Radio Frequency Integrated Circuit Design - Webs

322 Radio Frequency Integrated Circuit Design
would be first order in this case) with a single-sided bandwidth of 200 MHz.
Then the image is 600 MHz from the center, or at three times the corner
frequency; therefore, the gain is 1/3 or −9.5 dB. Equivalently, the number of
decades (log 3 = 0.477) could also be calculated and then multiplied by
−20 dB/decade (−20 � 0.477 =−9.54).
For good image rejection, a higher Q is needed or a higher order filter
must be employed. Note that for higher Q, the bandwidth gets narrower,
requiring the filter to be tuned to compensate for mismatch and process variation.
This filter will provide virtually no channel selection either. In comparison
to a typical wireless standard, in which a channel width is usually no more than
a few hundred kilohertz, or at most a few megahertz, and the whole frequency
spectrum available may be no more than 200 MHz, the bandwidth of this filter
is very wide by comparison. To make matters even worse, if a filter of this type
were actually to have such a narrow bandwidth, its gain would be huge.
Example 9.2 Determining Required Bandwidth
Determine the bandwidth required in the last example to get 60 dB of attenuation
at the image frequency.
Solution
Since in this case the image is 600 MHz away from the passband, a first-order
low-pass equivalent corner frequency at about 600 kHz is needed to get 3
decades of attenuation for the image at 600-MHz offset. The resulting bandwidth
is 1.2 MHz at a center frequency of 1.9 GHz and this implies a Q of 1,583
or about 0.06% accuracy. Such accuracy is not possible without some tuning
scheme, since the actual accuracy due to process variation will be more like
20%.
9.3.2 A Simple Bandstop Filter
Placing a parallel LC resonator in the collector of an LNA makes a crude bandpass
filter, but bandstop filters can be made as well. For example, a cascode
LNA with a series resonator attached to it as shown in Figure 9.3 forms a
bandstop filter [1].
In this case, the current that is produced at the collector of Q 1 is split
between Q 2 and the series resonant circuit made up of L, C, and R. Thus, it
can be shown that the gain in this circuit is given by
L�
s
A v =−gmR 2 + R 1
s +
L LC
s 2 R +
+
1
g m
s +
L
1
LC�
(9.4)