Successfully reported this slideshow.
We use your LinkedIn profile and activity data to personalize ads and to show you more relevant ads. You can change your ad preferences anytime.

Bjt+and+jfet+frequency+response

17,546 views

Published on

  • Be the first to comment

Bjt+and+jfet+frequency+response

  1. 1. BJT and JFET FrequencyResponse
  2. 2. Effects of Frequency on Operation of Circuits• The frequency of a signal can affect the response of circuits.• The reactance of capacitors increases when the signal frequency decreases,and its reactance decreases when the signal frequency increases.• The reactance of inductors and winding of transformers increases when thesignal frequency increases, and its reactance decreases when the signalfrequency decreases.• Devices such as BJTs, FETs, resistors, and even copper wires have intrinsiccapacitances, whose reactance at high frequencies could change the responseof circuits.• The change in the reactance of inductors and capacitors could affect the gainof amplifiers at relatively low and high frequencies.• At low frequencies, capacitors can no longer be treated as short circuits,because their reactance becomes large enough to affect the signal.• At high frequencies, the reactance of intrinsic capacitance of devicesbecomes low enough, that signals could effectively pass through them,resulting to changes in the response of the circuit.• At low frequencies, reactance of primary of transformers become low,resulting to poor low frequency response. Change in magnetic flux at lowfrequencies become low.• At high frequencies, the stray capacitance of transformer windings reducesthe gain of amplifiers.
  3. 3. Effects of Frequency on Operation of Circuits• Increase in the number of stages could also affect the frequency response of acircuit.• In general, the gain of amplifier circuits decreases at low and highfrequencies.• The cutoff frequencies are the frequencies when the power delivered to theload of the circuit becomes half the power delivered to the load at middlefrequencies.Voltage gainFrequency0.707 AVmidAVmidf1 f2BandwidthAvmid = voltage gain of amplifier at middle frequencies0.707 Avmid = voltage gain of amplifier at lower cutoff frequency and higher cutoff frequency(when output power is half the output power at middle frequencies)f1 = low cutoff frequency PO(HPF) = output power at higher cutoff frequency Vi = input voltagef2= high cutoff frequency PO(LPF) = output power at lower cutoff frequencyPomid= output power at middle frequencies    Omid2vmid2vmidO(LPF)(HPF) P0.5RoViA5.0RoVi0.707APPo Bandwidth = f2-f1
  4. 4. Frequency Response of Amplifier Circuits• f1 and f2 are called half power, corner, cutoff, band, break, or -3dbfrequencies.• f1 is the low cutoff frequency and f2 is the high cutoff frequency.• When the amplitude of a signal is 0.707 of its original amplitude, its powerbecomes half of its original power.PHP = PMF / 2 = power at half power frequencywhere: PHP = Power at half power point (f1 or f2)PMF = Power at middle frequencies• The bandwidth of the signal is equal to f2 – f1B = f2 – f1 = bandwidth
  5. 5. Effects of Frequency on Operation of Circuits• The 180 degrees phase shift of most amplifiers (Common emitter, commonsource) is only true at middle frequencies.• At low frequencies, the phase shift is more than 180 degrees.• At high frequencies, the phase shift is less than 180 degrees.Phase shiftbetween Voand ViFrequency18002700f1 f2Phase shift between Vo and Vi900
  6. 6. Frequency Response of Amplifier Circuits• The graph of the frequency response of amplifier circuits can be plottedwith a normalized gain. (gain is divided by the gain at middle frequencies.)Frequency0.707 AVmid1 AVmidf1 f2Normalized Gainin RatiofrequencymiddleatgainvoltageAffrequencyatgainvoltageA:whereAAGainNormalizedVmidVVmidVNormalized Plot of Voltage Gain Versus Frequency
  7. 7. Frequency Response of Amplifier Circuits• A decibel plot of the gain can be made using the following formula:VoltagegainFrequency0.707 AVmid1 AVmidf1 f2Normalized Gain in dbfrequencymiddleatgainvoltageAffrequencyatgainvoltageA:wheredbingainnormalizedAA20logAAVmidVVmidVVmidVdbDecibel plot of Normalized Voltage Gain Versus Frequency0 db-3 db-6 db-9 db
  8. 8. Capacitor Coupled Amplifier Circuit Frequency Response• For capacitor coupled (also called RC-coupled) amplifiers:– The drop in gain at low frequencies is due to the increasing reactance ofthe coupling capacitors (Cc), and bypass capacitors (Cb, CE, and Cs).– The drop in gain at high frequencies is due to the parasitic capacitanceof network and active devices, and frequency dependence of the gain ofBJTs, FETs, or vacuum tubes.VoltagegainFrequency0.707 AVmidAVmidf1 f2Drop in gain is due to increase in reactanceof coupling and bypass capacitorsAvmid = voltage gain of amplifier at middle frequencies0.707 Avmid = voltage gain of amplifier at lower cutoff frequency and higher cutoff frequency(when output power is half the output power at middle frequencies)f1 = low cutoff frequencyf2= high cutoff frequencyBandwidth
  9. 9. Transformer Coupled Amplifier Circuit Frequency Response• For transformer coupled amplifier circuits:– The drop in gain at low frequencies is caused by “shorting effect” of theinput terminals (primary) of the transformer at low frequencies. The reactanceof the primary of a transformer becomes very low at low frequencies andbecomes zero at 0 hertz.– At low frequencies, change in magnetic flux becomes low, resulting to loweroutput voltage.– The drop in gain at high frequencies is due to the stray capacitance at theprimary and secondary of a transformer, and frequency dependence of gain ofdevices. At high frequencies, the reactance of the stray capacitances becomeslow enough) that high frequency signals are also “shorted out”.VoltagegainFrequency0.707 AVmidAVmidf1 f2Drop in gain is due to “shorting” effectof primary of transformerat low frequencies.Drop in gain is due to straycapacitance at primary andsecondary of transformer andother components, andfrequency dependence of gainof active devices.Bandwidth
