Upload
others
View
7
Download
0
Embed Size (px)
Citation preview
EffectofTemperatureonI/Vcurves
Figure 4.21 The IV curves of the silicon PN diode shift to lower voltages with increasing temperature
Source: Hu
EffectofTemperatureonI/Vcurves
• ForagivenID wewanttounderstandhowthediodevoltageVDvarieswithtemperature
• Thediode’sI/Vequationcontainstemperatureintwoplaces:VT =KT/qandIS
• InsidetheexpressionofIS thetermsni andμalsodependsontemperature:
• Nv andNc arerespectivelytheeffectivedensityofenergystatesinvalencebandandconductionband
ID = IS eVDVT −1
⎛
⎝⎜⎜
⎞
⎠⎟⎟ IS = qAni
2 Dp
LpNd
+Dn
LnNa
⎛
⎝⎜⎜
⎞
⎠⎟⎟= qAni
2 µpVT
LpNd
+µnVTLnNa
⎛
⎝⎜⎜
⎞
⎠⎟⎟
n2i = NCNVe−EGKT = ACAVT
3e−EGKT
where:
µ∝T −3/2
NC = ACT3/2 and NV = AVT
3/2
EffectofTemperatureonI/Vcurves
• Ifforsimplicityweignoretheweaktemperaturedependenceoftheeffectivemassdensityoftheconductionbandelectrons(mn
*)andthevalencebandholes(mp
*)thevaluesofAC andAV areaboutconstant(sources:Pierret p.51):
• k=Boltzmann’sconstant=8.62×10−5 eV/K=1.38×10−23 Joule/K• ħ=ReducedPlanck’sconstant=1.055×10−34Joule-sec• mn
* =1.18× m0
• mp* =0.81× m0
• m0=electronrestmass=9.11×10−31Kg
n2i = ACAVT3e
−EGKT with :
AC = 2mn*k
2π!2⎡
⎣⎢
⎤
⎦⎥
3/2
≈ 6.2×1015 cm−3K −3/2
AV = 2mp*k
2π!2⎡
⎣⎢
⎤
⎦⎥
3/2
≈ 3.52×1015cm−3K −3/2
n2i = ACAV ×T3e
−EGKT
!
ni = ACAV ×T3/2e
−EG2KT = B×T 3/2e
−EG2KT
• Therefore≈:
• Ifforsimplicityweignoretheweaktemperaturedependenceofthebandgapenergy:
whereEG0 isthebandgapenergyat0K(EG0≈1.17eV)andthetemperatureTisexpressedinK
• WecanfinallywriteIS asfollows:
EffectofTemperatureonI/Vcurvesni = B×T
3/2e−EG2KT withB ≅ 5×1015 cm−3K −3/2
EG = EG0 −4.73×10−4 ×T 2
T + 636≈ EG0 −3×10
−3 ×T
IS = qAB2T 3e
−EGkT
κµ,pT−1.5kT / qLpNd
+κµ,nT
−1.5kT / qLnNa
⎛
⎝⎜⎜
⎞
⎠⎟⎟ ≡ ς ⋅T
2.5 ⋅e−EGkT
1.16 − 3×10()×𝑇source:Plummer
• Firstpass:
• thetemperaturedependencecausedbythepolynomialtermisweakcomparedtotheonecausedbytheexponentialterm,sowewillignoreit
• Let’sgetbacktoourobjective:givenafixedvalueofcurrentID wewanttofindoutthevoltagevariationoccurringΔVD whenthereisatemperaturevariationΔToccurring
EffectofTemperatureonI/Vcurves
IS = qAB2T 3e
−EGkT
κµ,pT−1.5kT / qLpNd
+κµ,nT
−1.5kT / qLnNa
⎛
⎝⎜⎜
⎞
⎠⎟⎟ ≡ ς ⋅T
2.5 ⋅e−EGkT
IS = ς ⋅T2.5 ⋅e
−EGkT ≡ζ ⋅e
−EGkT
TC ≡ dVDdT @ID=const
=ζ
• Let’sassumeforwardbias andfliptheI/Veq.:
EffectofTemperatureonI/Vcurves
ID ≈ IS eVDVT
⎛
⎝⎜⎜
⎞
⎠⎟⎟ ⇒
IDIS= e
VDVT ⇒ VD ≈VT ln
