Transcript 投影片 1

EXAMPLE 9.1
OBJECTIVE
Calculate the minority-carrier concentration at the edge of the space charge region of a
pn junction when forward-bias voltage is applied.
Consider a silicon pn junction at T = 300 K. Assume the n-type doping is Nd = 1016
cm-3 and assume that a forward bias of 0.60 V is applied to the pn junction. Calculate
the minority-carrier hole concentration at the edge of the space charge region.
 Solution
The thermal-equilibrium minority-carrier hole concentration in the n region is
pn 0

ni2
1.5 1010


Nd
1016

2
 2.25104 cm-3
Form Equation (9.11), we have

or

 eVa 
 0.60 
4
pn xn   pn 0 exp
  2.2510 exp

 0.0259
 kT 
pn(xn) = 2.59  1014 cm3
 Comment
The minority-carrier concentration can increase by many orders of magnitude when a
relatively small forward-bias voltage is applied. Low injection still applies, however, since
the excess electron concentration (equal to the excess hole concentration to maintain charge
neutrality) is much less than the thermal-equilibrium electron concentration.
EXAMPLE 9.2
OBJECTIVE
Determine the ideal reverse-saturation current density in a silicon pn junction at T = 300
K. Consider the following parameters in a silicon pn junction:
Na = Nd = 1016 cm-3
ni = 1.5  1010 cm-3
Dn = 25 cm2/s
p0 = n0 = 5  10-7 s
Dp = 10 cm2/s
r = 11.7
 Solution
The ideal reverse-saturation current density is given by
Js 
eDn n p 0
Ln
which can be rewritten as
 1
J s  en 
 Na

2
i

Dn
 n0
eDp pn 0
Lp
1

Nd
Substituting the parameters, we obtain Js = 4.15  10
 Comment
D p 
 p 0 
-11
2
A/cm .
The ideal reverse-bias saturation current density is very small. If the pn junction crosssectional area were A = 10-4 cm2, for example, then the ideal reverse-bias diode current
would be Is = 4.15  10-15 A.
EXAMPLE 9.3
OBJECTIVE
Calculate the forward-bias pn junction current.
Consider the pn junction described in Example 9.2 with a junction area of A = 10-4
cm2. Calculate the current for forward-bias voltages of Va = 0.5, 0.6, and 0.7 V.
 Solution
For forward-bias voltages, we can write

 Va
I  JA  J S Aexp
V
 t

For Va = 0.5 V, we obtain











 Va



1

J
A
exp

S

V

 t





 0.5 
I  4.151011 104 exp
  1.0A
 0.0259
For Va = 0.6 V, we find
 0.6 
11
4
I  4.1510 10 exp
  47.7 A
0
.
0259


For Va = 0.7 V, we have
 0.7 
I  4.151011 104 exp
  2.27mA
 0.0259
 Comment
We see, from this example, that significant pn junction currents can be induced for relatively
small forward-bias voltages even though the reverse-saturation current is very small.
EXAMPLE 9.4
OBJECTIVE
Design a pn junction diode to produce particular electron and hole current densities at given forwardbias voltage.
Consider a silicon pn junction diode at T = 300 K. Design the diode such that Jn = 20 A/cm2 and Jp =
5 A/cm2 at Va = 0.65 V. Assume the remaining semiconductor parameters are as given in Example 9.2.
 Solution
The electron diffusion current density is given by Equation (9.24) as
eDn n p 0 

Dn ni2
 eVa 
Jn 
exp kT   1  e   N
Ln




n0
a
Substituting the numbers, we have

19
20  1.6 10


25
1.5 1010

5 107
Na

2


 eVa 
exp

1




 kT 




 0.65 
exp 0.0259  1




Which yields
Na = 1.01  1015 cm-3
The hole diffusion current density is given by Equation (9.22) as
Jp
eDp nn 0 
D p ni2 


 eVa 
 eVa 

exp

1

e

exp

1







Lp
 p 0 N d 
 kT 
 kT 



Substituting the numbers, we have

19
5  1.6 10
Which yields


10
1.5 1010

5 107
Nd
Nd = 2.55  1015 cm-3

2


 0.65 
exp 0.0259  1




 Comment
The relative magnitude of the electron and hole current densities through a diode can be varied by
changing the doping concentrations in the device.
EXAMPLE 9.5
OBJECTIVE
To calculate the electric field required to produce a given majority-carrier drift current.
Consider a silicon pn junction at T = 300 K with the parameters given in
Example 9.2 and with an applied forward-bias voltage Va = 0.65 V.
 Solution
The total forward-bias current density ids given by

 eV
J  J s exp
 kT

 
  1
 
We determined the reverse saturation current density in Example 9.2, so we can write


