basic electronics instrumentation laboratory[基本的電子儀器實(shí)驗(yàn)室](41)

1、IVExperiment No. 7EE 312Basic Electronics Instrumentation Laboratory Wednesday, October 11, 2000Objectives:Measure dynamic impedance of a forward-biased diode & Zener diodeLearn about small-signal techniquesLearn about interference reduction through the use of proper grounding and twisted-pair techn

2、iquesBackground:What is dynamic impedance ?R=V Ird =dVd Idynamic resistanceresistanceV-I CharacteristicsVIRVIVIVIVIsloperddVdIR=VIdiodetransistorTubeOperating PointiDvDslope diD /dvDdiDrd =dvD=VDiD=VdIdIdVdID dc diode currentId ac amplitudeid ac diode currentiD total diode current+-dc circuitac circ

3、uitIDidiDidiDIDIdX d or DIEEE Standard NotationVD dc diode voltageVd ac amplitudevd ac diode voltagevD total diode voltagevdvDVDVd+-dc circuitac circuitVDvdvDID, VDId, VdSmall-Signal ConditionDynamic Resistance MeasurementiDvDIDVD2Id2Vdrd = Vd I diDvDIDVDInput Signal Too LargeiDvDIDVDInput Signal To

4、o SmallnoisenoiseMeasurement of rd looks simple.The problem is that vd in the millivolt range for forward bias.Thus, noise and stray pickup may cause trouble if you are not careful.Example:Questions: Where does come from ?!ABOscilloscopeRidStray magnetic fluxHow large is it ?Questions: Where does co

5、me from ?Answer: 1. Current iac in power lines on bench & drops from ceiling 2. fluorescent lights3. AC machinesrQuestion: How large is ? rrIABOscilloscopeR1 meterArea=1 m2Assume our experiment is about 2 meters from the power lines: r = 2 m100 amp. peak60 HZPeak value is 3.77 mV and this may be com

6、parable to signal amplitudes being measured!+-Must be concerned about in all parts of circuit.How is this problem avoided?rememberWe have control over A. We cant do much about r or I.So, we must minimize A.OSC.Step 1: Make the area smallStep 2: Twist wires togetherOSC.Twisting wires does two things,

7、1- Holds wires together2- voltages induced in adjacent sections cancelV1V212V1 -V2So induced signals cancelABOscilloscopeRKeep track of grounded leadsSingle Point GroundingUse Only One Ground Connection Such As CRO groundCan only one ground connection be realized? e. g. CRO ground. Not with BNCs bec

8、ause the each outer connector is another ground. Capacitive Coupling1. Assume 1 pF between your circuit and 120 VAC power lines.2. 60-Hz current I = jCV where = 377 rad/s at f = 60 Hz, C = 1 pF, and V = 120 VAC(rms)3. The voltage produced by I = ZxI where Z is the impedance I flows through. 4. Examp

9、le: CROZ = 1 MegVCRO = 377x1pFx120Vx1Meg = 45 mV(rms) = 130 mVppProcedures:I-Measure dynamic resistance of a Zener diode in the forward bias region.II- Simulation for Part I. (In Bell 242)III- Measure dynamic resistance in the Zener breakdown region.Components:Zener Diode 1N4742-12VDC-0.5 W2 Heathki

10、t Resistance Substitution Boxes1-kohm & 10 kohm ResistorsDecade Capacitor Box1- Dynamic Resistance in Forward RegionCH. 1CH. 2+-A0-20VR1R2dc circuitac circuitIDidCvD, iD+-A toR1dc circuitIDThe values of R1 and the voltage source are selected to control the dc bias current ID. Suppose we want ID = 10

11、 mA. Make the dc voltage across R1 = 10 VDC. Assume VD = 0.7 V. V=10.7 volts & ID =10 mAR1=1000 Ohms10 VDC+-AR2ac circuitidCR2 is selected so that ac current peak is 10% of dc current.R1=1000 OhmsR2=10,000By setting the dc power supply voltage to 10.7 VDC & the FG amplitude to 20 Vpp and R2 to 10R1,

12、 the ac current peak is 10% of dc current. I. E. ID =10 mA & id =1 mA .To obtain other values of ID & id change both R1 & R2 with R2/R1 = 10. The dc & ac voltage levels in the circuit change very little as R1 & R2 are changed to change the currents ID & id . 20Vpp1 kHzR1+-AR2ac circuitidCR2 is selec

13、ted so that ac current peak is 10% of dc current.R1=1000 OhmsR2=10,000C blocks dc current in the ac circuit & C should be large enough so that capacitance reactance is small compared with R2Note that R1 must be diode dynamic resistance so that most of the ac current goes through the diode & not the

14、dc circuit20Vpp1 kHzSelection of R2The values of R2 and the function generator voltage amplitude Vgen should be chosen to make the ac current amplitude id 10% to 20 % of ID. The corresponding diode peak ac voltage Vd will be 10% V to 20 % of nVT where VT = 25 mV at T = 290 K. (20 C). Thus Vd will be

15、 2.5 to 5 mV for n = 1 and the peak-to-peak diode ac voltage will be 5 to 10 mV. Fall 2000 Data Table For Forward rd Fall 2000 Data Table For Forward rd n=1 to 2rtheoretical ?1/Td(lnID)dVDslopegives nn?Examples: ID = 0.2 mAn = 1rd = 1X25mV/0.2mA = 125 n = 2rd = 2X25mV/0.2mA = 250 2- Simulationa- Sim

16、ulate Part 1 of experiment b- Plot I(D1) and V(2) on separate graphsc- Calculate dynamic impedance of the diode+-0-20VR1R2C01234D1DYNAMIC IMPEDANCEI1 0 1 PWL(0 .5M .00249 .5M .0025 1M .00499 1M .005 5M .00749 5M .0075 10M)D1 2 0 DIODE.MODEL DIODE D(RS=2 IS=2E-9 N=1.8)R2 3 2 15KC1 4 3 .22UV1 4 0 SIN(

17、0 5 1KHZ).TRAN .05M 10M 0 .05M.PRINT TRAN V(2) i(D1).END+-0-20VR1R2C01234D1time smA3- Dynamic Resistance of Zener in the Breakdown RegionCH. 1CH. 2+-A0-20VR1R2dc circuitac circuitIDidCvD, iDChoose values of dc bias current so that the dc power dissipation in the diode is less than 1/2 of its max rated power dissipation (1/2 Watt).Assume Zener Diode Breakdown Voltage VZ = 12VThe values of R1 and the dc voltage source are selected to control the dc bias current ID. Suppose we w