Electronics I - Diode Circuits - Gonzaga...
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Transcript of Electronics I - Diode Circuits - Gonzaga...
DiodeCircuits
• Applications:
– Rectifiers– LimitingCircuits(a.k.a.clippers)– Detectors– LevelShifters(a.k.a.clampers)– Regulators– Voltagedoublers– Switches
Warm-upexamplesExample#1: diodeandresistorinseries
(orVin <Vγ) (orVin <Vγ)
Vin
Vout
Vγ
Vγ
(e)
source:Razavi
Theinput/outputcharacteristicswithideal andconstant-voltage modelsyieldstwodifferentbreakpoints.Applyinganinappropriatediode’smodelcanbemisleading!
Sidenote:1. Whenthediodeisforward
biasedthecurrentthoughthediodeis≈Vin/R:wecannotmakemakeVin getsolargethatVin/R>IF,peakotherwisethediode“melts”
2. Whenthediodeisreversebiasedthevoltageacrossthediodeis≈−Vin:wecannotmakeVingetsosmallthat|−Vin|>VR,peakotherwisethediode“breaks”VR,peak isa.k.a.PIV(PeakInverseVoltage)
Warm-upexamples
Example#2: diodeandresistorinseries(half-waverectifier)
1
Vγ
Vin
Vin >0(orVin >Vγ)
Vin <0(orVin <Vγ)
source:Razavi
Warm-upexamples
Example#3: diodeimplementationofORgate
VA = 5V and VB = 4VLet's guess D1 is ON ⇒Vout (A− side) =VA −Vγ = 4.3VLet's guess D2 is ON ⇒Vout (B− side) =VB −Vγ = 3.7V ⇒ BAD GUESS !!Vout (A-side) must be the same as Vout (B-side) otherwise we violateKVL !!⇒ D2 is OFF
VA = 0.6V and VB = 0VLet's guess D1 is OFF Let's guess D2 is OFF
⎤
⎦⎥⇒Vout = 0V
VA = 3V and VB = 0VLet's guess D1 is ON ⇒Vout =VA −Vγ = 2.3VLet's guess D2 is OFF
CONSISTENCYMETHOD:Itissometimedifficulttocorrectlypredicttheregionofoperationofeachdiodebyinspection.Insuchcases,wemaysimplymakean“educated”guessproceedwiththeanalysis,andeventuallydetermineifthefinalresultagreesorconflictswiththeoriginalguess.
VA(V) VB(V) Vout (V) D1 D2
0 0 0 OFF OFF
0 5 ≈5 OFF ON
5 0 ≈5 ON OFF
5 5 ≈5 ON ON
Let’stryafewcases:
source:Razavi
Warm-upexamples
Example#4:source:Razavi
WhenthediodeisONwecanmodelthecircuitasfollow:
Vout =Vin −VD,onR1 + R2
R2 +VD,on =R2
R1 + R2Vin +VD,on
R1R1 + R2
FromthemodelofthecircuitiseasytoseethatifVin <VD,onwecannothavecurrentflowingthroughthediode=>thediodemustbeOFF Vout =Vinslope
IfweprefertousetheidealdiodemodelallwehavetodoistoassumeVD,on=0rather thanVD,on=0.7V.WedonotneedalotofdeepthinkingtogettheassociatedI/Ocharacteristic
@Vin =VD,on ⇒Vout =VD,on
@Vin =VD,on ⇒Vout =VD,on
Warm-upexamples
Example#5:source:Razavi
WhenthediodeisOFFwecanmodelthecircuitasaresistivedivider:
Vout =R2
R1 + R2Vin
WhenthediodegoesON=>Vout=VD,on =>
=>sotheturnonpointis VD,on =R2
R1 + R2Vin ⇒Vin =VD,on
R1 + R2R2
ForVin <0thediodeisdefinitelyOFF
Warm-upexamples
Example#6:source:Hambley 4kΩ 6kΩ
1.Let’sassumebothdiodesareON
ID1 =10−34k
=1.75mA
VR = 3V→ IR = 3 / 6k = 0.5mA
KCL :ID1 + ID2 = IR → ID2 = IR − ID1 = −1.25mA
Theresultisnotconsistent:currentcannotflowfromKtoA2.D1 isONandD2 isOFF
4kΩ 6kΩ
ID1 = IR =10
4k + 6k=1mA
IR
VRID1
ID2
VR = IR ×R =1m×6k = 6V
VD2 = 3V − 6V = −3V
Warm-upexamples
Example#7:source:Razavi
