BY PAUL FLORIAN
is specified for 10 pF to 35 pF loading.
The crystal should operate in the
fundamental mode. Adjusting Ctune will
compensate for the loading of capacitors C1 and C2. Furthermore, Ctune can
be set to make small changes in the
oscillator frequency. Q1 and its associated components form a Colpitts crystal oscillator and Q2 is used as a buffer.
The output impedance of Q2 is
about 20 ohms. Because there may be
distortion present at the output of this
buffer, it is followed by a resonant circuit composed of Lr and Cr (see Figure
1). The resonant circuit formed by Lr
and Cr requires a source and load resistance of 50 ohms. R7, R8, and Rp form
a 4 dB attenuator (more on this later).
■ PHOTO 1. Oscillator Board and Filter.
A Design Example
We will now design a 4.19304
MHz crystal oscillator. First, obtain a
crystal (X1) of this frequency. Table 1
lists the values for C1 and C2 corresponding to this frequency. For this
design, C1 = 220 pF and C2 = 100 pF.
Next, we can calculate the values for
the resonant circuit at the output of
the buffer. The design equations are
given in the sidebar Equations 1.
It is necessary to specify the following parameters for the equations: Fc
(center frequency), Q (loaded Q), Qp
(Q of the inductor), and R (source and
load resistance). Let Fc = 4.19304 MHz.
Qp for an airwound coil is
approximately 110 and R = 50.
A good choice of the loaded Q
for this resonant circuit is 5.
Higher Qs provide more
selectivity, however, at higher
frequencies the value of the
inductor Lr may be very small.
Examine the equations in
the sidebar. Solve for Xp (the
reactance at resonance) and
then solve for Lr and Cr.
For this example, Lr = 180
nH and Cr = 8 nF. The closest
standard value to Cr is 8.2 nF. I
used a 2% tolerance part for Cr.
Lr is a handwound coil. For information on how to wind this coil,
refer to Sources 1 and 2 listed in this
article. In some cases, the calculated
value of Cr may be between standard
values. If this is the case, two capacitors
can be used in parallel to achieve a
closer match to the calculated value.
Again, Lr is 180 nH. This is not a lossless
component. The coil has an inductance
with a series resistance (see Figure 2A).
There exists a transformation equation that will give us the equivalent
parallel resistance (see Figure 2B). The
design equations for Rp are given in the
Equations 2 sidebar. The inductor is
wound from 22 gauge wire. The resistance of this wire is 16.2 ohms per
1,000 feet. The stretched coil length is
six inches (. 5 feet). Rs is then calculated.
Qp of the air core inductor is 110. We
will now calculate Rp from Rs and Qp.
Rp = 98 ohms and Lp is approximately
equal to Ls. Note that Rp is a property
of the inductor and Rp on the schematic is not an installed component.
Rp is about 100 ohms. In order to
match the input and output of the resonant circuit to 50 ohms, we can make Rp
part of a 50 ohm T-Network attenuator
■ FIGURE 2. Series to Parallel
Equivalent of Inductor Resistance.
Design Equations for the Resonant
■ Fc = 4.19304 MHz
■ w = 2 * 3. 14 * Fc
■ Q = 5
■ Qp = 110
■ R = Rload = Rsource = 50 ohms
■ Xp = (R/2) * (Qp-Q) / (Qp*Q) = 4.773
■ Lr= Xp/ w = 180 nH
■ Cr= 1 / (Xp* w) = 7. 96 nF
Design Values Corresponding to
Design Equations for Rp.
■ Rs = ( 16.2)/(1000) * . 5 = .008 ohms
■ Qp = 110
■ Rp= Qp^2 * Rs= 98
August 2006 43