Radio Frequency Integrated Circuit Design - Webs

LNA Design
Z�
A v = (−g m R L )
R S + Z� =−20
10.9 − j73
50 + 10.9 − j73
= 13.26 − j8.08 = 15.52 ∠−31.34° (or 23.8 dB)
Input impedance is given by Z� , calculated as 10.9 − j73 (or 73.8 ∠
−81.5°). The output impedance for this simple example is infinity. However,
for a real circuit, the output impedance would be determined by the transistor,
integrated circuit, and package parasitic impedance.
We now turn to the case when the feedback is 500�. From (6.36), the
voltage gain is
A v = v o
v i
≈
−0.2 × 100
1 + 100
500
= 20
= 16.67 (or 24.4 dB)
1.2
(Note the exact expression from the same equation would have resulted
in 16.5, so the approximate expression is sufficient.) Thus, the gain has been
reduced from 26 dB to 24.4 dB.
From the source the gain is reduced much more, since the input impedance
is much reduced, so there is a lot of attenuation. The gain from the source,
instead of from vi , can be shown to be
where
A vsource =−5.487 + j1.290 = 5.637 ∠ 166.8° (or 15.02 dB)
Thus, gain is reduced from 23.8 to 15.02 dB by feedback.
Z in ≈ R f ||Z� || R f + R L
g m R L
Z out =
= 23.89 − j8.65
R f
1 + g m Z ip =
500 + 100
= 500||(10.9 − j73)||� 20
500
= 55.68 + j26.82
1 + 0.2 � (31.76 − j17.73)
Z ip = R S ||Rf ||Z� = 50||500||(10.9 − j73) = 31.76 − j17.73
We note that at low frequency without feedback, Z in would have been
500� due to r� , while with feedback it is 27�, so the impedance is much
steadier across frequency.
�
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