Post on 21-Jun-2018
transcript
Bipolar Junction Transistors
HO #4: ELEN 251 - Bipolar Transistors Page 1S. Saha
• Bipolar junction transistors (BJT) are active 3-terminal devices with areas of applications:– amplifiers, switch etc.– high-power circuits– high-speed logic circuits for high-speed computers.
• BJT structure:– sandwich of alternating type of Si-layers
♦ npn BJT: sequence of n-p-n♦ pnp BJT: sequence of p-n-p
– npn BJTs are most widely used.
N+ P
E B C
N- N+
A. A Typical NPN-BJT Fabrication
HO #4: ELEN 251 - Bipolar Transistors Page 2S. Saha
Antimony Implant
SiO2
N+ Buried Collector
P−
1. Form buried collector (BC)− grow oxide− mask #1: BC− implant Sb− drive-in
N+
P−
N-Epi 2. Form N− epitaxial collector− strip oxide− clean− grow single crystal epi layer
NPN-BJT Fabrication
HO #4: ELEN 251 - Bipolar Transistors Page 3S. Saha
Boron
N+
P−
N-Epi P+P+3. Form junction isolation
− grow oxide− mask #2: ISO− implant B− oxidation
N+
P−
N-EpiP+ P+N+
Phosphorus
4. Form deep collector (sinker)− mask #3: Collector− implant P− drive-in P+ and N+− oxidation
NPN-BJT Fabrication
HO #4: ELEN 251 - Bipolar Transistors Page 4S. Saha
5. Form Base region− mask #4: Base− implant B− oxidationN+
P−
N-EpiP+ P+N+P
B
N+
P−
N-EpiP+ P+N+P N+
As/P
6. Form Emitter region− mask #5: Emitter− implant As/P− drive-in emitter− oxidation
NPN-BJT Fabrication
HO #4: ELEN 251 - Bipolar Transistors Page 5S. Saha
7. Define contact area− mask #6: Contact− etch oxide
N+P−
N-EpiP+ P+N+P N+
N+
P−
N-EpiP+ P+N+P N+
Contacts
8. Interconnect− deposit Aluminum− anneal− mask #7: Metal
9. Deposit protective dielectric (SiO2 or Si3N4). Mask #8 to open bonding pads.
NPN-BJT: Active Region Doping Profile
HO #4: ELEN 251 - Bipolar Transistors Page 6S. Saha
a) Final 2d-cross-section of an npn BJT structure.
1.E+15
1.E+16
1.E+17
1.E+18
1.E+19
1.E+20
1.E+21
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8
Depth (um)
Con
cent
ratio
n (c
m-3
) N+ Emitter
P Base
N- Collector
N+ Burried layer
b) 1d-cross-section along the intrinsic device.
N+
P−
N-EpiP+ P+N+P N+
Contacts
Intrinsicdevice
NPN
NPN-BJT: A Typical Layout
HO #4: ELEN 251 - Bipolar Transistors Page 7S. Saha
Atypical layout for the vertical npn-BJTs described in the fabrication example:
CONTACTBase Emitter Collector
P+ IsolationN
−E
pita
xial
laye
r N+ Buried layerP+ Base
N+ Emitter
B. Basic Features of IC BJT Structures
HO #4: ELEN 251 - Bipolar Transistors Page 8S. Saha
1.E+15
1.E+16
1.E+17
1.E+18
1.E+19
1.E+20
1.E+21
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8
Depth (um)C
once
ntra
tion
(cm
-3) N+ Emitter
P Base
N- Collector
N+ Burried layer
• The base region is non-uniformly doped. This results in a built-in ε-field across the base which aids e− transport from E → C.
– this built-in ε-field is set up to establish an equilibrium between:
♦ mobile carriers attempt to diffuse away from high concentration regions
♦ mobile carriers pulled by the ε-filed (drift) of fixed ionized donors (ND
+) or acceptors (NA−) left behind by mobile carriers.
∴ ε-field is obtained from: diffusion = drift
Basic Features of IC BJT Structures
HO #4: ELEN 251 - Bipolar Transistors Page 9S. Saha
• The parasitic elements exist in a BJT structure include:– base resistance, RB from base contact to active area– collector resistance, RC (predominantly through n-layer).
