Current-Voltage Characteristics

Transcription

ECE-305: Spring 2015
BJTs:
Current-Voltage Characteristics
Professor Mark Lundstrom
Electrical and Computer Engineering
Purdue University, West Lafayette, IN USA
[email protected]
Pierret, Semiconductor Device Fundamentals (SDF)
pp. 371-399
Lundstrom ECE 305 S15
4/22/15
bipolar transistors
C: collector
IC
IC
(forward) active region
EB: FB, BC: RB
B: base
VBE1 , I B1
IB
E: emitter
saturation region
EB: FB, BC: FB
IE
cut-off region
EB: RB, BC: RB
VCE
NPN BJT
2
BJT operation: active region
FB
n+
emitter
RB
p
base
n
collector
n+
x
To understand this device, we should first draw an
Energy Band Diagram.
3
3
equilibrium
E
qVbi
EC
EC
EF
EV
EV
x
emitter
base
collector
4
VBE = 0, VCE > 0
E
qVbi
EC
Fn
EC
EV
Fn
“off”
“cut-off”
EV
x
emitter
base
collector
5
VBE > 0, VCE > 0
E
EC
(
q Vbi − V A
)
EV
I Cn = α T I En = I C
I En
IC = I0e
Fn
qVBE k BT
I B = I C β dc << I C
I Ep
EC
Fn
“active”
I B = I Ep
emitter
base
EV
x
collector
6
NPN BJT operation (active region)
(WB << Ln )
FB
IE
I nE
n+
emitter
RB
p
base
I Cn ≈ I En
n
collector
IC
n+
I pE
IB
I B = I Ep
I C ≈ I En
I E = I En + I Ep
7
Question 1)
1) For an NPN bipolar transistor biased in the forward active
region, which of the following is true?
a) 
b) 
c) 
d) 
e) 
VBE = 0, VCE = 0.
VBE > 0, VCE > 0.
VBE > 0, VCE < 0.
VBE < 0, VCE > 0.
VBE < 0, VCE < 0.
8
Question 2)
2) For a PNP bipolar transistor biased in the forward active
region, which of the following is true?
a) 
b) 
c) 
d) 
e) 
VBE = 0, VCE = 0.
VBE > 0, VCE > 0.
VBE > 0, VCE < 0.
VBE < 0, VCE > 0.
VBE < 0, VCE < 0.
9
Question 3)
3) How are the PN junctions biased in the saturation
region of an NPN BJT?
a) Emitter-base: forward biased.
forward-biased.
b) Emitter-base: forward biased.
reverse-biased.
c) Emitter-base: reverse biased.
forward-biased.
d) Emitter-base: reverse biased.
reverse-biased.
e) Emitter-base: forward biased.
biased breakdown.
Base-collector:
Base-collector:
Base-collector:
Base-collector:
Base-collector:
10
Question 4)
4) How are the PN junctions biased in the saturation
region of an PNP BJT?
a) Emitter-base: forward biased.
forward-biased.
b) Emitter-base: forward biased.
reverse-biased.
c) Emitter-base: reverse biased.
forward-biased.
d) Emitter-base: reverse biased.
reverse-biased.
e) Emitter-base: forward biased.
biased breakdown.
Base-collector:
Base-collector:
Base-collector:
Base-collector:
Base-collector:
11
outline
1) 
2) 
3) 
4) 
5) 
6) 
Review
