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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

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