• Isolation must be provided between adjacent devices:– reverse biased PN junctions– trench isolation.
• The N- collector region adjacent to the base:– reduces base-collector capacitance, CBC
– improves base-collector break down voltage, BVCB
– decreases base width (WB) modulation by the collector voltage
– but adds series resistance to the device.
C. Basic BJT Operation
HO #4: ELEN 251 - Bipolar Transistors Page 10S. Saha
B
E C
≈ 0.7 Vforward bias
WB
e− Mostof the
e−
N+ P N- N+
− + − +
5 Vreverse bias
hole
s• In a typical npn-BJT operation:
– an external potential (≈ 0.7 V) is applied across the E-B junction to forward bias it
– e− are injected into the base by the emitter. (Also, holes are injected into the emitter but their numbers are much smaller because of the relative values of NA, ND).
Basic Operation
HO #4: ELEN 251 - Bipolar Transistors Page 11S. Saha
– if WB << Ln (diffusion length) in the base, most of the injected e− get into the collector without recombining. A few do recombine; the holes necessary for this are supplied as base current.
– the e− reaching the collector are collected across the C-B junction depletion region.
ICIEElectrons flowingemitter to collector
Recombining electronsHoles
into emitter
E B C
IB
ICIB
IE
Since most of the injected e− reach the collector and only a few holes are injected into the emitter, the required IB << IC.
Therefore, the device has a substantial current gain.
Basic Operation - Derivation of Currents
HO #4: ELEN 251 - Bipolar Transistors Page 12S. Saha
In order to derive the basic relationship for e− current flowing between E → C, we assume that the device current gain is high.
∴ IB ≅ 0
or, Jp ≡ hole current density in the base ≅ 0
or, Jp ≅ 0 = qµppεx − qDp(dp/dx) (1)
For uniformly doped base, dp/dx = 0,∴εx = 0 and the e−
travelling through the base will move by diffusion only.
However, in IC-BJTs dp/dx ≠ 0 and εx ≠ 0. The direction of this field aids e− flow from E → C and retards e− flow from C → E.
dxdp
pqkT
dxdp
pD
p
px
11==∴
µε (2)
Basic Operation - Derivation of Currents
HO #4: ELEN 251 - Bipolar Transistors Page 13S. Saha
The current due to e− flow between E and C is given by: Jn = qµnnεx + qDn(dn/dx) (3)
Substituting the expression for aiding ε-field, εx from (2) in (3) we get:
We integrate (5) over the quasi-neutral base region, WB.
⎥⎦⎤
⎢⎣⎡ +=∴
dxdnp
dxdpnp
DqJ n
n
dxdn
Dqdxdp
pnkTJ nnn += µ (4)
( )dxpnd
pDq
J nn = (5)Or,
Basic Operation - Derivation of Currents
HO #4: ELEN 251 - Bipolar Transistors Page 14S. Saha
BE C
WB
N+ P N-xdbcxdeb
x = 0 x = WB
dxW
dxpnd
DdxW
qp
JB
on
B
on ∫∫
⎟⎠⎞⎜
⎝⎛
=∴ (6)
In (6) Jn is pulled outside the ∫ assuming no recombination of e- in the base region, i.e. Jn = constant.
