III. Gm-C Filtering - Epublications - Université de Limoges

this gives:
I
I
I
md
d1
d 2
= I
I
=
I
=
d1
0
0
− I
+ I
2
− I
2
d 2
Given that the differential input voltage is
Now, let’s consider
E md
1
md
md
2
- 213 -
(D.6)
(D.7)
(D.8)
= E − E , (D.9)
I 1 − I 2 = E md
1
β
2
1 2 2
I 1.
I 2 = ( I 0 − I md )
4
2
( I 1 I 2 ) = I 1 + I 2 − 2 I 1 I 2 = I 0 − 2 I 1 I 2
(D.10)
(D.11)
− (D.12)
Combining last three equations, one gets:
Then,
2
⎛ 1 ⎞
2 2
⎜ E ⎟
⎜ md β = I 0 − I 0 − I md
2 ⎟
(D.13)
⎝ ⎠
2
2
⎛
1
⎞
⎜ ⎛ ⎞
2 2
I 0 − ⎜ E ⎟ ⎟
md = I 0 − I md
⎜ ⎜
β (D.14)
2 ⎟ ⎟
⎝ ⎝ ⎠ ⎠
Developing this expression leads to:
1 4I
2
β Emd
(D.15)
2 β
0
I md = Emd
−