather small value. β= %=345 C =β B =345.5=38.7mA V out = C C =7.7V (hgher than before) B =80K B V CC =0V C =00 C That s why ths basng method s not used for transstor amplfers. On the other hand, due to the fact that ths method s very smple and costeffectve, t s wdely used when the transstor operates as a load swtch, for example as a relay or E drver. That s because the Q pont operates from cut-off to hard saturaton, and even large current gan changes have lttle or no effect at the output.

64 Chapter 3. Bpolar juncton transstor 3.. Emtter feedback bas Ths s the frst method that used to fx the problem of the unstable current gan f C s ncreased due to a temperature ncrement, E s also ncreased. The voltage drop across E (V E ) s ncreased whch eventually ncreases the base voltage (V B ). Ths base voltage ncrement has as a result the decrement of the voltage across the base resstor B, whch eventually decreases B. B =80K B V CC =0VC C =00 C Β=300 C VCC VBE mA / / 300 E B Suppose that the current gan s ncreased by 5%: E =00 E Fg.3.4. Emtter feedback bas

65 Chapter 3. Bpolar juncton transstor B =80K B E =00 V CC =0VC C =00 C Β=300 E Suppose that the current gan s ncreased by 5%: So, a 5% current gan ncrement caused a 5.% output current ncrement. By addng a 00 Ohms feedback resstor at the emtter, a 5% current gan ncrement caused a 0.6% output current ncrement. The ncrement s 4.5% less whch means that ths method works somehow, but stll the shftng of the Q-pont s too large to be acceptable.

66 Chapter 3. Bpolar juncton transstor 3..3 Collecter feedback bas (negatve feedback) Ths method used to stablze the Q pont. f the current gan s ncreased due to temperature ncrement, the current through the collector s ncreased as well, and ths decreases the voltage on the collector V C. But the base resstor s connected at ths pont, so less current wll go through the resstor n the base. ess current through the base means less current through the collector. C =00 B =80K B E C V CC =0V C VCC VBE / / 300 C B 5.3mA Β=300 Fg.3.5. Collector feedback bas

67 Chapter 3. Bpolar juncton transstor C =00 B =80K E C B V CC =0V Β=300 Fg.3.6. collector feedback bas When the current gan s ncreased by 5%: C = 8 ma The effectveness of ths method compared to the emtter resstor feedback bas shown before s exactly the same. The dfference s that, C s usually much larger than E, whch results n hgher stablty. Nevertheless, quescent pont Q cannot be consdered stable.

68 Chapter 3. Bpolar juncton transstor 3..3 Collecter feedback bas (cont.) The stablzaton s much better than before C VCC VBE mA / / 300 C E B Wth a 5% current gan ncrement: C =.3 ma B =80K B E =00 So, a 5% current gan ncrement causes a 7% output V CC =0VC current ncrement. Although t s better than the prevous C =00 crcuts, stll the Q pont s not stable enough. Add to ths C that h fe s extremely senstve to temperature changes and β=h fe =300 the transstor generates a lot of heat when t operates as a E power amplfer. So we need a much better stablzaton technque. Fg.3.7. Emtter & collector feedback bas

69 Chapter 3. Bpolar juncton transstor 3..4 Voltage dvder bas The most effcent and commonly used basng method for transstor amplfers, t s the voltage dvder bas (VB) C B C C B B B E C B E E BB EBB E V CC Fg.3.8. VB crcut M Fg.3.9. Thevenn equvalent crcut

70 Chapter 3. Bpolar juncton transstor 3..4 Voltage dvder bas The voltage dvder mantans a very stable voltage at the base, and snce the B s many tmes smaller than the current through the dvder, the base voltage remans practcally unchanged. The resstor E provdes the negatve feedback. ue to the base voltage remans unchanged, the negatve feedback works very effectvely and any unwanted ncrement of the current gan produces analmost-equal negatve feedback. The collector and emtter currents change just a few, and the Q pont remans practcally stable

71 Chapter 3. Bpolar juncton transstor C B B C B C C B BB E B E E EBB M E V CC Two transstors wth dfferent current gans can VB VCC V V C CC C C V V V E B BE V V CE CC C C E operate as amplfers wth exactly the same amplfcaton, only because they are based wth a voltage dvder. Moreover, snce V BE s many tmes smaller than V B and V B remans unchanged all the tme, the emtter voltage VE remans also unchanged, hence mantanng a stable emtter current.