Review of PN junctions under bias
IV Characteristics (Active region)
IV characteristics (Saturation region)
CE vs. CB
Wrap-up
12
NP junction in FB
Jn
Jn = q
Dn ni2 qVA
(e
WP N A
Jp = q
D p ni2 qVA
(e
WN N D
kBT
− 1)
kBT
− 1)
q (Vbi − VA )
Fp
Fn
WP
WN
Jp
13
quasi-neutral regions
WN
WP
q (Vbi − VA )
Fn
Fp
14
diffusion in the quasi-neutral regions
Δn ( 0 ) =
Δp ( 0 ′ ) =
ni2 qVA
(e
ND
− 1)
kBT
− 1)
Δn ( x )
Δp ( x )
WN << Ln
x′
kBT
ni2 qVA
(e
NA
Δn ( 0 ) = 0
WP << Ln
x
0′
WN
WP
0
15
NP junction in FB (N-region)
Δn ( 0 ) =
Δn ( x )
ni2 qVA
(e
NA
kBT
− 1)
J n = qDn
J n = −q
WB << Ln
Δn ( 0 ) = 0
Jn = q
dΔn ( x )
dx x=0
Dn
Δn ( 0 )
WP
Dn ni2 qVA
e
WP N A
(
kBT
)
−1
x
0
WP
16
N+P junction in FB
N D >> N A
Jn
J D (VA )
J n (VA ) = q
Dn ni2 qVA
e
WP N A
J p (VA ) = q
D p ni2 qVA
e
WN N D
(
kBT
(
kBT
)
−1
)
−1
“electron injection efficiency”
Jp
γ ≡
J n (VA ) + J p (VA )
γ ≡
J n (VA )
1
D p WP N A
1+
Dn WN N D
≤1
17
outline
1) 
2) 
3) 
4) 
5) 
6) 
Review
CE vs. CB
Wrap-up
18
NPN BJT operation (general)
FB/RB
I En
IE
n+
emitter
FB/RB
IC
I Cn
p
base
n
collector
n+
I Cp
I Ep
IB
I E = I En + I Ep
I C = I Cn + I Cp
I B = I E − IC
19
19
NPN BJT operation (active region)
(WB << Ln )
FB
IE
I nE
n+
emitter
RB
p
base
I Cn
n
collector
IC
n+
I pE
IB
I C ≈ I En
I B = I Ep
I E = I En + I Ep
20
NPN BJT operation (active)
I En
IE
n+
emitter
p
base
IC
I Cn
n
collector
n+
I Cp ≈ 0
I Ep
IB
I En (VBE ) = qAE
Dn ni2 qVBE
(e
WB N AB
I Ep (VBE ) = qAE
D p ni2 qVBE
(e
WE N DE
kBT
− 1) ???
kBT
− 1)
I Cn (VBE ) = I En (VBE )
αT = 1
(no recombination in the base)
21
diffusion in the quasi-neutral regions
Δn ( 0 ) =
ni2 qVBE
(e
N AB
kBT
− 1)
Δn ( x )
Δp ( 0 ′ ) =
2
i
n
( eqVBE
N DE
− 1)
Δn (WB ) =
Δp ( x )
WE << Ln
x′
kBT
ni2 qVBC
(e
N AB
kBT
− 1)
WB << Ln
x
WE
Emitter
0′
WB
0
Base
22
diffusion in the quasi-neutral base
Δn ( 0 ) =
ni2 qVBE
(e
N AB
kBT
− 1)
Δn ( x )
Δn (WB ) =
ni2 qVBC
(e
N AB
WB << Ln
kBT
− 1)
x
I En (VBE ) = qAE
Dn ni2 qVBE
(e
WB N AB
I En (VBE ) = qAE
Dn ni2 qVBE
(e
WB N AB
VBC << 0
kBT
− 1) ???
kBT
− eqVBC
kBT
)
(active region)
Δn (WB ) = −
ni2
≈0
N AB
WB
0
Base
23
Δn ( 0 ) =
ni2 qVBE
(e
N AB
kBT
− 1)
Δn ( x )
0
Base
Dn ni2 qVBE
(e
WB N AB
kBT
− eqVBC
VBC << 0
I En (VBE ) = qAE
WB << Ln
Δn (WB ) ≈ 0
I En (VBE ) = qAE
Dn ni2 qVBE
e
WB N AB
kBT
x
WB
24
kBT
)
I En
IE
n+
emitter
p
base
n
collector
I Ep (VA ) = qAE
n+
I Cp ≈ 0
I Ep
I En (VBE ) = qAE
IC
I Cn
Dn ni2 qVBE
e
WB N AB
D p ni2 qVBE
(e
WE N DE
IB
kBT
kBT
I Cn (VBE ) = I En (VBE ) (α T = 1)
− 1)
