III. Gm-C Filtering - Epublications - Université de Limoges
III. Gm-C Filtering - Epublications - Université de Limoges
III. Gm-C Filtering - Epublications - Université de Limoges
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g<br />
V<br />
m3<br />
in<br />
<strong>III</strong>.1.b Linearity Consi<strong>de</strong>rations<br />
The fully-differential equivalent topology of this circuit is consi<strong>de</strong>red. This<br />
configuration allows neglecting second or<strong>de</strong>r distortion as a first approximation. Furthermore,<br />
for each transconductor <strong>Gm</strong>,i, the output differential current Iout,i is given by:<br />
I = g V + g ′′ V . (<strong>III</strong>.7)<br />
out,<br />
i mi in mi<br />
3<br />
in<br />
The transfer function of the schematic can be computed, since the type of solution<br />
expected is:<br />
V = α V + βV<br />
, (<strong>III</strong>.8)<br />
out<br />
in<br />
These α and β coefficients can then be obtained. The differential current in the gyrator<br />
IinL gives the following relation:<br />
I<br />
+ g′<br />
′ V<br />
⎛<br />
⎜<br />
⎝<br />
r<br />
3<br />
in<br />
( ) ( ) 3<br />
3 ⎞ ⎛ r0<br />
3 ⎞<br />
g V + g′<br />
′ V ⎟ + g ′′ ⎜ g V + g ′′ V<br />
0<br />
inL = gm<br />
2⎜<br />
m1<br />
inL m1<br />
inL 2<br />
1<br />
1<br />
1 +<br />
⎟ m ⎜<br />
m inL m inL<br />
jr0C<br />
aω<br />
1 + jr0C<br />
aω<br />
3<br />
m3<br />
in<br />
Limiting to the third or<strong>de</strong>r terms, Equation (<strong>III</strong>.9) becomes:<br />
I<br />
inL<br />
gm1g<br />
m2r0<br />
= V<br />
1 + jr C ω<br />
0<br />
a<br />
inL<br />
⎛<br />
+ ⎜ g<br />
⎜<br />
⎝<br />
m2<br />
⎠<br />
⎝<br />
3<br />
g′<br />
′ m1r0<br />
⎛ gm1r0<br />
⎞ ⎞<br />
+ g ′′<br />
⎟<br />
m2<br />
V<br />
1 jr0C<br />
⎜<br />
a 1 jr0C<br />
⎟<br />
+ ω ⎝ + ⎟ aω<br />
⎠ ⎠<br />
- 85 -<br />
3<br />
inL<br />
⎟ ⎠<br />
(<strong>III</strong>.9)<br />
, (<strong>III</strong>.10)<br />
Keeping first and third or<strong>de</strong>r terms, the current of the gyrator can be written as:<br />
I<br />
Now, assuming<br />
this results in<br />
= I<br />
inL<br />
+ V<br />
out<br />
g<br />
g<br />
r<br />
g′<br />
′ r<br />
m1<br />
m2<br />
0<br />
m1<br />
0 3<br />
inL = VinL<br />
+ gm<br />
2 VinL<br />
, (<strong>III</strong>.11)<br />
1 + jr0C<br />
aω<br />
1 + jr0C<br />
aω<br />
V V = , (<strong>III</strong>.12)<br />
out<br />
inL<br />
3<br />
1+<br />
jr0C<br />
bω<br />
g 1g<br />
2r<br />
⎛<br />
0<br />
g 1r0<br />
g 1r<br />
⎞<br />
m m<br />
0<br />
3 1 jr0C<br />
V ⎜ ′′ m ⎛ m ⎞<br />
+ bω<br />
=<br />
out + g m2<br />
+ g ′′<br />
⎟<br />
m2<br />
Vout<br />
+ Vout<br />
r0<br />
1 jr0C<br />
aω<br />
⎜ 1 jr0C<br />
aω<br />
⎜<br />
1 jr0C<br />
aω<br />
⎟<br />
+<br />
+<br />
⎟<br />
r0<br />
⎝<br />
⎝ + ⎠ ⎠<br />
(<strong>III</strong>.13)<br />
Since the solution is of the type:<br />
V = α V + βV<br />
. (<strong>III</strong>.14)<br />
out<br />
in<br />
3<br />
in