FET - Summary

Introduction & review

Basic concepts (Review)

Review the 2 types of transistors

Calculation package

Modules for illustration

SemiParm={ Eg, me, gc, mh, ϵ} ;
NDOS0[SemiParm_]:= Module[{ Eg, me, gc, mh,ϵ},
{ Eg, me, gc, mh,ϵ}=SemiParm ;
4.829415887919019`*^15  {gc  (me^(3/2)) , (mh^(3/2))} 
];
Fermi[Dopant_,T_,SemiParm_]:=Module[{Eg, me, gc, mh,ϵ
, NC0,NV0,type, DopeDensity, kB=  8.61739*10^(-5)},
{ Eg, me, gc, mh,ϵ}=SemiParm  ;
{NC0,NV0}= NDOS0[SemiParm] ;
{type, DopeDensity}=Dopant ;
If[type>0  , 
kB T Log[DopeDensity/(NC0 (T^(3/2)))] 
,-kB T Log[DopeDensity/(NV0 (T^(3/2))) ] ]
] ;

DepletionW[T_,Vbias_,JuncDopant_, SemiParm_]:=Module[{Eg, me, gc, mh,ϵ
,Dopant1, Dopant2,
ContactV, V0, W,
,aB=5.29177*10^(-9)
,Ryd=13.6057},
{ Eg, me, gc, mh,ϵ}=SemiParm  ;
{Dopant1, Dopant2}=JuncDopant ;
ContactV=Eg+Sign[Dopant1[[1]] ]*Fermi[Dopant1,T,SemiParm]+Sign[Dopant2[[1]] ]*Fermi[Dopant2,T,SemiParm] ;
V0=ContactV-Vbias ;
W=Sqrt[(ϵ V0/(aB  Ryd))* (1/Dopant1[[2]] +1/Dopant2[[2]]) ]  ;
10^4* W {1, Dopant2[[2]]/(Dopant1[[2]] +Dopant2[[2]])
,Dopant1[[2]]/(Dopant1[[2]] +Dopant2[[2]])  }] ;

ϕJ[x_, T_,Vbias_,JuncDopant_, SemiParm_]:= Module[{Eg, me, gc, mh,ϵ
, ContactV, V0, W, x1,x2},
{ Eg, me, gc, mh,ϵ}=SemiParm  ; 
{W, x1,x2} =DepletionW[T,Vbias,JuncDopant, SemiParm] ;
ContactV=Eg+Sign[JuncDopant[[1,1]] ]*Fermi[JuncDopant[[1]],T,SemiParm]+Sign[JuncDopant[[2,1]] ]*Fermi[JuncDopant[[2]] ,T,SemiParm] ;
V0=ContactV-Vbias ;
Sign[JuncDopant[[1,1]] ]*If[x≤-x1,0.,If[x≤ 0., V0  (x+x1)^2 /(x1*W),
If[x≤ x2, V0*(1-(x-x2)^2/(x2*W)) ,V0]
]
]
] ;
BandDiagram[x_, T_,Vbias_,JuncDopant_, SemiParm_]:= Module[{Eg, me, gc, mh,ϵ
, ContactV, V0, W, x1,x2},
{ Eg, me, gc, mh,ϵ}=SemiParm  ; 
{W, x1,x2} =DepletionW[T,Vbias,JuncDopant, SemiParm] ;
ContactV=Eg+Sign[JuncDopant[[1,1]] ]*Fermi[JuncDopant[[1]],T,SemiParm]+Sign[JuncDopant[[2,1]] ]*Fermi[JuncDopant[[2]] ,T,SemiParm] ;
V0=ContactV-Vbias ;
w=Sign[JuncDopant[[1,1]] ]*If[x≤-x1,0.,If[x≤ 0., V0  (x+x1)^2 /(x1*W),
If[x≤ x2, V0*(1-(x-x2)^2/(x2*W)) ,V0]
]
] ;
w-Fermi[JuncDopant[[1]],T,SemiParm]
+{Sign[JuncDopant[[2,1]] ]*Eg,0.} 
] ;

FermiLine[x_, T_,Vbias_,JuncDopant_, SemiParm_]:=Module[{Eg, me, gc, mh,ϵ
, ContactV, V0, W, x1,x2},
{ Eg, me, gc, mh,ϵ}=SemiParm  ; 
{W, x1,x2} =DepletionW[T,Vbias,JuncDopant, SemiParm] ;
ContactV=Eg+Sign[JuncDopant[[1,1]] ]*Fermi[JuncDopant[[1]],T,SemiParm]+Sign[JuncDopant[[2,1]] ]*Fermi[JuncDopant[[2]] ,T,SemiParm] ;
V0=ContactV-Vbias ;
If[x≥- x1,-Sign[JuncDopant[[1,1]] ]*Vbias,  ]
];

CarrierEqui[x_, T_,JuncDopant_, SemiParm_]:= Module[{Eg, me, gc, mh,ϵ 
,NC0,NV0, F
, kB=  8.61739*10^(-5)},
{ Eg, me, gc, mh,ϵ}=SemiParm  ; 
{NC0,NV0}= NDOS0[SemiParm] ;
F=BandDiagram[x, T,0.,JuncDopant, SemiParm] ;
If[JuncDopant[[1,1]]>0, 
{NC0 (T^(3/2)) (E^(-F[[2]]/(kB T))), NV0 (T^(3/2)) (E^(F[[1]]/(kB T))), NC0 (T^(3/2)) (E^(-F[[2]]/(kB T)))  +  NV0 (T^(3/2)) (E^(F[[1]]/(kB T)))} 
, {NC0 (T^(3/2)) (E^(-F[[1]]/(kB T))), NV0 (T^(3/2)) (E^(F[[2]]/(kB T))), NC0 (T^(3/2)) (E^(-F[[1]]/(kB T)))  +  NV0 (T^(3/2)) (E^(F[[2]]/(kB T)))}
]
]

CarrierJ[x_, T_,Vbias_,JuncDopant_, SemiParm_,DiffParm_]:= Module[{Eg, me, gc, mh,ϵ
, μn,μp,τ,Ln, Lp 
,NC0,NV0
, W, x1,x2
,q1,q2,F1,F2,G
, kB=  8.61739*10^(-5)},
{ Eg, me, gc, mh,ϵ}=SemiParm  ; 
{ μn,μp,τ}=DiffParm  ;
Ln=10^4*Sqrt[μn kB T τ]; Lp=10^4*Sqrt[μp kB T τ];
{NC0,NV0}= NDOS0[SemiParm] ;
{W, x1,x2} =DepletionW[T,Vbias,JuncDopant, SemiParm] ;
F1=BandDiagram[-x1, T,0.,JuncDopant, SemiParm] ;
F2=BandDiagram[x2, T,0.,JuncDopant, SemiParm] ;
q1=If[JuncDopant[[1,1]]>0, NV0 (T^(3/2)) (E^(F1[[1]]/(kB T))), NC0 (T^(3/2)) (E^(-F1[[1]]/(kB T)))] ;
q2=If[JuncDopant[[1,1]]>0, NC0 (T^(3/2)) (E^(-F2[[2]]/(kB T))), NV0 (T^(3/2)) (E^(F2[[2]]/(kB T)))] ;
G=Exp[Vbias/(kB T)]-1;
If[JuncDopant[[1,1]]>0,
If[x≤ -x1, q1 G  (E^((x+x1)/Lp)),
If [x≥ x2, q2 G  (E^(-(x-x2)/Ln)), ]
]
, If[x≤ -x1, q1 G  (E^((x+x1)/Ln)),
If [x≥ x2, q2 G  (E^(-(x-x2)/Lp)), ]
]
]
]  ;

DiffJ[x_, T_,Vbias_,JuncDopant_, SemiParm_,DiffParm_]:= Module[{Eg, me, gc, mh,ϵ
, μn,μp,τ,Ln, Lp ,
,NC0,NV0
, W, x1,x2
,q1,q2,F1,F2,G
, kB=  8.61739*10^(-5)},
{ Eg, me, gc, mh,ϵ}=SemiParm  ; 
{ μn,μp,τ}=DiffParm  ;
Dn=10^4*Sqrt[μn kB T τ]; Lp=10^4*Sqrt[μp kB T τ];
Ln=10^4*Sqrt[μn kB T τ]; Lp=10^4*Sqrt[μp kB T τ];
{NC0,NV0}= NDOS0[SemiParm] ;
{W, x1,x2} =DepletionW[T,Vbias,JuncDopant, SemiParm] ;
F1=BandDiagram[-x1, T,0.,JuncDopant, SemiParm] ;
F2=BandDiagram[x2, T,0.,JuncDopant, SemiParm] ;
q1=If[JuncDopant[[1,1]]>0, NV0 (T^(3/2)) (E^(F1[[1]]/(kB T))), NC0 (T^(3/2)) (E^(-F1[[1]]/(kB T)))] ;
q2=If[JuncDopant[[1,1]]>0, NC0 (T^(3/2)) (E^(-F2[[2]]/(kB T))), NV0 (T^(3/2)) (E^(F2[[2]]/(kB T)))] ;
G=Exp[Vbias/(kB T)]-1;
If[JuncDopant[[1,1]]>0,
If[x≤ -x1,Lp q1 G  (E^((x+x1)/Lp))/τ,
If [x≥ x2,Ln q2 G  (E^(-(x-x2)/Ln))/τ, ]
]
, If[x≤ -x1,Ln q1 G  (E^((x+x1)/Ln))/τ,
If [x≥ x2,Lp q2 G  (E^(-(x-x2)/Lp))/τ, ]
]
]
]  ;

