Transcript EX4

EXAMPLE 4.1
OBJECTIVE
Calculate the drift current density induced in a semiconductor for a given applied electric field.
Consider a silicon semiconductor at T = 300 K with an impurity doping concentration of Nd =
1016 cm-3 and Na = 0. Assume electron and hole mobilities given in Table 4.1. Calculate the drift
current density for an applied electric field  = 35 V/cm.
 Solution
Since Nd > Na , the semiconductor is n type and, at room temperature, we can assume complete
ionization so that
n  Nd = 1016 cm-3
Then, the hole concentration is


10 2
n
1.5 10
p

n
1016
2
i
 2.25 104 cm 3
Since n >> p, the drift current density is
Jdrf = e (nn + pp)   enn
so
Jdrf = (1.6  10-19)(1350)(1016)(35) = 75.6 A/cm2
 Comment
Significant drift current densities can be obtained in a semiconductor with relatively small
applied electric fields. This result then implies that currents in the mA range can be generated in
very small semiconductor devices. (in chip max 10 mA/um2)
EXAMPLE 4.2
OBJECTIVE
Find the electron and hole mobilities in silicon at various temperatures.
(a) Find the electron mobility for (i) Nd = 1017 cm-3, T = 150C and
for (ii) Nd = 1016 cm-3, T = 0C. (b) Find the hole mobility for (i) Na =
1016 cm-3, T = 50C and for (ii) Na = 1017 cm-3, T = 150C.
 Solution
Using Figure 4.2, we find
(a) (i) For Nd = 1017 cm-3, T = 150C  n  500 cm2/V-s
(a) (ii) For Nd = 1016 cm-3, T = 0C  n  1500 cm2/V-s
(b) (i) For Na = 1016 cm-3, T = 50C  p  380 cm2/V-s
(b) (ii) For Na = 1017 cm-3, T = 150C  p  200 cm2/V-s
 Comment
We can see that the mobility decreases as the temperature increases.
EXAMPLE 4.3
OBJECTIVE
Determine the required impurity doping concentration in silicon at T = 300 K to produce a semiconductor resistor with
specified current-voltage characteristics.
Consider a bar of silicon uniformly doped with acceptor impurities and having the geometry shown in Figure 4.5. For an
applied voltage of 5 V, a current of 2 mA is required. The current density is to be no larger than Jdrf = 100 A/cm2. Find the
required cross-sectional area, length, and doping concentration.
 Solution
The cross-sectional area can be found as
I  J drf A  A 
I
J drf
2  10 3

 2  10 5 cm 2
100
The resistance of the device is
R
V
5
3


2
.
5

10
  2.5kΩ
3
I
2  10
From Equation (4.22b), the resistance, for the bar of semiconductor, is given by
R
L
L
L


A
e p pA
e p N a A
From this relation, we see that there is no unique solution for Na and L. If we choose a very small value of L, then Na might
be unreasonably small. On the other hand, if we choose a very large value of L, then Na might be unreasonably large. Hence,
we will choose a value of the doping concentration and then determine the required length.
Let Na = 1016 cm-3. From Figure 4.3, we find p  400 cm2/V-s. The device length is then found to be
L = AR = epNaAR = (1.6  10-19)(400)(1016)(2  10-5)(2.5  103)
or
L = 3.2  10-2 cm
 Comment
We must make sur that we use the mobility that corresponds to the doping concentration in our analysis and design.
EXAMPLE 4.4
OBJECTIVE
Design a p-type semiconductor resistor with a specified resistance to handle a given current density.
A silicon semiconductor at T = 300 K and with the geometry shown in Figure 4.5 is initially doped with donors at a
concentration of Nd = 5  1015 cm-3. Acceptors are to added to form a compensated p-type material. The resistor is to have a
resistance of R = 10 k, handle a current density of Jdrf = 50 A/cm2 when 5 V is applied, and have an applied electric field no
larger than  = 100 V/cm.
 Solution
For 5 V applied to a 10-k resistor, the total current is
I
V
5

 0.5mA
R 10
For a current density limited to 50 A/cm2, the cross-sectional area must be
I
0.5 10 3
A 
 10 5 cm 2
J
50
From the specified voltage and electric field, the device length is found to be
L
V