  10. 10. Direct Coupled Amplifier Circuit Frequency Response• For direct coupled amplifier circuits:– There are no coupling or bypass capacitors, or transformers to cause adrop in the gain at low frequencies. The gain at low frequencies istypically the same as that at middle frequencies.– The drop in gain at high frequencies is due to stray capacitance of thecircuit and the frequency dependence of the gain of active devices.VoltagegainFrequency0.707 AVmidAVmidf2Drop in gain is due to straycapacitance of the circuit, andthe frequency dependenceof the gain of active devices.Bandwidth
  11. 11. Frequency Analysis of High Pass Resistor Capacitor (RC) Circuit• A capacitor coupled circuit which acts as a high pass filter is shown below.• At middle and high frequencies, the capacitor C can be considered a shortcircuit because its reactance becomes low enough that the voltage appearingacross RL is almost equal to Vi (input voltage of combination of C and RL).• At low frequencies, the coupling capacitor C could no longer be treated as ashort circuit because its reactance becomes high enough that the voltageappearing at the load (RL) becomes significantly lower than Vi.• R can represent any resistance or resistance combination in a circuit.• At low frequencies, the RC combination of the coupling capacitor (C) and theresistance (R) determines the frequency response of the amplifier circuit.• The reactance of the coupling capacitor C can be computed as:RCcIRVi =Input voltageto RC networkCapacitor Coupled Circuit Which Acts As A High Pass FilterVo = Outputvoltage(Farad)CcofecapacitancC(Hz)signaloffrequencyf:whereffrequencyatCcofreactanceπfC21Xc
  12. 12. Frequency Analysis of High Pass Resistor Capacitor (RC) Circuit1ViVoRloadtheacrossvoltageViVosfrequenciehighatCcofreactance0fC21XcL• At high and middle frequencies, Xc becomes low enough that it can beassumed to be zero (0) and Cc is assumed to be a short circuit.– The voltage across R (Vo) can be assumed to be equal to the input voltageof the RC network (Vi).• If the frequency is equal to zero (0) such as when the signal is a DC voltage,the reactance of Cc is equal to infinity, and the capacitor Cc can be assumed tobe an open circuit.– The voltage across R (Vo) is equal to zero (0).• Between the two extremes, the ratio between Vo and Vi will vary betweenzero and one (1).0ViVoRloadtheacrossvoltage0Vohz0fwhenCcofreactancefC21XcL
  13. 13. Frequency Analysis of High Pass Resistor Capacitor (RC) Circuit2ViVowhensfrequenciemiddleatRatdissipatedpowersfrequenciemiddleatPowerRViRVi21R12ViRVoPXcRwhentageoutput vol0.707ViVo0.707Vi2ViRRRViRXcRVi)(R)I(VoXc,RWhenbelow.shownasXcRwhenoccursthisands,frequenciemiddleatpoweroutputtheofthathalfisRatpoweroutputthe(f1),frequencycutofflowthetoequalissignaltheoffrequencyWhen theRXcRVi)(R)I(Vo22222R2R2222• The magnitude of the output voltage can be computed as:
  14. 14. Frequency Analysis of High Pass Resistor Capacitor (RC) Circuit(Hz)frequencycutofflowerRC21f1f1C21XcR• When the frequency is equal to the low cutoff frequency (f1), R=XC and f1can be computed as follows:• The normalized voltage gain at lower cutoff frequency (f1) can becomputed as:• The normalized voltage gain at middle frequencies (fmid) can be computedas:3dbA0.707Alog20dbAAVmidVmidVmidcutoffVlowerdb0AAlog20dbAAVmidVmidVmidVmid
  15. 15. Frequency Analysis of High Pass Resistor Capacitor (RC) Circuit (Hz)frequencycutofflowerRC21f1:where(unitless)ffrequencyatgainvoltageff1j11AvfCR21j11RjXc11jXcRRjXcRIRIViVoAv• At frequency f, the voltage gain can be computed as:• In magnitude and phase form, the voltage gain at any frequency can becomputed as:fi/f/Tanff111Av 12
  16. 16. Frequency Analysis of High Pass Resistor Capacitor (RC) Circuit(db)ff111log20ViVolog20Av2db• In db (logarithmic form), the voltage gain at frequency f can be computedas:• When f=f1= lower cutoff frequency,db3-f1f111log20ViVolog20Av2db 
  17. 17. Frequency Analysis of High Pass Resistor Capacitor (RC) Circuit22/122dbff11log10ff11-20log(db)ff111log20ViVolog20Av• The voltage gain at frequency f can be written as:• When f<<f1, the above equation can be approximated by:• If we forget the condition f<<f1 and plot the right side of the aboveequation, the following points can be used. The plot is a straight line whenplotted in a log scale.db20-20log10-f1/10fAtdb12-20log4-f1/4fAtdb6-20log2-f1/2fAtdb020log1-f1fAtf1)f(whenff1log20ff1-20logAv2/12db ff1log20ff1-20logX2/12
  18. 18. Frequency Analysis of High Pass Resistor Capacitor (RC) Circuit• Using the points in the preceding slide, a bode plot can be made as shown below.• A Bode plot is a piecewise linear plot of the asymptotes and associatedbreakpoints.• A Bode plot for the low frequency region is shown below.• One octave is equivalent to a change in frequency by a factor of two (2).• One octave results to a 6 db change in the normalized gain.• One decade is a change in frequency by a factor of 10.• One decade results to a 20 db change in the normalized gain.Frequency (log scale)0.707 AV1 AVf1Normalized Gainin dbBode Plot for Low Frequency Region0 db-3 db-6 db-9 db-12 db-15 db-18 db-21 dbf1/2f1/4f1/10ff1log20X  (db)ff111log20Av2dbActual Response CurveAsymptoteAsymptote
  19. 19. Frequency Analysis of High Pass Resistor Capacitor (RC) Circuit• The plot in the preceding slide shows two asymptotes. One for f<< f1 (6 db /octave), and the other for f >> f1 (horizontal line – 0 db).• The plot of the line corresponding to f << f1 results to frequency response of6db per octave (6 db drop in gain for every reduction in the frequency by afactor of 2). The plot also corresponds to a response of 20 db per decade.• The decibel plot of voltage gain (Av) can be made by using the informationon the asymptotes and knowing that at f= f1, Avdb = -3db.