IDIS=kTqln ID
IS
⎛
⎝⎜
⎞
⎠⎟ d uv( )
dx= v du
dx+u du
dxddx
lnu( ) = 1ududx
ddx
xN( ) = N ⋅ xN−1
Andlet’skeepinmindacoupleofusefulmathrules:
dVDdT
=ddT
kTqln ID
IS
⎛
⎝⎜
⎞
⎠⎟
⎡
⎣⎢
⎤
⎦⎥=
kqln ID
IS
⎛
⎝⎜
⎞
⎠⎟+
kTq
ddT
ln ID − ln IS( ) ≅
≅kq⋅VDkTq
+kTq
ddT
ln ID − lnζ +EG
kT⎛
⎝⎜
⎞
⎠⎟ ≅
VDT+kTq
ddT
EG
kT⎛
⎝⎜
⎞
⎠⎟ ≅
≅VDT−EG
qTEG andkareconstantstakethemoutofthederivative
constant
constant
• AssumingVD=0.5VandT=300K:
• Atroomtemperature,adiode’sforwardvoltage-drophasathermalcoefficientofabout−2mVperdegree
• IfweknowthediodevoltageatsomereferencetemperatureT0wecanestimatethediodevoltageatanyothertemperatureTasfollows:
EffectofTemperatureonI/Vcurves
TC ≡ dVDdT
≅VDT−EG / qT
TC = 0.5−1.1300
= −2mV /K
VD (T ) ≅VD (T0 )− 2mV/degree× (T −T0 ) =VD (T0 )+TC × (T −T0 )
Idiode
Vdiode
ID
T1 >T0ΔV/ΔT=(VD1−VD0)/(T1−T0)≅ −2mV/degree
= -2.1 mV/K = -2.1 mV/degree0.5-1.12
Aside:secondpass• Ifwewanttoconsideralsothetemperaturedependencecausedbythe
polynomialterm,allwehavetodoisalittlemoreworktakingthederivative:
• ThetermVD/TresultsfromthetemperaturedependenceonVT.ThenegativetermsresultsfromthetemperaturedependenceofIS,anddoesnotdependonthevoltageacrossthediode
• AssumingVD=0.5VandT=300K:
• Soforconservativedesignitiscommontoassume:
dVDdT
=ddT
kTqln ID
IS
⎛
⎝⎜
⎞
⎠⎟
⎡
⎣⎢
⎤
⎦⎥=
kqln ID
IS
⎛
⎝⎜
⎞
⎠⎟+
kTq
ddT
ln ID − ln IS( ) ≅
≅VDT+kTq
ddT
ln ID − lnζ − lnT2.5 +
EG
kT⎛
⎝⎜
⎞
⎠⎟ ≅
≅VDT−2.5VTT
−EG / qT
IS = ς ⋅T2.5 ⋅e
−EGkT
ddT
lnTm( ) = 1Tm
ddT
T m( ) = mTm T
m−1 =mT
TC = 0.5− 2.5×26×10−3 −1.1
300≅ −2.2mV /degree
TC = dVDdT
≅ −2.5mV /degree-2.5 mV/degree
-2.3 mV/degree-1.12
• Inreversebias wefindoutthatthevariationofIS withTisabout14.6%percent/degree:
• Atroomtemperature:
• Since(1.152)5≈2weconcludethatthesaturationcurrentapproximatelydoublesforevery5degreesriseintemperature
EffectofTemperatureonI/Vcurves
ddT
ln IS( ) = 1ISdISdT
⇒dISdT
= IS ⋅d ln IS( )dT
ddT
ln ς ⋅T 2.5 ⋅e−EGkT
⎛
⎝⎜
⎞
⎠⎟
⎡
⎣⎢⎢
⎤
⎦⎥⎥=ddT
lnς + lnT 2.5 −EG
kT⎛
⎝⎜
⎞
⎠⎟=2.5T+EG
kT 2 =2.5T+EG
kT 2qq=2.5T+EG / qT ⋅VT
=IS
ddx
lnu( ) = 1ududx
Recalling:
dISdT
= IS2.5T+EG / qT ⋅VT
⎛
⎝⎜
⎞
⎠⎟
dISdT
= IS2.5300
+1.1
300 ⋅26 ⋅10−3⎛
⎝⎜
⎞
⎠⎟= IS ×
14.6100
[A/degree]1.12 15.2
• Inreversebias:
• Althoughstillquitesmall,realdiodesexhibitreversecurrentsthataremuchlargerthanIS.Alargepartofthereversecurrentisduetoleakageeffects.TheseleakageeffectsareproportionaltothejunctionareaAjustasIS is.
