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Unformatted text preview: 6.012  Microelectronic Devices and Circuits Lecture 10  MOS Caps II; MOSFETs I  Outline
S
c • Review  MOS Capacitor
The "DeltaDepletion Approximation" – vGS +
(= vGB) G SiO2 n+ (nMOS example) Flatband voltage: VFB ≡ vGB such pSi that φ(0) = φpSi: VFB = φpSi – φm B
Threshold voltage: VT ≡ vGB such that φ(0) = – φpSi: VT = VFB – 2φpSi + [2εSi qNA2φpSi ]1/2/Cox* Inversion layer sheet charge density: qN* = – Cox*[vGC – VT] • Charge stores  qG(vGB) from below VFB to above VT
Gate Charge: qG(vGB) from below VFB to above VT Gate Capacitance: Cgb(VGB) Subthreshold charge: qN(vGB) below VT • 3Terminal MOS Capacitors  Bias between B and C
Impact is on VT(vBC): 2φpSi → (2φpSi  vBC) • MOS Field Effect Transistors  Basics of model
Gradual Channel Model: electrostatics problem normal to channel
drift problem in the plane of the channel
Clif Fonstad, 10/15/09 Lecture 10  Slide 1 S
C The nMOS capacitor – vGS +
(= vGB) G SiO2 n+ pSi Right: Basic device
with vBC = 0 B
Below: Onedimensional structure for depletion approximation analysis* vGB
+ –
SiO 2 G tox 0
Clif Fonstad, 10/15/09 pSi B
x * Note: We can't forget the n+ region is there; we
will need electrons, and they will come from there. Lecture 10  Slide 2 MOS Capacitors: Where do the electrons in the inversion
layer come from?
Diffusion from the ptype substrate?
If we relied on diffusion of minority carrier electrons from
the ptype substrate it would take a long time to build up
the inversion layer charge. The current density of electrons flowing to the interface is just the current across a
reverse biased junction (the psubstrate to the inversion
layer in this case):
D
2
2 Je = q ni e N A w p ,eff [Coul/cm  s] The time, τ, it takes this flux to build up an inversion charge is the the increase in the charge, Δqn, divided by Je: !
s o τ is Clif Fonstad, 10/15/09 ! #ox
"q =
" (vGB $ VT )
t ox
*
N # q* $ox N A w p ,eff
N
"=
=
# (vGB % VT )
2
Je
q n i De t ox
Lecture 10  Slide 3 Diffusion from the ptype substrate. cont.?
Using NA = 1018 cm3, tox = 3 nm, wp,eff = 10 µm, De = 40 cm2/V
and Δ(vGBVT) = 0.5 V in the preceding expression for τ we
find τ ≈ 50 hr!
Flow from the adjacent n+region?
As the surface potential is increased, the potential energy
barrier between the adjacent n+ region and the region
under the gate is reduced for electrons and they readily
flow (diffuse in weak inversion, and drift and diffuse in
strong inversion) into the channel; that's why the n+
region is put there:
G
S There are many electrons
here and they don't have
far to go once the barrier
is lowered.
Clif Fonstad, 10/15/09 – vGS +
(= vGB) SiO2 n+ pSi B Lecture 10  Slide 4 Electrostatic potential and net charge profiles  regions and boundaries
φ(x) φ(x) φ(x)
tox
vGB φp x φm
ρ(x) vGB tox
φm qNAxd tox  Cox*(vGB  VFB)
Cox*(vGB  VFB) Acccumulation
vGB < VFB x vGB
xd x φp qNAXDT +
Cox*(vGB  VT) ρ(x) t o x xd −qNA
vGB φ p
tox
φm XDT 2φp 
φp
ρ(x) t o x x XDT −qNA qD* = qNAxd x qN* qD* = qNAXDT
=  Cox*(vGB  VT) x Strong Inversion
VT < vGB Depletion ( Weak Inversion )
when φ(0) > 0
VFB < vGB < VT vGB
Threshold Voltage
VT = VFB+2φp+(2εSi2φpqNA)1/2/Cox* Flat Band Voltage
VFB = φp – φm φ(x) φ(x) vGB tox
φm
φp x φp
tox
φm qNAXDT ρ(x)
tox vGB x tox
−qNA Clif Fonstad, 10/15/09 XDT 2φp  x φp
ρ(x)
XDT
qD* = qNAXDT x
Lecture 10  Slide 5 MOS Capacitors: the gate charge as vGB is varied qG [coul/cm2]
*
"
2Cox2 (vGB $ VFB ) (
#SiqN A %
"
' 1+
qG =
$ 1*
"
'
*
Cox &
#SiqN A
) "
"
qG = Cox (vGB # VT ) Inversion
Layer
Charge + qN AP X DT qNAPXDT ! Depletion
Region
Charge VFB q =C
"
G "
ox (vGB # VFB ) The charge expressions:
"
, Cox (vGB # VFB )
.