72 Chapter 3. Bpolar juncton transstor 3.3. Problems of bas transstor C Problem C 3k C (ma) C V BB 3V B B 0k E V CC V 3 0 Q B =40A B =30A B =0A B =0A Soluton B V BB V B BE V 0.6V 0A 0K C E. B 00 0A ma V CE (V) ma 3K V VCE VCC C. C V 6 P VCE. C 6V ma mw

73 Chapter 3. Bpolar juncton transstor Problem C C (ma) C V BB 3V B B 70k C E 0.5k.5k E V CC V 3 0 Q B =40A B =30A B =0A B =0A Soluton V BB B U CE (V) VBB VBE 3V 0.7V. B VBE E. E B. 70K K C E. B 00 0A ma VE E. E ma 0.5K V VB VE VBE V 0.6V. 6V B E ma.5k V VC VCC C. C V 7 V V V ma.5k 0.5K V CE CC C C E 6 0A

74 Chapter 3. Bpolar juncton transstor Problem 3 B 50k C.5k B C V CC V E 0.5k E Soluton V CC B VCC VBE V 0.6V. B VBE E. E B. 50K K C E. B 00 0A ma V. ma 0.5K V E E E VB VE VBE V 0.6V. 6V B E ma.5k V VC VCC C. C V 7 0A

75 Chapter 3. Bpolar juncton transstor Problem 4 C 56k 0k B B C.5k B 0.5k E C E Vcc V V BB.8V B B C 8.5k E 0.5k.5k E V CC V Soluton B 0K VBB VCC. V. 8V 56K 0K B B B B // B 8. 5K V BB B VBB VBE.8V 0.7V. B VBE E. E B. 8.5K K B E 0A

76 Outlne Chapter Chapter Chapter 3 Chapter 4 Chapter 5 Chapter 6 Chapter 7 Materal structure Semconductor and ode Bpolar juncton transstor H parameter & equvalent crcut Feld effect transstor C coupled amplfer Opamp & applcaton crcuts

77 U near two-port Network U du U U d U du Í du U d d Í ; Í U h h r ; U h h U ; f Í h h 0 h h U u h h u h h u 4.. Parameter H Chapter 4. Parameter H and transstor equvalent crcut Fg.4.. Two-port network ), ( ), ( U f U f U Í Í

78 Chapter 4. Parameter H and transstor equvalent crcut u h h h h u u. h h. U / h U h U.. Fg.4.. Hybrd equvalent

79 Chapter 4. Parameter H and transstor equvalent crcut 4.. Transstor bass crcuts 4... Common emtter crcut Vcc B C C s C B C s Vs B V E E C E V Equvalent crcut u h h u u h h u u h u h h h u h h u h r EB b e rb CB EB B E EB B E f o C fb e ob CB C fb E C. E

80 Chapter 4. Parameter H and transstor equvalent crcut 4... Common emtter crcut Equvalent crcut S M B B C C h fe B S B B V h E C V V S N Fg.4.3. Common emtter & equvalent crcut E Total nput mpedance Z // h B // B E h E h fe 5( mv) ( ma) E Total output mpedance Z 0 C // h 0E C

81 Chapter 4. Parameter H and transstor equvalent crcut Voltage amplfcaton coeffcent V V A ve BE B h E S fe B B ( C // V h // h V V h fe ( Full voltage amplfcaton coeffcent h C E // ) // B E ) V S S Z Z V V A V full S S Z Z h V S fe V B S ( C // ) V ( / / ) fe( C / / ) C h Z hfe V h Z h S E S S E Z Z B h E S Z Z

82 Chapter 4. Parameter H and transstor equvalent crcut 4... Common base crcut S C C S c B V V S E C B Vcc B V Equvalent crcut u h h u u h h u u h h h u h h u h r BC C B rc EC BC e B f o E fc B oc EC E fc B

83 Chapter 4. Parameter H and transstor equvalent crcut 4... Common base crcut Equvalent crcut S E E C V S S h B h fb E C E V B C Z / / h h E B B Z0 C // h0 B C Fg.4.4. Common base & equvalent crcut h B he h fe h h E fe

84 Chapter 4. Parameter H and transstor equvalent crcut 4... Common base crcut V h ( / / ), h fb E C fb V V h ( / / h ) A A EB E B S E B vb full V hfb( C / / ) ( C / / ) V h h B B Z Z Z V V V V h S S S S E B Z S Z Z V ( / / ) Z h fb V ( Z ) h C S S B C E C / / C / / A h h B fb fb S C E S

