Radio Frequency Integrated Circuit Design - Webs

Voltage-Controlled Oscillators
noise floor have not been included in this equation but are straightforward to
add (see discussion about Figure 8.34).
The term A added to Leeson’s formula is usually referred to in the literature
as the excess small-signal gain. However, A has been shown, for most operating
conditions, to be equal to 3 dB, independent of coefficients in the power series
used to describe the nonlinearity, the magnitude of the noise present, the
amplitude of oscillation, or the excess small-signal gain in the oscillator.
8.18 Making the Oscillator Tunable
Varactors in a bipolar process can be realized using either the base-collector or
the base-emitter junctions or else using a MOS varactor in BiCMOS processes
[10, 11]. However, when using any of these varactors, there is also a parasitic
diode between one side and the substrate. Unlike the base-collector or baseemitter
junctions, which have a high Q, this parasitic junction has a low Q due
to the low doping of the substrate. This makes it desirable to remove it from
the circuit. Placing the varactors in the circuit such that the side with the
parasitic diodes tied together at the axis of symmetry can do this.
Example 8.9 VCO Varactor Placement
Make a differential common-base oscillator tunable. Use base-collector junctions
as varactors and choose an appropriate place to include them in the circuit.
Solution
The most logical place for the varactors is shown in Figure 8.38. This gives a
tuning voltage between power and ground and prevents the parasitic substrate
diodes from affecting the circuit.
As can be seen from the last section, low-frequency noise can be very
important in the design of VCOs. Thus, the designer should be very careful
how the varactors are placed in the circuit. Take, for instance, the −Gm oscillator
circuit shown in Figure 8.39. Suppose that a low-frequency noise current was
injected into the resonator either from the transistors Q 1 and Q 2 or from the
current source at the top of the resonator. Note that at low frequencies the
inductors behave like short circuits. This current will see an impedance equal
to the output impedance of the current source in parallel with the transistor
loading (two forward-biased diodes in parallel). This would be given by
R Load = rcur //re1 //re2 ≈ re
2
295
(8.102)
where rcur is the output impedance of the current source and the transistors
are assumed to be identical.