Vout =Vin +VD,on
Vin +VD,on =R2
R1 + R2Vin ⇒VD,on = −
R2R1 + R2
Vin ⇒Vin = −VD,onR1 + R2
R2
ForVin <0thediodeisdefinitelyON.WhenthediodeisON:
(straightlinewithslope1andcrossingxaxisat–VD,on)
WhenthediodeisOFF,thecircuitcanbemodeledasavoltagedivider:
Vout =R1
R1 + R2Vin (straightlinepassingthroughtheoriginandwithslopeR1/(R1+R2)
ThebreakpointbetweenONandOFFiswhenVout =Vin +VD,on
−VD,on
−+−
Warm-upexamples
Example#8:source:Razavi
Vout =Vin +VD,on
Vin +VD,on =R2
R1 + R2Vin ⇒VD,on = −
R2R1 + R2
Vin ⇒Vin = −VD,onR1 + R2
R2
ForVin <0thediodeisdefinitelyON.WhenthediodeisON:
(straightlinewithslope1andcrossingxaxisat–VD,on)
WhenthediodeisOFF,thecircuitcanbemodeledasavoltagedivider:
Vout =R1
R1 + R2Vin (straightlinepassingthroughtheoriginandwithslopeR1/(R1+R2)
ThebreakpointbetweenONandOFFiswhenVout =Vin +VD,on
−VD,on
−+−
Half-waverectifiers
vIvS=N1N2
source:Neamen
VS
Vγ
Vγ
vOsource:Sedra &Smith
WhenthediodeisONthemaxcurrentflowingthroughthediodeis:
ID,max =VS −VγR
≅VSR
WhenthediodeisOFFthePIVacrossthediodeis:PIV =VS
RuleofthumbitisgoodpracticetoselectadiodewithVBR≥1.5×PIVandIF,max ≥1.5xID,max
Halfwaverectifierasasignalstrengthindicator
Vout (t) =Vp sinωt for 0≤ t≤T/20 for T/2≤ t≤T
⎧⎨⎩
Vout,avg =1T
Vout (t)dt0
T
∫ =1T
Vp sinωt dt0
T /2
∫ =1TVp
ω−cosωt[ ]0
T /2=Vp
π
source:Razavi
Vp
Vout(t)
𝑉"#$,&'( =𝑉*2 2�
Halfwaverectifierasabatterycharger
source:Neamen
• ifVB <VB,nominal thebatterygetrecharged(diodeisONfromt1tot2)
• otherwisethebatteryisleftalone(thediodeisOFFallperiodT)
ωT
𝑉- > 𝑉0,1"'2134 + 𝑉6
Precisionhalf-waverectifier
• a
Thetransferfunctionisalmostideal:itdoesn’tsufferfromhavingoneortwodiodedrops
• IfvI >0thediodeisON.WiththediodeONthecircuitbecomesafollower.
• IfvI <0thediodeisOFFwiththediodeOFFtheloadisatground
• Fortheo.a.tostarttooperateandturn-onthediode,vI hastoexceedonlyanegligiblysmallvoltageequaltoVγ/Ad
source:Sedra &SmithPIV = −VSS
IF,max =VO,maxRL
=VI ,maxRL
Full-waverectifiers
vI=
vO
Vγ
VS
ID,max =VS −VγR
≅VSR
PIV =VS −Vγ − (−VS ) = 2VS −Vγ
D1ON
D2ON
D1ON
D1ON
D2ON
source:Neamen
source:Sedra & Smith
2vS
Diode-bridgefullwaverectifiera.k.a.Grätz bridge
VS
2Vγ
vO
(a) whenvS ispositive,D1 andD2 areturnedON
(b) whenvS isnegative,D3 andD4 areturnedON
IneithercasecurrentflowsthroughRinthesamedirection
(a)
(b)
vI=
D3
D4
source:Sedra & Smith
source:Neamen
ID,max =VS − 2Vγ
R≅VSR
PIV =VS −Vγ
VX
vD4=vs−vD1vD3=vD1+vOvO =vs−2Vγ
Clippers(a.k.a.Limiters)• Theideabehindclippersisquitesimple.Wehavealreadybuiltoneinthe
past
• Allwehavetodotoshifttheclippingthresholdtoadifferentvalueistoaddabattery
Vin
VinPositive-cyclelimitingcircuit source:Razavi
Averycommonclipper’sapplication
• Protectioncircuitry:keepthesignalsbelowcertainthresholds
Vin
R VX
VB1=VDD VB2=0
inputpin
DiodeLimiters
CMOSIC
VDD
VXD1
D2
source:Harris&Weste
source:Razavi
Non-idealitiesinlimitingcircuits
source:Razavi
TheclippingregionisnotexactlyflatsinceasVinincreases,thecurrentsthroughdiodeschange,andsodoesthevoltagedrop.