"
. %SiqN A &
2Cox2 (vGB # VFB ) )
"
( 1+
qG (vGB ) = # 1+
"
(
+
%SiqN A
. Cox '
*
. C " (v # V ) + qN X
/ ox GB
T
A DT
Clif Fonstad, 10/15/09 ! vGB [V] VT Accumulation
Layer Charge "
Cox # for vGB $ VFB for $ox
t ox VFB $ vGB $ VT
!
VT $ vGB for Lecture 10  Slide 6 MOS Capacitors: the small signal linear gate capacitance, Cgb(VGB)
$
#qG
Cgb (VGB ) " A
#vGB Cgb(VGB) [coul/V] vGB = VBG Cox
Accumulation Depletion VFB
"
& A Cox
(
(
"
Cgb (VGB ) = ' A Cox
(
"
( A Cox
) ! VGB [V] VT for VGB # VFB for VFB # VGB # VT for "
2Cox2 (VGB $ VFB )
1+
%SiqN A This expression can also be written
)1
as:
# t ox x d (VGB ) &
Cgb (VGB ) = A% +
(
"ox
"Si '
$ Clif Fonstad, 10/15/09 Inversion ! VT # VGB G
εox Aεox/tox [= Cox] εSi AεSi/xd(VGB)
B Lecture 10  Slide 7 MOS Capacitors: How good is all this modeling?
How can we know?
Poisson's Equation in MOS
As we argued when starting, Jh and Je are zero in steady
state so the carrier populations are in equilibrium with
the potential barriers, φ(x), as they are in thermal
equilibrium, and we have: n ( x ) = n ie q" ( x ) kT and p( x ) = n ie# q" ( x ) kT Once again this means we can find φ(x), and then n(x) and
p(x), by solving Poisson's equation: ! d 2" ( x )
q
= # n i (e# q" ( x ) / kT # e q" ( x ) / kT ) + N d ( x ) # N a ( x )
dx 2
$ [ ! This version is only valid, however, when φ(x) ≤ φp.
When φ(x) > φp we have accumulation and inversion layers,
and we assume them to be infinitely thin sheets of charge,
i.e. we model them as delta functions.
Clif Fonstad, 10/15/09 Lecture 10  Slide 8 Poisson's Equation calculation of gate charge
Calculation compared with depletion approximation
model for tox = 3 nm and NA = 1018 cm3: tox,eff ≈ 3.2 nm tox,eff ≈ 3.3 nm Clif Fonstad, 10/15/09 We'll look in
this vicinity
next. Lecture 10  Slide 9 Plot courtesy of Prof. Antoniadis MOS Capacitors: Subthreshold charge
Assessing how much we are neglecting
Sheet density of electrons below threshold in weak inversion:
In the depletion approximation for the MOS we say that the
charge due to the electrons is negligible before we reach
threshold and the strong inversion layer builds up:
*
qN ( inversion ) (vGB ) = " Cox (vGB " VT ) But how good an approximation is this? To see, we calculate the electron charge below threshold (weak inversion): qN ( sub " threshold ) (vGB ) = " q ! 0 $ ne x i ( vGB ) i q# ( x ) / kT dx This integral is difficult to do because φ(x) is nonlinear "( x) = " p +
! qN A
2
x  xd )
(
2#Si but if we use a linear approximation for φ(x) near x = 0,
where the term in the integral is largest, we can get a very
good approximate analytical expression for the integral.
Clif Fonstad, 10/15/09 ! Lecture 10  Slide 10 Subthreshold electron charge, cont. We begin by saying 2qN A [" (0) % " p ]
d" ( x )
" ( x ) # " (0) + ax where a $
=%
where
dx x = 0
&Si With this linear approximation to φ(x) we can do the integral
and find ! ! qN ( sub " threshold ) (vGB ) # q kT n (0)
kT
="q
qa
q $Si
n ie q% ( 0 ) kT
2qN A [% (0) " % p ] To proceed it is easiest to evaluate this expression for various
values of φ(0) below threshold (when its value is φp), and to
also find the corresponding value of vGB, from vGB " VFB = # (0) " # p + t ox
2$SiqN A [# (0) " # p ]
$ox This has been done and is plotted along with the strong
inversion layer charge above threshold on the following foil.
