11/18/2010
1
LECTURE 5: Bipolar Junction
Transistor (BJT)
EECS 170A
11/18/2010
1
Summary
•
In previous lectures we understood how the most basic electronic device (p
‐
n
junction diode) works.
•
We started basic assumptions to idealize the diode
•
The diode is operating under steady state condition: There are no transient effects, d/dt
0
•
The semiconductor is a non
‐
degenerately doped semiconductor, E
C
‐
E
F
>3kT, & E
V
‐
E
F
>3kT,
•
One dimensional diode structure, (variations are always along x direction)
•
Low level injection: Carrier density diffusing from across the junction is much lower than the majority carriers,
∆
n<<p in p side,
and
∆
p<<n in n side,
•
The generation process in the quasi neutral regions is negligibly smal , G
L
=0
•
Total current flowing through the device is constant at any point along the device, J
total
(x)=FIXED
We said that everything is triggered by diffusion of carriers from one side to
11/18/2010
2
V
A
+V
A
q(Vbi)
q(VbiVA)
q(VbiVA)
Holes with enough
energy to overcome built
in potential
electrons with enough
energy to overcome builtin
potential
•
other. Applying voltage regulates the amount of carriers diffusing across the
junction by lowering or
increasing the potential barrier
seen by carriers
Summary (2)
•
As a result we create a minority carrier gradient at quasineutral regions
11/18/2010
3
Under Forward Bias V
A
>0
•
Diffusion of these carrier
will generate diffusion current
V
A
<0
Summary (3)
•
Mathematically you can link the total current passing through the diode by using
continuity equations
11/18/2010
4
P
Quasi
‐
NeutralRegion
Space Charge
(Depletion)
Region
N
QuasiNeutral Region
V
A
X=xp
X=0
X=xn
2
2
If you are after calculating electron or hole current at any point you can us
1
1
()
@
(
:
1
)
e
p
n
pN
A
A
xx L
qV
kT
i
q
Np P
n
N
P
xx
x
Vk
T
i
x
dn
dp
JJ x J
n
d
J
xq
D
q
D
dx
n
qD
e
e
qD
e
e
x
x
x
d
N
N
p
A
P
N
AN
dx N
NL
J
2
2
2
If you are after calculating the total current you can evaluate electron and hole currents at
and :
1
1
1
[email protected]
nP
A
P
A
n
A
qV
kT
i
P
n
qV
kT
i
P
DP
qV
k
pn
T
D
i
n
n
d
qD
n
J
n
qD
e
e
ee
x
dx N
JJ
x
q
e
2
NiP
Dn
D
NL NL
J
0
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document11/18/2010
2
Summary (4)
•
The same calculation can be applied to any diode but we specifically introduced
narrow base diode where N region is narrow
•
Where the math is slightly different due to
different boundary conditions
•
We can apply the same continuity equation and we get solution as:
11/18/2010
5
N

∞
P
X=x
n
X=x
p
X=x
c
2
2
()
( )
cosh
1
A
Pn
N
p
cn
P
i
qV
kT
Ni
P
Total current J
J
x
J
x
xx
L
n
J
Dn
D
qq
e
A
N
D
I
A
J
A
a
r
e
a
•
If the diode is P
+
‐
N diode, i.e. N
A
>>N
D
'
sinh
.
This is the end of the preview.
Sign up
to
access the rest of the document.
 Fall '10
 OzdalBoyraz
 lb

Click to edit the document details