  20. 20. Frequency Analysis of High Pass Resistor Capacitor (RC) Circuit• Example: For the RC network shown below, Determine the break frequency(cutoff frequency), sketch the asymptotes and the frequency response curve.Frequency (log scale)0.707 AV1 AVf1 =99.47Normalized Gainin dbBode Plot for Low Frequency Region0 db-3 db-6 db-9 db-12 db-15 db-18 db-21 dbf1/2=49.74f1/4=24.87f1/10=9.947RL =8 kohmCc = 0.2 microfaradIRLVi =Input voltageto RC networkVo = Outputvoltage Hz99.47)0x12.0)(000,8(21CR21f1 6L AsymptoteAsymptote-3db point
  21. 21. Low Frequency Analysis of Capacitor Coupled BJT Amplifier• A capacitor coupled (also called RC coupled) BJT amplifier circuit is shownbelow.• At middle and high frequencies, the capacitors Cc, Cs, and Ce can beconsidered short circuits because their reactance become low enough, that thereare no significant voltage drops across the capacitors.• At low frequencies, the coupling capacitors Cc, Cs, and Ce could no longer betreated as short circuits because their reactance become high enough that thethere are significant voltage drops across the capacitors.RcQ1VccRB2CcVo = OutputvoltageRLViVsRsigCsRB1RECeZi ZoVs = Signal sourceRsig = internal resistance of signal sourceCs =coupling capacitor for VsCc= coupling capacitor for RLCe= bypass capacitor for RE
  22. 22. Low Frequency Analysis of Capacitor Coupled BJT Amplifier• The frequency analysis of high pass RC network can be used for capacitorcoupled BJT amplifier circuits. The values of R and C are taken from theequivalent resistances and capacitances in the BJT amplifier circuit.• For the portion of the circuit involving the coupling capacitor Cs, theequivalent circuit is shown below.– Equivalent circuit assumes that the input impedance of the amplifier (Zi) ispurely resistive and is equal to Ri.CsZi = RiViEquivalent Circuit of Vs, Cs and ZiIiCsZi = RiViEquivalent Circuit of Vs, Cs and ZiIiRB1//RB2 hie = rereZi = RiVsRsigVsRsig
  23. 23. Low Frequency Analysis of Capacitor Coupled BJT AmplifierCsjXRsigRiRiVsVi• The value of the input impedance (resistance) of the amplifier can be computedas:Zi = Ri = RB1 // RB2 // hie= RB1 // RB2 // re• The voltage Vi can be computed using voltage divider rule.• The voltage Vi at middle frequencies (Cs can be considered short circuit) canbe computed as:• The lower cutoff frequency can be computed as:RsigRiRiVsVi midCsinvolvingcircuittheofportionfor thefrequencyoffcutlowerRi)Cs(Rsig21fLs 
  24. 24. Low Frequency Analysis of Capacitor Coupled BJT Amplifier• For the portion of the circuit involving the coupling capacitor Cc, theequivalent circuit is shown below.– Equivalent circuit assumes that the output impedance of the transistor ispurely resistive and is equal to ro.CcZo= RoVRLEquivalent Circuit of Circuit Portion Involving CcIRLRc RLib roCcZo= Ro = Rc // roVRLIRLRc // roRL
  25. 25. Low Frequency Analysis of Capacitor Coupled BJT Amplifier• The value of the output impedance (resistance) of the amplifier can becomputed as:Zo = Ro = RC // ro• The lower cutoff frequency can be computed as:Ccinvolvingcircuittheofportionfor thefrequencyoffcutlower)CR(Ro21fCLLC 
  26. 26. Low Frequency Analysis of Capacitor Coupled BJT Amplifier• For the portion of the circuit involving the bypass capacitor Ce, the equivalentcircuit is shown below.– The equivalent circuit uses the re model.• The resistance (Re) seen looking into RE from the output side can be computedas:Ce(Rs’/ + reEquivalent Circuit of Portion of Circuit Involving RE and CERE(Ampere)currentquiescentEmitter(Ampere)currentDCEmitterI(ohms)I10X26r//RRsig//RRs:Where(ohms)sideoutputthefromRintolookingseenimpedancerRs//RReEE3-eB2B1EeE eE rRs//RRe
  27. 27. Low Frequency Analysis of Capacitor Coupled BJT Amplifier• The lower cut-off frequency of the portion of the circuit involving thebypass capacitor Ce can be computed as:• The voltage gain of the amplifier without considering the effects of thevoltage source resistance Rsig can be computed as:• At middle frequencies, RE is shorted out because the reactance of Ce isvery low. Voltage gain can be computed as:• At low frequencies, the reactance of Ce becomes high and RE should beconsidered in the computation of the voltage gain.side.outputthefromRintolookingresistanceequivalentRe:wherecircuittheofportionfor thefrequencyoffcutlowerCeRe21fELE)considerednot(rosfrequencielowatamplifiertheofgainvoltageRrRc//RViVViVoAvEeLRLsfrequenciemiddleatamplifiertheofgainvoltagerro//Rc//RViVViVoAveLRL
  28. 28. Low Frequency Analysis of Capacitor Coupled BJT Amplifier• Overall, the effects of the capacitors Cs, Cc, and Ce must be considered indetermining the lower cutoff frequency of the amplifier.• The highest lower cutoff frequency among the three cutoff frequencies willhave the greatest impact on the lower cutoff frequency of the amplifier.• If the cutoff frequencies due to the capacitors are relatively far apart, thehighest lower cutoff frequency will essentially determine the lower cutofffrequency of the amplifier.• If the highest lower cutoff frequency is relatively close to another lower cutofffrequency, or if there are more than one lower cutoff frequencies, the lowercutoff frequency of the amplifier will be higher than the highest lower cutofffrequency due to the capacitors.fLT = overall lower cutoff frequency of amplifierfLT > fLSfLT > fLCFlt > fLE
  29. 29. Low Frequency Analysis of Capacitor Coupled BJT Amplifier• Example: A voltage divider BJT amplifier circuit has the parameters listed below.Determine the low cutoff frequency of the amplifier and sketch the low frequencyresponse.Cs=12 uF Rsig = 2 kohm RL= 2 kohm RB1=50 kohm RB2= 10 kohmCc=2 uF RE = 2 kohm RC= 4 kohm =100 Vcc= 20 voltsCE=20 uF Assume that output resistance of transistor to be infinite.742.1910x1.31710x26reCurrentQuiescentEmitterA10x1.31720007.0333.3RVVRVIgroundtorelativebaseatvoltageDCvolts3.33350,00010,0000)(20)(10,00RRRVVdone.becanionsapproximatfollowingtheandRohms20,0000)(100)(2,00Rβ333EBEBEREEB1B2B2CCBB2E