DiodeC[ T_,Vbias_,JuncDopant_, SemiParm_,DiffParm_]:= Module[{Eg, me, gc, mh,ϵ
, μn,μp,τ,Ln, Lp ,
,NC0,NV0
, W, x1,x2
,q1,q2,F1,F2,G
, kB=  8.61739*10^(-5)},
{ Eg, me, gc, mh,ϵ}=SemiParm  ; 
{ μn,μp,τ}=DiffParm  ;
Ln=10^4*Sqrt[μn kB T τ]; Lp=10^4*Sqrt[μp kB T τ];
{NC0,NV0}= NDOS0[SemiParm] ;
{W, x1,x2} =DepletionW[T,Vbias,JuncDopant, SemiParm] ;
F1=BandDiagram[-x1, T,0.,JuncDopant, SemiParm] ;
F2=BandDiagram[x2, T,0.,JuncDopant, SemiParm] ;
q1=If[JuncDopant[[1,1]]>0, NV0 (T^(3/2)) (E^(F1[[1]]/(kB T))), NC0 (T^(3/2)) (E^(-F1[[1]]/(kB T)))] ;
q2=If[JuncDopant[[1,1]]>0, NC0 (T^(3/2)) (E^(-F2[[2]]/(kB T))), NV0 (T^(3/2)) (E^(F2[[2]]/(kB T)))] ;
G=Exp[Vbias/(kB T)]-1;
1.602176487`*^-19*If[JuncDopant[[1,1]]>0,
( Lp q1 +Ln q2 ) G  /τ ,( Ln q1 +Lp q2 ) G /τ]
]  ;

DiodeC1[ T_,Vbias_,JuncDopant_, SemiParm_,DiffParm_]:= Module[{Eg, me, gc, mh,ϵ
, μn,μp,τ,Ln, Lp ,
,NC0,NV0
, W, x1,x2,ContactV,Ef1,Ef2
,q1,q2,F1,F2,G,r1,r2,H
, kB=  8.61739*10^(-5)},
{ Eg, me, gc, mh,ϵ}=SemiParm  ; 
{ μn,μp,τ}=DiffParm  ;
Ln=10^4*Sqrt[μn kB T τ]; Lp=10^4*Sqrt[μp kB T τ];
{NC0,NV0}= NDOS0[SemiParm] ;
Ef1=  Fermi[JuncDopant[[1]],T,SemiParm];
Ef2=  Fermi[JuncDopant[[2]],T,SemiParm];
ContactV=Eg+Sign[JuncDopant[[1,1]] ]*Fermi[JuncDopant[[1]],T,SemiParm]+Sign[JuncDopant[[2,1]] ]*Fermi[JuncDopant[[2]] ,T,SemiParm] ;
{W, x1,x2} =DepletionW[T,Vbias,JuncDopant, SemiParm] ;
F1=BandDiagram[-x1, T,0.,JuncDopant, SemiParm] ;
F2=BandDiagram[x2, T,0.,JuncDopant, SemiParm] ;
q1=If[JuncDopant[[1,1]]>0, NV0 (T^(3/2)) (E^(F1[[1]]/(kB T))), NC0 (T^(3/2)) (E^(-F1[[1]]/(kB T)))] ;
q2=If[JuncDopant[[1,1]]>0, NC0 (T^(3/2)) (E^(-F2[[2]]/(kB T))), NV0 (T^(3/2)) (E^(F2[[2]]/(kB T)))] ;
r1=  If[JuncDopant[[1,1]]>0,(mh/me)^3/2  *Exp[(F1[[1]]+F1[[2]])/(kB T)] , (me/mh)^3/2  *Exp[-(F1[[1]]+F1[[2]])/(kB T)] ] ;
r2=   If[JuncDopant[[1,1]]>0,(me/mh)^3/2  *Exp[-(F2[[1]]+F2[[2]])/(kB T)] , (mh/me)^3/2  *Exp[(F2[[1]]+F2[[2]])/(kB T)] ] ;
G=Exp[Vbias/(kB T)]-1; H=1-Exp[-2(ContactV-Vbias)/(kB T)] ;
1.602176487`*^-19*If[JuncDopant[[1,1]]>0,
( Lp q1 (1+r1 (G+1)) +Ln q2  (1+ r2 (G+1))) G  /(τ H)
,( Ln q1 (1+r1 (G+1)) +Lp q2 (1+ r2 (G+1)) ) G /(τ H)]
]  ;

BandDiagramBJT[x_, T_,dbase_,Vbias_,JuncDopant_, SemiParm_]:= Module[{Eg, me, gc, mh,ϵ
, ContactV1, ContactV2, V01,V02, W, x1,x2, Y, y1,y2, },
{ Eg, me, gc, mh,ϵ}=SemiParm  ; 
{W, x1,x2} =DepletionW[T,Vbias[[1]],Take[JuncDopant,{1,2}], SemiParm] ;
{Y, y1,y2} =DepletionW[T,Vbias[[2]],Take[JuncDopant,{2,3}], SemiParm] ;
ContactV1=Eg+Sign[JuncDopant[[1,1]] ]*Fermi[JuncDopant[[1]],T,SemiParm]+Sign[JuncDopant[[2,1]] ]*Fermi[JuncDopant[[2]] ,T,SemiParm] ;
ContactV2=Eg+Sign[JuncDopant[[2,1]] ]*Fermi[JuncDopant[[2]],T,SemiParm]+Sign[JuncDopant[[3,1]] ]*Fermi[JuncDopant[[3]] ,T,SemiParm] ;
V01=ContactV1-Vbias [[1]];
V02=ContactV2-Vbias [[2]];
w=Sign[JuncDopant[[1,1]] ]*If[x≤-x1,0.,If[x≤ 0., V01  (x+x1)^2 /(x1*W),
If[x≤ x2, V01*(1-(x-x2)^2/(x2*W)) ,
If[x≤ dbase-y1,V01, 
If[x≤ dbase, V01-V02*(x-dbase+y1)^2/(y1*Y),
If[x≤ dbase+y2,V01- V02*(1-(x-dbase-y2)^2/(y2*Y)), V01- V02]
]
]
]
]
] ;
w-Fermi[JuncDopant[[1]],T,SemiParm]
+{Sign[JuncDopant[[2,1]] ]*Eg,0.} 
] ;

BandDiagramMODFET[y_,x_, T_,dbase_,Vbias_,JuncDopant_, SemiParm_]:= Module[{Eg, me, gc, mh,ϵ
, ContactV1, ContactV2, V01,V02, W, x1,x2, Y, y1,y2, },
{ Eg, me, gc, mh,ϵ}=SemiParm  ; 
{W, x1,x2} =DepletionW[T,Vbias[[1]],Take[JuncDopant,{1,2}], SemiParm] ;
{Y, y1,y2} =DepletionW[T,Vbias[[2]],Take[JuncDopant,{2,3}], SemiParm] ;
ContactV1=Eg+Sign[JuncDopant[[1,1]] ]*Fermi[JuncDopant[[1]],T,SemiParm]+Sign[JuncDopant[[2,1]] ]*Fermi[JuncDopant[[2]] ,T,SemiParm] ;
ContactV2=Eg+Sign[JuncDopant[[2,1]] ]*Fermi[JuncDopant[[2]],T,SemiParm]+Sign[JuncDopant[[3,1]] ]*Fermi[JuncDopant[[3]] ,T,SemiParm] ;
V01=ContactV1-Vbias [[1]];
V02=ContactV2-Vbias [[2]];
w=Sign[JuncDopant[[1,1]] ]*If[x≤-x1,0.,If[x≤ 0., V01  (x+x1)^2 /(x1*W),
If[x≤ x2, V01*(1-(x-x2)^2/(x2*W)) ,
If[x≤ dbase-y1,V01, 
If[x≤ dbase, V01-V02*(x-dbase+y1)^2/(y1*Y),
If[x≤ dbase+y2,V01- V02*(1-(x-dbase-y2)^2/(y2*Y)), V01- V02]
]
]
]
]
] ;
w-Fermi[JuncDopant[[1]],T,SemiParm]
+{Sign[JuncDopant[[2,1]] ]*Eg,0.} 
] ;

Application

Overview:

Note: there are many types of FET, for this course, we need to learn only MOSFET 
(MODFET is also useful for FET concept illustration but not essential).

