模拟电路分析与设计英文pptch.ppt
Huazhong University of Science and TechnologyThe Department of Electronics and Information Engineering,First Term 08/09,Electronic Circuit Analysis and Design,Dr.Tianping Deng,PART 1SEMICONDUCTOR DEVICES AND BASIC APPLICATIONSChapter 1Semiconductor Materials and DiodesChapter 2Diode CircuitsChapter 3The Field-Effect TransistorChapter 4Basic FET AmplifiersChapter 5The Bipolar Junction TransistorChapter 6Basic BJT AmplifiersChapter 7Frequency ResponseChapter 8Output Stages and Power AmplifiersPART 2ANALOG ELECTRONICSChapter 9Ideal Operational Amplifiers and Op-Amp CircuitsChapter 10Integrated Circuit Biasing and Active LoadsChapter 11Differential and Multistage AmplifiersChapter 12Feedback and StabilityChapter 13Operational Amplifier CircuitsChapter 14Nonideal Effects in Operational Amplifier CircuitsChapter 15Applications and Design of Integrated Circuits,Contents,Ch1.Semiconductor Materials and Diodes,1.1 Semiconductor Materials and Properties,1.2 The PN Junction,1.4 Diode Circuits AC Equivalent Circuit,1.3 Diode Circuits DC Analysis and Models,1.5 Other Diode Types,1.6 Design Application:Diode Thermometer,1.7 Summary,1.1 Semiconductor Materials and Properties,1.1.1 Intrinsic Semiconductor,1.1.2.Extrinsic Semiconductor,N-type semiconductor,P-type semiconductor,Ch1.Semiconductor Materials and Diodes,1.1 Semiconductor Materials and Properties,Materials,ConductorInsulatorSemiconductor,:conduction electrons_electrical conductivity,:electrons in bonding mechanism _cannot move,:silicon,germanium,gallium arsenide,1.1.1.Intrinsic Semiconductor,Silicon Valley,1.1.1 Intrinsic Semiconductor,1.Silicon,germanium_single-crystal structure,At temperature T=0oK,silicon is an insulator.,1.1.1 Intrinsic Semiconductor,Each Si atom shares one electron with each of its four closest neighbors so that its valence band will have a full 8 electrons.,1.1.1 Intrinsic Semiconductor,2.When T increases,free electrons and“holes”are created,In pure semiconductor,the concentration of electrons and holes are equal,and very small,so it has very small conductivity.,Increasing temperature,generating electron-hole pairs,Free electron-produced by thermal ionization.It can move freely in the lattice structure so as to form current.Hole-empty position in broken covalent bond.It can be filled by free electron(recombination)and can also“move”freely to form current.,Carriers,A free electron is negative charge and a hole is positive charge.,1.1.1 Intrinsic Semiconductor,A hole can be regarded as a positive charge carrier,Does the hole can move through the crystal freely?,1.1.2.Extrinsic Semiconductor,1.phosphorus+silicon=N-type semiconductor,Holes present because of thermal energy,What are the majority carriers in n-type materials?What are the minority carriers in n-type materials?,N-type semiconductor material(phosphorus)donor:provide free electrons majority carrier electrons minority carrier holes,Doped Semiconductorn type,Donor,Free E,bound charge,Donor-pentavalent impurity provides free electrons,usually entirely ionized.Positive bound charge-impurity atom donating electron gives rise to positive bound charge.Majority carriers-free electrons(mostly generated by ionized donor and a very tiny portion by thermal ionization).Minority carriers-holes(only generated by thermal ionization).,n type Semiconductor,1.1.2.Extrinsic Semiconductor,What are the majority carriers in P-type materials?What are the minority carriers in P-type materials?,2.boron+silicon=P-type semiconductor,P-type semiconductor material(Boron)acceptor:accept an extra electrons majority carrier holes minority carrier electrons,Doped Semiconductorp type,Boundcharge,Hole,Acceptor,Acceptor-trivalent impurity provides holes,usually entirely ionized.Negative bound charge-impurity atom accepting hole give rise to negative bound chargeMajority carriers-holes(mostly generated by ionized acceptor and a tiny small portion by thermal ionization)Minority carriers-free electrons(only generated by thermal ionization.),p type Semiconductor,1.1.2.Extrinsic Semiconductor,2.boron+silicon=P-type semiconductor,1.phosphorus+silicon=N-type semiconductor,Positive