I C (VBE ) = I 0 eqVBE
kBT
I C = I En (VBE )
I 0 = qAE
Dn ni2
WB N AB
25
NPN BJT in active region
IC
I C = I 0 eqVBE
kBT
VBE, I B
What base current
produced this collector
current?
VCE
26
26
NPN BJT (active region base current)
IE
I En
n+
emitter
p
base
IC
I Cn
n
collector
n+
I Cp ≈ 0
I Ep
IB
I Ep (VBE ) = qA
D p ni2 qVBE
(e
WE N DE
kBT
− 1)
I B (VBE ) = qAE
D p ni2 qVBE
e
WE N DE
kBT
27
IC
I C = I 0 eqVBE
I C (VBE ) = qAE
D p ni2 qVBE
e
WE N DE
I B (VBE ) = qAE
D p ni2 qVBE
e
WE N DE
kBT
kBT
VBE, I B
VCE
I C Dn N DE WE
=
= β dc
I B D p N AB WB
28
kBT
(forward) active region summary
IC
I C = I 0 eqVBE
VBE, I B
I 0 = qAE
kBT
Dn ni2
WB N AB
I B = I C β dc
VCE
β dc =
Dn N DE WE
D p N AB WB
29
outline
1) 
2) 
3) 
4) 
5) 
6) 
Review
CE vs. CB
Wrap-up
30
bipolar transistors
IC
C: collector
EB: FB, BC: RB
IC
B: base
VBE1 , I B1
IB
saturation region
EB: FB, BC: FB
IE
E: emitter
VCE
NPN BJT
31
Δn ( 0 ) =
ni2 qVBE
(e
N AB
kBT
− 1)
Δn ( x )
I En (VBE ) = qAE
Δn (WB ) =
ni2 qVBC
(e
N AB
kBT
Dn ni2 qVBE
(e
WB N AB
kBT
− eqVBC
− 1)
WB << Ln
y
WB
0
Base
32
kBT
)
NPN BJT operation (saturation)
FB
I En
IE
n+
emitter
FB
IC
I Cn
p
base
n
collector
n+
IB
I C = I 0 eqVBE
kBT
(1− e− qVCE
kBT
)
I 0 = qAE
Dn ni2
WB N AB
33
outline
1) 
2) 
3) 
4) 
5) 
6) 
Review
CE vs. CB
Wrap-up
34
NPN bipolar transistor
IC
saturation
active
I C = β dc I B
VCE
IB
VBE
inverted
active
IE
BE: FB VBE > 0
BC: RB VCB = VCE − VBE > 0
VCE (V )
cut-off
Pierret, Fig. 10.4
35
common base (active region)
IC
IE
I C = α dc I E
VCB
IB
VCE
VBE
VCB > 0
VEB < 0
I B = IC β
IE
36
IV characteristics
common base (active region)
saturation
IC
IE
active
I C = α dc I E
VEB
IB
VCB
BE: FB VEB < 0
BC: RB VCB > 0
VCB (V )
cut-off
Pierret, Fig. 10.4
37
outline
1) 
2) 
3) 
4) 
5) 
6) 
Review
CE vs. CB
Wrap-up
38
bipolar transistors
IC
C: collector
IC
I C = I 0 eqVBE
B: base
I B = I C β dc
kBT
VBE1 , I B1
IB
I C = I 0 eqVBE
IE
E: emitter
kBT
(1− e
− qVCE kBT
)
I B > I C β dc
D n2
I 0 = qA n i
WB N AB
NPN BJT
VCE
39
NPN BJT operation (general)
FB/RB
IE
I En
n+
emitter
FB/RB
p
base
IC
I Cn
n
collector
n+
I Cp
I Ep
IB
I E = I En + I Ep
I C = I Cn + I Cp
I B = I E − IC
40
FB/RB
IE
I En
n+
emitter
FB/RB
p
base
I Cn ≈ I En
n
collector
IC
n+
I Ep
IB
I B = I Ep
I C ≈ I En
41
NPN BJT operation (saturation)
FB/RB
IE
I En
n+
emitter
FB/RB
p
base
IC
I Cn
n
collector
n+
I Cp
I Ep
IB
I C = I En − I Cn
I B = I Ep + I Cp
42

Current-Voltage Characteristics

Transcription

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