Zener diode
Zener diodesymbol
−VZT
iD
vD=−IBV
=−BVVD,on
−VZT
K standsforknee
VZT =zener voltage
constantvoltageI-Vmodelofbreakdownregion
Oftenisconvenienttomodelthebreakdownregion withapiece-wiselinearmodel: iD
vD
−IZT
−VZT −VZ0
VZ =VZ 0 + rZ × IZ
ThemodelisvalidforIZ >IZK andVZ >VZ0:
Q 1rZ=
ΔIZΔVZ
⎛
⎝⎜
⎞
⎠⎟@Q
ThelowerthevalueofrZthemoreconstantthezener’s voltageremain
source:Sedra &Smith
Zener diode• Azener diodeisadiodespecificallymanufacturedtobebeusedinbreakdownregion.
Thezener’s I-Vcurveinbreakdownregionisverysteep(morethanusual)
• Diodebreakdownisnormallynotdestructive,providedthepowerdissipatedinthediodeislimitedtoasafelevel
• ThefactthatthediodeI/Vcharacteristicinbreakdownisalmostaverticalline(justlikeabattery)enablesittobeusedinvoltageregulation(moretocomesoon!)
• Therearetwomechanismcausingthebehaviorwehaveinbreakdownregion(…despitethemechanismtheendresultisthesame)
– Avalanche:occurswhentheminoritycarrierssweptbytheelectricfieldindepletionregionhaveenoughkineticenergytobeabletobreakcovalentbondsinatomswithwhichtheycollide
– Zener:occurswhentheelectricfieldinthedepletionregionincreasestothepointthatitcantearoutaboundelectronfromitscovalentbond
Zener diode:datasheetexampleOnSemiconductor:Zener VoltageRegulatorwithVZ,nom=2.4V
IfthecurrentexceedacertainlimitthepowerdissipatedPD=VD×ID risesthejunctiontemperaturetoomuch(>150°Cinourcase)andthedevicemaygetdamaged
TJ = TA +PD ×RΘJAAdevicemaygetdamagedalsointhecasethejunctiontemperaturebecomestoosmall(<−65°Cinourcase)
Zener diode:datasheetexample
Themaxpowerratingofthediode(PD,max =300mW)goesdownof2.4mW/°Cfortemperaturesabove25°C
ClippingwithZener diodes
source:NeamenBasicidea:ReplacingbatterieswithZener diodes
MoreclippingwithZener diodes
• ForlargepositivevI thediodeDZ1 isforwardbiasedandDZ2 isbiasedinzener region(vI >Vz2 +0.7)
• ForlargenegativevI thediodeDZ1 isbiasedinzener regionandDZ2 isforwardbiased(vI <−VZ1−0.7)
• Intherange−(VZ1+0.7)<vI <VZ2+0.7oneofthediodesisinforwardregionandtheotheroneinreverseregion(thereforevO=vI)
source:Sedra &Smith
Anotherapplicationofclippers:softlimiters
cellphonefarfrombasestation cellphonenearabasestation
source:Razavi
source:Hambley
Dissectingthepeakdetectoralittlemore
Note: thevoltageacrossthediode(VD1)isjustlikeVin,onlyshifteddown
source:Razavi
Detectors:AMdemodulator
Modulatedinputsignal
Detectorcircuit
Demodulatedoutputsignal
source:Neamen
AMDemodulator
RC>>TCperiodcarrier
Clampers(a.k.a.levelshifters)
• ClampersshifttheentiresignalappliedattheinputbyaDClevel.
