Radio Frequency Integrated Circuit Design - Webs

LNA Design
Extracting only the small-signal components from this equation gives
vs = v be + R E i c
173
(6.73)
Also, from the basic properties of the junction,
VBE + v be VBE v be v be
v
IC + ic = IS e T v = IS e T e vT v = IC e T (6.74)
where, from Chapter 3, v T = kT /q. Solving for v be gives
v be = vT ln�1 + ic IC�
Now making use of the math identity
(6.75)
ln (1 + x) = x − 1
2 x 2 + 1
2 x 3 . . . (6.76)
and expanding (6.75) using (6.76) and substituting it back into (6.73), we get
vs = R E i c + v T� ic
I C
− 1
2� ic
IC� 2
+ 1
3� ic
IC� 3
...�
Noting that vT /IC = re and rearranging, we get
vs = (R E + r e )i c − 1
2 re
i 2 c
IC
This can be further manipulated to give
vs
(R E + re ) = ic − 1
2
re
(R E + re )
i 2 c
IC
+ 1
+ 1
3
3 re
i 3 c
I 2
C
re
(R E + re )
(6.77)
... (6.78)
i 3 c
I 2
C
... (6.79)
This is the equation we need, but it is in the wrong form. It needs to be
solved for ic . Thus, a few more relationships are needed. Given
y = a1x + a 2x 2 + a 3x 3 + ... (6.80)