  30. 30. Low Frequency Analysis of Capacitor Coupled BJT Amplifier0.4442,0001,596.081,596.08RsigRiRiVsViamplifierofimpedanceinputohms1,596.082.974,11000,101000,5011r////RRRiZiRs)gconsiderin(notsfrequenciemidatgainvoltage538.67742.19000,2000,4)000,2)(000,4(rRc//RViVoAv:sfrequenciemiddleAt2.974,1)742.19(100)(βreB2B1eLmide
  31. 31. Low Frequency Analysis of Capacitor Coupled BJT AmplifierCcinvolvingcircuitoffrequencycutofflowerHz13.26310X2)000,2(4,00021)CcR(Rc21f:CcofeffectsthegConsiderinCsinvolvingcircuittheofportionfor thefrequencyoffcutlowerHz3.68810X21,596.08)1(2,00021Ri)Cs(Rsig21famplifierofimpedanceinputohms1,596.082.974,11000,101000,5011r////RRRiZi:CsofeffectsthegConsiderinVs)ofresistance(internalRsgconsideringainvoltage9.98720.444)(-67.538)(VsViViVoVsVoAvs6-LLC6-LSeB2B1mid
  32. 32. Low Frequency Analysis of Capacitor Coupled BJT Amplifiercircuit.amplifierwholetheoffrequencycutoffaffect thetlypredominanwillfrequencycutoffloweritsCc,andCstoduefrequeciescutoffthetocomparedhighrelativelyisCinvolvingcircuittheofportiontheoffrequencycutofflowertheBecauseHz225.82210X205.239)3(21CRe21fohms5.2393742.911001612.903//2000reβR//RReohms1612.903000,101000,501200011//RRsig//RRs:CofeffectsthegConsiderinE6-ELESEB2B1E
  33. 33. Low Frequency Response of JFET Common Source AmplifierCcID = Drain current(ac)VGSVDSIG (GateCurrent)= 0Drain (D)SourceGate (G)VO=Output voltageRG1Vi =InputvoltageCGRDVDDZi ZoCsRs• The analysis of low frequency response of FET amplifiers is similar to that ofBJT amplifiers.• At middle and high frequencies, the capacitors Cc, Cs, and CG can beconsidered short circuits because their reactance become low enough, thatthere are no significant voltage drops across the capacitors.• At low frequencies, the coupling capacitors Cc, Cs, and CG could no longerbe treated as short circuits because their reactance become high enough thatthe there are significant voltage drops across the capacitors.Vs =SourcevoltageRsigIiRLRG2
  34. 34. Low Frequency Response of JFET Common Source Amplifier• The frequency analysis of high pass RC network can be used for capacitorcoupled FET amplifier circuits. The values of R and C are taken from theequivalent resistances and capacitances in the FET amplifier circuit.• For the portion of the circuit involving the coupling capacitor CG, theequivalent circuit is shown below.– Equivalent circuit assumes that the input impedance of the amplifier (Zi) ispurely resistive and is equal to Ri.CGZi = Ri= RG1 // RG2ViEquivalent Circuit of Vs, CG and ZiIiCGViEquivalent Circuit of Vs, CG and ZiIiRG1 // RG2VsRsigVsRsigZi = Ri= RG1 // RG2Zi = Ri= RG1 // RG2
  35. 35. Low Frequency Response of JFET Common Source AmplifierCGjXRsigRiRiVsVi• The value of the input impedance (resistance) of the amplifier can be computedas:Zi = Ri = RG1 // RG2 Zi = RG2 if RG1 is not present (RG1 = infinity)• The voltage Vi can be computed using voltage divider rule.• The voltage Vi at middle frequencies (when CG can be considered as shortcircuit) can be computed as:• The lower cutoff frequency (half power frequency) can be computed as:RsigRiRiVsVi midGGLGCinvolvingcircuittheofportionfor thefrequencyoffcutlowerRi)C(Rsig21f 
  36. 36. Low Frequency Response of JFET Common Source Amplifier• For the portion of the circuit involving the coupling capacitor Cc, theequivalent circuit is shown below.– Equivalent circuit assumes that the output impedance of the transistor ispurely resistive and is equal to Ro.rdDrain (D) IDRDZo = RoRLgmVgs+- ---+ +gmVgsIrd IRD IRLCc
  37. 37. Low Frequency Response of JFET Common Source Amplifier• The value of the output impedance (resistance) of the amplifier can becomputed as:Zo = Ro = RD // rd Zo = Ro = RD if rd is equal to infinity• The lower cutoff frequency can be computed as:Ccinvolvingcircuittheofportionfor thefrequencyoffcutlower)CR(Ro21fCLLC 
  38. 38. Low Frequency Response of JFET Common Source Amplifier• For the portion of the circuit involving the bypass capacitor Cs, the equivalentcircuit is shown below.– The resistance (Req) seen looking into Rs from the output side can becomputed as:CsEquivalent Circuit of Portion of Circuit Involving RS and CSRSgm1Rs//Reqbewillaboveequationtheinfinity,rwhen(ohms)sideoutputthefromRintolookingseenimpedance//RRrrgm1Rs1RsReqdSLDddReqSystem
  39. 39. Low Frequency Response of JFET Common Source Amplifier• The low cut-off frequency of the portion of the circuit involving the bypasscapacitor Cs can be computed as:side.outputthefromRintolookingresistanceequivalentReq:wherecircuittheofportionfor thefrequencyoffcutlowerCsReq2π1fSLs• Overall, the effects of the capacitors CG, Cc, and CS must be considered indetermining the low cutoff frequency of the amplifier.• The highest lower cutoff frequency among the three cutoff frequencies willhave the greatest impact on the low cutoff frequency of the amplifier.• If the cutoff frequencies due to the capacitors are relatively far apart, thehighest low cutoff frequency will essentially determine the low cutofffrequency of the amplifier.• If the highest lower cutoff frequency is relatively close to another lower cutofffrequency, or if there are more than one lower cutoff frequency, the low cutofffrequency of the amplifier will be higher than the highest lower cutofffrequency due to the capacitors.involving Cs
  40. 40. Low Frequency Response of JFET Common Source Amplifier• Example: Given a common source FET amplifier with the followingparameters, determine the lower cutoff frequency of the amplifier.CG =0.02F Cc = 0.6 F Cs = 2 FRsig = 12 K RG= 1 Mohm RD= 5 K Rs= 1 K RL = 2 KIDSS= 9 mA Vp= -7 volts rd= infinity VDD = 20 voltsSince RG1 is not present, configuration is self bias FET. negative)moregoesVwhenreachedfirstisvalue(ThisA10x2.9806IChoose1016.49xIA10x2.9806I2(183))10x4(183)(93.5653.565III183I565.310x90I183I565.210x920,408II285110x97-)(1000)(I-110x9I(Ampere)currentDrainV))(R(I-1IVV1IIIGS3DQ3DQ3DQ32DQDDD3DD3DD32D3D2PSDDSS2PGSDSSDQD222
