Download BJT breakdown

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# 10
BJT breakdown
• Possible mechanisms
Avalanche breakdown of the collector-base junction
Base pinch-through
SDM 2, ©Michael Shur 1999-2009
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# 10
Breakdown for Common Base
For a common-base configuration, the
avalanche breakdown voltage, BVcb, can
be found by equating the maximum electric
field at the collector-base interface to the
breakdown field, Fbr. Using the depletion
approximation, we obtain:
2
ε s Fbr
BVcb ≈
2q
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⎛ 1
⎞
F
ε
1
s br
+
⎜
⎟≈
⎝ N ab N dc ⎠ 2qN dc
2
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# 10
Breakdown Field
•
•
•
•
300 kV/cm for Si
400 kV/cm for GaAs
> 2,500 kV/cm or more for SiC
> 2,500 kV/cm or more for GaN
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# 10
More accurate approach
then using a breakdown field
Ic = Mcb (Icbo + αIe )
where
Mcb =
1
1 − (Vcb / BVcb )mb
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Common Emitter Configuration
Ic = Mcb (Icbo + αIe )
Substituting Ic = Ie, we find
Mcb
Ic = Icbo
1 − αMcb
Hence, the breakdown occurs when
αM cb = 1
BVce = BVcb (1 − α)
1/ mb
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# 10
Example
Find the ratio BVce/BVcb for a transistor with α = 0.99 and mb = 3.
Solution:
From
BVce = BVcb (1 − α)
1/ mb
BVce = 0.215 BVcb.
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Multiplication rate for CE
configuration
Since
We define Mce as
I c = M ce I ceo
Icbo
Iceo =
1− α
1− α
Mce = Mcb
1 − αMcb
BVce = BVcb (1 − α )1/ mb
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# 10
Breakdown for
Common Emitter Configuration
Example
Find the ratio BV ce /BV cb for a transistor with α - 0.99
and m b = 3.
Solution
BV ce - 0.215 BV cb.
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Multiplication coefficient
6
V bc +
Mce
5
+
V ce
–
4
3
2
–
Mcb
1
0
0
10
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Voltage (V)
40
50
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Leakage Currents
250
open
base
200
150
open
emitter
100
50
0
0
10
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30
Voltage (V)
40
50
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# 10
BJT
Operation
Look at
the depletion
regions
From
http://www.research.ibm.com/journal/rd/501/zutic7.gif
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# 10
Punch-through breakdown
Merging depletion regions
Emitter
Base
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Collector
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Punch-through equations
V pth ≈
2⎛
qN ab W
2ε s
2
2
⎞
qN abW
N ab
⎜1 +
⎟≈
Ndc ⎠ 2ε s N dc
⎝
BVcb ⎛ ε sFbr ⎞
≈⎜
⎟
Vp th ⎝ qnG ⎠
For Si
SDM 2, ©Michael Shur 1999-2009
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⎛
BVcb ⎜ 2×1012
≈
Vp th ⎜ nG cm −2
⎝
(
⎞
⎟
⎟
⎠
2
)
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Example
# 10
For silicon, the breakdown field, Fbr - 3x10 7 V/m, the dielectric permittivity,
εs - 1.05x10 -10 F/m. At what base thickness will a silicon BJT with the base
doping level Nab = 1017 cm-3 have equal breakdown voltages BV cb and Vpth ?
Solution:
The ratio of the avalanche breakdown voltage for the common-base
configuration, BV cb (see eq. (5-3-1)), and the punch-through voltage, Vpth , is
given by
BVcb ⎛ ε s Fbr ⎞
≈⎜
⎟
Vpth ⎝ qnG ⎠
2
where n G = NabW is called the Gummel number . For the silicon BJT
we find
2
BVcb
⎛ 2 x 1012 ⎞
≈⎜
⎟
Vpth
⎜ n (cm − 2) ⎟
⎝ G
⎠
level Nab = 1017 cm -3,
For the base doping
Vpth is equal to BV cb
for the base width W = 0.2 µm. For thinner bases, punch through breakdown occurs at voltages smaller than BV cb. For
thicker bases, avalanche breakdown occurs at voltages smaller
than Vpth .