• AsaruleofthumbIR doublesforevery10degreeriseintemperature
EffectofTemperatureonI/Vcurves
IS (T ) = IS (T0 )×2(T−T0 )/5
IR (T ) = IR (T0 )×2(T−T0 )/10
Idiode
Vdiode
T1 >T0
DifferentModels
Figure 5–23 Streetmanapproximations of junction diode characteristics: (a) the ideal diode; (b) ideal diode with an offset voltage; (c) ideal diode with an offset voltage and a resistance to account for slope in the forward characteristic.
Vγ Vγ
Vγ Vγ0
Vγ=0.7V Vγ0
VD VDVD
ID ID ID
VD VD VD
dIDdVD
= gD
ID ID IDrD=1/gD
Idealdiodewithvoltageoffset• Constantvoltagemodel:
– theconstantvoltageiscalledcut-involtage,turn-onvoltage,orthresholdvoltageandisusuallydenotedasVD,onorVγ
• Thismodelisbasedontheobservationthataforward-conductingdiodehasavoltagedropthatvariesinarelativelynarrowrange(e.g.0.6to0.8).We’llassumeVγ ≈0.7V
• BelowVΥ thecurrentisverysmall(lessthan1%ofthemaximumratedvalue).
source:Razavi
for T > 25 !C ⇒ PD,max = 500−1.68× (T − 25) [mW ]
1N4148Forwardcontinuouscurrent:IF=300mA
forVF=0.7V,IF≈3mATA=25°C
TJ = TA + RΘJA ×PD
Idealdiodewithvoltageoffset
• I/VcharacteristicofatheoreticaldiodewithIS=10−14A=10fA
• Forward-biaspartofthecharacteristicswithcurrentplottedonalogscale
Thediodecurrentincreasesafactorof10every60mVincreaseinvoltage
NOTE:realdiodesexhibitreversecurrentsthatareconsiderablylargerthanIS(thisismainlyduetoholesandelectronsbeinggeneratedwithinthespacechargeregion).Atypicalvalueofreverse-biascurrentmaybe1nA(stillsmallandnegligibleinmostcases)
source:Neamen
0.20.40.60.8
• Shouldweworryaboutthefactthatthediodehasresistance?
• ForwardRegion:
Example:IS ≈10fA =>ID≈ISexp(VD/VT)=10−14exp(0.7/26m)≈4.9mA=>VT/ID =26/4.9≈5.3Ω
• ReverseRegion:
Isthemodelgoodenough?
rdiode =dIDdVD
⎛
⎝⎜
⎞
⎠⎟
−1
=ddVD
IS eVDVT −1
⎛
⎝⎜⎜
⎞
⎠⎟⎟
⎡
⎣⎢⎢
⎤
⎦⎥⎥
−1
=ISVTeVDVT
⎛
⎝⎜⎜
⎞
⎠⎟⎟
−1
=ISe
VDVT − IS + ISVT
⎛
⎝
⎜⎜⎜
⎞
⎠
⎟⎟⎟
−1
=VT
ID + IS
rdiode =VT
ID + IS≅VTID
ID >> IS
VT/ID isonlyafewΩ
rdiode =VT
ID + IS≅∞ [Ω]
ID ≅ −IS
IdealDiode
• Inapplicationsthatinvolvevoltagesmuchgreaterthanthediodevoltagedrop(0.6V-0.8V)wemayneglectthediodevoltagedropaltogether.