 0.65 
2
J  4.151011 exp
  1  3.29A/cm
 0.0259




The total current far from the junction in the n region will be majority-carrier electron drift
current, so we can write
J = Jn  enNd
The doping concentration is Nd = 1016 cm-3, and if we assume n = 1350 cm2/V-s, then the
electric field must be
Jn
3.29


 1.52V/cm
19
16
en N d 1.6 10 135010 
 Comment
We assumed, in the derivation of the current-voltage equation, that the electric field in the neutral p and n regions was zero. Although the electric field is not a zero, this example shows
that the magnitude is very small-thus, the approximation of zero electric field is very good.
EXAMPLE 9.6
OBJECTIVE
To determine the change in the forward-bias voltage of fa pn junction with a change in
temperature.
Consider a silicon pn junction initially forward biased at 0.60 V at T = 300 K. Assume the
temperature increases to T = 310 K. Calculate the change in forward-bias voltage required to
maintain a constant current through the junction. Neglect the temperature effects on the density of
states parameters.
 Solution
The forward-bias current can be written as follows:
  Eg 
 eV 
 exp a 
J  exp
 kT 
 kT 
If the temperature changes, we can take the ratio of the diode currents at the two temperatures.
This ratio is
exp  E / kT expeV / kT 

J2
g
2
a2
2

J1
exp Eg / kT1 expeVa1 / kT1 
If current is to be held constant, then J1 = J2 and we must have
Eg / eVa 2
Eg  eVa1

kT2
kT1
Let T1 = 300 K, T2 = 310 K, Eg = 1.12 eV, and Va1 = 0.60 V. Then, solving for Va2, we obtain Va2 =
0.5827 V.
 Comment
The change in the forward-bias voltage is – 17.3 mV for a 10C temperature change.
EXAMPLE 9.7
OBJECTIVE
Calculate the reverse-saturation current density in a silicon Schottky diode.
Assume the barrier height is Bn = 0.67 V and the temperature is T = 300 K.
 Solution
We have
J sT
or
  eBn 
  0.67 
2
 A T exp
  120300 exp

 0.0259
 kT 
*
2
JsT = 6.29  10-5 A/cm2
 Comment
In general, the reverse-saturation current density in a Schottky barrier diode is several
orders of magnitude larger than the reverse-saturation current density in a pn junction
diode. This result is actually an advantage in several applications of Schottky diodes.
EXAMPLE 9.8
OBJECTIVE
Calculate the forward-bias voltage required to generate a forward-bias current density
of 25 A/cm2 in a Schottky diode and a pn junction diode.
 Solution
For the Schottky diode, we have
  eVa  
J  J sT exp
  1
  kT  
Neglecting the (1) term, we can solve for the forward-bias voltage. We find
 kT   J
Va  
 ln
 e   J sT

 J
  Vt ln

 J sT

20 
  0.0259 ln
 0.334V
5 
 5 10 

For the pn junction diode, we have
 J
Va  Vt ln
 Js

20
  0.0259 ln 11   0.734V
 10 

 Comment
A comparison of the two forward-bias voltages shows that the Schottky barrier diode
has a turn-on voltage that, in this case, is approximately 0.4 V smaller than the turn-on
voltage os the pn junction diode.
EXAMPLE 9.9
OBJECTIVE
To calculate the small-signal admittance of a pn junction diode.
This example is intended to give an indication of the magnitude of the diffusion
capacitance as compared with the junction capacitance considered in Chapter 5. The
diffusion resistance will also be calculated. Assume that Na >> Nd so that pn0 >> np0. This
assumption implies that Ip0 >> In0. Let T = 300 K, p0 = 10-7 s, and Ip0 = IDQ = 1 mA.
 Solution
The diffusion capacitance, with these assumptions, is given by
 1
Cd  
 2Vt