85 Chapter 4. Parameter H and transstor equvalent crcut Common collector crcut B C Q C 0 r s V CC v s E Equvalent crcut u h h u u h h u u h h h u h h u h r BC C B rc EC BC C B f o E fc B oc EC E fc B

86 Chapter 4. Parameter H and transstor equvalent crcut Common collector crcut B B C B B h C E r s hc h fc B r s h fc B v s B E E B E v s C Fg.4.5. Common collector & equvalent crcut Z B // r VBC VBE VEC r C hc hfc E // B B 5mV hc he mh. fe eq Z // r C B C B C

87 Chapter 4. Parameter H and transstor equvalent crcut 0 //r 0 Z E ) // ( 0 fc B S C B fc BC EB E EC h h h V V V r ) // ( ) ( ) // ( E B fc E E h V B E fc C B BC h h r V V ) // ( )( ) // )( ( ) // )( ( E fc C E fc v h h h V V A full S V A V C E C E S B B E E S h h A // // E fc C h A ) // ( ( ) Common collector crcut

88 Chapter 4. Parameter H and transstor equvalent crcut Problems Vcc=VC etermne the resstors, of the followng crcut to C reach the Max-swng. Calculate the output current? h fe =80, V BE =0.7V Vn C snt k k 3 k Q C3 C 6 k Soluton: 4 0.5k 5 0.k C resstor: AC resstor: C 3 4.5k / / / / 0.75 k AC V CC CQ max swng 3.7 C AC 0 4 BB hfe 4k ma

89 Chapter 4. Parameter H and transstor equvalent crcut VBB BBB VBE C4.735V VCC BB 7.5 k V BB BB 5.k VBB VCC 3 max CQ max swng.47 ma 3 6 t t ma max sn

90 Outlne Chapter Chapter Chapter 3 Chapter 4 Chapter 5 Chapter 6 Chapter 7 Materal structure Semconductor and ode Bpolar juncton transstor H parameter & equvalent crcut Feld effect transstor C coupled amplfer Opamp & applcaton crcuts

91 Chapter 5. Unjuncton transstor (UJT) 5.. Classfcaton JFET type N type P Self-conductor UJT MOSFET EMOSFET MOSFET type N type P type N type P Self-break Fg.5.. UJT Classfcaton

92 Chapter 5. Unjuncton transstor (UJT) fference between JFET and MOSFET JFET Juncton Feld Effect Transstor t only operates n depleton mode t has less nput mpedance than a MOSFET (about 0 9 ohm) Manufacturng process s smple and less sophstcated whch makes them cheaper t s less susceptble to damage because of the hgh nput capactance t s manly used n low nose applcatons MOSFET Metal Oxde Semconductor Feld Effect Transstor t operates n both depleton mode and enhancement mode t offers hgher nput mpedance than JFET (about 0 4 ohm) Manufacturng process s complex plus the addtonal metal oxde layer adds to the hgh cost The metal oxde nsulator reduces the nput capactance makng t more susceptble to damage t s mostly used n hgh nose applcatons

93 Chapter 5. Unjuncton transstor (UJT) 5.. Symbol G G G G G G S S S S S S type N type P type N type P type N type P JFET epleton mode MOSFET Enhancement mode MOSFET Fg.5.. Symbol

94 Chapter 5. Unjuncton transstor (UJT) 5.3. Structure G P N P G N P N S JFET type N S JFET type P Fg.5.3. Structure

95 Chapter 5. Unjuncton transstor (UJT) V characterstc curve V V GS SS A V GS = 0 Saturaton regon near regon Breakdown regon B 0 V p V S Fg V characterstc curve

96 Chapter 5. Unjuncton transstor (UJT) SS V GS = 0 3 V GS = - V GS = - V GS = -3 0 V S V p V p V GS Fg V characterstc curve

97 Chapter 5. Unjuncton transstor (UJT) 5.5. FET transstor basng C bas of a FET devce needs settng of gate-source voltage V GS to gve desred dran current Fxed bas Z n G 0 V V V V 0 V GS G S GG GG Vout VS V G -V GG V V S V p SS V GS V V GS p Fg.5.5. Fxed basng crcut for N-channel JFET