Illustration with two types of FET:
1- Channel normally on: Generic MODFET
	- Understand channel pinch-off concept
	- Illustration channel current saturation and IV characteristics
2- Channel normally off: MOSFET
	- MOS junction and inversion
	- C-V characteristics and switching efficiency
	- IV curve

1. Concept illustration "Generic" FET - MODFET (skip)

Can one make a FET with a channel that is already OPEN?
YES - If a channel is normally on - then the effect of the field is to "pinch-off" the channel
MODFET or Generic FET
This is a different from MOSFET - which is normally off 
MODFET is a very useful model to illustrate the pinch-off concept

- Start out with a conducting channel
- Use the depletion effect to pinch off
- the consequence is current saturation

SemiParm={ 1.12, (0.19^2 0.98)^(1/3.),6, 0.49,11.7 } ;   (* this is Si  *)
VG0=. ;
T=300. ;  
Manipulate[ VBC=VG0  ; VBE=VG0  ;
d1=18. ; d2=17.5; d3=18. ;dbase=0.95 ;
xR1=0.5 ; xR2=1.5  ; dtype=1  ;
Show[Plot[BandDiagramBJT[x, T,dbase,{VBE,VBC},{ {dtype,10^d1},{-dtype,10^d2},{dtype,10^d3} }, SemiParm],{x,-xR1,xR2}, PlotRange-> All, PlotStyle-> Thick]
,Graphics[{If[dtype<0,Red,Blue], Rectangle[{0,0}
,{DepletionW[T,VBE,{ {dtype,10^d1},{-dtype,10^d2} }, SemiParm][[3]],-Sign[dtype]* 10^d2/(10^d1+10^d2)}]} ]
,Graphics[ {If[dtype<0,Blue,Red],Rectangle[
{-DepletionW[T,VBE,{ {dtype,10^d1},{-dtype,10^d2} }, SemiParm][[2]]
,Sign[dtype]*10^d1/(10^d1+10^d2)},{0,0}]}
]
,Graphics[{If[dtype<0,Blue,Red], Rectangle[{dbase,0}
,{dbase+DepletionW[T,VBC,{ {-dtype,10^d2},{dtype,10^d3}}, SemiParm][[3]],Sign[dtype]* 10^d3/(10^d3+10^d2)}]} ]
,Graphics[ {If[dtype<0,Red,Blue],Rectangle[
{dbase-DepletionW[T,VBC,{ {-dtype,10^d2},{dtype,10^d3}}, SemiParm][[2]]
,-Sign[dtype]*10^d2/(10^d3+10^d2)},{dbase,0}]}
]
, Plot[BandDiagramBJT[x, T,dbase,{VBE,VBC},{ {dtype,10^d1},{-dtype,10^d2},{dtype,10^d3} }, SemiParm][[2]],{x,-xR1,xR2}, PlotRange-> All, PlotStyle-> {Thick, Purple}]
, Plot[BandDiagramBJT[x, T,dbase,{VBE,VBC},{ {dtype,10^d1},{-dtype,10^d2},{dtype,10^d3} }, SemiParm][[1]],{x,-xR1,xR2}, PlotRange-> All, PlotStyle-> {Thick, Hue[0.4,1,0.5]}
,PlotRange-> {{-0.5,1.5},All}] 
 ], {{VG0,0},0,-5}]

Pinch-off can be created by using depletion- Increasing reverse bias voltage.

SemiParm={ 1.12, (0.19^2 0.98)^(1/3.),6, 0.49,11.7 } ;   (* this is Si  *)
T=300. ;   VG=. ; 
VG[x_,VG0_,L_]:=VG0*(1-x/L)  ; 
Manipulate[  d1=18. ; d2=17.5; d3=18. ;dbase=0.95 ; L=5.  ;
xR1=0.5 ; xR2=1.5  ; dtype=1  ;
Plot3D[BandDiagramMODFET[x,y, T,dbase,{VG[x,VG0,L],VG[x,VG0,L]},{ {dtype,10^d1},{-dtype,10^d2},{dtype,10^d3} }, SemiParm],{y,-xR1,xR2},{x,0,L}, PlotRange->{{-xR1,xR2},{0,L},{-1.,6}}, BoxRatios-> {1,1,1}]   , {{VG0,0},0,-5}]

3D illustration of band diagram with pinch off

1.1 Pinch-off effect on ID

Derivation of drain current: ID.
 ({{ρ, channel resistivity}, {(Subscript[D, 0]), channel depth}, {(Subscript[h, 0]), channel original width (zero (Subscript[V, G]))}, {(Subscript[I, D]), drain current}, {(Subscript[V, G]) or (Subscript[V, g]), gate voltage}, {(Subscript[V, D]) or (Subscript[V, d]), drain voltage}, {(Subscript[V, S]) or (Subscript[V, s]), source voltage, which can be used as reference = 0}, {V[x], voltage along the channel length}, {W[x], channel opening}, {(Subscript[V, PO]), pinh-off voltage (a function of channel structure) }})

ID= .; Vg=. ;

ΔVx:=ID(ρ/(D0  h[x]))Δx

h[x_]:=h0 - 2 W[x]

W[x_]:=(h0/2)(Sqrt[((V[x]-Vg)/VPO)])

ΔVx

ID(ρ/(D0  h0))Δx==ΔVx(1- (Sqrt[((V(x)-Vg)/VPO)]))
Solve the differential equation by integrating both side.

Solve for Vx:

∫ID(ρ/(D0  h0))\[DifferentialD]x== ∫(1-(Sqrt[((Vx-Vg)/VPO)]))\[DifferentialD]Vx

We must include a constant and solve for the boundary condition V[0]= Vs

((ID x ρ)/(D0 h0))+C1==Vx-(2/3) (Vx-Vg) (Sqrt[((Vx-Vg)/VPO)]) /. x->0 /. Vx-> 0

Hence:     ((ID x ρ)/(D0 h0))==-(2/3) Vg (Sqrt[-(Vg/VPO)])+Vx-(2/3) (Vx-Vg) (Sqrt[((Vx-Vg)/VPO)]) 
                           ==VPO ((Vx/VPO)-(2/3) (((((Vx-Vg)/VPO)) )^(3/2))+(2/3) (((-(Vg/VPO) ))^(3/2)))  
We can also define: (((Subscript[D, 0])(Subscript[h, 0]))/(ρ (Subscript[I, D])))=(1/Ec): (~ channel effective electric field)
If we care to solve for V[x] as function of x.

Solve[Ec  x ==  VPO   ((Vx/VPO)-(2/3)  (((((Vx-Vg)/VPO)) )^(3/2))+ (2/3)  (((-(Vg/VPO) ))^(3/2))) ,Vx]

The solution is above.

Do we care? No.. this is just a simplistic model, no need for so much algebra.
Usually, we care only numerical solution.

Vc[x_,L_,VG_,VD_,VPO_]:= Module[{Ec, y} ,
Ec=VPO     ;
Print["  EC=", Ec]   ] ;

Vc[x_,L_,VG_,VD_,VPO_]:= Module[{Ec, y} ,
Ec=VPO  ( ((Min[VD,VG+VPO])/VPO)-(2/3) (((((Min[VD,VG+VPO]-VG)/VPO)))^(3/2))+(2/3)  (((-(VG/VPO)))^(3/2))) /L    ;
y=Vx  /. FindRoot[Ec  x ==  VPO   ((Vx/VPO)-(2/3)  (((((Vx-VG)/VPO)))^(3/2))+ (2/3)  (((-(VG/VPO)))^(3/2))) ,{Vx, VD*x/L}] ;
y] ;

Vc[0.2,3,-2,5,VPO=2.5]

L=3.  ;
Manipulate[VPO=2.5;
Plot[Vc[x,L,VG,VD,VPO]/VD,{x,0,L},PlotRange-> {0,1}]
,{VG,0,-1},{VD,0.1,1.5}]

Solve for ID:

We solve for drain current by integrating the entire channel:

(∫_0^L)   ((ID ρ)/(D0 h0))  \[DifferentialD]x== (∫_0^VD) (1-(Sqrt[((Vx-VG)/VPO)]))\[DifferentialD]Vx

-(2/3) VD (Sqrt[((VD-VG)/VPO)])+(2/3) VG (Sqrt[((VD-VG)/VPO)])+VD-(2/3) VG (Sqrt[-(VG/VPO)]) /. VD-> 0

((ID L ρ)/(D0 h0))== FullSimplify[-(2/3) VD (Sqrt[((VD-VG)/VPO)])+(2/3) VG (Sqrt[((VD-VG)/VPO)])]+VD-(2/3) VG (Sqrt[-(VG/VPO)])

Thus, the solution is:
((ID L ρ)/(D0 h0))==VD-(2/3) VPO (((((VD-VG)/VPO)))^(3/2))-(2/3) VG (Sqrt[-(VG/VPO)])
Or:

ID=((D0 h0)/(L ρ))VPO  ( (VD/VPO)-(2/3) (((((VD-VG)/VPO)))^(3/2))+(2/3)  (((-(VG/VPO)))^(3/2)))

ID=.  ;
ID[VD_,VG_,VPO_,G0_]:=G0  VPO  ((VD/VPO)-((2 )/3)((((((VD-VG)/VPO)))^(3/2))-(((-(VG/VPO)))^(3/2))))

Take a look at the differential conductance: ((∂ (Subscript[I, D]))/(∂ (Subscript[V, D])))

(∂_VD)ID[VD,VG,VPO,G0]

The limit is: 1-(Sqrt[((VD-VG)/VPO)])≥ 0  Or: VD-VG=VPO:
The differential conductance: ((∂ (Subscript[I, D]))/(∂ (Subscript[V, D])))=0 at VD-VG=VPO. This means current DOESNOT increase anymore vs VD: current saturation. VPO is a function of the channel (structure of the channel).