  41. 41. Low Frequency Response of JFET Common Source Amplifier3voltsvolts2.9806)1000)(10x-(2.9806))(R-(IV 3SDGSQ  ctancetransconduSiemensx1033.19-3-19)x10(92VV1VI2gm 3--3PGSPDSSHz7.8610x)0.02000,000,1(12,00021)CR(Rsig21Ri)C(Rsig21f 6-GGGLG Hz89.730.6x10)000,2(5,00021)CR(R21)CR(Ro21f 6-CLDCLLC amplifiertheoffrequencycutofflowon theimpacthighestthehasits,frequenciecutofflowerthreetheoflargesttheisfSinceHz185)x10(429.18)(221CsReq2π1fLS6-Ls ohms18.29410x33.1/11000)10x33.1/1)(1000(gm1Rs//Reqinfinity,rdSince 3--3
  42. 42. Low Frequency Response of JFET Common Source Amplifiersfrequenciemidlleatgainvoltage9.12,0005,000000)(5,000)(2,)10x(1.33)//Rgm(RViVoAvmid 3-LDAv / Avmid (db)NormalizedVoltagegainFrequency0.707 AVmid1 AVmidfLSfLCNormalized Gain in dbLow Frequency Response (Normalized Voltage Gain Versus Frequency0 db-3 db-5 db-25 db-20 db-15 db-10 db1 10 100 1K 10K 100K 1M 10M 100MfLG fHi fHof- 20 db / decade
  43. 43. High Frequency Response of Low Pass RC Network• At the high frequency end, the frequency response of a low pass RC networkshown below is determined by the decrease in the reactance of the capacitor asfrequency of operation increases.• Because of the decrease in the capacitance, there is a “shorting” effect acrossthe terminals of the capacitor at high frequencies, and the voltage drop acrossthe capacitor decreases as frequency increases.RIRVi =Input voltageto RC networkVo = OutputvoltageFrequency (log scale)0.707 AV1 AVf2Normalized Gainin dbBode Plot for High Frequency Region0 db-3 db-6 db / octaveCAv = Vo / Vi
  44. 44. High Frequency Response of Low Pass RC Network• The voltage gain of the low pass RC network can be computed as: sfrequenciemidatgainvoltagethetimes0.707isgaingewhen voltafrequency(Hz)frequencycutoffhighRC21f2:where(unitless)ffrequencyatgainvoltagef2fj11Avff21j11fRC211j11R1R1fC21Rj11fC21Rj11jXcR11jXcRjXc-jXcRI(-jXc)IViVoAv• The above equation results to plot that drops off at 6db per octave withincreasing frequency.
  45. 45. Miller Effect Capacitance• When the frequencies being processed by an amplifier are high, the frequency responseof the amplifier is affected by:• Interelectrode (between terminals) capacitance internal to the active device• Wiring capacitance between leads of the network• The coupling and bypass capacitors are considered short circuits at mid and highfrequencies because their reactance levels are very low.• The diagram below shows the existence of a “feedback” capacitance whose reactancebecomes significantly low at high frequencies, that it affects the performance of anamplifier.• The input and output capacitance are increased by a capacitance level sensitive to theinterelectrode (between terminals) capacitance (Cf) between the input and outputterminals of the device and the gain of the amplifier.• Because of Cf, an equivalent capacitance, called Miller capacitance, is produced at theinput and output.Vo+-Vi+-ZiAv =Vo / ViCfI2I1IiRi
  46. 46. Miller Effect Capacitance• The value of the Miller effect input capacitance can be computed as:CMifMiCMff2121X1Ri1Zi1ecapacitancinputeffectMillerCAv)(1CecapacitancinputeffectMillerofReactanceXfC2Av)(11CAv)(11Av)(1XcfAv)(1Xcf1Ri1Zi1XcfAv)(1Ri1Zi1XcfAv)Vi(1RiViZiViIIIiXcfAv)Vi(1XcfAvViViXcfVoViIRiViIZiViIi
  47. 47. Miller Effect Capacitance• The equivalent circuit due to the Miller Effect Capacitance is shown below.• Above results show that for any inverting amplifier (negative AV), the inputcapacitance will be increased by a Miller effect capacitance, which is afunction of the gain of the amplifier and the interelectrode (parasitic)capacitance between the input and output terminals of the active device.• If the voltage gain is negative (with phase reversal), Miller Effect capacitance(CM) is positive and higher than the interelectrode capacitance.• If the voltage gain is positive (no phase reversal) and greater than 1, MillerEffect capacitance (CM) is negative.+-ViZiIiCM i= (1-AV)CfRi
  48. 48. Miller Effect Capacitance• At high frequencies, the voltage gain Av is a function of the Miller effectcapacitance (CM).• There is difficulty in solving the value of the Miller effect capacitance (CM)since it is a function of the gain AV which in turn is a function of the Millereffect capacitance.– In general, the midband value of the voltage gain is used for AV, to getthe worst case scenario for the Miller effect capacitance, since thehighest value of Av is the midband value.• The Miller effect also increases the level of the output capacitance, and itmust also be considered in determining the high cutoff frequency.• The diagram below shows the “feedback” capacitor as seen in the output sideof the amplifier.Vo+-Vi+-ZoAv =Vo / ViCf I2I1IoRo
  49. 49. Miller Effect Capacitance• The Miller effect output capacitance can be determined as follows:ecapacitancoutputeffectMillerofReactanceC1CAv111Av11XcfIoVoXcfAv11VoIoXcfAv11VoXcfAvVoVoXcfViVoIo:byedapproximatbecanIolargelysufficientusuallyisRobecausesmallryusually veisRoVobutXcfViVoRoVoIoXcfViVoIRoVoIZoVoIoIIIoMof2121
  50. 50. Miller Effect CapacitanceecapacitancoutputeffectMillerCfC:byedapproximatbecanecapacitancoutputeffectMillerthe1,angreater thmuchisAvWhenecapacitancoutputeffectMillerCfAv11CMoMo
  51. 51. BJT High Frequency Response• At the high frequency end, the high cutoff frequency (-3 db) of BJT circuits isaffected by:– Network capacitance (parasitic and induced)– Frequency dependence of the current gain hfe• At high frequencies, the high cutoff frequency of a BJT circuit is affected by:– the interelectrode capacitance between the base and emitter, base andcollector, and collector and emitter.– Wiring capacitance at the input and output of the BJT.• At high frequencies, the reactance of the interelectrode and wiring capacitancebecome significantly low, resulting to a “shorting” effect across thecapacitances.• The “shorting” effect at the input and output of an amplifier causes a reductionin the gain of the amplifier.• For common emitter BJT circuits, Miller effect capacitance will affect the highfrequency response of the circuit, since it is an inverting amplifier.
