EE 432/532 ion implantation – 1

0.01

0.1

1

10 100 1000

boronphosphorusantimonyarsenic

rang

e (µ

m)

implant energy (keV)

B

P

As

Sb

0.001

0.01

0.1

10 100 1000

boronphosphorusantimonyarsenic

stra

ggle

(µm

)

implant energy (keV)

B

P

As

Sb

Range and straggle for implants into silicon

EE 432/532 ion implantation – 2

Ion implant – example 1A silicon wafer with n-type background doping is subjected to a boron implant. The implant energy is 80 keV and the dose is 1014 cm–2. The background doping of the wafer is 2x 1016 cm–3. Find the peak concentration and the junction depth of the implanted layer.

First, find the range and straggle for a boron implant at 80 keV. From the graph, RP ≈ 0.24 µm and ∆RP ≈ 0.063 µm. Then

13 =4�

���53=

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��.� � ����FP

� = �.� � ����FP��

[M = 53 ±���53

�ln

�131%

�� ��

= �.��µP ±�� (�.���µP)

�ln

��.� � ����FP��

� � ����FP��

�� ��

= –0.026 µm or 0.45 µm. In this case, only the positive value is real

EE 432/532 ion implantation – 3

You want to do an phosphorus implant into a p-type silicon wafer (background doping of 1016 cm–3). You choose to do the implant at 100 keV. What dose would be need to get a junction depth of 0.3 µm? What peak concentration will result?

Again, start by finding the range and straggle for a phosphorus implant at 100 keV. From the graph, RP ≈ 0.12 µm and ∆RP ≈ 0.045 µm. Then

Example 2

1% = 13 exp

��

�[M � 53

��

��5�3

�13 = 1% exp

��[M � 53

��

��5�3

�

→

13 =�����FP��

�exp

�(�.�µP � �.��µP)�

� (�.���µP)�

�= � � ����FP��

=���

��.� � ����FP

���.� � ����FP��

�= �.� � ����FP��

4 =����5313

EE 432/532 ion implantation – 4

A silicon wafer with n-type background doping of 1016 cm–3 is subjected to a boron implant. The implant energy is 100 keV and the dose is 1016 cm–2. Then the wafer is annealed for 30 minutes at 1000°C. Find the peak concentration and junction depth(s) immediately after implantation and then after annealing.

First, find the range and straggle for a boron implant at 100 keV. From the graph, RP ≈ 0.3 µm and ∆RP ≈ 0.07 µm. Before annealing,

Example 3

13 =4����53

=����FP��

���

��.� � ����FP

� = �.� � ����FP��

[M = 53 ±���53

�ln

�131%

�� ��

= �.��µP ±�� (�.��µP)

�ln

��.� � ����FP��

����FP��

�� ��

= +0.1 µm or 0.5 µm.

EE 432/532 ion implantation – 5

1014

1015

1016

1017

1018

0 0.1 0.2 0.3 0.4 0.5 0.6

NC

N (c

m–3

)

x (µm)

NB

after implant

after anneal

For the heat treatment, Dt = 2.5x10–11 cm2

13 =4

���

�(�53)�� +'W

�

=����FP��

��

��(������FP)�

� +��.� � �����FP�

��

= 4.7x10–17 cm–3

[M = 53 ±

������

(�53)�

� +'W�

ln

�131%

�

= � � ����FP ±

�������

�

��

�� � ����FP

��

� +��.� � �����FP�

��

�� ln

�� � ����FP��

����FP��

�

= +0.03 µm or 0.57 µm.