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 cm -3

T

= 300 K. Assume the n-type doping is

N d

= 10 16 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

p n

0 

n i

2 

N d

Form Equation (9.11), we have  1 .

5  10 10 10 16  2  2 .

25  10 4 cm 3

p n

 

n



p n

0 exp

eV a

  2 .

25  10 4  exp 0.60

or

kT

0.0259

p n

(

x n

) = 2.59  10 14 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:

N a

=

N d

= 10 16 cm -3

n i

= 1.5  cm -3

D D p n

= 25 cm 2 /s = 10 cm 2 /s 

p

0 

r

= 

n

= 11.7

0 = 10 10 5  10 -7 s

Solution

The ideal reverse-saturation current density is given by

J

which can be rewritten as

J s s

 

eD en i

2  

n L

1

N n n a p

0 Substituting the parameters, we obtain

J s

 

eD p L p p D n

 1

N n

0 = 4.15 

d

10 -11 

n

0

D p p

0   A/cm 2 .



Comment

The ideal reverse-bias saturation current density is very small. If the pn junction cross sectional area were A = 10 -4 would be

I s

= 4.15  10 -15 A.

cm 2 , for example, then the ideal reverse-bias diode current

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

cm 2 . Calculate the current for forward-bias voltages of

V a



Solution

= 0.5, 0.6, and 0.7 V.

= 10 -4 For forward-bias voltages, we can write

A

   exp  

V a V t

For

V a I



JA

= 0.5 V, we obtain 

J S I

    4 .

15  10  11   exp  1    For

V a

= 0.6 V, we find

I

  4 .

15  10  11   exp  0 .

5 0 .

6

J S

0 .

0259 0 .

0259

A

exp  

V a V t

   1 .

0  A  47 .

7  A For

V a

= 0.7 V, we have

I

  4 .

15  10  11   exp 0 .

7  2 .

27 mA 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 forward bias voltage.

Consider a silicon pn junction diode at 5 A/cm 2 at

V a T

= 300 K. Design the diode such that

J n

= 20 A/cm 2 and

J p

= 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

J n



eD n n L n p

0   exp

eV a kT

 1   Substituting the numbers, we have Which yields 20   1 .

6  10

N a

 19  5  25 10 = 1.01   7 10 15    1 .

5 cm -3

e

 10 10

N a



D n n

0  2 

n i

2

N a

  exp The hole diffusion current density is given by Equation (9.22) as

J p



eD L p p n n

0   exp

eV a kT

 1   

e D



p p

0 

n i

2

N d

  exp

eV a kT

0 .

65 0 .

0259   exp

eV a kT

Substituting the numbers, we have Which yields 5   1 .

6  10  19

N d

 10 5  10 = 2.55   7  10 15  1 .

5  10 10 cm -3

N d



Comment

 2   exp 0 .

65 0 .

0259  1    1    1    1   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

V a



Solution

= 0.65 V.

The total forward-bias current density ids given by

J



J s

  exp

eV kT

 1   We determined the reverse saturation current density in Example 9.2, so we can write

J

  4 .

15  10  11    exp 0 .

65 0 .

0259   1   3 .

29 A/cm 2 The total current far from the junction in the n region will be majority-carrier electron drift current, so we can write

J

=

J n



e



n N d

 The doping concentration is

N d

= 10 16 cm -3 , and if we assume 

n

= 1350 cm 2 /V-s, then the electric field must be  

J e



n n N d





1 .

6  10  19 3 .

29



 1350 

 

 1 .

52 V/cm 

Comment

We assumed, in the derivation of the current-voltage equation, that the electric field in the n eutral 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:

J

 exp   

E kT g

  exp

eV a kT

If the temperature changes, we can take the ratio of the diode currents at the two temperatures. This ratio is

J J

2 1  exp    

E g g

/

kT

2

kT

1   exp  

eV a eV a

1 2 / /

kT

2

kT

1   If current is to be held constant, then Let

T

1 = 300 K,

T

2 = 310 K,

E g E g J

/ 1 =

J

2

eV a

and we must have 2 

E g



eV a

1

kT

2 = 1.12 eV, and

V a

1

kT

1 = 0.60 V. Then, solving for

V a

2 , we obtain

V a

2 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



A

*

T

2 exp    

e



kT Bn

   

  

2 exp     0 .