  52. 52. BJT High Frequency Response• The figure below shows the RC network which affects the frequency response ofBJT circuits at high frequencies.CcIRCCwoIB CEBVO= OutputvoltageREVi =Input voltageCSRCVCCRB2RB1IRB2IRB1CeZi ZoZix ZoxIRLCwiVs =SourcevoltageRsigIiCbcCceCbeCbe = capacitance between the base and emitter of transistorCce = capacitance between collector and emitter of transistorCbc = capacitance between base and collector of transistorCwi = wiring capacitance at input of amplifierCwo = wiring capacitance at output of amplifier
  53. 53. BJT High Frequency Response• The figure below shows the ac equivalent circuit of the BJT amplifier in thepreceding slide.• At mid and high frequencies, Cs, Cc, and Ce are assumed to be short circuitsbecause their impedances are very low.• The input capacitance Ci includes the input wiring capacitance (Cwi), thetransistor capacitance Cbe, and the input Miller capacitance CMi.• The output capacitance Co includes the output wiring capacitance (Cwo), thetransistor parasitic capacitance Cce, and the output Miller capacitance CMo.• Typically, Cbe is the largest of the parasitic capacitances while Cce is thesmallestECiIcIbVo=VceRi  re= reroZix ZoxVi IbERLBIiCoRCRiRB1// RB2VsRsigCi = Cwi + Cbe + CMi Co = Cwo + Cce + CMoThi Tho
  54. 54. BJT High Frequency Response• The Thevenin equivalent circuit of the ac equivalent circuit of the BJT amplifieris shown below.• For the input side, the -3db high cutoff frequency can be computed as:Vo=VceViCoRiRThi = Rsig // RB1// RB2 // RiVThiThi ThoVThoCiRTho= Rc // RL// rocircuitofecapacitancinputAv)C-(1CCCCCCsideinputatresistanceequivalentThevenin1)re(//R//R//RsigRi//R//R//RsigRfrequency)(-3dbsideinputfor thefrequencyoffcuthigherCiR21fbcbewiMibewiiB2B1B2B1THiThiHi
  55. 55. BJT High Frequency Response• At the high frequency end, the reactance of capacitance Ci will decrease asfrequency increases, resulting to reduction in the total impedance at the inputside.– This will result to lower voltage across Ci, resulting to lower base current,and lower voltage gain.• For the output side, the -3db high cutoff frequency can be computed as:• At the high frequency end, the reactance of capacitance Co will decrease asfrequency increases, resulting to reduction in the total impedance at the outputside.– This will result to lower output voltage Vo, resulting to lower voltage andpower gain.circuitofecapacitancoutputCAv1-1CCCCCCsideoutputatresistanceequivalentThevenin//roR//RcRfrequency)(-3dbsideoutputfor thefrequencyoffcuthigherCoR21fbccewoMocewooLTHoThoHo
  56. 56. BJT High Frequency Response• The Hybrid or Giacolletohigh frequency equivalent circuit for commonemitter is shown below.• The resistance rb includes the base contact resistance (due to actual connectionto the base) , base bulk resistance (resistance from external base terminal to theactive region of transistor), and base spreading resistance (actual resistancewithin the active region of transistor).• The resistances r, ro, and ru are the resistances between the indicated terminalswhen the BJT is in the active region.• Cbe and Cbc are the capacitances between the indicated terminals.ECu = CbcIcIbro =1 / hoeZix ZoxEBC = Cber  rerbCru Ib =hfe IbHybrid  High Frequency Equivalent Circuit (Common Emitter)
  57. 57. BJT High Frequency Response• At the high frequency end, hfe of a BJT will be reduced as frequency increases.• The variation of hfe (or ) with frequency can approximately be computed as:    rtransistoofcurrentemitterDCII26mVr))(r(hferrCuCr211fCuCr21hfe1CuCr21ffsheet)specsatgivenusuallyone(thefrequencymiddleathfehfe:whereffrequencyathfeffj1hfehfeEEeemidemidmidβmidhfeβmidmidβmidee
  58. 58. BJT High Frequency Response• Since re is a function of the DC emitter current IE, and f is a function of re, f is afunction of the bias condition of the circuit.• hfe will drop off from its midband value with a 6 db / octave slope.• For the common base configuration:• It has improved frequency response compared to common emitter configuration.• Miller effect capacitance is not present because of its non-invertingcharacteristics.• f is higher than f.Frequency (log scale)fNormalized hfein dbBode Plot for hfe () in the High Frequency Region0 db-3 db-6 db / octave (for f)hfe / hfe mid
  59. 59. BJT High Frequency Response• The relationship of f (-3db high cutoff frequency for ) and f db highcutoff frequency for ) is shown below.• The upper cutoff frequency of the entire system (upper limit for the bandwidth)is lower than the lowest upper cutoff frequency (lowest among fHi, fHo, and f)• The lowest upper cutoff frequency has the greatest impact on the bandwidth ofthe system. It defines a limit for the bandwidth of the system.• The lower is the upper cut off frequency, the greater is its effect on thebandwidth of the entire system. forfrequencyoffcutdb3)1(ff Frequency (log scale)fNormalized hfbin dbBode Plot for hfb () in the High Frequency Region (Common Base)0 db-3 db-6 db / octave (for f)hfb / hfb mid f
  60. 60. BJT High Frequency Response• The gain-bandwidth product of a transistor is defined by the following condition:  productbandwidthgainCuCr21fCuCr211)())(f(fbandwidthffbandwidthfandgainsinceproduct,bandwidthgain))(f())(fhfe(f1ffhfeff1hfe:ascomputedis)f(fwhenhfeofmagnitudetheandfbydenotedis0dbtoequalishfeat whichfrequencyThedb01log20ffj1hfelog20hfeand1ffj1hfehfeTmidmidβmidTmidTββmidβmidβmidTβTmid2βTmidTT,dbβmiddbβmidee