1.2 Saturation current

We now define:

IDsat[VG_,VPO_] :=ID[VD,VG,VPO,G0] /. VD->VG+VPO

IDsat[VG,VPO]

Simplify[%]

1.3 Illustration of ID

ID[VD_,VG_,VPO_,G0_]:=G0 VPO    (((Min[VD,VG+VPO])/VPO)-((2 )/3)((((((Min[VD,VG+VPO]-VG)/VPO)))^(3/2))-(((-(VG/VPO)))^(3/2))))

Plot[Release[  Table[ ID[VD,VG,2.5,0.05]  ,{VG,-1,0,0.25}]
                                                  ],{VD,0,10}
     , PlotRange->{{0,10.},{0,0.05}}
	, AxesLabel->{Drain voltage,Drain current}
     , AspectRatio-> 1 
     , PlotStyle-> {                    {RGBColor[0,0,1], Thickness[0.02]}
		                                                  , {RGBColor[1,0,0], Thickness[0.02]}
			                                                , {RGBColor[0,0.5,0], Thickness[0.02]}  }]

Plot3D[  ID[VD,VG,2.,0.05] 
				                           ,{VD,0,10}
	                                         ,{VG,-1,0}
                                                 
     , PlotRange->{{0,10.},{-1,0},{0,0.05}}
,BoxRatios->{1,1,2}
        ]

1.4 Illustration of pinch-off

Set up

-(1/3) 2 (Sqrt[((VD-VG)/VPO)]) VD+VD+(2/3) VG (Sqrt[((VD-VG)/VPO)])-(2/3) VG (Sqrt[-(VG/VPO)]) /. VD->Vx

((ID L ρ)/(D h0)) /. L->x /. ID->((D h0 (-2 VPO (((((VD-VG)/VPO)))^(3/2))+3 VD+2 (((-(VG/VPO)))^(3/2)) VPO))/(3 L ρ))

x /.  Solve[(((-2 VPO (((((VD-VG)/VPO)))^(3/2))+3 VD+2 (((-(VG/VPO)))^(3/2)) VPO) x)/(3 L))==-(2/3) (Sqrt[-(VG/VPO)]) VG+(2/3) (Sqrt[((Vx-VG)/VPO)]) VG+Vx-(2/3) Vx (Sqrt[((Vx-VG)/VPO)]), {x}]

ChannelWidth[x_,VD_,VG_,VPO_,h0_,L_]:=Module[{Vx ,Vm=Min[VD,VPO+VG] },
		               
		           Vx= y /.    FindRoot[Min[x,L] ==-(3 L ((2/3) (Sqrt[-(VG/VPO)]) VG-(2/3) (Sqrt[((y-VG)/VPO)]) VG-y+(2/3) y (Sqrt[((y-VG)/VPO)])))/(-2 VPO (((((Vm-VG)/VPO)))^(3/2))+3 Vm+2 (((-(VG/VPO)))^(3/2)) VPO),
			                                      {y,0.,Max[Vm,0.001]   }];
		             h0 * (1-(Sqrt[((Vx-VG)/VPO)]))   
	                                                     ]

Illus for channel opening as a function of Drain voltage and Gate voltage

Manipulate[ VPO=10. ;VD=Drainvoltage;
VG=Gatevoltage ;
 Plot[{ ChannelWidth[x,VD,VG,VPO,0.5,3.],
					                 - ChannelWidth[x,VD,VG,VPO,0.5,3.] 
				                    },{x,0.,3.}, Filling-> {1-> {2}}, PlotRange-> {-0.5,0.5}]
,{Drainvoltage,0.01,9},{Gatevoltage,0,-VD}]

Illustration of channel opening

Manipulate[VPO=Pinchoffvoltage ;VD=Drainvoltage;
VG=Gatevoltage ;
 Plot[{ ChannelWidth[x,VD,VG,VPO,0.5,3.],
					                 - ChannelWidth[x,VD,VG,VPO,0.5,3.] 
				                    },{x,0.,3.}, Filling-> {1-> {2}}, PlotRange-> {-0.5,0.5}]
,{Drainvoltage,1,9},{Gatevoltage,0,-VD},{Pinchoffvoltage,4,10}]

Illustration of channel opening

2. Element of MOSFET channel: MOS junction, low field and depletion

MOSFET:
- Channel normally closed
- Gate voltage is to open (create) channel - A threshold voltage.
- A similar "pinchoff" effect (but a bit more complicated) not horizontally, but vertically.
- The effect leads to current saturation

2.1 Metal-Oxide-Semiconductor junction

We will see that semiconductor can be:
- insulator-like or very-poor conductor-like
- metal-like
- the transition happens with a “threshold”

In the illustration below, we see what happens to the charge at the semiconductor junction as a function of gate voltage.

2.2 Band bending and electrostatic

When we apply a field to MOS, what happens in the semiconductor? what is the charge profile in the semiconductor?

We need to calculate the electrostatic potential and charge density at the channel beneath the oxide (or insulating layer). V and Q.
This is done by solving the Poisson's equation.
We will see that there are roughly three conditions:
- low field: charge screening
- stronger field: depletion
- even stronger field: inversion

2.1.1 Charge density expression

Charge density is
                ρ[x]=q ((SuperPlus[(Subscript[N, D])])-(SuperMinus[(Subscript[N, A])])+p[x]-n[x])
At x->∞, the charge is neutral: ie. ρ[x=∞]=0=  q ((SuperPlus[(Subscript[N, D])])-(SuperMinus[(Subscript[N, A])])+p0-n0) where p0 and n0 are hole and electron density at equilibrium (assume ND and NA are uniform). At equilibrium, (see Chapter 3)

(Subscript[n, 0])=(Subscript[n, i]) (e^(((Subscript[E, F]) - Ei)/((Subscript[k, B]) T)))=(Subscript[n, i]) (e^(-(Subscript[ϕ, F])/((Subscript[k, B]) T)))  

where we denote: (Subscript[E, i])-(Subscript[E, F])=(Subscript[ϕ, F])

(Subscript[p, 0])=((((Subscript[n, i])^2))/(Subscript[n, 0]))=(Subscript[n, i])(e^((Subscript[ϕ, F])/((Subscript[k, B]) T)))
At any point with a potential q ϕ[x] added to the band:

n[x]=ni (e^((EF - (Ei-q ϕ[x]))/((Subscript[k, B])T)))=ni (e^((EF - Ei+q ϕ[x])/((Subscript[k, B]) T)))=n0 (e^(q ϕ[x]/((Subscript[k, B]) T)))
p[x]=(((ni^2) )/(n[x]))=((ni^2)/n0)(e^(-q ϕ[x]/((Subscript[k, B]) T)))=p0 (e^(-q ϕ[x]/(k T)))

Detailed algebra for   ρ[x]=q ((SuperPlus[(Subscript[N, D])])-(SuperMinus[(Subscript[N, A])])+p[x]-n[x])

ρ[x]=q (NDi-NAi+p[x]-n[x]);
ρ[x] = ρ[x]/. p[x]->p0 (E^(-q ϕ[x]/(k T)))  /. n[x]->n0 (E^(q ϕ[x]/(k T)))

ρ[x] /.-NAi+NDi->n0-p0

FullSimplify[%]

ρ[x] = (n0(1-(E^((q ϕ(x))/(k T))) ) +(-1+(E^(-((q ϕ(x))/(k T))))) p0) q

2.1.2 Maxwell-Poisson's equation of electrostatic and charge density (Poisson)

Maxwell-Poisson's equation:
(((d^2)ϕ[x])/(d (x^2)))=-((ρ[x])/ϵ)
From the detailed algebra result above:

(((d^2)ϕ[x])/(d (x^2)))=-(q/ϵ)(n0(1-(e^((q ϕ(x))/(k T))) ) +(-1+(e^(-((q ϕ(x))/(k T))))) p0) 
Or: 
2(((d^2)ϕ[x])/(d (x^2)))((dϕ[x])/(d x))=-((2 q)/ϵ)(n0(1-(e^((q ϕ(x))/(k T))) ) +(-1+(e^(-((q ϕ(x))/(k T))))) p0)((dϕ[x])/(d x))

This equation is transcendental and cannot be solved. But we can get reasonable approximation.
But first, we can to transform it into 1st order differential eq. for approximation.