98 Chapter 5. Unjuncton transstor (UJT) Fxed bas wth resstor feedback V S S S S G S V V V V V 0 V GS G S GG GG S V p SS -V GG Fg.5.6. Fxed basng crcut for JFET wth resstor feedback

99 Chapter 5. Unjuncton transstor (UJT) Self-bas Ths s the most common method for basng a JFET 0 V 0 G G G G V V S S S S V GS V V 0 V G S S S JFET N V V S S G S S V p SS Fg.5.7. Fxed basng crcut for JFET wth resstor feedback

100 Chapter 5. Unjuncton transstor (UJT) Potental-vder Basng The crcut s so desgned that S s greater than V G so that V GS s negatve. Ths provdes correct bas voltage. VG V V VGS VG VS VG S 0 V V V S S S V V V GS S S S Fg.5.8. Potental-dvder basng crcut for N-channel JFET

101 Chapter 5. Unjuncton transstor (UJT) 5.6. ow-frequency small-sgnal equvalent crcut model SS V V GS P g V GS SS V V V P GS P V GS V GS V S V S gv GS V r S G G = 0 G G = 0 r G r gv GS gv GS S S

102 Chapter 5. Unjuncton transstor (UJT) 5.7. Common Source JFET Amplfer V s C C 0 r s v s S C S s G G r s r G r v s g.v GS S

103 Chapter 5. Unjuncton transstor (UJT) s G G = 0 r s r v s g.v GS Z // rg // // GS V gv r // // 0 S Z // r 0 V V GS V0 A V gr // // V V / / A g r / / / / 0 full vs rs / / V V GS v s r s // // v s V GS r s // // A V // // GS g // s VGS s r

104 Chapter 5. Unjuncton transstor (UJT) 5.8. Common ran JFET Amplfer V C s r s C 0 v s S s G G r s r G r g.v GS v s S S

105 Chapter 5. Unjuncton transstor (UJT) s G G S s G G = 0 S r s r G r s r G r S r S v s g.v GS vs g.v GS Z // Z S // r 0 V gv r // V // 0 S GS S A full V v V S 0 s gv GS r // S // V V V V V gv r // // G GS S GS GS S V GS s // A V V V 0 V GS gv GS gv r GS // r // // // A s g r // S // //

106 Chapter 5. Unjuncton transstor (UJT) 5.8. Common Gate JFET Amplfer V C n JFET N C out + V n s C G - V out n S s gv GS v n Vout + G _

107 Chapter 5. Unjuncton transstor (UJT) 5.8. Common Gate JFET Amplfer v n v GS v gv o GS ( // ) v A vg 0 g // vn v v gvgs gv S S v S Z g Z // o r ds S S // v g S g

108 Chapter 5. Unjuncton transstor (UJT) 5.9. Problems When we desgn the crcuts wth FET, we can choose: V SS VS 3 V GS V 3 P Problem : esgn the bellow crcut. SS =6mA, V p =-4V, A V =0 V =V s C C 0 r s v s S C S

109 Chapter 5. Unjuncton transstor (UJT) Soluton: SS SS 3mA 3 V VGS VP, V VS 6V 3 SS V GS.6, g,( ma / V ) V p V p 4 4 V V S S V VS 6 S k 3 A V / / / /0 k / /0k 0 / g /,,5 S

110 Chapter 5. Unjuncton transstor (UJT).0k 0k k (5 0 )( 0 k) k selected 6.35k k 0, k choose 0 S S 3 VG VGS S,, V

111 Chapter 5. Unjuncton transstor (UJT) V G V choose / / 00K G G 00 06k choose 00k VG 0.74 V G 00 V 6k choose.5m V 0.74 G

112 Chapter 5. Unjuncton transstor (UJT) Problem : The crcut s gven as follows. V =30V, =k, V GG =V, po =0mA, V po =-6V, U =0.5snt(V) etermne the quescent pont Q etermne the voltage gan V C C Vo G U V G G

113 Chapter 5. Unjuncton transstor (UJT) V GSQ VGSQ VGG V SQ po 7mA V po V V 6V Q 7 ma,6v Soluton: SQ SQ U U 0,5sn t V Vgs U max mn 0,5V 0,5V VGS max V GSQ U max 0,5V VGS mn V GSQ U mn,5v V GS max 0,5 S max po 0 8, 4mA V po 6 V GS mn,5 S mn po 0 5, 6mA V po 6