  61. 61. BJT High Frequency Response• Example: Given a common emitter BJT amplifier with the following parameters,determine the following:a. High cutoff frequency for the input of the circuit (fHi)b. High cutoff frequency for the output of the circuit (fHo)c. High cutoff frequency for fd. Gain bandwidth product (fT) e. Sketch the frequency response for the low and high frequency range Specs similar to example on BJT low frequency response:Cs=12 uF Rsig = 2 kohm RL= 2 kohm RB1=50 kohm RB2= 10 kohmCc=2 uF RE = 2 kohm RC= 4 kohm Vcc= 20 voltsCE=20 uF  = hfemid = 100 ro = infiniteAdditional specs:C = Cbe= 35 pF Cu = Cbc= 3 pF Cce = 1 pFCwi = 5 pF Cwo = 6 pF
  62. 62. BJT High Frequency Responseresponse)frequencylow(foramplifierofimpedanceinputohms1,596.08RiZiRs)gconsiderin(notsfrequenciemidatgainvoltage538.67Av742.19rresponsefrequencyhighforRi2.974,1)742.19(100)(βrresponse,frequencylowonexampleprevioustheFrommideeCeinvolvingcircuitoffrequencycutofflowerHz225.822fCcinvolvingcircuitoffrequencycutofflowerHz263.13fCsinvolvingcircuittheofportionfor thefrequencyoffcutlowerHz3.688fVs)ofresistance(internalRsgconsideringainvoltage9.9872AvsLELCLSmidohms678.8872.974,11000,101000,501000,211Ri//R//R//RsigRpF614.452pF3(-67.538))-(1pF35pF5CbcAv)-(1CbeCwiCi:responsefrequencyhighFor theB2B1Thi
  63. 63. BJT High Frequency ResponseHz359,884,11)10x)(10.04433.333,1(21)(Co)R(21fpF044.01pF3538.6711pF1pF6CbcAvmid11CceCwoCCceCwoCoohms33.333,12,0004,000000)(4,000)(2,R//RcRHz981,729)10x614.45)(2678.887(21)(Ci)R(21f12-ThoHoMoLTho12-ThiHi    1hfewhenfrequencyHz800,092,215)892,150,2)(001())(fhfe(fHz892,150,2x103x103519.472)(21(100)1CuCr21hfe1CuCr21ffβmidT1212midhfeβ  e
  64. 64. BJT High Frequency Response• In the low frequency region, the lower cutoff frequency due to the emitter capacitor(fLE) has the highest value. Consequently, it has the greatest impact on the bandwidth ofthe system, among the three lower cutoff frequencies.• In the high frequency region, the high cutoff frequency due to the input capacitors andresistors (fHi) has the lowest value. Consequently, it has the greatest impact on thebandwidth of the system, among the three high cutoff frequencies.Av / Avmid (db)NormalizedVoltagegainFrequency0.707 AVmid1 AVmidfLEfLCNormalized Gain in dbFull Frequency Response (Normalized Voltage Gain Versus Frequency)0 db-3 db-5 db-25 db-20 db-15 db-10 db1 10 100 1K 10K 100K 1M 10M 100MfLS fHi fHofBandwidth- 20 db / decade(-6 db / octave)+20 db / decade(6 db/octave)
  65. 65. FET High Frequency Response• The high frequency response analysis for FET is similar to that of BJT.• At the high frequency end, the high cutoff frequency (-3 db) of FET circuits isaffected by the network capacitance (parasitic and induced).• The capacitances that affect the high frequency response of the circuit arecomposed of:– the interelectrode capacitance between the gate and source, gate and drain,and drain and source.– Wiring capacitance at the input and output of the circuit.• At high frequencies, the reactance of the interelectrode and wiring capacitancebecome significantly low, resulting to a “shorting” effect across thecapacitances.• The “shorting” effect at the input and output of an amplifier causes a reductionin the gain of the amplifier.• For common source FET circuits, the Miller effect will be present, since it is aninverting amplifier.
  66. 66. FET High Frequency Response• The figure below shows the RC network which affects the frequency response ofFET circuits at high frequencies.CcIRDCwoIG DSGVO= OutputvoltageRSVi =Input voltageCGRDVCCRG2RG1IRG2IRG1CSZi ZoZix ZoxIRLCwiVs =SourcevoltageRsigIiCgdCdsCgsCgs = capacitance between the gate and source of transistorCds = capacitance between drain and source of transistorCgd = capacitance between gate and drain of transistorCwi = wiring capacitance at input of amplifierCwo = wiring capacitance at output of amplifierCommon Source FET Amplifier Circuit
  67. 67. FET High Frequency Response• The figure below shows the ac equivalent circuit of the FET amplifier.• At mid and high frequencies, CG, CS, and Cc are assumed to be short circuits becausetheir impedances are very low.• The input capacitance Ci includes the input wiring capacitance (Cwi), the transistorcapacitance Cgs, and the input Miller capacitance CMi.• The output capacitance Co includes the output wiring capacitance (Cwo), thetransistor parasitic capacitance Cds, and the output Miller capacitance CMo.• Typically, Cgs and Cgd are higher than Cds.• At high frequencies, Ci will approach a short-circuit and Vgs will drop, resulting toreduction in voltage gain.• At high frequencies, Co will approach a short-circuit and Vo will drop, resulting toreduction in voltage gain.SCiIdVo=VdsrdZix ZoxVi = Vgsgm VgsSRLGIiCoRDRG1// RG2VsRsigCi = Cwi + Cgs + CMi Co = Cwo + Cds + CMoThi ThoDIRL
  68. 68. FET High Frequency Response• The Thevenin equivalent circuit of the ac equivalent circuit of the FET amplifieris shown below.• For the input side, the -3db high cutoff frequency can be computed as:Vo=VdsViCoRThi = Rsig // RG1// RG2VThiThi ThoVThoCiRTho= RD // RL// rd scenariocaseworstget thetoAvforusedisAvmidwheresideinputatecapacitanceffectMillerCgdAv1CcircuitofecapacitancinputCCgsCCsideinputatresistanceequivalentTheveninR//R//RsigRfrequency)(-3dbsideinputfor thefrequencyoffcuthighCiR21fMiMiwiiG2G1THiThiHi
  69. 69. FET High Frequency Response• For the output side, the -3db high cutoff frequency can be computed as:scenariocaseworstget thetoAvforusedisAvmidsideoutputat theecapacitanceffectMillerCgdAv11CcircuitofecapacitancoutputCCdsCCsideoutputatresistanceequivalentThevenin//rdR//RRfrequency)(-3dbsideoutputfor thefrequencyoffcuthighCoR21fMoMowooLDTHoThoHo