• Insteadystate,theoutputsignalisanexactreplicaoftheinputwaveform,buttheoutputsignalisshiftedbyaDCvalue
• Commonapplication:
• Supposethereisastage(e.g.anamplifier)thatdoesnotoperateproperlywiththeDClevelprovidedatitsinput,theissuecanbesolvedbyputtingalevelshifterinfrontofthestage
PositivePeaksClamper
Thisisthe“same”circuitofthepeakdetector,butnowwetaketheoutputacrossthediode!
vO = vI − vC
D1ON
Cgetscharged(atT/4vC≈VM)
D1OFF
WhenthediodegoesOFFthereisnopathtogroundsothecapacitorcannotdischarge:vC staysconstantatVM (soafterT/4:vO =vI – VM)
source:Neamen
Thepositivepeaksoftheoutputvoltageareclampedat0V
Assumingrf≅0Ωand Vγ=0V
NegativeDClevel
Levelshifterwithpeakat-2VM
Positivepeaksclamper
Vout =Vin −VC1D1OFF D1ON
C1chargesveryquicklytoVC1=4V
C1blocksVin=−6VDC
Vout =0 Vout =4– 4==0
D1OFF
C1remainschargedatVC1=4V
Vout =–6– 4==–10
D1ON
C1alreadychargedatVC1=4V
Vout =4– 4==0
Thepositivepeaksoftheoutputvoltageareclampedat0V
NegativeDCLevelShifter
PositivepeaksclamperwithBattery
source:Neamen
steadystate(ittakesT/4toreachit)
steadystate(ittakesTtoreachit)Superpositionof +
ThepositivepeaksoftheoutputvoltageareclampedatVB
Positivepeaksclamperwithbattery• Ifwetakethecircuitwejustanalyzed,andreversethepolarityofthe
batteryweclampthepositivepeaksofthesignaltoanegativevoltagevalue.
• Thisisnosurprise:westillclampthepositivepeakstoVB(butnowVBhappenstobenegative)
ThecircuitisdesignedtohaveRC>>T
NOTE:whenvin isat+5VthediodeisONandthecapischargedtoVC=10V
source:Hambley
Negativepeaksclamper
source:Razavi
ThediodeturnsONandthecap.chargesto−VP
vC1
vout = vin − vC1 = vin
ThediodeturnsOFFforgood:
vout = vin − (−VP ) = vin +VP
initiallythecapisuncharged:vC1(0)=0
Thenegativepeaksoftheoutputvoltageareclampedat0V
tt4t2t1PositiveDClevel
Levelshifterwithpeakat+2VP
Negativepeaksclamper
Vout =Vin −VC1D1ON
C1chargesveryquicklytoVC1=−6V
D1OFF
C1remainschargedatVC1=−6V
Vout =–6– (–6)==0
Vout=4– (–6)==10
D1ON
C1alreadychargedatVC1=−6V
Vout =–6– (–6)==0
Thenegativepeaksoftheoutputvoltageareclampedat0V(itisbettertousePWLratherthanPULSE)
PositiveDCLevelShifter
PositiveDClevelshifter:effectofload
• Inpracticetheclamperwillbedrivingaload.
• weneedtomakesurethatR1C1 >>T/2,otherwisewhenD1 isOFFthecap.C1losestoomuchchargeontheload
ExampleshowingtheeffectofhavingR1C1toosmall(R1C1 =T/2)
Negativepeaksclamperwithbattery
• Example:ThiscircuitclampsthenegativepeaksofanACsignalto+6V
source:Hambley
VB =6ifweassume:Vγ ≈0VorVB =6.7ifweassume:Vγ ≈0.7V
vin =VP sinωt
VB
Negativepeaksclamperwithbattery
idealdiodeVγ ≈0VVB
vin =VP sinωt
Vp =2V,fin=1KHz
VB =6VVB =−1V
WhataboutreplacingbatterieswithZeners ?
• Itkindofworks,butweneedtokeepinmindthat(differentlyfromwhathappenedwiththelimiters)herethezener mustworkinzener regionatalltime.Soitmustbebiasedinzener regionatalltime!!