ID >0VD=0“ashort”(R=0)inforwardbias
VD<0ID =0,“anopen”(R=∞)inreversebias
source:Razavi
Piecewiselinearmodel• Thismodelisusefulwhenthereisasmallvaryingsignalsuperimposedto
thebiasingvoltage
• Let’ssaywewanttoforwardbiasadiodesothatitoperatesatagivenvalueofID,weneedtofindthecorrespondingVD (Q=operatingpoint=(ID,VD)
ID =VDD −VD
R⇔ ID =
VDDR
−VDR
KVL=topologicaleq.
constitutiveeq.ofdeviceID ≈ ISeVDVT
Thisistheeq,ofaline(ID vs.VD)withslope−1/R
Piecewiselinearmodel
ExampleWehaveVDD=5VandthediodeatVD0=0.7VhasacurrentofID0=1mA.WewantID ≅ 4.3mA.
Let’sassumeVD ≅ 0.7V(anditerateuntilwefindtherightvaluecorrespondingtoID)
R = VDD −VDID
=5− 0.74.3m
=1KΩ
VD1 −VD0 = 2.3VT × LogID1ID0
⎛
⎝⎜
⎞
⎠⎟⇒VD1 =VD0 + 60mV × Log
ID1ID0
⎛
⎝⎜
⎞
⎠⎟= 0.7+ 60mV × Log
4.3m1m
⎛
⎝⎜
⎞
⎠⎟ ≈ 0.738V
ID2 =VDD −VD1
R=5− 0.7381K
= 4.262mA VD2 =VD1 + 60mV × LogID2ID1
⎛
⎝⎜
⎞
⎠⎟ ≈ 0.738+ 60mV × Log
4.262m4.3m
≅ 0.738V
(iteration1)
(iteration2)
Nofurtheriterationsarenecessary.ThecircuitusedgivesID≅4.262mAandVD≅0.738V
𝑁𝑂𝑇𝐸: ln x = 𝛼 5 𝐿𝑜𝑔 𝑥 → ln e = 1 = 𝛼 5 𝐿𝑜𝑔 𝑒 = 𝛼 5 0.43 → 𝛼 ≈ 2.3
Avoltageincreaseof60mVcorrespondstoa10xcurrentincrease
Piecewiselinearmodel
• aa
VD
Vγ0
vD =VD + vd ⇒ vd = vD −VD ≡ ΔVD ⇒ vD =VD +ΔVD
iD ≅ ISevDVT = ISe
VD+ΔVDVT = ISe
VDVT e
ΔVDVT = IDe
ΔVDVT
WehaveasignalsuperimposedtothebiasvoltageVD:
Ifthemaxexcursionofthesignalissmall(thatisΔVD/VT<<1):
iD = IDeΔVDVT ≈ ID 1+
ΔVDVT
⎛
⎝⎜
⎞
⎠⎟ ex =1+ x + x
2
2!+x3
3!+...
for x<<1: ex ≅1+ xΔVDVT
<<1
iD ≈ ID +IDVTΔVD = ID + gdΔVD = ID +ΔID
wefindoutthattheresponseofthediodeisaboutlinearanditmakesensetoapproximatethediodecharacteristicwiththe tangentlineatQ:
1rd= gd =
ΔIDΔVD
=idvd=diDdvD @VD
=gd =id=ΔID
*
Piecewiselinearmodel
vD
iD
VγVγ0
Q=(ID,VD)ID
VDgd =
IDVT
⇒VT =VD −Vγ 0 ⇒Vγ 0 =VD −VT
iD = IS eVDVT −1
⎛
⎝⎜⎜
⎞
⎠⎟⎟ ≈ ID + gd vD −VD( ) = ID +
IDVD −Vγ 0
vD −VD( )
slope=ΔI/ΔV
slope:gd
Equationoflinegiventwopoints:
y− y0 =m x − x0( )
m =y1 − y0x1 − x0
=ΔyΔx
*Howsmallisasmallsignal?ex ≅1+ xTheerroroftheapprox.shouldbe≤10%Error = ex −1− x ≤ 0.1Solvingnumericallywesethatforx≤0.4theerroris≤0.09ΔVDVT
≤ 0.4⇒ΔVD ≤ 26mV ×0.4 ≈10.4mV
y =mx + n⇒ iD ≈vDrd−Vγ 0rd
alternatively:
y(x ) ≡ 0 =mx + n⇒n = −mx
DiodeCircuits
• …Finallylet’sstartbuildingsomecircuit• Applications:
– Rectifiers– LimitingCircuits(a.k.a.Clippers)– LevelShifters(a.k.a.Clampers)– Detectors– Voltagedoublers– Regulators– Switches