0 .

67 0259    or

J sT

= 6.29  10 -5 A/cm 2 

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/cm 2 in a Schottky diode and a pn junction diode.



Solution

For the Schottky diode, we have

J



J sT

  exp

eV kT a

 1   Neglecting the (  1) term, we can solve for the forward-bias voltage. We find

V a



kT e

ln  

J J sT

  

V t

ln  

J J sT

    0 .

0259  ln 20 5  10  5  0 .

334 V For the pn junction diode, we have

V a



V t

ln  

J J s

    0 .

0259  ln 20 10  11  0 .

734 V 

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

N a

assumption implies that

I p

0 >>

I n

0 . Let

T

= 300 K, 

p

0 = 10 -7 >>

N d

so that

p n

0 s, and

I p

0 =

I DQ

>>

n p

0 . This = 1 mA.



Solution

The diffusion capacitance, with these assumptions, is given by

C d

   2 1

V t

 



I p

0 

p

0



 1   0 .

0259    10  7   1 .

93  10  9 F The diffusion resistance is

r d



V t I DQ

 0 .

0259 V 1 mA  25 .

9  

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 

Solution

= 

p

0 = 

n

0 = 5  10 -7 s.

The ideal reverse saturation current density was calculated in Example 9.2 and was found to be

J s

= 4.15  10 -11 A/cm 2 . The generation current density is again given by Equation (9.61) as and the depletion width is given by

J

gen 

en i

2 

W

0     1 / 2

W

    2 

e s

 

N a



N d N a N d

  

V bi



V R

If we assume, for example, that

V bi

Example 9.2 we find that

W

= 1.14  density to be

J

gen +

V

10 -4

R

= 5 V, then using the parameters given in cm, and then calculate the generation current = 2.74  10 -7 A/cm 2 

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

V a

the ratio of

J

rec calculated in part (

a

= 0.3 V. (

b

) determine ) to the ideal diffusion current density at

V a

= 0.3 V.



Solution

(a) We find that and

V bi



V t

ln  

N a N d n i

2     0 .

0259  ln   

   

1 .

5  10 10



2     0 .

695 V

W

 



   1 / 2    2 

e s

    

N a N a



N d N d

 



V bi

2



11 .

7





8 .

85 1 .

6  10  14  10  19





V a

  10 16 10 16  10 16  



0 .

695  0 .

30



 1 /   2 or

W

= 0.320

 m

EXAMPLE 9.11



Solution

Then

J

rec  



eWn i

2  0 1 .

6 exp  10    19 2

V t



V a

  0 .

32 2



5  10   10  7



4



1 .

5  10 10 or

J

rec = 2.52  (b) From Example 9.2, we found that

J S

10 -5 = 4.15 



exp A/cm 2   2



0 .

30 0 .

0259 10 -11 A/cm 2 . So or

J D



J S

exp  

V a V t J D

  



4 .

15 = 4.45   10  11 10 -6



exp   A/cm 2 0 .

30 0 .

0259  



   Then

J J

rec

D

 2 .

52  10  5 4 .

45  10  6  5 .

66 

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.

3  Consider a silicon pn junction diode at 10 18

T

= 300 K. Assume that

N d

cm -3 . Design the diode such that the breakdown voltage is

V B

100 V.

= = 

Solution

From Figure 9.30, we find that the doping concentration in the low doped side of a one-sided abrupt junction should be approximately 4  10 15 cm -3 for a breakdown voltage of 100 V.

For a doping concentration of 4  10 15 from Figure 9.29, is approximately 3.7  cm 10 5 -3 , the critical electric field, 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.