114 Chapter 5. Unjuncton transstor (UJT) VS mn V S max V VS max V S mn 30 5, 6 8,8V V0max VS max VSQ 8,8 6,8V V0mn VS mn VSQ 3, 6,8V V0 t,8sn t,8sn t V V0,8 Av 5,6 V 0,5

115 Chapter 5. Unjuncton transstor (UJT) Problem 3: The crcut s gven as follows. V =0V, V SQ =0V, G =M, V GG =V, po =6mA, V po =-4V, U =0.snt(V), r S. etermne etermne the voltage gan V C C Vo G U V GG

116 Chapter 5. Unjuncton transstor (UJT) V GSQ VGSQ VGG V SQ po.5ma V po V VSQ 0 V VSQ SQ k 3 Soluton: G SQ G gv G r S V 0 V S S d S po V GSQ 3 g ms dvgs V po V po V gv 0 GS rs / / Av g 0 V V GS

117 Chapter 5. Unjuncton transstor (UJT) Problem 4: The crcut s gven as follows. V =0V, V GG =V, G =M, =k, po =0mA, V po =-5V, r S. etermne S subject to SQ =0mA etermne the voltage gan V C C Vo G U V GG S C

118 Chapter 5. Unjuncton transstor (UJT) Soluton: V GSQ SQ po 0mA V po VGS SS V GSQ S SQ G S 46 ( selected ) S S 853 G gv GS r S V 0 g A v V S d S po V GSQ 5,66mS dvgs V po V po V gv 0 GS rs / / g 5,66 5,66 V V GS

119 Chapter 5. Unjuncton transstor (UJT) Wthout capactor C: gv GS G S r S G S V 0 U G S gv GS G S V 0 U

120 Chapter 5. Unjuncton transstor (UJT) U V V V gv g V GS S GS GS S S GS V0 gvgs A v V0 g 3, V g S

121 Chapter 5. Unjuncton transstor (UJT) Problem 5: The crcut s gven as follows. V =0V, V GG =V, =0k, =30k, =70k, =k, r S =0k, V T =V, K=mA/V. etermne quescent pont Q wth S =0.5k etermne the voltage gan V C C V 0 U S C

122 Chapter 5. Unjuncton transstor (UJT) Soluton: V V GS S S 30 V V 0 6V K V V V S GS T GS V GS S S 6 V V V V 6 GS GS GS GS VGSQ SQ,79V 6, 4mA 4 V V V SQ S SQ

123 Outlne Chapter Chapter Chapter 3 Chapter 4 Chapter 5 Chapter 6 Chapter 7 Materal structure Semconductor and ode Bpolar juncton transstor H parameter & equvalent crcut Feld effect transstor C coupled amplfer Opamp & applcaton crcuts

124 Chapter 6. C coupled amplfer Example : The crcut s gven as follows. Calculate the resstors value. Calculate total nput mpedance, output mpedance, voltage gan, current gan of each stage and entre of crcut. V S (t) s small and medum frequency, capactes wth large capactance VCE VCE 3 V; VBE VBE 0. 7V Cmax Cmax 3 ma; hre hre 0; h0 E h0 E 0; V CC =4V C 4 C C C O C n s 0 =00 =00 K V s E C E 3 E C E

125 Chapter 6. C coupled amplfer Calculate from the fnal stage to the frst Stage 4 Choose VE V CC VCC. 4V VE VE.4V E 0.8K K 3mA E C V V 4V 3.4 VCC CE E V C 6.K 6. 8K 3mA C VCC VB VBE VE 0.7V.4V 3. V

126 Chapter 6. C coupled amplfer Choose BB // E K Choose = 0K.48K = 68K.. 00 K K Smlar to stage Thus C = C = 6.8K E = E = K 3 = = 0K 4 = = 68K

127 Chapter 6. C coupled amplfer How to choose the rght capactor sze? n order for a couplng or bypassng capactor to operate effectvely, t must have the rght sze. The capactor acts lke a resstor n AC current. The resstance of a capactor s called "mpedance". Unlke resstors, capactors do not have a fxed mpedance. nstead, the mpedance s determned by the frequency of the AC sgnal. The equaton to calculate the mpedance s the followng: X C C fc Where, C s the capactance n Farads, and f s the AC sgnal frequency n Hertz. So, a 0uF capactor connected n seres wth a KHz sgnal generator, wll present an mpedance of: XC=5.9Ohms There are two factors that must be taken nto account: The frequency and the crcuts total resstance

128 Chapter 6. C coupled amplfer How to choose the rght capactor sze? The optmum capactor value that we choose must be 0 tmes smaller than the total crcut resstance, calculated for the worst case scenaro C s nput r Example. A crcut wth a couplng capactor, s =50, r =.K, and f= 0Hz0KHz. The total resstance : = = 50Ohms The worst case scenaro s 0Hz (lowest frequency), so the resstance of capactor must be less than 5 (=/0) : C = / ( * 3.4 * 0 * 5) = 35.3uF So, the capactor must have at least 35uF capactance. Ths makes sure that the capactor wll have less than % effect on the total resstance of the crcut. We choose the value s 47uF.