Details: transforming into 1st order DE

(∫_0^(ϕ[x]))((- 2 q)/ϵ) n0(1-(E^((q y)/(k T))) )\[DifferentialD]y + (∫_0^(ϕ[x]))((- 2 q)/ϵ) p0(-1+(E^(-((q y)/(k T)))))\[DifferentialD]y

(∫_0^(El[x]))2 El \[DifferentialD]El

((El(x))^2) ==-((2 k n0 T)/ϵ)-((2 k p0 T)/ϵ)+((2 n0 ((E^((q ϕ(x))/(k T))) k T-q ϕ(x)))/ϵ)+((2 p0 ((E^(-((q ϕ(x))/(k T)))) k T+q ϕ(x)))/ϵ)

((2 k  T p0)/ϵ) ((E^(-((q ϕ(x))/(k T))))+((q ϕ(x))/(k T)) -1)+((2 k  T n0)/ϵ)((E^((q ϕ(x))/(k T))) -((q ϕ(x))/(k T))-1)

ϕ'[x]==Sqrt[ ((2 k  T p0)/ϵ) ((E^(-((q ϕ[x])/(k T))))+((q ϕ[x])/(k T)) -1)+((2 k  T n0)/ϵ)((E^((q ϕ[x])/(k T))) -((q ϕ[x])/(k T))-1)]

Hence, we have a differential equation:   
ϕ'[x]==Sqrt[ ((2 k  T p0)/ϵ) ((e^(-((q ϕ[x])/(k T))))+((q ϕ[x])/(k T)) -1)+((2 k  T n0)/ϵ)((e^((q ϕ[x])/(k T))) -((q ϕ[x])/(k T))-1)]
As mentioned above, this cannot be solve analytically, but can be approximated for different conditions below.

2.3 Approximated solutions for the electrostatic potential: low field

2.3.1 Low field region - detailed algebra

T=. ;  k =. ;

First, let it be p-doped material. And low field, then: n0<<p0: we can simplify:

ϕ'[x]==Sqrt[ ((2 k  T p0)/ϵ) ((E^(-((q ϕ[x])/(k T))))+((q ϕ[x])/(k T)) -1)+((2 k  T n0)/ϵ)((E^((q ϕ[x])/(k T))) -((q ϕ[x])/(k T))-1)] /.  n0-> 0

ϕ'[x]==Sqrt[ ((2 k  T p0)/ϵ) ((e^(-((q ϕ(x))/(k T))))+((q ϕ(x))/(k T)) -1)]
What if ϕ[x] is small? we have:
(e^(-((q ϕ(x))/(k T))))+((q ϕ(x))/(k T)) -1       --->  1 -((q ϕ(x))/( k T)) +(1/2)(((((q ϕ(x))/( k T))))^2)+((q ϕ(x))/( k T))-1
(1/2)(((((q ϕ(x))/( k T))))^2)

Sqrt[ ((2 k  T p0)/ϵ) (1/2)(((((q ϕ[x])/( k T))))^2)]

ϕ'[x]==  ϕ[x]  (Sqrt[((p0 (q^2))/(k T ϵ))])

2.3.2 Low field solution

Define (Sqrt[((p0 (q^2))/(k T ϵ))]) = (1/LD): where LD has the unit of length. This is called Debye length, NOT DIFFUSION length.

DSolve[-ϕ'[x]==((ϕ[x])/LD),ϕ[x],x]

Expplot=Plot[Exp[-x],{x,0.,5}, PlotStyle->{RGBColor[0.,0.,0.7], Thickness[0.01]}]

So, at low field, the potential has the exponential decay profile with a characteristic length LD = (Sqrt[((k T ϵ)/(p0 (q^2)))]). LD is called the Debye length. It signifies the length dimension for which the charges screen out the potential.

2.4. Approximated solutions for the electrostatic potential: large field region (depletion)

2.4.1 Detailed algebra

Consider now that ((q ϕ(x))/(k T)) is large, >>1 and so (e^(-((q ϕ(x))/(k T)))) is very small, then 
ϕ'[x]==Sqrt[ ((2 k  T p0)/ϵ) ((e^(-((q ϕ(x))/(k T))))+((q ϕ(x))/(k T)) -1)]=Sqrt[ ((2 k  T p0)/ϵ) (((q ϕ(x))/(k T)) )]

FullSimplify[DSolve[ϕ'[x]==(Sqrt[((2 k  T p0)/ϵ) (((q ϕ[x])/(k T)) )]), ϕ[x],x]]

FullSimplify[%]

2.4.2 Solution

Hence, the electrostatic is parabolic, similarly to the depletion regions at a p-n junction, and it is so for exactly the same reason: depletion of carriers.
        ϕ(x)= (((Subscript[p, 0]) q)/(2 ϵ)) (((x-(Subscript[c, 1])))^2)

Quadplot=Plot[(x-3.)^2/9.,{x,0,3}
  , PlotStyle->{RGBColor[0.3,0.3,0.], Thickness[0.01]}]

Show[Quadplot, Expplot]

This is a simple parabolic profile: equivalent with constant charge. This is the case since by dropping the exponent term (e^(-((q ϕ(x))/(k T)))) which represents the charge density variation, the residual charge must be constant. The charge is:

ϵ  (∂_x)((∂_x)(((p0 q  (((x-(Subscript[c, 1])))^2))/(2 ϵ))))

Indeed this is similar to what we have seen at p-n junction depletion.

3. Channel inversion

Introduction: Let’s take a look at this:

3.1 Discussion

ϕ'[x]==Sqrt[ ((2 k  T p0)/ϵ) ((e^(-((q ϕ[x])/(k T))))+((q ϕ[x])/(k T)) -1)+((2 k  T n0)/ϵ)((e^((q ϕ[x])/(k T))) -((q ϕ[x])/(k T))-1)]

In the above approximation, the potential is just that of  the constant charge (Subscript[p, 0]) q from doping.
But notice something: we always assume that (e^(-((q ϕ(x))/(k T)))) is very small. If so,  (e^((q ϕ(x))/(k T))) is very large and at some point, there is a term that we neglect:   (Subscript[n, 0]) (e^((q ϕ(x))/(k T))) only because the minority carrier concentration (Subscript[n, 0]) is small.
At some point, because (e^((q ϕ(x))/(k T))) is so large,   (Subscript[n, 0]) (e^((q ϕ(x))/(k T))) is nolonger negligible, (EVEN as n0 is small as minority carriers)
To deal with this, we have to modify our approximation.

3.2 Surface charge density, inversion

The above simple approximation gives us some insight into the behavior of the MOS junction. With the concept of Debye length and charge screening, we can rewrite the equation: 
Efield= - ϕ'[x]==-Sqrt[ ((2 k  T p0)/ϵ) ((e^(-((q ϕ[x])/(k T))))+((q ϕ[x])/(k T)) -1)+((2 k  T n0)/ϵ)((e^((q ϕ[x])/(k T))) -((q ϕ[x])/(k T))-1)]
  = - (Sqrt[((2 k  T p0)/ϵ)]) Sqrt[  ((e^(-((q ϕ[x])/(k T))))+((q ϕ[x])/(k T)) -1)+(n0/p0)((e^((q ϕ[x])/(k T))) -((q ϕ[x])/(k T))-1)]
  We can simplify further by replacing (n0/p0) --> (e^(-2 q ϕF/(k T)))
  We we use Gauss' law, we know that the net charge accumulation is:
  (Subscript[Q, S]) = -ϵ Efield   at the surface. So, the surface charge is:
   (Sqrt[2 k  T p0 ϵ]) Sqrt[  ((e^(-((q (Subscript[ϕ, S]))/(k T))))+((q (Subscript[ϕ, S]))/(k T)) -1)+(e^(-2 q ϕF/(k T)))((e^((q (Subscript[ϕ, S]))/(k T))) -((q (Subscript[ϕ, S]))/(k T))-1)]
  q p0 (Sqrt[((2 k  T  ϵ)/(p0  (q^2)))])Sqrt[  ((e^(-((q (Subscript[ϕ, S]))/(k T))))+((q (Subscript[ϕ, S]))/(k T)) -1)+(e^(-2 q ϕF/(k T)))((e^((q (Subscript[ϕ, S]))/(k T))) -((q (Subscript[ϕ, S]))/(k T))-1)]
 q  p0  LD Sqrt[ 2 ((e^(-((q (Subscript[ϕ, S]))/(k T))))+((q (Subscript[ϕ, S]))/(k T)) -1)+2 (e^(-2 q ϕF/(k T)))((e^((q (Subscript[ϕ, S]))/(k T))) -((q (Subscript[ϕ, S]))/(k T))-1)]
 Solution is:

Plot[ - Sign[y]*(Sqrt[2 (E^(-2 a)) (-y+(E^y)-1)+2 (y+(E^(-y))-1)])  /. a->10. ,{y,-20,40},
               PlotStyle->{RGBColor[1,0,0], Thickness[0.01]}
	              ,AspectRatio->1
            ,  ImageSize->{300,300}]

We see what happens here:

Illustration on log scale

Plot[Log[10.,(Sqrt[2 (E^(-2 a)) (-y+(E^y)-1)+2 (y+(E^(-y))-1)])] /. a->10. ,{y,-20,40},
               PlotStyle->{RGBColor[1,0,0], Thickness[0.01]}
	              ,AspectRatio->1
            ,  ImageSize->{300,300}]

Exercise: where is the majority carrier accumulation? where is the depletion? where is the threshold? and where is the inversion?