5
 5  10  2 cm
100
The semiconductor conductivity, from Equation (4.22b) is
L
5 10 2
1


 


0
.
50
Ω

cm
RA
10 4 10 5


The conductivity of a compensated p-type semiconductor is

 = epp = ep(Na - Nd)
where the mobility p is a function of the total ionized impurity concentration Na + Nd.
Using trial and error, if Na = 1.25  1016 cm-3, then Na + Nd = 1.75  1016 cm-3, and the hole mobility, from Figure 4.3 is p
 410 cm2/V-s. The conductivity is then
= ep(Na- Nd) = (1.6  10-19)(410)(1.25  1016  5  1015) = 0.492 (-cm)-1
which is very close to the value we need.
 Comment
Since the mobility is related to the total ionized impurity concentration, the determination of the impurity concentration to achieve a particular
conductivity is not straightforward.
EXAMPLE 4.5
OBJECTIVE
Determine the carrier density gradient to produce a given diffusion current density.
The hole concentration in silicon at T = 300 K varies linearly from x = 0 to x = 0.01 cm. The
hole diffusion coefficient is Dp = 10 cm2/s, the hole diffusion current density is Jdif = 20 A/cm2,
and the hole concentration at x = 0 is p = 4  1017 cm-3. Determine the hole concentration at x =
0.01 cm.
 Solution
The diffusion current density is given by
J dif  eD p
or

dp
p
 p 0.01  p 0  
 eD p
 eD p 

dx
x
0.01  0


20   1.6 10
which yields
19
 p0.01  4 1017
10
0.01  0






p(0.01) = 2.75  1017 cm-3
 Comment
We can note that, since the hole diffusion current is positive, the hole gradient must be negative,
which implies that the value of hole concentration at x = 0.01 must be smaller than the
concentration at x = 0.
EXAMPLE 4.6
OBJECTIVE
Determine the induced electric field in a semiconductor in thermal equilibium given a linear
variation in doping concentration.
Assume that the donor concentration in an n-type semiconductor at T = 300 K is given by
Nd(x) = 1016  1019 x (cm-3)
where x is given in cm and ranges between 0  x  1 m.
 Solution
Taking the derivation of the donor concentration, we have
dN d  x 
19
-4
 10 cm
dx


The induced electric field is given by Equation (4.42), so we have

19
 kT   1  dN d x   0.0259  10
  



16
19


e
N
x
dx
10

10
x

 d


At x = 0, for example, we find


 = 25.9 V/cm
 Comment
We can recall from our previous discussion of drift current that fairly small electric fields can
produce significant drift current densities, so that an induced electric field from nonuniform
doping can significantly influence semiconductor device characteristics.
EXAMPLE 4.7
OBJECTIVE
Determine the diffusion coefficient given the carrier mobility.
Assume that the mobility of a particular carrier is  = 1200 cm2/V-s at
T = 300 K.
 Solution
Using the Einstein relation, we have that
 kT 
2
D
   0.0259 1200   31.1cm / s
 e 
 Comment
Although this example is simple and straightforward, it is important to
keep in mind the relative orders of magnitude of the mobility and
diffusion coefficient. The diffusion coefficient is approximately 40 times
smaller in magnitude than the mobility at room temperature.
EXAMPLE 4.8
OBJECTIVE
Determine the majority-carrier concentration and mobility, given Hall effect parameters.
Consider the geometry shown in Figure 4.20. Let L = 10-1 cm, W = 10-2 cm, and d = 10-3 cm.
Also assume that Ix = 1.0 mA, Vx = 12.5 V, B = 500 gauss = 5  10-2 tesla, and VH = 6.25 mV.
 Solution
A negative Hall voltage for this geometry implies that we have an n-type semiconductor. Using
Equation (4.70), we can calculate the electron concentration as



 103 5 102
21
3
n

5

10
m
1.6 1019 105  6.25 103




or
n = 5  1015 cm-3
The electron mobility is then determined from Equation (4.74) as
n 
or
10 10 
 0.10m
5 10 12.510 10 
3
1.6 10
19
21
3
4
5
2
/V - s
n = 1000 cm2/V-s
 Comment
It is important to note that the MKS units must be used consistently in the Hall effect equations to
yield correct results.