129 Chapter 6. C coupled amplfer V CC =4V s C n 0 C 4 C =00 C =00 C O K V s E C E 3 E C E s B C B C r s h fe B h fe B h e C 3 4 h e C V s

130 Chapter 6. C coupled amplfer Stage : h Z E 5mV 5 hfe E 3 // 4 // h 0//68// E Z C 6. 8K A A h 0 // 6.8// C V hfe he h E AV Stage : Z E 87 5mV 5 hfe E 04 // // h 0//68// E

131 Chapter 6. C coupled amplfer Z C 6. 8K A A 0 // Z 6.8// C V hfe he h E AV Z Both of two stages: Z Z 760 Z Z 6. 8K A A V A V AV A A

132 Chapter 6. C coupled amplfer Example : The crcut s gven as follows. Calculate the resstors value. Calculate total nput mpedance, output mpedance, voltage gan, current gan of each stage and entre of crcut. V S (t) s small and medum frequency, capactes wth large capactance g g 3mA/ V r r 30K r G r G V =V C 3 K 6 0K C C 0 s 0 M K 4 M 5 K V s C C 5 0K 7

133 Chapter 6. C coupled amplfer The equvalent of FET V =V C 3 K 6 0K C C 0 s 0 M K 4 M 5 K V s C C 5 0K 7 r s v s g V GS g V GS 4 0k 7

134 Chapter 6. C coupled amplfer Stage Z M 4 Z // 0K //30K 7. 5K 0 6 V gv. GS // 6 // 7. 7 V VGS V AV g // 6 7 V A // 3 30//0//0 3 V // // GS g4 AV VGS 7 7 Stage Z M Z // K //30K 8. 6K

135 Chapter 6. C coupled amplfer V 0 gv. GS // 3 // V VGS V0 AV g // 3 // ////000 V A 5 V // // GS 3 4 g AV VGS 4 4 Both of two stages Z Z M Z Z 7. 5K A V A V AV A A A

136 Outlne Chapter Chapter Chapter 3 Chapter 4 Chapter 5 Chapter 6 Chapter 7 Materal structure Semconductor and ode Bpolar juncton transstor H parameter & equvalent crcut Feld effect transstor C coupled amplfer Opamp & applcaton crcuts

137 Chapter 7. Opamp & applcaton crcuts 7.. C amplfer The C and the transformer coupled amplfer only response to the AC sgnal wth small ampltude, and low frequency at least greater than Hz. Although the drect coupled amplfer responses to the C sgnal, but the number of stages s lmted and complexty calculaton The C amplfer response to the small frequency sgnal (less than Hz, consderng as C sgnal) that s the reason we want to amplfy. C amplfer characterstc: The sgnal wth small ampltude (V) and low frequency (<Hz) Symmetry nput sgnal arge amplfcaton factor Good ant-nterference ablty nput and output bas voltage equals zeros Stablty basng, unmovng the quescent pont under change temperature

138 Chapter 7. Opamp & applcaton crcuts 7.3. nvertng Operatonal Amplfer V VA VA V V V GN 0V A B V V V A V V The mnus sgn ( ) ndcates a 80 o phase shft because the nput sgnal s connected drectly to the nvertng nput termnal of the Op-amp

139 Chapter 7. Opamp & applcaton crcuts 7.4. Non-nvertng Operatonal Amplfer 0 VB V0 V B V V V B A V V AV V 0

140 Chapter 7. Opamp & applcaton crcuts 7.5. Op-amp fferental Amplfer The dfferental amplfer amplfes the voltage dfference present on ts nvertng and non-nvertng nputs f 3 V V V a V b UA V out 4 V V V V V V,, a b a out f 3 4 Va Vb, Vb V 4