We plot with the following transformed unitless variable:   y = ((q (Subscript[ϕ, S]))/(k T)) ;  a = ((q ϕF)/(k T)). 

The turning point is: y = 2 a or: (Subscript[ϕ, S]) = 2 ϕF.
We observe that there are 2 regions. A region with  (Subscript[ϕ, S]) < 2 ϕF and a region with (Subscript[ϕ, S]) > 2 ϕF.
What happens when (Subscript[ϕ, S]) < 2 ϕF?  We can plot the carrier density:  
						
n[x]=ni (e^((EF - (Ei-q ϕ[x]))/(k T)))=ni (e^((EF - Ei+q ϕ[x])/(k T)))=ni (e^((q ϕ[x]-ϕF) /(k T)))
p[x]=(((ni^2) )/(n[x]))=((ni^2)/n0)(e^(-q ϕ[x]/(k T)))=p0 (e^(-q ϕ[x ]/(k T)))
n[x] 
For simplicity, we can use the approximation  (Subscript[ϕ, S]) (e^(-(x/LD)))

3.3 Carrier density (mobile charges)

3.3.1 Illustrations

There are two types of charge (no, not holes and electrons).
1- fixed charges associated with ionized acceptors in p-type (or donors in n-type): these are NOT relevant, because these charges do NOT conduct electricity
2- mobile charges associated with carriers (either holes or electrons, depending on the channel type). For the channel, the inversion mobile charge is the key thing that makes a MOSFET work.

See illustration of carrier density

Illustration of charge density and mobile carrier (electrons) density

Manipulate[
If[start==1,carrierconc={
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,}  ;
 start=0 ,
] ;
Animate[carrierconc[[i]],{i,1,Length[carrierconc],1}]
,
{{start,1,"Click to start"},{0,1}} ]

Illustration of hole and electron density on log scale

Manipulate[
If[start==1,logcarrierconc={
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,};
 start=0 ,
] ;
Animate[logcarrierconc[[i]],{i,1,Length[logcarrierconc],1}]
,
{{start,1,"Click to start"},{0,1}} ]

This is a plot of hole (blue) and electron (red) density on log scale as a function of distance into the semiconductor

The point where carrier type crosses over each other is called the inversion point. The regime where there is a strong electron concentration is called the inversion regime. MOSFET channel is formed by the inversion region underneath a gate.

3.3.2 Net charge density

ρ[x] = (n0(1-(e^((q ϕ(x))/(k T))) ) +(-1+(e^(-((q ϕ(x))/(k T))))) p0) q
           = q p0  ((e^(-((2  q ϕF)/(k T))))(1-(e^((q ϕ[x])/(k T))) ) +(-1+(e^(-((q ϕ[x])/(k T))))))

Manipulate[Plot[  {((E^(-Vb  (E^(-x )))) -1)+(E^(-20))  (1-(E^(Vb   (E^(-x ))))) },{x,0.,10}, 
			                             Filling -> Axis, FillingStyle->  RGBColor[0.4,0.4,1.]
			                            ,PlotRange->{0,-5}
,Frame-> True, GridLines-> Automatic
,FrameLabel->{"Position inside semiconductor (a. u.)","Charge density (a. u.)"}
,LabelStyle->Directive[14,FontFamily-> "Arial"] ],{{Vb,0,"Gate bias voltage (a. u.)"},0,25}]

4. MOS junction C-V and capacitance characteristics

Critical for inversion is the voltage inside semiconductor.

It is crucial for MOSFET to potential in semiconductor  (Subscript[ϕ, s])>> (Subscript[V, oxide])

4.1 Capacitance concept

MOS junction is a capacitor. In the above we study the field (or potential) profile inside the semiconductor. Externally, MOS can be described interms of capacitance and accumulated charge 
The bias voltage is a sum of voltage across different region:
VB = VOxide + (Subscript[ϕ, S])
VOxide= -((Subscript[Q, S])/(Subscript[C, oxide])) where (Subscript[Q, S]) is the charge in metal and semiconductor (equal and opposite sign).
Thus:
(Subscript[Q, S]) = (VB-(Subscript[ϕ, S])) (Subscript[C, oxide])
But using Gauss' law, we know that:
(Subscript[Q, S]) = ϵ E[x=0], or = -ϵ ((Subscript[dϕ, S])/dx)[x=0]
Thus, the charge balance equation is:
(VB-(Subscript[ϕ, S])) (Subscript[C, oxide]) = -ϵ ((Subscript[dϕ, S])/dx)[x=0]
This is a fundamental relation of MOS junction.
The full equation is:
(VB-(Subscript[ϕ, S])) (Subscript[C, oxide])=  q p0 LD Sqrt[ 2 ((e^(-((q (Subscript[ϕ, S]))/(k T))))+((q (Subscript[ϕ, S]))/(k T)) -1)+2 (e^(-2 q ϕF/(k T)))((e^((q (Subscript[ϕ, S]))/(k T))) -((q (Subscript[ϕ, S]))/(k T))-1)]
Or:
(VB-(Subscript[ϕ, S])) =((q p0 LD Sqrt)/(Subscript[C, oxide]))[ 2 ((e^(-((q (Subscript[ϕ, S]))/(k T))))+((q (Subscript[ϕ, S]))/(k T)) -1)+2 (e^(-2 q ϕF/(k T)))((e^((q (Subscript[ϕ, S]))/(k T))) -((q (Subscript[ϕ, S]))/(k T))-1)]

See charge plot again:

Manipulate[Plot[  {((E^(-Vb  (E^(-x )))) -1)+(E^(-20))  (1-(E^(Vb   (E^(-x ))))) },{x,0.,10}, 
			                             Filling -> Axis, FillingStyle->  RGBColor[0.4,0.4,1.]
			                            ,PlotRange->{0,-5}
,Frame-> True, GridLines-> Automatic
,FrameLabel->{"Position inside semiconductor (a. u.)","Charge density (a. u.)"}
,LabelStyle->Directive[14,FontFamily-> "Arial"] ],{{Vb,0,"Gate bias voltage (a. u.)"},0,25}]

We can solve the above equation numerically. But it is more intuitive to see what's going on.

First, we observe that there are 2 types of charges in the semiconductor. A charge due to the exposed ionized acceptor as the holes move away from the junction, called depletion charge. The other is the electrons especially at the inversion. The first charge is NOT mobile, so they are not contributing to the drain current. The mobile electron charge is what makes FET work.

From above discussion (Section 2 and the plottings), we see that charge behavior is quite complicated. There are weak field, strong depletion, and inversion conditions. Since accumulated charge behavior directly determines the capacitance, it is clear that the capacitance of a MOS is very complicated but understanding this capacitance is essential. In fact, measurement and analysis of MOS becomes an indispensible tool for experimental characterization of a MOS channel. Also important is the high frequency characteristic for fast switching applications.

The differential capacitance is defined as:
(Subscript[C, S]) = ((d QS)/(d V))=((d QS)/(d (Subscript[ϕ, S])))
This is what we need to analyze.

4.2 Depletion region capacitance

For this condition:
 (Subscript[C, S]) = (ϵ/W), where W=((Sqrt[2 ϵ (Subscript[ϕ, s ])])/(Sqrt[p0 q ]))  is a depletion width. But this is practically the capacitance of the depletion width itself. Hence, the capacitance decreases with voltage. 
This is the reason to consider MOS as a series of 2 capacitor: the oxide and the semiconductor. 
The net capacitor is (1/(Subscript[C, T]))=(1/(Subscript[C, oxide]))+(1/(Subscript[C, depletion ]))
The larger the oxide capacitance, the closer the total capacitance to (1/(Subscript[C, depletion ])): more charge for the channel.

4.3 Inversion condition and threshold voltage

4.3.1 Discussion

When inversion happens, the mobile electron charge can build up very fast next to the oxide-semiconductor interface. So intuitively, the capacitance must increase again.

The capacitance initially increases with voltage exponentially. Obviously it can not increase forever. What happens is that (Subscript[ϕ, S]) is "clamped" just above 2 ϕF so that increasing the total bias voltage will result mostly increase in voltage across the oxide.