  70. 70. FET High Frequency Response• Example: Given a common source FET amplifier with the following parameters,determine the following:a. High cutoff frequency for the input of the circuit (fHi)b. High cutoff frequency for the output of the circuit (fHo)c. Sketch the frequency response for the low and high frequency range Specs similar to example on FET low frequency response:CG =0.02mF Cc = 0.6 mF Cs = 2 mFRsig = 12 K RG= 1 Mohm RD= 5 K Rs= 1 K RL = 2 KIDSS= 9 mA Vp= -7 volts rd= infinity VDD = 20 voltsSince RG1 is not present, configuration is self bias FET.Additional specs:Cgd= 3 pF Cgs = 5 pF Cds = 1 pFCwi = 5 pF Cwo = 6 pF
  71. 71. FET High Frequency ResponseCsinvolvingcircuittoduefrequencycutofflowerHz851fCcinvolvingcircuittoduefrequencycutofflowerHz37.89fCinvolvingcircuittoduefrequencycutofflower7.86HzfRs)gconsiderin(notsfrequenciemidatgainvoltage9.1Avresponse,frequenclowonexampleprevioustheFromLSLCGLGmidHz800,717)10x8.71)(857,11(21)(Ci)R(21fohms857,11000,000,1000,12)000,000,1)(000,12(R//RsigRpF7.18pF3(-1.9))-(1pF5pF5CgdAv)-(1CgsCwiCi:responsefrequencyhighFor the12-ThiHiGThi
  72. 72. FET High Frequency ResponseHz605,621,9)10x579.1)(157.428,1(21)(Co)R(21fpF579.11pF39.111pF1pF6CgdAvmid11CdsCwoCCdsCwoCoohms57.428,12,0005,000000)(5,000)(2,R//RR12-ThoHoMoLDTho
  73. 73. FET High Frequency Response• In the low frequency region, the lower cutoff frequency due to the source capacitor (fLS)has the highest value. Consequently, it has the greatest impact on the bandwidth of thesystem, among the three lower cutoff frequencies.• In the high frequency region, the high cutoff frequency due to the input capacitors andresistors (fHi) has the lowest value. Consequently, it has the greatest impact on thebandwidth of the system, among the two high cutoff frequencies.Av / Avmid (db)NormalizedVoltagegainFrequency0.707 AVmid1 AVmidfLSfLCNormalized Gain in dbFull Frequency Response (Normalized Voltage Gain Versus Frequency)0 db-3 db-5 db-25 db-20 db-15 db-10 db1 10 100 1K 10K 100K 1M 10M 100MfLG fHi fHoBandwidth-20 db / decade(-6 db / octave)+20 db / decade(6 db / octave)
  74. 74. Frequency Response of Multistage (Cascaded) Amplifiers• If there are several stages in a cascaded amplifier system, the overall bandwidthof the system will be lower than the individual bandwidth of each stage.• In the high frequency region:• The output capacitance Co must now include the wiring capacitance (Cwi),parasitic capacitance (Cbe or Cgs), and input Miller capacitance (CMi) of thenext stage.• The input capacitance Ci must now include the wiring capacitance (Cwo),parasitic capacitance (Cce or Cds), and input Miller capacitance (CMO) of thepreceding stage.• The lower cutoff frequency of the entire system will be determined primarily bythe stage having the highest lower cutoff frequency.• The upper cutoff frequency of the entire system will be determined primarily bythe stage having the lowest higher cutoff frequency.• For n stages having the same voltage gain and lower cutoff frequency (f1), theoverall lower cutoff frequency (f1’) can be computed as:stagesofnumbernstageeachoffrequencycutofflowerf1:whereamplifierentiretheoffrequencycutoffloweroverall12f1f11/n
  75. 75. Frequency Response of Multistage (Cascaded) Amplifiers• For n stages having the same voltage gain and higher cutoff frequency (f2),the overall higher cutoff frequency (f2’) can be computed as:stagesofnumbernstageeachoffrequencycutoffhigherf2:whereamplifierentiretheoffrequencycutoffhigheroverall12f2f2 1/n
  76. 76. Square Wave Testing• A square wave signal can be used to test the frequency response of singlestage or multistage amplifier.• If an amplifier has poor low frequency response or poor high frequencyresponse, the output of the amplifier having a square wave input will bedistorted (not exactly a square wave at the output).• A square wave is composed of a fundamental frequency and harmonicswhich are all sine waves.• If an amplifier has poor low or high frequency response, some low or highfrequencies will not be amplified effectively and the output waveform will bedistorted.
  77. 77. Square Wave Testing• The figures below show the effect of poor frequency response of anamplifier using a square wave input.VtVtttVVNo distortion(Good Frequency Response)Poor High FrequencyResponsePoor Low FrequencyResponsePoor High and LowFrequency Responsetiltlong rise timetiltVttVVery Poor High FrequencyResponseVery Poor Low FrequencyResponsetiltvery long rise time
  78. 78. Square Wave Testing• The high cutoff frequency can be determined from the output waveform bymeasuring the rise time of the waveform.• Rise time is between the point when the amplitude of the waveform is 10 %of its highest value up to the point when the amplitude is 90 % of itshighest value.• The high cutoff frequency can be computed as:• The lower cutoff frequency can be determined from the output waveformby measuring the tilt of the waveform.(seconds)timerisetr:where)ftoequalelyapproximatis(bandwidthamplifierofBandwidthtr0.35fBW(Hz)amplifieroffrequencycutoffuppertr0.35fHHH(Hz)wavesquareoffrequencyfs(unitless)tiltVV-VP:where(Hz)amplifieroffrequencycutofflowerfsPfLOtV tiltVV’Rise time (tr)
  79. 79. Square Wave Testing• Example: The output waveform of an amplifier with a 4 Khz square waveinput has the following characteristics:Rise time = 15 microseconds Maximum amplitude (V) = 40 millivoltsMinimum voltage of tilt (V’) = 30 millivoltsDetermine: high cutoff frequency, bandwidth, low cutoff frequency.Hz23,333.3315x100.35fBWHz23,333.3315x100.35tr0.35f6-H6-HfrequencycutofflowerHz318(4,000)0.25fsPf(unitless)tilt0.2510x4010x30-10x40VV-VPLO3--3-3

×