Charges+holes=electrons,Negative Charges+electrons=holes,majority carrier,minority carrier,Doping,Temperature,majority carrier,Doping,minority carrier,Temperature,Majority carrier is only determined by the impurity,but independent of temperature.Minority carrier is strongly affected by temperature.If the temperature is high enough,characteristics of doped semiconductor will decline to the one of intrinsic semiconductor.,Conclusion on the doped semiconductor,1.1.2.Extrinsic Semiconductor,Diffusion:a concentration of charge carriers tends to spread with time,1.1.2.Extrinsic Semiconductor,Diffusion:a concentration of charge carriers tends to spread with time,Drift:The average motion of the charge carriers due to an applied electric field.,1.2 The PN Junction,1.2.1 The Equilibrium PN Junction,1.2.2 Forward-Biased PN Junction,1.2.3 Reverse-Biased PN Junction,1.2.4 PN Junction Diode,1.2 The PN Junction,1.2.1 The Equilibrium PN Junction,N,P,hole,Electron,Holes diffuse from P-to N-region.Electrons diffuse from N-to P-region,1.2 The PN Junction,1.2.1 The Equilibrium PN Junction,1.Depletion region(space-charge region)_PN junction,Electric field,2.Electric field,N,P,1.2.1 The Equilibrium PN Junction,3.Drift current IDR=diffusion current IDF _Balance,1.Depletion region(space-charge region)_PN junction,2.Electric field,IDF,IDR,N,P,1.2.1 The Equilibrium PN Junction,The procedure of forming pn the dynamic equilibrium of drift and diffusion movements for carriers in the silicon.In detail,there are 4 steps:DiffusionSpace charge regionDriftEquilibrium,1.2.2 Forward-Biased PN Junction,Positive voltage is applied to P,forward-biased PN.,1.Depletion region is reduced,low resistance.,2.Majority carriers flow across PN junction more easily.IDFIDR,iD=IDFIDR,vD,The two fields are opposite.The results are:,1.2.2 Forward-Biased PN Junction,1.2.2 Forward-Biased PN Junction,3.The current-voltage(v-i)characteristic,The current is an exponential function of voltage.,4.V _turn-on,cut-in voltage,IS_reverse-bias saturation current,VT _thermal voltage at room temperature VT=0.026V,vD,1.2.3 Reverse-Biased PN Junction,Positive voltage is applied to N,reverse-biased PN.,1.Depletion region is increased,high resistance.,2.Majority carriers cannot cross the junction.Minority carriers sweep across PN easily.IDFIDR,iD=IDR,The two fields are in the same direction.The results are:,1.2.3 Reverse-Biased PN Junction,3.The reverse current flow will be due to the drift current IDR,known as the reverse saturation(or leakage)current IS.,1.2.3 Reverse-Biased PN Junction,4.When the voltage is increased to some point,breakdown occurs and current increases rapidly.The voltage at this point is called the breakdown voltage.,5.Avalanche breakdown Zener diode operate in the breakdown region,1.2.3 Reverse-Biased PN Junction,6.Junction capacitance Cj,VR_applied voltageVbi_build-in voltage,VR,1.2.4 PN Junction Diode,1.Diode_ has one PN junction in it,circuit symbol,Photo of real diode,1.2.4 PN Junction Diode,2.Temperature Effect,T1 T0,T1 T0,IS and VT are functions of T.,T increasesrequired forward-bias voltage decrease,1.2.4 PN Junction Diode,3.Switching Transient,Storage time ts Turn-off time tf,When forward-bias,excess carrier is stored in both regions.When switching from forward to reverse,it need time to remove.,1.3 Diode Circuits DC Analysis and Models,1.3.1 Models,1.3.2 DC Analysis,1.3 Diode Circuits_DC Analysis and Models,1.3.1 Models,1.Ideal model,Describing i-v characteristic,when analyzing circuit,Case 2,Case 1,1.3.1 Models,2.Piecewise model,Two linear approximations,1.3.2 DC Analysis,Determine the diode voltage and current for the circuit shown in figure,Assume,Sol:For diode,For R and VPS branch,VPS=5V,R=2k,Find two special points(5V,0mA)and(0V,2.5mA),(1)using the graphical analysis.,Intersection_quiescent point or Q-point,VD,ID,1.3.1 Models,3.Small signal model_AC model,When a diode is operating in the small range，it can be a small-signal incremental resistance.,That is:,small-signal conductance,Room T（T=300K）,1.3.2 DC Analysis,(2)using the model analysis.,Ideal Model,Piecewise model_1,（Silicon Diode）,Piecewise