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# 10
Example (temperature coefficient)
Assuming that the collector current is kept
constant,
estimate the temperature coefficient of the emitterbase voltage, dVbe/dT, for a typical silicon
transistor.
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Constants and parameters (SI units)
(*Constants*)
q = 1.602 10^-19;
kB = 1.38 10^-23;
me = 9.109 10^-31;
hbar=1.055 10^-34;
(*Silicon material parameters*)
mneff = 1.28; (*density of states mass at T = 300 K*)
mpeff = 0.81; (*density of states mass at T = 300 K*)
EG =1.12;
mob = 0.08;
(*Device parameters*)
W = 1. 10^-7;
Nab = 1. 10^23;
A = 1. 10^-6; (*area in m^2*)
Ic = 1.;
T = 300.;
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# 10
Solution
2
i
n
V be 1
jc = qDn
exp
N ab
Vth W
EG
n = N c N v exp−
Vth
2
i
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Solution (continued)
Nc = 2 (q mn Vth/(2 Pi hbar^2))^1.5;
Nv = 2 (q mp Vth/(2 Pi hbar^2))^1.5;
ni2 = Nc Nv Exp[-EG/Vth];
VBE= Vth Log[Ic Nab W/(q ni2 A Vth mob)];
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dVbe/dT (mV/degree K) versus Ic (A)
-1.60
-1.65
-1.70
-1.75
-1.80
-1.85
2
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6
8
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Safe Operating Area
Saturation
Maximum current
limitation
Safe
Operating
Area
Maximum power
limitation
Break down voltage
limitation
Log(V ce)
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SiC Diode Breakdown
# 10
From P. Rabkin, R. Cottle, P. A. Blakey, and M. S. Shur, 2D Simulation of DC, AC, and Breakdown Characteristics of Bipolar and
Unipolar Silicon Carbide Devices, in Proceedings of International Semiconductor Device Research Symposium,
Charlottesville, VA, Dec. 1-3, pp. 569-572 (1993)
SDM 2, ©Michael Shur 1999-2009
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Cross-section of SiC thyristor
Anode
Anode
Gate
p+ SiC
p+ SiC
n SiC
p- SiC
n+ SiC
n+ SiC substrate
Cathode
From M. S. Shur, SiC Transistors, in "SiC Materials and Devices", ed. Y. S. Park, Academic Press, Semiconductors and
Semimetals, Vol. 52, pp. 161-193 (1998)
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# 10
Forward conduction and blocking I-V characteristics
of a 30A, 1kV 4H-SiC epitaxial emitter BJT with
an active area of 3.4x3.4mm2
35
IB = 1 A
IC (A)
30
25°C
800 mA
25
600 mA
20
400 mA
15
10
BVCEO = 1000 V
@ 50 μA
200 mA
5
0 mA
0
0
2
4
6
8
10
12
VCE (Volts)
From S. Krishnaswami, A.K. Agarwal, S.-H. Ryu, J. Richmond, C. Capell, J.W. Palmour, S. Balachandran, T.P. Chow,
B. Geil, S. Bayne, C. Scozzie, and K.A. Jones, “1000V, 30A SiC Bipolar Junction Transistors and Integrated Darlington
Pairs,” Paper ThP2-28, European Conference on Silicon Carbide and Related Materials, Bologna, Italy,
August 31-September 4, 2004
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Collector-base
voltage
Summary
breakdown
Increase in collector current
for voltages higher than
BVcb
Multiplication factor due to
avalanche breakdown in the
collector-base junction
Collector current under
avalanche
multiplication
(open base)
Collector-emitter breakdown
voltage
Base punch-through voltage
BV
cb
≈
2
εs F
br ⎛
2q
1
⎜N
⎝
# 10
2
εs F
br
1 ⎞
≈
N ⎟⎠ 2qN
dc
dc
+
ab
Ic = Mcb (Icbo + αIe )
Mcb =
1
1 − (Vcb / BVcb )mb
1− α
Ic = Mce Iceo where Mce = Mcb
1 − αMcb
1/ m
b
BV ce = BV cb (1 − α )
2
2
2
qN ab W ⎛
N ab ⎞ qN abW
V pth ≈
⎜1 +
⎟≈
2ε s ⎝
Ndc ⎠ 2 ε s N dc
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