Physics 160Lecture 4R. JohnsonAC (linear) Analysis with PSpice• See the PSpice tutorial and HW #1.0V40.00mVR16R24V_V1 N00013 0 DC 0.1Vdc AC 1VacPSpice Netlist13V100.0mVIV11Vac0.1VdcC120uFL12mHR_R1 N00013 N00019 6 R_R2 N00019 N00022 4 L_L1 0 N00022 2mHCC10N0001920uF19 22000VC_C1 0 N00019 20uF “Capture” Schematic0Hz7961PSpice Output(AC Analysis)Hz 7962 LCApril 10, 2014 Physics 160 2(AC Analysis)Nice Alternative: www.circuitlab.comThe syntax and methods are the same as in other Spice circuit simulators, but very easy to use.The free student version includes tfth t l blmost of the parts, or else reasonable substitutes, that we use in the lab. You have to register online as a student to use the free version and to be able to save your work.April 10, 2014 Physics 160 3yButterworth Low-Pass Filter• Filter that is – flat at low frequency–falls very rapidly at the 3dB point–falls very rapidly at the 3dB point•(6dB times the number of “poles”) per octave– flat and near zero above the transition •A5pole example is tested in the lab•A 5-pole example is tested in the lab– Be sure to look at the output with a ×10 probe, so that you do not load down the output and distort the response!•Demonstration example withPSpice•Demonstration example with PSpice• In the complex frequency plane, the poles (zeroes in the complex impedance) are evenly distributed on a semicircle above the real axis and centered on zero (See Wikipedia )above the real axis and centered on zero. (See Wikipedia.)April 10, 2014 Physics 160 43-Pole Butterworth FilterPlot of H in the complex frequency plane.C2=4/3 F, R4=1 , L1=3/2 H, L3=1/2 HApril 10, 2014 Physics 160 53-Pole Butterworth FilterPlot of H in the complex frequency plane.C2=4/3 F, R4=1, L1=3/2 H, L3=1/2 2413jsssssVsVsHiO where2211)()()(32Transfer function:April 10, 2014 Physics 160 6ssssVi221)(3-Pole Butterworth FilterC2=4/3 F, R4=1 , L1=3/2 H, L3=1/2 HPlot of H in the complex frequency plane.1I2I1out)()(ZRVHApril 10, 2014 Physics 160 7214inout)()(ZRVH3-Pole Butterworth Filter ResponseCalculated by solving 2 complex equations for two unknowns in MathCad.Easier way: usePSpice!April 10, 2014 Physics 160 8Easier way: use PSpice!Pspice:April 10, 2014 Physics 160 9Butterworth Frequency Response5-pole filter will fall off at 5(6)= 30 dB/octave.April 10, 2014 Physics 160 10Spice Simulation of 5-Pole ButterworthApril 10, 2014 Physics 160 11Other 5-Pole Linear FiltersApril 10, 2014 Physics 160 12LC Notch Filter (LCR application)LCf210LCRLffQ dB30• The resonance frequency is given by L and CfdB3qyg y• High “Q” (sharp dip) is obtained by a low driving impedance (small R in this case)•The phase shift is zero well off resonance but changes by 180The phase shift is zero well off resonance but changes by 180 degrees while passing through the resonance, corresponding to an inversion.April 10, 2014 Physics 160 13LC Notch Filter10Magnitude of Notch Filter Transfer FunctionR=100 ohmsL1 H0.11h ()12L= 1 mHC= 1 F0= 31.6 kHzMagnitude of Notch Filter Transfer Function1103 1104 1105 11060.01Q= 0.32110Magnitude of Notch Filter Transfer Functionh ()R=10 ohmsL= 1 mHC1F0010.112C= 1 F0= 31.6 kHzQ= 3.2April 10, 2014 Physics 160 141103 1104 1105 11060.01LC Bandpass FilterLCf210LCRCffQ dB30• The resonance frequency is given by L and CHi h“Q”( h k)i bt i db hi hdi i i dfdB3•High “Q” (sharp peak) is obtained by a high driving impedance (large R in this case)– This is desirable for selecting one radio frequency for reception, as an examplean example.– Resistance in the inductor will lower Q• The phase shift is zero at resonanceThi i h fi i i k’ l b iApril 10, 2014 Physics 160 15•This is the first experiment in next week’s lab session.LC Bandpass FilterR=100 ohmsL1 H110Magnitude of Bandpass Filter Transfer Functionh()L= 1 mHC= 1 F0= 31.6 kHz0.11h()12Q= 3.21103 1104 1105 11060.01Magnitude of Bandpass Filter Transfer FunctionR=1000 ohmsL= 1 mHC=1F110gph ()C 1 F0= 31.6 kHzQ= 320010.112April 10, 2014 Physics 160 161103 1104 1105 11060.01Scope Probes• After lab 2, you should always use ×10 probes to measure signals in your circuits at any high-impedance point.–Probesshouldbe calibrated to match the scope input using theProbes should be calibrated to match the scope input, using the calibration square wave provided on the front of the scope.– Set the scope channel to ×10, so that your voltage readings will be correct.Probe tipProbe tipAdjustable capacitor on the BNC plug end fth byes9Mof the probeno9MThe input capacitance of the diff fno1Mscope can differ from one instrument to another, so the probe capacitor is adjusted by hand until the attenuation is 1/10 at all frequencies. Then April 10, 2014 Physics 160 17Scopeqthe square wave is not distorted.Semiconductor DiodesExponential curveExponential curve for forward biasSmall (~20 nA) reverse “leakage” currentcurrent1N4148 small signal diode~75 VApril 10, 2014 Physics 160 18Lab 3 Diode Applications• Half-wave and full-wave rectifiers, as used in power supplies.– Diode drops–Filtering and ripple voltage–Filtering and ripple voltage• Voltage clamps• Rectified differentiatorDi d li it•Diode limiter• Zener diode voltage reference– This is discussed in the Lab-3 chapter, but you’ll see it in practice in Lab 12.April 10, 2014 Physics 160 19Diode RectifiersHalf WaveFull WaveApril 10, 2014 Physics 160 20Output SmoothingNote that there is no ground connection on The load (shown here as a 1 kohmresistance) draws current from thecapacitor when a diode is not supplyinggthis side of the rectifier!the current.=R1C1 decay time constantIf R1 is small (high load; i.e. high current), then C1 must be large inorder for the time constant to be long enough to minimize the ripple.LaterinthequarterwewillseehowtoimproveonthisusingactiveApril 10, 2014 Physics 160 21Laterinthequarterwewillseehowtoimproveonthisusingactivevoltage regulators to eliminate ripple and hold the output voltageconstant even with a changing