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Radio Frequency Integrated Circuit Design - Webs

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334 <strong>Radio</strong> <strong>Frequency</strong> <strong>Integrated</strong> <strong>Circuit</strong> <strong>Design</strong><br />

some simple formulas for the image rejection they provide. For the circuit of<br />

Figure 9.9, the gain in the stop band G SB is given by<br />

R −<br />

G SB =−gm1RL�R −<br />

=−g m1RL�<br />

R −<br />

g m3<br />

� 2 C 1C2<br />

g m3<br />

� 2 C 1C2<br />

Isharp R −<br />

+ 1<br />

g m2�<br />

� 2 C 1C2vT<br />

Isharp<br />

� 2 C 1C2vT<br />

and the gain in the passband G PB is approximately<br />

G PB =−g m1RL<br />

+ 1<br />

g m2�<br />

(9.25)<br />

(9.26)<br />

It is assumed that the passband is sufficiently far away that the series<br />

resonator is essentially an open circuit there. Therefore, the image rejection (IR)<br />

provided by this filter is<br />

IR = 20 log | G |<br />

R −<br />

PB<br />

G SB | = 20 log<br />

I sharp<br />

� 2 C 1C 2v T<br />

Isharp<br />

R −<br />

� 2 C 1C2vT |<br />

+ 1<br />

g m2<br />

(9.27)<br />

If Isharp can be tuned to perfectly cancel the loss in the resonator, this<br />

circuit can have perfect image rejection. In practice, the accuracy to which Isharp<br />

can be set will limit the image rejection available from this circuit.<br />

Example 9.4 Effect of Process Tolerance on Image Rejection<br />

Assume that the current Isharp in the circuit of Figure 9.9 can be set to an<br />

accuracy of 1%. Determine the image rejection that can be expected from this<br />

circuit. Assume that Q 1 and Q 2 are biased at 3 mA and that the loss in the<br />

resonator is 5�.<br />

Solution<br />

In this case, we assume that Isharp will take on a value of<br />

Isharp = � 2 C 1C2vT R �

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