Circuitthatclampsthenegativepeaksto−5V
Ifwetakeoffthe−15VbiasvoltageandreturnRdirectlytogroundthediodeneverturnsONandthecircuitdoesn’twork
whenthevin isat2VthediodeisONandthecapchargesat2−5=−3V
−4.3V
Vγ=0.7V
source:Hambley
ReplacingbatterieswithZeners• Exampleofcircuitforclampingpositivepeaks
source:Hambley
Circuitthatclampsthepositivepeaksto+6V
5.3V
Example:anotherclamper
fin=1/20ns
Circuit’selements:
C=1nFR=100KΩIdealdiode
VR=11V VR=2V
Alternativewaysofclamping• InsideCMOSICsDClevelshiftingisusuallyachievedusingcurrentsources
(i.e.MOStransistors)andcascadeofdiodes(ordiodeconnectedMOStransistors)
Assumption:thecurrentpulledbythenextstageisnegligible(oratleastconstant),sothatthecurrentthroughthediodeestablishadropofVD,on acrossthediode
shiftdowncircuit
shiftupcircuit
source:Razavi
• InsideCMOSICs,anothercommonwayofaachievingDClevelshiftingisbyusingaCommonDrainstage
Alternativewaysofclamping
source:B.Murmann &R.Dutton
Application:DCPowersupply• Let’stakealookathowtobuildaDCpowersupply(AC-DCpower
converter)
Figure 4.22 Block diagram of a dc power supply.
source:Sedra &Smith
detectthepeak(smoothingcaporfiltercap)
smooththerectifiedwaves
Rectifier+FilterCapacitor+Load
The following circuit (peak rectifier or peak detector) provides a DC voltage equal to the peak of the input sine wave
So at a first glance it would seem a reasonable solution to use it as a DC power supply to drive a load.
source:Sedra &Smith
vS(t)
Vm
However, once we connect the load if we look at the circuit a little harder we realize it presents some issues
iC(t)
source:Hambley
Rectifier+FilterCapacitor+Load
Waveformsforhalf-rectifierwithsmoothingcapacitor
Thevoltageacrosstheloadisnotconstant.Thereisa ripple.Aripplemorethan5%−10%ofVm(orVm−Vγ withtheconstantvoltagemodel)isaproblem!
Wehavetwodesigndecisions:- pickC(howmuch
ripplewewanttotolerate)
- pickD(howmuchcurrentthediodemustwithstandinforwardregion,andwhatisthePIVinreverseregion)
= vO(t)/RL
exponentialdecaywithtimeconstantτ=RLC
iLRL
source:Hambley
source:Neamen
vD(t) = vS(t) − vO(t)
Ripplefordifferentcapacitorvalues
• ForVout tohavesmallrippleweneedlargeC
• Theamplitudeoftherippleisgivenbythedecayingexponential
source:Razavi
PIV≈2×VMVD=Vin−Vout
iLRL RippleandID,max
AssumingRLC>>T:
VmRLC
≅VRT⇒VR ≅
VmRL
×TC
slopeofexponentialdecayatt=0
0
Vm
Vme−
tRLC
vS
Thediode’scurrentismaxatthebeginningoftheconductionintervalanditgoesdownasthediodetendstoturnsoff
Thisisalsowhenthecurrentthroughthecap.ismax(thisisbecausetheslopeofVout ismax)
ThecurrentsuppliedtoRLisalmostconstantandisboundedbyVm/RL
ID,max =CdVoutdt t=−Δt
+VmRL
=IC,max
whendiodeisOFFvO =
WeneedtofindIC,max !
slopebyusingsimplegeometry
𝑑𝑣9𝑑𝑡 ;$<9
=𝑑 𝑉'𝑒>$/(AB C)
𝑑𝑡E$<9
PIV
IC,max =CdVoutdt t=−Δt
=C ddt
Vm cosωt( )t=−Δt
=CVmω[−sin(−ωΔt)]=CVmω sinωΔt
cosωΔt ≈1− 12ωΔt( )2
RippleandID,max
Lookingatthe“geometry”ofVoutweseethat:
vout(t)
Vm cos −ωΔt( ) =Vm cosωΔt =Vm −Vr ⇒ cosωΔt =1− VrVm
Thediodeconductscurrentonlyasmallportionoftheperiod(Δt/T<<1)thereforeωΔt isasmallangleandsin(ωΔt)≈ωΔt
IC,max ≈ωCVm ωΔt( )
Taylorforsmallangles
1− 12ωΔt( )2 ≅1− Vr
Vm⇒ωΔt ≈ 2Vr
VmIC,max ≈ωCVm ωΔt( ) ≈ 2π
TCVm
2VrVm
ΔtT≈12π
2VrVm
Finally
%oftimethediodeisON
0
RippleandID,max
ID,max =2πTCVm
2VrVm
+VmRL
=VmRL
1+ 2π CRLT
2VrVm
⎛
⎝⎜⎜
⎞
⎠⎟⎟ ≈
VmRL
1+ 2π 2VmVr
⎛
⎝⎜⎜
⎞
⎠⎟⎟
VmRLC
≅VrT← slope of exponentialdecay at t=0
ID,avg ≈1T×VmRL
1+ 2π 2VmVr
⎛
⎝⎜⎜
⎞
⎠⎟⎟×
Δt2=12×VmRL
×ΔtT1+ 2π 2Vm
Vr
⎛
⎝⎜⎜
⎞
⎠⎟⎟ ≅
VmRL
×14π
2VrVm
+1⎛
⎝⎜⎜
⎞
⎠⎟⎟
areatriangleΔtT≈12π
2VrVm
Δt Δt
0 T
Canwefurtherreducetheripple?• Yesitis.Insteadofusingasimpledioderectifierwecanuseabridge
• SinceCdischargesonlyfor½period,theripplevoltageisdecreasedbyafactorof2
• AlsoeachdiodeisapproximatelysubjectedtoonlyoneVm reversebiasdrop(versusthe2Vm wehadwiththehalf-waverectifier).