Clif Fonstad, 10/15/09 ! Lecture 10  Slide 11 Subthreshold electron charge, cont. 6 mV Neglecting this charge results in a 6 mV error in the threshold
voltage value, a very minor impact. We will see its impact on
subthreshold MOSFET operation in Lecture 12.
Clif Fonstad, 10/15/09 Lecture 10  Slide 12 MOS Capacitors: A few more questions you might have about our model Why does the depletion stop growing above threshold?
A positive voltage on the gate must be terminated on negative
charge in the semiconductor. Initially the only negative charges
are the ionized acceptors, but above threshold the electrons in the
strong inversion layer are numerous enough to terminate all the
gate voltage in excess of VT. The electrostatic potential at 0+ does
not increase further and the depletion region stops expanding. How wide are the accumulation and strong inversion layers?
A parameter that puts a rough upper bound on this is the extrinsic
Debye length
2 LeD " kT #Si q N When N is 1019 cm3, LeD is 1.25 nm. The figure on Foil 11 seems to
say this is ~ 5x too large and that the number is nearer 0.3 nm.* Is n, p = nie±qV/kT valid in those layers? ! It holds in Si until φ ≈ 0.54 V, but when φ is larger than this Si
becomes "degenerate" and the carrier concentration is so large
that the simple models we use are no longer sufficient and the
dependence on φ is more complex. Thinking of degenerate Si as
a metal is far easier, and works extremely well for our purposes. Clif Fonstad, 10/15/09 * Note that when N = 1020 cm3, LeD ≈ 0.4 nm. Lecture 10  Slide 13 Bias between n+ region and substrate, cont. Reverse bias applied to substrate, I.e. vBC < 0 C
vBC < 0
–
vBC
+ – vGC + G SiO2 n+ pSi B
Soon we will see how this will let us electronically adjust MOSFET threshold voltages when it is convenient for us to do so. Clif Fonstad, 10/15/09 Lecture 10  Slide 14 φ(x) With voltage between substrate
and channel, vBC < 0 Threshold: vGC = VT(vBC) with vBC < 0 vGB =
VT(vBC) φp – vBC
tox φ p 2φp – vBC
XDT(vCB = 0) X (v < 0)
DT BC φm x φp
qNAXDT ρ(x) VT(vBC) = VFB + 2φp + [2εSi(2φpvBC)qNA]1/2/Cox*
{This is vGC at threshold} XDT(vBC < 0) = [2εSi(2φpvBC)/qNA]1/2
XDT(vBC < 0)
tox −qNA Clif Fonstad, 10/15/09 qN* = qNAxDT qN* = [2εSi(2φpvBC)qNA]1/2 x Lecture 10  Slide 15 Bias between n+ region and substrate, cont.  what electrons see The barrier confining the electrons to the source is lowered by the
voltage on the gate, until high level injection occurs at threshold.
Clif Fonstad, 10/15/09 Lecture 10  Slide 16 Bias between n+ region and substrate, cont.  what electrons see The barrier confining the electrons to the source is lowered by the
voltage on the gate, until high level injection occurs at threshold.
When the sourcesubstrate junction is reverse biased, the barrier is
higher, and the gate voltage needed to reach threshold is larger. Clif Fonstad, 10/15/09 Lecture 10  Slide 17 An nchannel MOSFET capacitor: reviewing the results of the
depletion approximation,
now allowing for vBC ≠ 0. C
vBC < 0 vGC – + G SiO2 n+ –
vBC
+ pSi B Flat  band voltage : VFB " vGB at which # (0) = # p $ Si
VFB = # p $ Si $ # m
Threshold voltage : VT " vGC at which # (0) = $ # p $ Si + v BC
VT (v BC ) = VFB
Accumulation [ Depletion VFB ! { 1
$ 2# p $ Si + * 2%Si qN A 2# p $ Si $ v BC
Cox
Inversion 0 Inversion layer sheet charge density : VT vCG *
q* = " Cox [vGC " VT (v BC )]
N Accumulation layer sheet charge density :
Clif Fonstad, 10/15/09
} 1/ 2 *
q* = " Cox [vGB " VFB )]
P
Lecture 10  Slide 18 An nchannel MOSFET S – vGS G
+ iG n+ vDS D
iD
n+ pSi
vBS +
B iB Gradual Channel Approximation: There are two parts to the problem:
the vertical electrostatics problem of relating the channel charge to the
voltages, and the horizontal drift problem in the channel of relating the
channel charge drift to the voltages. We will assume they can be worked
independently and in sequence.