( ) ( )0pnWpnDdxW
qp
J Bno
n
B
−=∴ ∫ (7)
pn-Product at the Depletion Edge
HO #4: ELEN 251 - Bipolar Transistors Page 15S. Saha
In thermal equilibrium, we define: npo ≡ # e− in neutral P-region
≅ ni2/NA
pno ≡ # holes in neutral N-region ≅ ni
2/ND
nno ≡ # e− in N region ≅ ND
ppo ≡ # holes in P region ≅ NA
nn
qkT
nNN
qkT
pnp
n
i
AD
bio
olnln:junction- of potential-inBuilt 2 ==−∴ φ
N P+ ++++ +
+ ++++
−−−−−−
−−−−−
nno
pno
ppo
npo
(8)
eq
pp
eq
nn
kTbi
nop
kTbi
pono
o
φ
φ
=
=: thatshowcan weThen,
(9)
pn-Product at the Depletion Edge
HO #4: ELEN 251 - Bipolar Transistors Page 16S. Saha
Under the applied bias, VA we replace φbi by (φbi − VA) in (8) & (9) to obtain the non-equilibrium carrier density as:
We now assume low-level injection i.e. the injected carrier densities are small compared to the background concentrations so that nn = nno and pp = ppo. Then from (10) and (8),
N P+ ++++ +
+ ++++
−−−−−−
−−−−−
nn
pn
pno
pp
np
npoe
Vpp
eV
nn
kTAbi
q
np
kTAbi
q
pn
⎟⎠⎞⎜
⎝⎛ −
⎟⎠⎞⎜
⎝⎛ −
=
=
φ
φ (10)
(11)
⎟⎟⎠
⎞⎜⎜⎝
⎛==⎟
⎟
⎠
⎞
⎜⎜
⎝
⎛=
⎟⎠⎞⎜
⎝⎛ −
eq
nneVq
nn kTbi
pnkTAbi
pn oo
φφ(12)
pn-Product at the Depletion Edge
HO #4: ELEN 251 - Bipolar Transistors Page 17S. Saha
Therefore, under low-level injection we get from (12):
Similarly from (9) and (11) we can show under low level injection:
Therefore, for low level injection we get from (13) and (14):
(13) & (14) define the minority carrier densities at the edge of the space charge region under applied bias while (15) defines the pn-product of carriers at the depletion edge.
nininpipip nnnnppnpnnoooo
// and ;// ,know We 2222 ====
(14)eVq
pp kTA
nn o=
(13)eVq
nn kTppA
o=
eVq
npn kTiA2= (15)
Basic Operation - Derivation of Currents
HO #4: ELEN 251 - Bipolar Transistors Page 18S. Saha
Thus, the pn-products at the edge of the depletion regions within the base of an npn-BJT are given by:
Now, the total charge in the un-depleted base region is given by:
( )
( )
(7)) Eq (from
0
2
2
2
∫
⎥⎦⎤
⎢⎣⎡ −
=∴
=
=
W
Dpdx
eenqJ
enWpn
enpn
B
o n
kTV BEq
kTV CBq
i
n
kTV CBq
iB
kTV BEq
i (16)
(17)
(18)
∫=W
pdxqAQB
oB (19)
Basic Operation - Base Gummel Number
HO #4: ELEN 251 - Bipolar Transistors Page 19S. Saha
This is an extremely important result. Note that:1 Usually, only one of the two exponential terms is important
since one junction is typically reversed biased. When the deviceis in saturation, both junctions are forward biased and both terms must be included.
2 The quantity, is called the base Gummel number.
It is the total integrated base charge (atoms/cm2). Since I ∝ 1/QB, it is important to minimize QB. ∴ IC transistors use low base doping levels.
⎥⎦⎤
⎢⎣⎡ −=∴ eeII kT
V BEqkTV CBq
sn (20)
(21)Q
DnAqIB
nis
222
where, =
∫=W
dxxNqAQ B
oA
B )(
IC Vs. VBE
HO #4: ELEN 251 - Bipolar Transistors Page 20S. Saha
Since BC junction is reversed biased the eqVBC/kT term is negligible and we can show from (20) and (21):
(22) predicts that IC vs. VBE is related exponentially. Slope = (kT/q)ln(10)
= 60 mV/(decade I) (@ 25 oC) Relationship holds extremely well for IC
transistors over many decades of JC. Generally, QB is obtained by integration
over the base region. Therefore, QB is typically well controlled to ~ 1012 cm-2 to give high IC for a given VBE.
eQ
nDAqI kT
BEVq
B
inn
222
−=∴ (22)
1.E-10
1.E-09
1.E-08
1.E-07
1.E-06
1.E-05
1.E-04
1.E-03
1.E-02
1.E-01
0.0 0.2 0.4 0.6VBE (V)
J C (A
/cm
2 )
ExptIdeal
Decadechangein JC
60mVJS
D. Current gain
HO #4: ELEN 251 - Bipolar Transistors Page 21S. Saha
• Let us consider the factors that can contribute to base current in a BJT: 1 Recombination in the neutral base region2 Hole injection into the emitter3 Recombination in the E-B space charge region.