source:Razavi
Vm
VD,on
B
Vout
VAB =VD,on +Vout
slopeofexponentialdecayatt=0
VmRLC
≅VrT / 2
⇒Vr ≅12×VmRL
×TC
TreplacedbyT/2
ID,max ≅VmRL
× 1+π 2VmVr
⎛
⎝⎜⎜
⎞
⎠⎟⎟
ID,avg ≈VmRL
×1π
Vr2Vm
+1⎛
⎝⎜⎜
⎞
⎠⎟⎟
BridgeRectifier+FilterCapacitor+Load
Vm
source:Sedra andSmith
Δ𝑡𝑇 ≃
1𝜋
2𝑉&𝑉'
�
%oftimethediodeisON
VoltageRegulator
• Theripplecreatedbytherectifiercanbeunacceptabletosensitiveloads.Therefore,aregulatorisrequiredtoobtainamorestableoutput.
source:Razavi
source:Hambley
Vss
VariationsinVss mayberippleduetotherectifierbutalsofluctuationsintheACline
VoltageRegulator
• Aslongasrd <<R1,theuseofaZener diodeprovidesarelativelyconstantoutputdespiteinputvariations
Regulator
R1
ΔVss ΔVout
ΔVout =rd
rd + R1ΔVss
Example: rd=5Ω,R1=1KchangesinVss areattenuatedbyabout200timesattheoutput
source:Razavi
Vss
Vss
VoltageRegulationwithZener Diode
• ExampleDesignavoltageregulatortopoweracarradioatVout=9Vfromanautomobilebatterywhosevoltagemayvarybetween11Vand13.6V.Thecurrentintheradiowillvarybetween0(off)to100mA(fullvolume).
• TheresistanceRi limitsthecurrentthroughthezener diodeanddropsthe“excess”voltagebetweenVss andthenominalvoltagewewantontheloadVout,nom =VZ,T=VZ,nom (inotherwordsitsetsthediodeoperatingpointQT)
Vss = Initially,weneedtofindouttheproperinputresistanceRi.
VoutVss,nom=12V,VSS,middle=12.3V
source:Neamen
VoltageRegulationwithZener Diode
63
VZT
IZT
Vss,minVss,max
QT
Ri =VSS,nom −VZ ,nomIZ ,nom + IL,nom
IL,nom =VZ ,nomRL,nom
Initially,assumeidealdiode:
Morethoroughly,forthecircuittoworkproperly,thediodemustremaininzener regionandthepowerdissipationofthediodemustnotexceeditsratedvalue(PD).Inotherwords:
• ThecurrentinthediodeisaminimumIZ,minwhentheloadcurrentisamaximumIL,max,andthesourcevoltageisaminimumVss,min
• ThecurrentinthediodeisamaximumIZ,max,whentheloadcurrentisaminimumIL,min andthesourcevoltageisamaximumVss,max
Thereforewecanimposethetwofollowingconstraints: Ri =VSS,min −VZ ,nomIZ ,min + IL,max
Ri =VSS,max −VZ ,nomIZ ,max + IL,min
and
VZ (V)
IZ (A)
VoltageRegulationwithZener Diode
64
Reasonably,wecanassumethatweknowtherangeofinputvoltage,therangeofoutputloadcurrent,andtheZener voltage.Further,itisreasonabletosettheminimumzener currenttobeIZ,min ≈0.1× IZ,maxMorestringentdesignrequirementsmayrequiretheminimumzenerdiodecurrenttobe20or30percentofthemaximumvalue.TheimportantpointinsettingIZ,min istomakesureisfarenoughfromtheknee!!