141 Chapter 7. Opamp & applcaton crcuts 7.5. Op-amp fferental Amplfer f 3 V V V a V b UA V out f V V V 3 0, then out ( a) 4 3 f V 0, then Vout ( b) V Vout Vout ( a) Vout ( b) V V 4 when ;, then V V V out 4

142 Chapter 7. Opamp & applcaton crcuts 7.6. Summng Amplfer Crcut V V V V

143 Chapter 7. Opamp & applcaton crcuts 7.6. Summng Amplfer Crcut V V V / / V A 4 4 / / / / / / V / / V V / / V V V V

144 Chapter 7. Opamp & applcaton crcuts 7.7. Op-amp ntegrator Crcut n f C + - V c V n n X UA -V out Q Vc, Vc VX Vout 0 Vout C dvout dq dq, but dq/dt s electrc current dt Cdt C dt Vn 0 Vn n n n dv dq dq dvout C dt Cdt dt dt out f C C

145 Chapter 7. Opamp & applcaton crcuts 7.7. Op-amp ntegrator Crcut n V V n out n f n n V V dt V out n n nc 0 0 To smplfy the math, ths can be rewrtten as: V out dt C V dvout C dt V j C n t n t dt C n V n n n f X C + - V c UA -V out The mnus sgn ( ) ndcates a 80 o phase shft because the nput sgnal s connected drectly to the nvertng nput termnal of the Op-amp

146 Chapter 7. Opamp & applcaton crcuts 7.8. Op-amp fferentator Crcut VA A 0 du d V C V V C C dt dt V V GN 0V A B dv V 0 C dt V 0 dv C dt The mnus sgn ( ) ndcates a 80 o phase shft because the nput sgnal s connected drectly to the nvertng nput termnal of the Op-amp

147 Chapter 7. Opamp & applcaton crcuts 7.9. Op-amp dode log amplfer qu q VA V0 V KT V A KT S e S e V V GN 0V A B V0q KT S e KT V V V0 ln q S

148 Chapter 7. Opamp & applcaton crcuts 7.0. Op-amp exponental Crcut qu q V V A V KT A V 0 KT S e S e V V GN 0V A B qv V KT 0 S e V 0 e S q V KT

149 Chapter 7. Opamp & applcaton crcuts Problems Problem : V V V3 K.5K 3 K X 00K +0VC UA -0VC Vout V 00sn t( mv ); V 80( mv ); V 50( mv ) 3 etermne the output voltage V o, Keepng V, V3. etermne the V that make postve and negatve sgnal dstorton

150 Chapter 7. Opamp & applcaton crcuts 7.. Problems Soluton V K 4 00K +0VC V V3.5K 3 K X UA -0VC Vout 00 V V mv 0 V V V mv 6.4 V V V mv.5 V V V V V 8.9V

151 Chapter 7. Opamp & applcaton crcuts 4 V K 00K +0VC Postve sgnal dstorton V V3.5K 3 K X UA -0VC Vout 00V 80V 50V 0V 00V max 3 max 8.9 Vmax V 89mV 00 Negatve sgnal dstorton 00V 80V 50V 0V 00V max 3 max. Vmax V mv 00

152 Chapter 7. Opamp & applcaton crcuts 7.. Problems Problem : V V K K 3 +0VC 8 UA + - Vout V3 3 3K 4-0VC 5 4 0K 00K V ( V ); V ( V ); V 3( V ) 3 etermne the output voltage V o, Keepng V, V3. etermne the V that make postve and negatve sgnal dstorton

153 Chapter 7. Opamp & applcaton crcuts Soluton 5 / / 3 00 / /3 V0 V V 6V 6V 4 / / 3 0 / /3 5 / / 3 00 / /3 V0 V V 3V 6V 4 / / 3 0 / /3 5 / / 00 / / V03 V3 V3 V 3 6V 4 3 / / 0 3 / / V0 V0 V0 V V

154 Chapter 7. Opamp & applcaton crcuts Postve sgnal dstorton 6V 3V V 0V 6V 0 V max 3 max max 8 V 6 Negatve sgnal dstorton 6V 3V V 0V 6V 0 V max 3 max max 3 6 V

155 eferences. PGS-TS. ê Ph Yến, Kỹ thuật đện tử. PGS-TS. ê Tến Thường, Mạch đện tử & 3. KS. Nguyễn Tấn Phước, nh kện đện tử 4. onald.schllng Electronc Crcuts Some webste:. Software:. Proteus 8.6. Orcard