ICIEElectrons flowingemitter to collector
Recombining electronsHoles
into emitter
E B C
IB
1. Recombination in the Neutral Base Region
HO #4: ELEN 251 - Bipolar Transistors Page 22S. Saha
Typically, some of the e− traveling the base will recombine with majority carrier holes. (This is not important for modern IC BJTs).
If we assume that the base is uniformly doped so that εx = 0, the e− current and continuity equations are:
N+ Nnp
npo
P
QBpn
pno
pno
02
2
=−
−
=
τ n
ppn
pnn
n pn
xdnd
D
dxnd
DqAI
o
(23)
(24)
Recombination in the Neutral Base Region
HO #4: ELEN 251 - Bipolar Transistors Page 23S. Saha
Similar to PN-junction, the general solution of (24) is: np − npo = K1e−x/Ln + K2ex/Ln (25)
The appropriate boundary conditions (from Eq 13) are: np(x = 0) = npoe
qVBE/kT
np(x = WB) ≅ 0
Using these boundary conditions we get from (25):
Substituting (26) in (23) we get the emitter (InE) and collector (InC) e− currents.
⎟⎠
⎞⎜⎝
⎛
⎟⎠
⎞⎜⎝
⎛ −
=
LWL
xW
eV
nn
n
B
n
B
kT
q
ppBE
o
sinh
sinh(26)
Base Transport Factor
HO #4: ELEN 251 - Bipolar Transistors Page 24S. Saha
The ratio of these two currents is defined as the base transport factor and is given by: αT ≡ InC/ InE = sech(WB/Ln) (29)
In modern IC BJTs, WB << Ln and the recombination in the neutral base region is significantly low.
If WB << Ln, we can use series expansion of sech(WB/Ln) from (29) to get: αT ≅ 1 − WB
2/2Ln2
) (@ csc1
0) (@ coth1
Bn
BkT
q
n
n
C
n
BkT
q
n
n
E
Wx L
WheV
L
n pDqAIn
xL
WeV
L
n pDqAIn
BEo
BEo
=⎟⎟⎠
⎞⎜⎜⎝
⎛⎟⎠
⎞⎜⎝
⎛ −=
=⎟⎟⎠
⎞⎜⎜⎝
⎛⎟⎠
⎞⎜⎝
⎛ −= (27)
(28)
Base Transport Factor
HO #4: ELEN 251 - Bipolar Transistors Page 25S. Saha
In a typical advanced BJT, WB ≤ 1 µm and Ln ≥ 30 µm ∴αT ≅ 0.9994.
This value of αT would imply a forward current gain:
which is higher than the normally observed value of βF in IC BJTs with WB = 1 µm. Thus, αT is NOT usually a limiting factor in current gain.
The base current due to αT is: (InE - αTInE = (1 - αT)InE)
where τn is e− lifetime in the base.
⎟⎠⎞⎜
⎝⎛ −=∴ 1
2
2
eN
WnAqI kT
V BEq
nA
BiEBREC τ
(30)
16001
>−
=−
≅≡T
T
nCnE
nET
B
CF II
III
αααβ
2. Hole Injection into the Emitter
HO #4: ELEN 251 - Bipolar Transistors Page 26S. Saha
The dominant mechanism in limiting β in modern BJTs is hole injection from B to E. This process must occur because VBE reduces E → B e-flow barrier and also B → E hole flow the barrier.
The injected hole currents in each case come directly from the analysis of long base and short base diodes (E denotes emitter properties):
⎟⎠
⎞⎜⎝
⎛ −=<<
⎟⎠
⎞⎜⎝
⎛ −=>>
1:
1:
2
2
eV
xND
nqAILx
eV
LND
nqAILx
kT
q
EDE
pEipEpEE
kT
q
pEDE
pEipEpEE
BE
BE
(31)
(32)
E B CxE WB
nppn pn
xE >> LpE
xE
xE << LpE
E B C
nppn pn
Emitter Efficiency
HO #4: ELEN 251 - Bipolar Transistors Page 27S. Saha
The emitter injection efficiency is defined as:
Then from (27) and (32) we get:
(If WB >> Ln or XE >> Lp, then LB diode approximations replace WBand/or XE with Ln or Lp.)
