diffusion example 1 - iowa state...

13
EE 432/532 diffusion examples – 1 A constant-source boron diffusion is performed into an n-type silicon wafer. The diffusion temperature is 1050°C and the diffusion time is 1 hour. Assume that the surface concentration is limited by the solid- solubility limit. If the n-type background doping of the silicon is 5x10 16 cm –3 , find the junction depth. Start by finding Dt for this diffusion: So Dt will be Diffusion example 1 ' = ' R exp - ( $ N % 7 = . FP V exp - .H9 . × - H9 . (.) = . × - FP V 'W = . × - FP V (V)= . × - FP

Upload: others

Post on 08-Apr-2020

5 views

Category:

Documents


0 download

TRANSCRIPT

Page 1: Diffusion example 1 - Iowa State Universitytuttle.merc.iastate.edu/ee432/topics/doping/diffusion... · 2019-07-08 · EE 432/532 diffusion examples – 9 Design a constant-dose diffusion

EE 432/532 diffusion examples – 1

A constant-source boron diffusion is performed into an n-type silicon wafer. The diffusion temperature is 1050°C and the diffusion time is 1 hour. Assume that the surface concentration is limited by the solid-solubility limit. If the n-type background doping of the silicon is 5x1016 cm–3, find the junction depth.

Start by finding Dt for this diffusion:

So Dt will be

Diffusion example 1

' = 'R exp

�� ($N%7

=

��.�FP

V

�exp

�� �.�H9��.��� � ���� H9

.

�(����.)

� = �.�� � ����� FP�

V

'W =

��.�� � ����� FP�

V

�(����V) = �.�� � �����FP�

Page 2: Diffusion example 1 - Iowa State Universitytuttle.merc.iastate.edu/ee432/topics/doping/diffusion... · 2019-07-08 · EE 432/532 diffusion examples – 9 Design a constant-dose diffusion

EE 432/532 diffusion examples – 2

Next, look up the surface concentration, which is the solid-solubility limit at 1050°C. Interpolating from the table,

Ns ≈ 2.4x1020 cm–3.

Finally, apply the equation that was derived for junction depth for a constant source diffusion:

So that

with 1– NB/No = 0.999792. Consultation with a table of error function values (next page) shows that

[M = ��'W

�HUI��

�� � 1%

1V

��

erf( 2.63 ) ≈ 0.9998

[M = ���.��[�����FP� (�.��) = �.� � ����FP = �.��µP

Page 3: Diffusion example 1 - Iowa State Universitytuttle.merc.iastate.edu/ee432/topics/doping/diffusion... · 2019-07-08 · EE 432/532 diffusion examples – 9 Design a constant-dose diffusion

EE 432/532 diffusion examples – 3

Page 4: Diffusion example 1 - Iowa State Universitytuttle.merc.iastate.edu/ee432/topics/doping/diffusion... · 2019-07-08 · EE 432/532 diffusion examples – 9 Design a constant-dose diffusion

EE 432/532 diffusion examples – 4

An arsenic, constant-dose diffusion is performed. The initial dose is 1014 cm–2. The diffusion temperature is 1100°C and the diffusion time is 2 hr. The starting wafer had a p-type background doping of 1017 cm–3. Find the concentration of the As at the surface and find the junction depth.

Again, we need to find Dt. From the table

Example 2

' = 'R exp

�� ($N%7

So Dt will be

=

��.��FP

V

�exp

�� �.��H9��.��� � ���� H9

.

�(����.)

� = �.�� � ����� FP�

V

'W =

��.�� � ����� FP�

V

�(����V) = �.�� � �����FP�

Page 5: Diffusion example 1 - Iowa State Universitytuttle.merc.iastate.edu/ee432/topics/doping/diffusion... · 2019-07-08 · EE 432/532 diffusion examples – 9 Design a constant-dose diffusion

EE 432/532 diffusion examples – 5

Using the previously derived junction depth formula for Gaussian diffusions:

Find the surface concentration

1 (�) =4��'W

[M =

�'W�ln

�1 (�)1%

��

=

�������.�� � �����FP�

� �ln

��.� � ����FP��

����FP��

��= �.� � ����FP = �.��µP

=����FP��

��

��.�� � �����FP�

� = �.� � ����FP��

Page 6: Diffusion example 1 - Iowa State Universitytuttle.merc.iastate.edu/ee432/topics/doping/diffusion... · 2019-07-08 · EE 432/532 diffusion examples – 9 Design a constant-dose diffusion

EE 432/532 diffusion examples – 6

In a two-step process, phosphorus was diffused into a p-type silicon wafer (NB = 1016 cm–3). In the deposition step, the temperature was 900°C and the diffusion time was 45 minutes. In the drive step, the temperature was 1100°C and the time was 60 minutes. Find the surface concentration and junction depth.

Example 3

Find (Dt)1 for the first step

' = 'R exp

�� ($N%7

=

��.�� FP

V

�exp

�� �.��H9��.���� ���� H9

.

�(����.)