We can write the charge equation as:
(VB-(Subscript[ϕ, S])) (Subscript[C, oxide])=  (Subscript[Q, D ])+ (Subscript[Q, n]) 
where (Subscript[Q, D]) represents the non-mobile charge of the depletion region, and (Subscript[Q, n]) is the electron charge of the MOS channel.
How to express (Subscript[Q, D ])and (Subscript[Q, n])?
(Subscript[Q, D]) is the depletion charge: p0 q W = (Sqrt[2 p0 q ϵ  (Subscript[ϕ, s ])])
By definition: (Subscript[Q, n]) = (Subscript[Q, S ])- (Subscript[Q, D]).

At the onset of inversion, (Subscript[Q, S]) --> q (Subscript[N, A]) W, and (Subscript[ϕ, S]) -> 2 ϕF.

We can thus define a bias votage for the onset of inversion, called threshold voltage:
(Subscript[V, T])  =  ((q (Subscript[N, A]) W)/(Subscript[C, oxide])) + 2 ϕF

To see this, let's look at the condition right around (Subscript[V, T]).

4.3.2 Some background algebra (no need to open unless wanting to know details)

Remember the equation for charge balance is:
(VB-(Subscript[ϕ, S])) (Subscript[C, oxide])=  q p0 LD Sqrt[ 2 ((e^(-((q (Subscript[ϕ, S]))/(k T))))+((q (Subscript[ϕ, S]))/(k T)) -1)+2 (e^(-2 q ϕF/(k T)))((e^((q (Subscript[ϕ, S]))/(k T))) -((q (Subscript[ϕ, S]))/(k T))-1)]
For large field condition: (e^(-((q (Subscript[ϕ, S]))/(k T)))) ~ 0; ((q (Subscript[ϕ, S]))/(k T))>> 1; and (e^((q (Subscript[ϕ, S]))/(k T))) >>((q (Subscript[ϕ, S]))/(k T))+1

Co (VB-(Subscript[ϕ, S]))==LD p0 q (Sqrt[((2 q (Subscript[ϕ, S]))/(k T))+2 (E^(((q (Subscript[ϕ, S]))/(k T))-((2 q ϕF)/(k T))))]) /. ((q (Subscript[ϕ, S]))/(k T))-((2 q ϕF)/(k T))-> χ

Co (VB-(Subscript[ϕ, S]))==LD p0 q (Sqrt[((2 q (Subscript[ϕ, S]))/(k T))+2 (E^χ)])  /.  (Subscript[ϕ, S])->  2 ϕF+ k T χ/q

Simplify[%]

Co (VB-2 ϕF-((k T χ)/q))==(Sqrt[2]) LD p0 q (Sqrt[((2 q ϕF)/(k T))+(E^χ)+χ]) /. (( q ϕF)/(k T))->f

((Co k T)/(LD p0  q)) (((q VB)/(k T))-2 (( q ϕF)/(k T))- χ)==(Sqrt[2])  (Sqrt[2 f+(E^χ)+χ]) /. (( q ϕF)/(k T))->f /. p0->k T ϵ /((LD^2) (q^2))

What is ((Co LD)/ϵ)? Remember that the capacitance of the depletion region is ~ ϵ/W
W->(((Sqrt[2]) LD (Sqrt[q]) (Sqrt[(Subscript[ϕ, S])]))/((Sqrt[k]) (Sqrt[T])))
((Co LD)/ϵ) is the ratio of the oxide capacitance vs. a unit capacitance of the semiconductor when ((2 q ϕS)/(k T))=1. We define a ratio

((Co LD (-2 f+((q VB)/(k T))-χ))/ϵ)==(Sqrt[2]) (Sqrt[2 f+(E^χ)+χ])  /. ((Co LD)/ϵ)-> η

η (-2 f+((q VB)/(k T))-χ)==(Sqrt[2]) (Sqrt[2 f+(E^χ)+χ])/. ((q VB)/(k T))->B

B == χ+2 f+((Sqrt[2])/η) (Sqrt[2 f+(E^χ)+χ]) /. χ->0

We call this the threshold terms, and express B as the difference between B and this condition:

β == χ+2 f+((Sqrt[2])/η) (Sqrt[2 f+(E^χ)+χ]) -2 f-(((Sqrt[2]) (Sqrt[2 f+1]))/η)

β==χ+((Sqrt[2])/η)( (Sqrt[2 f+(E^χ)+χ])- (Sqrt[2 f+1]))

β==χ+((Sqrt[2])/η)( (Sqrt[2 f+(E^χ)+χ])- (Sqrt[2 f+1]))   ((( (Sqrt[2 f+(E^χ)+χ])+ (Sqrt[2 f+1])))/(( (Sqrt[2 f+(E^χ)+χ])+ (Sqrt[2 f+1]))))

Expand[( (Sqrt[2 f+(E^χ)+χ])- (Sqrt[2 f+1]))   ( (Sqrt[2 f+(E^χ)+χ])+ (Sqrt[2 f+1]))]

χ+((Sqrt[2])/η)(((χ+(E^χ)-1))/(( (Sqrt[2 f+(E^χ)+χ])+ (Sqrt[2 f+1]))))

η=.; f = . ;

Solve[t==χ+(((Sqrt[2]) (χ-1))/(η ((Sqrt[2 f+1])+(Sqrt[2 f+χ])))),{χ}]

Solve[t== χ+(((Sqrt[2]) (2 χ))/(η ((Sqrt[2 f])+(Sqrt[2 f+2 χ])))),{χ}]

{{χ->FullSimplify[((t (η^2)+2 (Sqrt[f]) η-2 (Sqrt[f (η^2)+t (η^2)+2 (Sqrt[f]) η+1])+2)/(η^2))]},{χ->((t (η^2)+2 (Sqrt[f]) η+2 (Sqrt[f (η^2)+t (η^2)+2 (Sqrt[f]) η+1])+2)/(η^2))}}

4.3.3 Plot of (Subscript[ϕ, S]) clamping and effect of oxide capacitance

We are ready to see the relationship between the oxide capacitance, bias voltage, and bias potential in semiconductor.
To simplify the discussion, we will define the bias voltage (Subscript[ϕ, S]) in semiconductor in terms of a unitless quantity:

χ=((q )/((Subscript[k, B]) T)) ((Subscript[ϕ, S])-2 (Subscript[ϕ, F]) )    or:   (Subscript[ϕ, S])=(((Subscript[k, B]) T)/q)χ+2 (Subscript[ϕ, F])   ;
It is relative to the Fermi level (Subscript[ϕ, F]). Remember: (Subscript[ϕ, F]) is simply the Fermi level relative to the intrinsic level and it is a function of doping. 
In turn, we also want to use a unitless quantity for (Subscript[ϕ, F]):
  f=(( q (Subscript[ϕ, F]))/((Subscript[k, B]) T))         or:       (Subscript[ϕ, F])=(( (Subscript[k, B]) T)/q) f      

For capacitance areal density (which is capacitance per unit area), we use the quantity:
η=(Subscript[C, oxide])(( (Subscript[L, Debye]))/ϵ)   or:    (Subscript[C, oxide])=η (ϵ/(Subscript[L, Debye]))  =η (Subscript[C, S0])  ; we simply define (ϵ/(Subscript[L, Debye]))≡ (Subscript[C, S0]). 
It is just an quantity that is a property of the semiconductor having the dimension of capacitance areal density.
Below is a code to calculate:

Chi[t_,f_, η_]:=χ /. FindRoot[t ==  χ+(((Sqrt[2]) (χ+(E^χ)-1))/(η ((Sqrt[2 f+1])+(Sqrt[2 f+(E^χ)+χ])))) ,{χ,0.}]
(* t is bias voltage - unitless *)

For VB above threshold, (Subscript[ϕ, S]) -> 2 ϕF + small and very slow increase: this is called clamping. What happens to the rest of the voltage?

Manipulate[
If[start==1,Chi[t_,f_, η_]:=χ /. FindRoot[t ==  χ+(((Sqrt[2]) (χ+(E^χ)-1))/(η ((Sqrt[2 f+1])+(Sqrt[2 f+(E^χ)+χ])))) ,{χ,0.}] ;
 start=0 ,
] ;
Manipulate[{Plot[ {x,Chi[x,20.,Co],x- Chi[x,20.,Co]}
		                  
			,{x,-15,40}
	, PlotRange->{-20,40}
         ,PlotStyle->{{Hue[0.35,1,0.7],Thickness[0.01]},{RGBColor[1,0,0] , Thickness[0.01]} ,{RGBColor[0,0,1], Thickness[0.01]}}
,Frame-> True, GridLines-> Automatic
,FrameLabel->{"Bias Voltage - rel. to threshold (unitless)","VBias (green) - ϕ Sem (red) - V oxide (blue)"}
,LabelStyle->Directive[14,FontFamily-> "Arial"]]
,Plot[(Chi[x,f,Co]+2 f) /(x+ 2 f+(((Sqrt[2]) (Sqrt[2 f+1]))/Co)),{x,-15,40}
	, PlotRange->{0,1}
 ,PlotStyle->{{RGBColor[1,0,0] , Thickness[0.01]}}
,Frame-> True, GridLines-> Automatic
,FrameLabel->{"Bias Voltage (rel to threshold) (unitless)"," ϕ Sem /V bias"}
,LabelStyle->Directive[14,FontFamily-> "Arial"]]}
,{{f,20,"Fermi Level (a. u.)"},10,30}
,{{Co,0.1,"Oxide capacitance (a. u.)"},0.1,10}],
{{start,1,"Click to start"},{0,1}} ]

Critical for inversion is the voltage inside semiconductor.