Clif Fonstad, 10/15/09 Lecture 10  Slide 19 An nchannel MOSFET showing gradual channel axes S G
+ vGS –
n+ 0
0 iG D
iD
n+ x pSi
vBS vDS +
B iB L y Extent into plane = W Gradual Channel Approximation:
 We first solve a onedimensional electrostatics problem in the x direction
to find the channel charge, qN*(y)
 Then we solve a onedimensional drift problem in the y direction to find
the channel current, iD, as a function of vGS, vDS, and vBS.
Clif Fonstad, 10/15/09 Lecture 10  Slide 20 Gradual Channel Approximation iv Modeling
(nchannel MOS used as the example) We restrict voltages to the following ranges:
v BS " 0, v DS # 0
This means that the sourcesubstrate and drainsubstrate
junctions are always reverse biased and thus that:
! iB (vGS , v DS , v BS ) " 0 The gate oxide is insulating so we also have:
iG (vGS , v DS , v BS ) " 0 With the back current, iB, zero, and the gate current, iG, zero,
!
the only current that is not trivial to model is iD.
The drain current, iD, is also zero except when when vGS > VT.
!
The inplane problem: (for just a minute so we can see where we're going)
Looking at electron drift in the channel we write iD as
iD = " W sey ( y ) q* ( y ) = " W µe q* ( vGS , v BS , vCS ( y ))
n
n dv cs
dy (at moderate E  fields) This can be integrated from y = 0 to y = L, and vCS = 0 to vCS =
vDS, to get iD(vGS, vDS,vBS), but first we need qn*(y).
! Clif Fonstad, 10/15/09 (derivation continues on next foil) Lecture 10  Slide 21 Gradual Channel Approximation, cont.
We get qn*(y) from the normal problem, which we do next.
The normal problem:
The channel charge at y is qn*(y), which is:
*
*
q* ( y ) = " Cox {vGC ( y ) " VT [vCS ( y ), v BS ]} = " Cox {vGS " vCS ( y ) " VT [vCS ( y ), v BS ]}
n {
} [ with VT [vCS ( y ), v BS ] = VFB " 2# p " Si + 2$SiqN A 2# p " Si " v BS + vCS ( y ) ! 1/ 2 *
Cox We can substitute this expression into the iD equation we
just had and integrate it, but the resulting expression is
deemed too algebraically awkward:
iD (v DS , vGS , v BS ) = ( 1
W
v'
3
+
*$
µe Cox 2& vGS " 2# p " VFB " DS )v DS +
2*SiqN A  2# p + v DS " v BS
,
L
2(
2
3% ) 3/2 ( " 2# p " v BS ) 3/2 .4
05
/6 A simpler, more common approach is to simply ignore the
dependence of VT on vCS, and thus to say { [ VT [vCS ( y ), v BS ] " VT (v BS ) = VFB # 2$ p # Si + 2%SiqN A 2$ p # Si # v BS
} 1/ 2 *
Cox With this simplification we have:
*
q* ( y ) " # Cox [vGS # VT (v BS ) # vCS ( y )]
n ! lif Fonstad, 10/15/09
C (derivation continues on next foil) ! Lecture 10  Slide 22 Gradual Channel Approximation, cont.
The drain current expression (the inplane problem):
Putting our approximate expression for the channel charge
into the drain current expression we obtained from
considering the inplane problem, we find:
iD = " W µe q* (vCS )
n dvCS
dv
*
# W µe Cox {vGS " VT (v BS ) " vCS ( y )} CS
dy
dy This expression can be integrated with respect to dy for
y = 0 to y = L. On the lefthand the integral with respect
to y can be converted to one with respect to vCS, which
ranges from 0 at y = 0, to vDS at y = L: ! L "i D 0 dy = v DS "W µ e *
Cox {vGS # VT (v BS ) # vCS ( y )} dvCS 0 Doing the definite integrals on each side we obtain a
relatively simple expression for iD:
! Clif Fonstad, 10/15/09 ! v&
*#
iD L = W µe Cox $vGS " VT (v BS ) " DS ' v DS
%
2(
(derivation continues on next foil) Lecture 10  Slide 23 Gradual Channel Approximation, cont.
