Radio Frequency Integrated Circuit Design - Webs

264 Radio Frequency Integrated Circuit Design
Figure 8.18 (a) Colpitts oscillator circuit; and (b) equivalent series model.
Example 8.3 Negative Resistance for Series and Parallel Circuits
Assume that, as before, L = 10 nH, R p = 300�, C 1 = 2.5 pF, C 2 = 10 pF,
and the transistor is operating at 1 mA, or re = 25� and g m = 0.04. Using
negative resistance, determine the oscillator resonant frequency and apparent
frequency shift.
Solution
As before, C T is 2 pF and
� =
1
√ LC T
1
=
7.07 Grad/sec
√10 nH × 2pF=
or frequency f o = 1.1254 GHz. The series negative resistance is equal to
g m
rs =−
� 2 C 1C2
Then Q can be calculated from
Q =−
1
=
�rs C T
−0.04
(7.07107 GHz) 2 � 2.5 pF � 10 pF =−32.0�
=
−1
7.07107 GHz � 32 � 2pF =−2.2097
rpar = rs (1 + Q 2 ) =−32(1 + 2.2097 2 ) =−188�
We note that the parallel negative resistance is smaller in magnitude than
the original parallel resistance, indicating that the oscillator should start up
successfully.
The above is sufficient for a hand calculation; however, to complete the
example, the equivalent parallel capacitance can be determined to be