(34) is only approximately correct in IC structures because NA and ND are not constant. Note that γ is made close to 1 by: (1) making NDE >> NAB; (2) WB small; (3) XE large (prevent hole recombination at E contact).
⎟⎟⎠
⎞⎜⎜⎝
⎛+==
+=
II
II
III
nE
pE
TOT
nE
pEnE
nE 1/1γ (33)
DNDN
xW
nBDE
pEAB
E
B+=
1
1γ (34)
3. E-B Space Charge Recombination
HO #4: ELEN 251 - Bipolar Transistors Page 28S. Saha
αT and γ are independent of VBE imply that the ratio of collector to base current is a constant, independent of VBE i.e. current level.
In practice, the ratio of the two currents is NOT independent of IC. At low levels the dominant reason is recombination in the E-B depletion region.
From PN junction theory, we find that some recombination of the carriers moving through the depletion region will occur, causing a recombination current.
E B Ce−
injectionholeinjection
*recombination
E-B Space Charge Recombination
HO #4: ELEN 251 - Bipolar Transistors Page 29S. Saha
exnqAI kTVq
o
EiREC
A
22τ
= (35)
where τo is the lifetime in the depletion region.
Note:– This current flows in the EB circuit and does not directly affect IC.
Thus, as IREC becomes important, the ratio of IC/IB will change.– eqVBE/2kT dependence is important at low current levels.
Summarizing our discussion of current gain:
II
III
II
II
nE
REC
nE
nCnE
nE
pE
C
B +−
+≅≡β1
eV
nDxWN
LW
DxNDWN
kT
q
oin
EBA
n
B
nED
PBA BE22
2
221 −++≅
τβ(36)
E. Heavy Doping Effect in BJTs
HO #4: ELEN 251 - Bipolar Transistors Page 30S. Saha
The modern BJTs are limited in β due to emitter efficiency. From simple analysis we found for SB-diodes Eq (34):
Eq (37) ⇒ γ↑ as NDE↑
∴ β[≅ γ/[1−γ) ]↑ as NDE↑ for the heavily doped emitters
DNDN
xW
nBDE
pEAB
E
B+=
1
1γ(37)
1019 1020 1021
β max
NDE (cm−3)
Simple theory (Eq 37)
Actual
Heavy Doping Effect: Bandgap Narrowing
HO #4: ELEN 251 - Bipolar Transistors Page 31S. Saha
β↓ in actual devices is due to heavy doping effect in the emitter.
The effective ionization energy for impurities decreases at high doping (> 1018 cm−3) level which can be described as bandgap narrowing (BGN) in the semiconductor. Assume,
∆Eg = reduction in energy gap, Eg due to heavy doping effect Ego = bandgap for the lightly doped value of semiconductors
Due to BGN, the intrinsic carrier concentration is given by: nie
2 = CT3e−(Ego
− ∆Eg)/kT = nio
2e∆Eg)/kT (38)
where, nio2 = CT3e−Ego/kT is the lightly doped value of ni.
BGN is directly responsible for γ↓ as well as β↓ in actual devices.
BGN due to Heavy Doping Effect
HO #4: ELEN 251 - Bipolar Transistors Page 32S. Saha
For an NPN transistor in the active mode of operation, the injected e− current into base is:
Here, QB = base Gummel number; WB = width of the neutral base region.