The drain current expression, cont:
Isolating the drain current, we have, finally:
iD (vGS , v DS , v BS ) = W
v&
*#
µe Cox %vGS " VT (v BS ) " DS ( v DS
$
L
2' Plotting this equation for increasing values of vGS we see
that it traces inverted parabolas as shown below.
inc.
!
iD
vGS vDS
Note,however, that iD saturates at its peak value for larger
values of vDS (solid lines); it doesn't fall off (dashed lines).
Clif Fonstad, 10/15/09 Lecture 10  Slide 24 Gradual Channel Approximation, cont.
The drain current expression, cont:
The point at which iD reaches its peak value and saturates is
easily found. Taking the derivative and setting it equal to
" iD
zero we find:
=0
when
v DS = [vGS # VT (v BS )]
" v DS
What happens physically at this voltage is that the channel (inversion) at the drain end of the channel disappears: ! *
q* ( L) " # Cox {vGS # VT (v BS ) # v DS }
n =0 when v DS = [vGS # VT (v BS )] For vDS > [vGSVT(vBS)], all the additional draintosource voltage
appears across the high resistance region at the drain end of
the!
channel where the mobile charge density is very small,
and iD remains constant independent of vDS:
iD (vGS , v DS , v BS ) =
Clif Fonstad, 10/15/09 ! 1W
2
*
µe Cox [vGS " VT (v BS )] for
2L v DS > [vGS " VT (v BS )]
Lecture 10  Slide 25 iD Gradual Channel Approximation, cont.
Derived neglecting the variation of the depletion layer charge with y. ! iG (vGS , v DS , v BS ) =
#
)
)
)
iD (vGS , v DS , v BS ) = $!
)
)W
)L
%
Linear or Triode Region Clif Fonstad, 10/15/09 0 iD + B vBS vGS
– S for [vGS " VT (v BS )] < 0 < v DS
0 < [vGS " VT (v BS )] < v DS for 0 < v DS < [vGS " VT (v BS )] for 1W
2
*
µe Cox [vGS " VT (v BS )]
2L
v&
*#
µe Cox $vGS " VT (v BS ) " DS ' v DS
%
2( iB G+ iB (vGS , v DS , v BS ) = 0 0 + vDS iG The full model: D Cutoff
Saturation
Linear or
Triode increasing vGS
Saturation or
Forward Active
Region Cutoff Region vDS Lecture 10  Slide 26 The operating regions of MOSFETs and BJTs: Comparing an nchannel MOSFET and an npn BJT
iD MOSFET D
+ Linear
iD or
Triode iG Saturation (FAR) vDS G+ iD ! K [vGS  V T(vBS)]2/2! vGS
– iC BJT C – S + Cutoff vDS iB
vCE B+ Saturation
iC i vBE
– B – E FAR iB ! IBSe qV BE /kT
vCE > 0.2 V Cutoff 0.6 V Input curve
Clif Fonstad, 10/15/09 vBE Forward Active Region
iC ! !F iB 0.2 V Cutoff vCE Output family
Lecture 10  Slide 27 6.012  Microelectronic Devices and Circuits Lecture 10  MOS Caps II; MOSFETs I  Summary • Quantitative modeling (Apply depl. approx. to MOS cap., vBC = 0)
Definitions: VFB ≡ vGB such that φ(0) = φpSi
VT ≡ vGB such that φ(0) = – φpSi Results and expressions Cox* ≡ εox/tox (For nMOS example) 1. Flatband voltage, VFB = φpSi – φm
2. Accumulation layer sheet charge density, qA* = – Cox*(vGB – VFB)
3. Maximum depletion region width, XDT = [2εSi2φpSi/qNA]1/2
4. Threshold voltage, VT = VFB – 2φpSi +
[2εSi qNA2φpSi]1/2/Cox*
5. Inversion layer sheet charge density, qN* = – Cox*(vGB – VT) • Subthreshold charge
Negligible?: Yes in general, but (1) useful in some cases, and (2) an
issue for modern ULSI logic and memory circuits • MOS with bias applied to the adjacent n+region
Maximum depletion region width: XDT = [2εSi(2φpSi – vBC)/qNA]1/2 Threshold voltage: VT = VFB – 2φpSi + [2εSi qNA(2φpSi – vBC)]1/2/Cox* • MOSFET iv (the Gradual Channel Approximation) (vGC at threshold) Vertical problem for channel charge; inplane drift to get current
Clif Fonstad, 10/15/09 Lecture 10  Slide 28 MIT OpenCourseWare
http://ocw.mit.edu 6.012 Microelectronic Devices and Circuits
Fall 2009 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms. ...
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