4.3.4 Effects of oxide capacitance

One way to think about this is the ratio of ((Subscript[ϕ, S])/(Subscript[V, bias])).
Obviously, the higher oxide capacitance, the better.

Advanced oxide materials

4.4 Mobile charge

We are interested in the mobile charge, which is what makes the channel to work.

(Subscript[Q, n]) = (Subscript[Q, S ])- (Subscript[Q, D]) = (VB-(Subscript[ϕ, S])) (Subscript[C, oxide]) -(Sqrt[2 p0 q ϵ  (Subscript[ϕ, s ])])

Illustration of mobile charge density vs gate bias voltage

Manipulate[
If[start==1,Chi[t_,f_, η_]:=χ /. FindRoot[t ==  χ+(((Sqrt[2]) (χ+(E^χ)-1))/(η ((Sqrt[2 f+1])+(Sqrt[2 f+(E^χ)+χ])))) ,{χ,0.}] ;
Qn2 [  VB_,Ef_,η_,kT_]:=   Module[{ B,f,χ },
		  
		B=VB/kT ; f=Ef/kT ; 
		                                   χ=Chi[B,f,η];
	                                             ((E^χ)/((Sqrt[ (2 f+χ)+2 (E^χ)])+ (Sqrt[2 f+χ])))] ;
 start=0 ,
] ;
Manipulate[η=Oxidecapacitance ; T=Temperature ;
Plot[Qn2[VB,f,η,T], {VB,-10,40}, PlotRange-> {0,20}
,PlotStyle->{{RGBColor[1,0,0] , Thickness[0.01]}}
,Frame-> True, GridLines-> Automatic
,FrameLabel->{"Bias Voltage (rel to threshold) (unitless)","Mobile charge (rel unit)"}
,LabelStyle->Directive[14,FontFamily-> "Arial"]
]
,{{f,20,"Fermi Level (a. u.)"},10,30},{{Temperature,1},0.5,2},{{Oxidecapacitance,1},0.1,5}],
{{start,1,"Click to start"},{0,1}} ]

5. MOSFET simple I-V

5.1 Basic

If we know the mobile charge (Subscript[Q, n]), we are ready to find the drain current as a function of gate voltage and drain-source voltage.
What determines the drain current?
In earlier example, we write the voltage-current differential equation in terms of local resistivity:

ΔVx:=ID(ρ/(D  h[x]))Δx

In the example in earlier section, we use a model of MODFET, which allows us to visuallize the change of the channel as a function of gate voltage. It is intuitively easy to comprehend the pinch-off and the channel opeing and closing. BUT the essence of any FET is really in the LOCAL RESISTIVITY, whether the local resistivity is affected by the channel size or by the conductivity DOES NOT matter.

Thus, we can have another perspective of MOSFET by describing it in terms of conductivity:
(Subscript[I, D])=-(Subscript[μ, n;S]) (Subscript[Q, n])[x] D (Subscript[E, f])[x] 
where (Subscript[Q, n])[x] is the mobile charge density of the MOS channel as discussed above. (Subscript[μ, n;S]) is the channel carrier (n-type) mobility in the channel, which is at the interface beween the gate and the oxide. D is the effective channel depth, which, for simplicity, we approximate as a constant. (Discussion in class about modification to this approximation). We do not need to specify channel width and the current is implicitly the current density per unit of channel width.

(Subscript[I, D])=(Subscript[μ, n;S]) (Subscript[Q, n])[x] D ((dV)/(dx)) 

(Subscript[I, D]) Δx=(Subscript[μ, n;S]) (Subscript[Q, n])[x] D ΔV[x] 

To solve this equation, we need a relation between (Subscript[Q, n])[x] and ΔV[x].

Both (Subscript[ϕ, S]) and (Subscript[Q, D]) are functions of ((Subscript[V, G]) - V[x]).

(Subscript[Q, n]) == (Subscript[C, oxide]) ((Subscript[V, G])-V[x]-(Subscript[ϕ, S]))-(Subscript[Q, D])  /. (Subscript[Q, D]) -> (Sqrt[2 (Subscript[p, 0]) q ϵ  (Subscript[ϕ, s]) ])

5.2 Simple plot

The channel is functional above threshold voltage. At that point  (Subscript[ϕ, S]) ~ 2 (Subscript[ϕ, F]) . But we recall the definition: 
(Subscript[V, T])  =  ((q (Subscript[N, A]) W)/(Subscript[C, oxide])) + 2 (Subscript[ϕ, F]) for V threshold,
where ((q (Subscript[N, A]) W)/(Subscript[C, oxide])) ~ (Subscript[Q, D]), and 2  (Subscript[ϕ, F]) ~ (Subscript[ϕ, S]), the charge equation can be written: 
(Subscript[Q, n])->((Subscript[V, G])-V[x]) (Subscript[C, oxide])-(Subscript[C, oxide]) (Subscript[V, T]) 
Or:
(Subscript[Q, n])->((Subscript[V, G])-(Subscript[V, T])-V[x]) (Subscript[C, oxide])

Hence:
(Subscript[I, D]) Δx=(Subscript[μ, n;S]) (Subscript[C, oxide])  ((Subscript[V, G])-(Subscript[V, T])-V[x])  D ΔV[x] 
We can integrate both sides:
(∫_0^L)(Subscript[I, D]) dx=(Subscript[I, D]) L 
and:
(∫_0^(Subscript[V, D]))((Subscript[V, G])-(Subscript[V, T])-V)  dV=(Subscript[V, D]) ((Subscript[V, G])-(Subscript[V, T]) )-(1/2) ((Subscript[V, D])^2)

VD=.  ; VG=.  ; VT= . ; L=.  ;
Solve[ID L == D0  (VD (VG-VT)-((VD^2)/2)) (Subscript[C, oxide]) (Subscript[μ, S]),{ID}]

Or: ID->((D0 (Subscript[C, oxide]) (Subscript[μ, S]) )/(2 L))(2 VD VG-(VD^2)-2 VD VT)

and ((D (Subscript[C, oxide]) (Subscript[μ, S]))/L) is defined as the transconductance (Subscript[k, n]). Saturation of drain current occurs when d ID/d VD =0:

Solve[ (∂_VD) (2 VG VD - (VD^2)-2 VT VD)  == 0,{VD}]

Notice the similarity of VT and VPO in the generic FET.

Now, we can plot

ID[VD_,VG_,VT_]:=If[VG<VT,0,2 VG Min[VD,VG-VT] - ((Min[VD,VG-VT])^2)-2 VT Min[VD,VG-VT]]

Manipulate[ Plot[ID[5,x,VT],{x,0,6}, PlotRange->{0,15}],{VT,1,3}]

Plot[ Release[Table[ (2 iG Min[VD,iG-VT] - ((Min[VD,iG-VT])^2)-2 VT Min[VD,iG-VT]) /. VT->1 , {iG,1.,3.,0.5}]],
               
{VD,0,5}
, PlotStyle->{{RGBColor[1,0,0],Thickness[0.01]},{RGBColor[1,0,0],Thickness[0.01]},{RGBColor[1,0,0],Thickness[0.01]},{RGBColor[1,0,0],Thickness[0.01]},{RGBColor[1,0,0],Thickness[0.01]},{RGBColor[0,0,1],Thickness[0.01]}} ]

We see that this model gives the same I-V characteristics of typical FET or transistor.

Manipulate[ iG=gatevoltage ;VT=thresholdvoltage  ;
Show[Plot[2 iG Min[VD,Max[0,iG-VT] ] - ((Min[VD,Max[0,iG-VT] ])^2)-2 VT Min[VD,Max[0,iG-VT] ]  ,       
{VD,0,5}, PlotRange-> {0,4}
,PlotStyle->{{RGBColor[1,0,0] , Thickness[0.01]}}
,Frame-> True, GridLines-> Automatic
,FrameLabel->{"Drain source voltage (a. u.)","Drain current (rel. unit)"}
,LabelStyle->Directive[14,FontFamily-> "Arial"] ]
,Graphics[{Hue[0.05,1,1,1],Rectangle[{4.5,0},{5,thresholdvoltage}]
,Hue[0.35,1,1,0.8],Rectangle[{4.75,0},{5,gatevoltage}] }]
]
,{gatevoltage,0.,3.}, {thresholdvoltage,1, 2}]