ByequatingtheconstrainsonRi andsettingIZ,min ≈0.1× IZ,maxwecanwrite:
IZ ,max =IL,max ⋅ Vss,max −VZ ,nom( )− IL,min ⋅ Vss,min −VZ ,nom( )
Vss,min − 0.9×VZ ,nom − 0.1×Vss,max
Ri =VSS,max −VZ ,nomIZ ,max + IL,min
Ri =VSS,min −VZ ,nomIZ ,min + IL,max
K standsforknee
ThemaximumpowerdissipatedintheZener diodeisapproximately:
PZ ,max ≈ IZ ,max ×VZ ,nom
Therefore: Ri =Vss,max −VZ ,nomIZ ,max + IL,min
IZ ,min =Vss,min −VZ ,nom
Ri− IL,maxand
VZK=BV
IZK=IBV
VZTVZ0
VZ (V)
IZT
IZ (A)
…andfinally makesurePZ,max <PDandIZ,min <IZK
Let’snowgobacktotheexampleandpluginsomenumbers:
VoltageRegulationwithZener Diode
Vss =
IL,max =100mA
IZ ,max =IL,max ⋅ Vss,max −VZ ,nom( )− IL,min ⋅ Vss,min −VZ ,nom( )
Vss,min − 0.9×VZ ,nom − 0.1×Vss,max=100 ⋅ (13.6− 9)− 011− 0.9 ⋅9− 0.1⋅13.6
≅ 300mA
source:Neamen
PZ ,max ≈ IL,max ×VZ ,nom = 300mA×9V = 2.7W Ri =Vss,max −VZ ,nomIZ ,max + IL,min
=13.6− 9300m+ 0
≅15.3Ω
PRi,max =Vss,max −VZ ,nom( )2
Ri=13.6− 9( )2
15.3≅1.4W
𝐼L,'21 = 0.1×𝐼L,'3P ≅ 30𝑚𝐴
Regulator’sfiguresofmerit
Sourceregulation(a.k.a.lineregulation)Itisameasureofhowmuchtheoutputvoltagechangesasthesourcevoltagechange(assumingno-loadconditionRL=∞)
• Inrealitythezener isnotideal.Ithassomenonzeroresistance,thereforeifthesourcevoltageortheloadcurrentfluctuates,sodoestheVout=VZ
Vss Vout
source:Neamen
sourceregulation≡ ΔVoutΔVss
×100% abilitytomaintainaconstantoutputvoltagelevelontheoutputdespitechangestotheinputvoltagelevel
Lineregulationexample
Example:Findthelineregulationforthepreviousexample,assumingrZ=2Ω
ForVss=13.6V: IZ =Vss −VZR1 + rZ
=13.6− 915.3+ 2
≅ 265.9mA⇒Vout = IZ × rZ +VZ = 9.532V
Alternativelybyconsideringjustthevariations(smallsignalcircuit)
rZ
ΔVoutΔVss
ΔVoutΔVss
=rZ
R1 + rZ=
215.3+ 2
≅15.6%
ForVss=11V: IZ =Vss −VZR1 + rZ
=11− 915.3+ 2
≅115.61mA⇒Vout = IZ × rZ +VZ = 9.231V
ΔVoutΔVin
=9.532− 9.23113.6−11
≅15.6%
R1
ΔIZ
Vout
source:Neamen
source:Razavi
Regulator’sfiguresofmerit
LoadregulationItisameasureofthechangeinoutputvoltagewithachangeinloadcurrent
VssVout
load regulation≡Vout,noload −Vout, fullload
Vout, fullload×100%
capabilitytomaintainaconstantvoltageontheoutputdespitechangesintheload(suchasachangeinresistancevalueconnectedacrossthesupplyoutput
where:- Vout,noload istheloadvoltageforzeroloadcurrent- Vout,fullload istheloadvoltageforthemaximumratedloadcurrent
Inpractice,thereareacoupleofotherwaysofdefiningloadregulation.