Similarly, the injected hole current into emitter is:
Here, QE = emitter Gummel number; LpE = hole diffusion length into E. If width of the neutral E, xE < LpE, we use xE in (40).
eW dxxN
nDqe
QnDq
J kTV BEq
AB
ibnBkT
V BEq
B
ibnBnE
B∫=≅
0
222
)((39)
eL dxxN
nDqeQ
nDqJ kT
V BEq
DE
ienBkT
V BEq
E
ienBpE
pE∫=≅
0
222
)((40)
BGN due to Heavy Doping Effect
HO #4: ELEN 251 - Bipolar Transistors Page 33S. Saha
(41)
∫∫+
=+
=∴
L dxxNnD
W dxxNnDJJ
pE
B
DEibnB
ABiepE
nE
pE
0
2
0
2
)(
)(1
1
1
1γ
Assuming no BGN in the lightly doped (< 1018 cm−3) base region, we get from (38): nie
2/nib2 = e∆Eg/kT
Therefore, If NAB and NDE are constants, we get from (41):
If xE < LpE, replace LpE by xE in (42).
(42)kTE
pE
B
DE
AB
nB
pE
nE
pEg
eLW
NN
DD
JJ ∆
+
=+
=
1
1
1
1γ
BGN due to Heavy Doping Effect
HO #4: ELEN 251 - Bipolar Transistors Page 34S. Saha
(43)kTE
pE
B
DE
AB
nB
pE
nE
pEg
eLW
NN
DD
JJ ∆
+
=+
=
1
1
1
1γ
• Eq (34) shows that γ↓ by the term e∆Eg/kT
• In order to estimate the reduction in emitter efficiency we need to know how ∆Eg varies with NDE.
• Typically ∆Eg ≈ 100 − 200 meV for NDE ≈ 1019 − 1020 cm−3.
• Note the validity of (43):– NDE and NAB are not constants but f(x)– in heavily doped regions, the hole lifetime and hence LpE
depends strongly on doping level.
Home Work 2: Due April 14, 2005
HO #4: ELEN 251 - Bipolar Transistors Page 35S. Saha
1) The cross-section of a double-poly bipolar transistor fabricated as part of a silicon IC is shown below. Design a plausible process flow to fabricate such a structure following the idea of junction isolated npn-bipolar process flow discussed in class. Your answers should be given in terms of a series of 2d-sketches of the structure after each major process steps.
Briefly explain your reasoning for each step and the order you choose to do things. Clearly, label all the sketches.
Trench-Isolated double-poly NPN
Home Work 2: Due April 14, 2005
HO #4: ELEN 251 - Bipolar Transistors Page 36S. Saha
2) An npn-transistor has a cross-sectional area of 10−5 cm2 and a quasi-neutral base region that is uniformly doped with NA = 4x1017 cm−3, Dn = 18 cm2/sec, and WB = 0.5 µm.
(a) If a close emitter contact is used so that the recombination in the neutral emitter can be neglected, find the emitter injection efficiency γ. Assume that the total emitter Gummel number is 8x109 atoms and Dp = 2 cm2/sec.
(b) If the minority-carrier lifetime in the base is 10−6 sec, estimate the base transport factor αT.
(c) What is β for this transistor?
3) Consider a uniformly doped p-base region of an npn-BJT device with base width = WB, emitter area = AE, base doping = NA under active normal bias.
Use electron drift-diffusion Eq. to show that the injected electron current InE from n-emitter at the edge of the emitter depletion region in the p-base region can be approximately given by:
⎟⎠
⎞⎜⎝
⎛ −= 12
eVq
WNnDqA
In kT
A B
inEE
BE
Home Work 2: Due April 14, 2005
HO #4: ELEN 251 - Bipolar Transistors Page 37S. Saha
4) A bipolar transistor is fabricated by growing a P-type epitaxial layer on an N-substrate. The P-type base is uniformly doped at 5x1017 cm−3. The epi-layer is 1.0 µm thick. The emitter is formed by ion implantation of As (Q = 5x1015 @ 50 KeV). The emitter depth of ≅ 0.28 µm.
(a) Calculate the base Gummel number.
(b) Estimate the transistor current gain β assuming it is limited by emitter efficiency. Neglect heavy doping effects and assume Dn = 2Dp.
(c) If a reduction in energy gap due to heavily doped emitter is ~ 200 mV, re-calculate β to include heavy doping effects. Assume µn = 1350 cm2/v.sec, τpE =0.2 nsec, and LpE = 0.2 µm. State any assumptions you make.
(d) Comment on your results in part (b) and (c).