1
I p 0 p 0  
103 107  1.93109 F
20.0259




The diffusion resistance is
Vt
0.0259V
rd 

 25.9
I DQ
1 mA
 Comment
The value of 1.93 nF for the diffusion capacitance of a forward-biased pn junction is 3 to 4
orders of magnitude larger than the junction or depletion capacitance of the reverse-biased
ph junction that we found in Chapter 5. Typically, we found junction capacitances on the
order of a few tenths of a pF. The forward-bias diffusion capacitance will also become
important in bipolar transistors covered in Chapter 10.
EXAMPLE 9.10
OBJECTIVE
Determine the relative magnitudes of the ideal reverse-saturation current density and the
generation current density in a silicon pn junction at T = 300 K.
Consider the silicon pn junction described in Example 9.2 and let 0 = p0 = n0 = 5  10-7 s.
 Solution
The ideal reverse saturation current density was calculated in Example 9.2 and was found to
be Js = 4.15  10-11 A/cm2. The generation current density is again given by Equation (9.61)
as
eniW
J gen 
and the depletion width is given by
2 0
1/ 2
 2 s  N a  N d 



Vbi  VR 
W 


 e  Na Nd 

If we assume, for example, that Vbi + VR = 5 V, then using the parameters given in
Example 9.2 we find that W = 1.14  10-4 cm, and then calculate the generation current
density to be
Jgen = 2.74  10-7 A/cm2
 Comment
Comparing the solutions for the two current densities, it is obvious that, for the silicon pn
junction diode at room temperature, the generation current density is approximately four
orders of magnitude larger than the ideal saturation current density. The generation current is
the dominant reverse-bias current in a silicon pn junction diode.
EXAMPLE 9.11
OBJECTIVE
Determine the recombination current density.
Consider a silicon pn junction with the same parameters as considered in Example
9.10. (a) Determine the recombination current density for Va = 0.3 V. (b) determine
the ratio of Jrec calculated in part (a) to the ideal diffusion current density at Va = 0.3
V.
 Solution
(a) We find that
and
N N
Vbi  Vt ln a 2 d
 ni
 2 s
W 
 e

 Na  Nd 

Vbi  Va 
 Na Nd 


or





 1016 1016 

  0.0259 ln 
  0.695V
10 2
 1.5 10


 211.7  8.8510

19
1
.
6

10

14
1/ 2
  10  10 0.695 0.30






10
10



16
16
W = 0.320m
16
16
1/ 2
EXAMPLE 9.11
 Solution
Then
J rec
 Va
eWni

exp
2 0
 2Vt



1.6 10 0.3210 1.5 10  exp 0.30 

 20.0259
25 10 


19
4
10
7
or
Jrec = 2.52  10-5 A/cm2
(b) From Example 9.2, we found that JS = 4.15  10-11 A/cm2. So
 Va 
 0.30 
11


J D  J S exp   4.1510 exp

 0.0259
 Vt 

or
Then

JD = 4.45  10-6 A/cm2
J rec 2.52105

 5.66
6
J D 4.4510
 Comment
For a low value of forward-bias voltage, the recombination current dominates the total
forward-bias current.
EXAMPLE 9.12
OBJECTIVE
Design an ideal one-sided n+p junction diode to meet a breakdown
vo0ltage specification.
Consider a silicon pn junction diode at T = 300 K. Assume that Nd =
3  1018 cm-3. Design the diode such that the breakdown voltage is VB =
100 V.
 Solution
From Figure 9.30, we find that the doping concentration in the lowdoped side of a one-sided abrupt junction should be approximately 4 
1015 cm-3 for a breakdown voltage of 100 V.
For a doping concentration of 4  1015 cm-3, the critical electric field,
from Figure 9.29, is approximately 3.7  105 V/cm. Then from Equation
(9.83), the breakdown voltage is 110 V, which correlates quite well with
the results from Figure 9.30.
 Comment
As Figure 9.30 shows, the breakdown voltage increases as the doping
concentration decreases in the low-doped region.