Radio Frequency Integrated Circuit Design - Webs

308 Radio Frequency Integrated Circuit Design
Solution
We can first compute R p in the usual way, assuming that the inductor is the
sole loss in the system. It is easy to see that at 2 GHz this will be 628.3�.
Next, we also need to know the value of the two capacitors in the resonator.
Noting that there are two capacitors,
C var =
2
� 2 oscL =
1
(2� � 2 GHz) 2 = 2.53 pF
5nH
Now since the transistor limiters will turn on for a voltage of about 0.85V
in a modern bipolar technology, we can assume that the VCO amplitude will
end up being close to this value. Therefore, we can compute the final value for
the resonator current from (8.68).
I tank = V tank
0.635R p
=
1.7V
= 4.26 mA
(0.635)628.3�
We will choose to use the 10:1 ratio for the mirror in order to get the
best dc gain of 10 from this stage. Thus, the current through the mirror transistor
will be 426 �A. We can now compute the values of transconductance and
resistance in the model:
g m5 = I bias
vT
g m6 = I bias
vT
= 426 �A
25 mV
= 4.26 mA
25 mV
= 17.04 mA
V
= 170.4 mA
V
r�5 = � 100
= = 5.87 k�
g m4 17.04 mS
r�6 = � 100
= = 587�
g m4 170.4 mS
Once we have chosen the gain for the current mirror, we also know the
gain from the resonator:
A 2(s) = V tank(0)
I tank(0) =
2 1
� C var
1
s +
R p C var
= 2
� R p = 399 V
A
and since the gain from the mirror will be 10, we can now find the gain required
from the limiters.