load regulation≡Vout,noload −Vout, fullload
Vout,nomload×100%
load regulation≡Vout,noload −Vout, fullloadIL,noload − IL, fullload
×100%=ΔVoutΔIL
×100%(W)
Loadregulationexample
Example:Findtheloadregulationfortheusualexample. AssumerZ=2Ω
ForIL=0A: IZ =Vss,max −VZR1 + rZ
=13.6− 915.3+ 2
≅ 265.9mA⇒Vout = IZ × rZ +VZ = 9.532V
ForIL=100mA: IZ =VR1R1
− IL =Vss,max − VZ + rZ × IZ( )
R1− IL
⇒ IZ =Vss,max −VZ − IL ×R1
R1 + rZ=13.6− 9−100m×15.3
15.3+ 2≅177.46mA⇒Vout = IZ × rZ +VZ = 9.355V
Vss Vout
Vout,noload −Vout, fullloadVout, fullload
×100%= 9.532− 9.3559.355
×100%≅1.89%
Note:Whenmeasuringtheloadregulationthesourceisassumedconstant.SincethefullloadcurrentisreachedforVss =Vss,maxforloadregulationcomputationswemustassumeVss =Vss,max=const
Loadregulationexample
Alternativelybyconsideringjustthevariations(smallsignalcircuit)
Vss Vout
rZ
ΔIL
ΔVss=0 ΔVout = R1 || rZ( )ΔIL ⇒ΔVoutΔIL
= R1 || rZ( ) =15.3 || 2 ≅1.77Ω
ForaΔILof100mAwehavethatΔVout≅ 177mV
(AsexpectedthisisthesameresultwegotbeforeΔVout=9.532−9.355=177mV)
Vout,noload Vout,fullload
EvolutionofanAC-DCconverter
source:Razavi
Ideallywewantbothlineregulationandloadregulationtobeascloseaspossibleto0%
Regulator
Voltagedoubler
clamper peakdetectorD1ON
VC1 fol-lowsVinallthewayto−VP
VX =Vin−VC1=0
FromT/4onD1isOFF
C1stayschargedto−VP
VX =Vin−(−VP)==Vin+VP
=Vin−VC1 Vout
clamper
X
peakdetector
Ifwetaketheclamperjustdesignedandattachapeakdetectoratitsoutputwegetavoltagedoubler
Tin =1ms,VP=2V
Voltagedoubler:detailedanalysis
IntherealitythechargingofC1andC2(VC2 =Vout)isnotassimpleasassumedinthepreviousslide(slide52isasnapshotofthesteadystate)
Voltagedoubler:detailedanalysis
Eachinputcycle,theoutputincreasesbyVp ,Vp/2,Vp/4, etc.,eventuallysettlingto2Vp
Vout =VP 1+12+14+18+...
⎛
⎝⎜
⎞
⎠⎟=VP
12⎛
⎝⎜⎞
⎠⎟
n=0
∞
∑n
=VP1
1−1/ 2= 2VP
t10 t6
DiodesasSwitches
76
Iftheswitchissuddenlyopened,avoltagespikedevelopsacrosstheinductor vL = L
diLdt
Inductorsdonotlikeabruptchanges incurrent
Motor
Inrealitytheswitchisatransistor
Thetransistorgoesin“smoke”
DiodesasSwitches
Adiodeplacedacrosstheinductiveload,willgivethevoltagespikeasafepathtodischarge,loopingover-and-overthroughtheinductoranddiodeuntiliteventuallydiesout.
SpecialDiodes• Schottky-BarrierDiode(SBD)
• Varactors
• Photodiodes
• [email protected] 78
• SBDarebuiltusingametal-semiconductorjunction• currentisconductedbymajoritycarriers(electrons).
ThusSBDdonotexhibittheminoritycarrierchargestorageeffect.AsaresultSBDcanbeswitchedfromontooffandviceversamuchfaster
• Theforwardvoltagedropislower(0.3Vto0.5Vforsilicon)
Thewavelengthofthelightemitted,andthusthecolor,dependsonthebandgapenergyofthematerialsformingthep-njunction.(Example.RedLED:λd=630nm,VF=2.1VIF=50mA,luminousIntensityIV=7500mcd)
Voltagedoubler modeledwithswitches
C2ismovingsomechargebacktoC1(ΔV=Vp/4)
C2ismovingsomechargebacktoC1(ΔV=3Vp/8)
noinitialchargemoving(VC1=VC2=VP)