� = �.��� ����� FP�V

('W)� =

��.�� � ����� FP�

V

�(����V) = �.�� � �����FP�

Page 7: Diffusion example 1 - Iowa State Universitytuttle.merc.iastate.edu/ee432/topics/doping/diffusion... · 2019-07-08 · EE 432/532 diffusion examples – 9 Design a constant-dose diffusion

EE 432/532 diffusion examples – 7

The dose introduced during the first step is

Ns is the solid-solubility of phosphorus in silicon at 900°C. From the table, Ns = 7x1020 cm–3, giving

Next, calculate (Dt)2,

'� =

��.��FP

V

�exp

�� �.��H9��.��� � ���� H9

.

�(����.)

� = �.�� � ����� FP�

V

4 =�1V��

�('W)�

4 =�

�� � ����FP��

��

��.�� � �����FP� = �.� � ����FP��

('W)� =

��.�� � ����� FP�

V

�(����V) = �.�� � �����FP�

Page 8: Diffusion example 1 - Iowa State Universitytuttle.merc.iastate.edu/ee432/topics/doping/diffusion... · 2019-07-08 · EE 432/532 diffusion examples – 9 Design a constant-dose diffusion

EE 432/532 diffusion examples – 8

Now, we can find the surface concentration

and the junction depth

1 (�) =4��'W

=�.� � ����FP��

��

��.�� � �����FP�

� = �.�� � ����FP��

[M =

�'W�ln

�1 (�)1%

��

=

�������.�� � �����FP�

� �ln

��.�� � ����FP��

����FP��

��= �.�� � ����FP = �.��µP

Page 9: Diffusion example 1 - Iowa State Universitytuttle.merc.iastate.edu/ee432/topics/doping/diffusion... · 2019-07-08 · EE 432/532 diffusion examples – 9 Design a constant-dose diffusion

EE 432/532 diffusion examples – 9

Design a constant-dose diffusion of antimony into p-type silicon that gives a surface concentration of 5x1018 cm–3 and a junction depth of 1 µm. The background p-type doping in the silicon is 5x1016 cm–3. In this case, the “design entails finding suitable values for Dt and Q for the diffusion.

Example 4

Start with the constant-dose equation

We can use the two requirement (junction depth and surface concentration) in the above equation to find Dt

1 ([, W) =4��'W

exp

�� [��'W

= 1 (�) exp

�� [��'W

'W =[�M

ln�1(�)1%

� =

�����FP

��

� ln�������FP��

������FP��

� = �.�� � �����FP�

Page 10: Diffusion example 1 - Iowa State Universitytuttle.merc.iastate.edu/ee432/topics/doping/diffusion... · 2019-07-08 · EE 432/532 diffusion examples – 9 Design a constant-dose diffusion

EE 432/532 diffusion examples – 10

We finish by finding the required Q,

Note that we didn’t use the information about the type of dopant. That wouldn’t show up until we try to use pick specific temperature and time to give the specific Dt we calculated. At that point, we would have to use the diffusion parameters for antimony.

4 = 1 (�)��'W

=�� � ����FP��

� ��

��.�� � �����FP�

�= �.�� � ����FP��

Page 11: Diffusion example 1 - Iowa State Universitytuttle.merc.iastate.edu/ee432/topics/doping/diffusion... · 2019-07-08 · EE 432/532 diffusion examples – 9 Design a constant-dose diffusion

EE 432/532 diffusion examples – 11

Design a two-step diffusion of boron into n-type silicon that will result in a surface concentration of 5x1017 cm–3 and a junction depth of 2 µm. The background n-type doping in the silicon is 1x1015 cm–3. (In this case, design means to find suitable values for (Dt)1 and (Dt)2 for the diffusion.)

Example 5

This will be similar to example 4. Start by using the junction depth to find (Dt)2.

1 ([, W) = 1 (�) exp

�� [�� ('W)�

('W)� =[�M

ln�1(�)1%

=

�� � ����FP

��

� ln�������FP��

����FP��

� = �.�� � ����FP�

Page 12: Diffusion example 1 - Iowa State Universitytuttle.merc.iastate.edu/ee432/topics/doping/diffusion... · 2019-07-08 · EE 432/532 diffusion examples – 9 Design a constant-dose diffusion

EE 432/532 diffusion examples – 12

Then we can use the surface concentration requirement to find (Dt)1.

We need a value for Ns. Assume that the deposition will be done at 900°C. From the solid-solubility data, we see that for boron at 850°C, Ns = 9.5x1019 cm–3. Then

1 (�) =�1V�

�('W)�('W)�

('W)� =

��1 (�)�1V

��('W)�

=

��

�� � ����FP��

���.� � ����FP��

���.�� � ����FP�

�= �.� � �����FP�

Page 13: Diffusion example 1 - Iowa State Universitytuttle.merc.iastate.edu/ee432/topics/doping/diffusion... · 2019-07-08 · EE 432/532 diffusion examples – 9 Design a constant-dose diffusion

EE 432/532 diffusion examples – 13

Since we assumed a temperature of 900°C for the first step, we can calculate the corresponding time.

'� =

��.�FP

V

�exp

�� �.�H9��.��� � ���� H9

.

�(����.)

� = �.�� � ����� FP�

V

W� =('W)�'�

=�.� � �����FP�

�.�� � ����� FP�

V= ���VHFRQGV

To find (Dt)2, we would first need to choose either a time or a temperature for the second step and then calculate the other quantity.