model_2,Assume,（Silicon Diode）,VD=V+ID rf=1.09V,1.4 Diode Circuits_AC Equivalent Circuit,1.4.1 Sinusoidal Analysis,1.4.1 Sinusoidal Analysis,E.g.1.2.Determine the output voltage in Fig.1.31.Assume circuit and parameter of VPS=5V,R=2k,Vr=0.6V,and vi=0.1sinwt(V).,Sol:because,(1)For the DC analysis,we set vi=0,then,(2)For the AC analysis,we set VPS=0,then,1.5 Other Diode Types,1.5.1 Solar Cell,1.5.2 Photodiode,1.5.3 Light-Emitting Diode(LED),1.5.4 Schottky Barrier Diode,1.5.5 Zener Diode,1.5 Other Diode Types,1.5.1 Solar Cell,It gets power from the solar array and can be used eitherto power an electric motor or to charge a battery pack.,1.5 Other Diode Types,1.5.2 Photodiode,It converts optical signals into electrical signals.,155Mbps,622Mbps,2.5Gbps,10Gbps,40Gbps,Trunk-line,Metro-core,Metro-access,Data-com,Access,DFB/MOD,CW+Locker,Coaxial LD&PD,APD+amp,OSA,10G-APD+TIA10G-PIN+TIA,Uncooled 10G-LD,40G-PIN(R&D),Can LD,1.5 Other Diode Types,1.5.3 Light-Emitting Diode(LED),It converts current into light.E/O,1.5 Other Diode Types,1.5.4 Schottky Barrier Diode,There is no minority carrier storage in the Schottky diode,so the switching time from a forward bias to a reverse bias is very short compared to the pn junction diode.,1.5.5 Zener Diode,1.Symbol and I-V characteristic,(a)symbol,(b)I-V characteristic,Zener diodes can be operated in the breakdown region by limiting the current to the capabilities of the diodes.,1.5.5 Zener Diode,2.Parameters,VZ _ Zener breakdown voltage,IZ _ Reverse-bias current when the diode is operating in the breakdown region.,3.Zener Diode Circuits,Assume Zener diode VZ=5.6V,rZ=0 and IZ max=3mA,find the value of resistance.,Sol:,Question:,Can we eliminate the R?Why?If VPS is replaced by sinusoidal input voltage vi=15sinwt(V),plot output voltage vO versus vi.,If VI is the sinwave,and VomVZ.VO=?,Think about:Does the output voltage can equal 6V?,Sol:,Which is bigger between UO and UZ?,Assume,The output voltage cant equal 6V,UO UZ,If we want the output is 6V,R,4.Ideal Voltage Reference Circuit,(1)Zener diode is useful in a voltage regulator,or a constant-voltage reference circuit.,(3)When IL is a max,and VPS is a min,IZ is a min.When IL is a min,and VPS is a max,IZ is a max.Due to,We obtain,and,(2)Ri limits Iz and drops the“excess”voltage between VPS and VZ.,4.Ideal Voltage Reference Circuit,Equating these two expressions,we know VPS(min)and VPS(max),and assume IZ(min)=0.1IZ(max).,Then we can determine Ri and max power rating PZ(max)of Zener diode.,(page 66),Sol:from Equation(3),IL(min)=0,IL(max)=100mA,VPS=11-13.6V,VZ=9V.Design Ri and PZ(max).,from Equation(2),Max power dissipated in Ri is,Diode Characteristic,Diode Characteristic,Intrinsic Semiconductor Doped Semiconductor Carriers Diffusion,Drift,Summary,Gain a basic understanding of semiconductor material properties Two types of charged carriers Two mechanisms that generate currentsDetermine the properties of a pn junction Examine dc analysis techniques for diode circuits using various models to describe the nonlinear diode characteristicsDevelop an equivalent circuit for a diode that is used when a small,time-varying signal is applied to a diode circuit,Homework:,P56 1.42P58 1.53P59 D1.61,1、You must sketch the circuit of the problem.,2、After the second week,we submit the homework at the second lesson of each week,please submit it on time.,P56 1.42,Consider the circuit shown in Figure P1.42.Determine the diode current ID and diode voltage VD for(a)V=0.6V and(b)V=0.7V.,First,determine if the diode is on or off.,Figure P1.42,P58 1.53,Consider the Zener diode circuit shown in figure P1.53.The Zener breakdown voltage is VZ=5.6V at IZ=0.1mA,and the incremental Zener resistance is.,rZ=10W,(a)Determine VO with no load.,(b)Find the range in the output voltage if VPS changes by 1V.,(c)Find VO if VPS=10V and.,Notice:rZ 0W,Figure P1.53,P59 1.53,Figure P1.61,Design a circuit to produce the characteristics shown in Figure P1.61,where iD is the diode current and vI is the input voltage.Assume the diode has piecewise linear parameters of V=0.7V and rf=0.,Exercise：（not submit),P57 1.44P57 1.47P59 D1.62,QuestionsandAnswers,
