Impedance Inverter Cohn Crystal Filter.
A block diagram of a superhet receiver is shown below. Recall
in the superhet receiver that the RF signal is mixed with the
VFO signal by the RF Mixer down to the IF. In the NorCal 40A,
the IF is approximately 4.9 MHz.
Stage 1:
After this, the IF signal is mixed with the BFO signal by the
Product Detector down to audio frequencies (approximately 620
Hz in the NorCal 40A).
Stage 2:
One difficulty here is that the BFO image is only about 1.2 kHz
away from the IF. For a center frequency of 4.9 MHz, we need a
bandpass IF filter with a Q of approximately
That’s a large Q! The Q’s we’ve seen for discrete elements (and
TL resonators) have been approximately 200 or less. That’s 50x
too small.
Quartz crystals will be used instead to achieve this high Q.
However, we need both series and parallel resonant elements to
realize bandpass ladder filters. While crystals have both these
resonances, they occur at different frequencies. We need these
resonant frequencies to be the same.
So, how do we make a bandpass filter with identical quartz
crystals? We will couple them together in a special way using
impedance inverters. Impedance Inverters
An impedance inverter is a device or circuit that has an input
impedance inversely proportional to the load impedance. More
specifically, the normalized input impedance equals the
normalized load admittance.
Actually, we’ve already seen an example of an impedance
inverter already: a ?/4 length of transmission line.
Recall from Lecture 9 that for an open circuit load ( ZL = ?), V (?/ 4) = 0 and I (?/ 4) =maximum. Therefore, Z (?/ 4) = 0.Consequently, this TL has “inverted” the load impedance.This is a general fact true of any load impedance connected to a ?/4 length of TL. As shown in the text (Section 4.11)
We make the following definitions.
Z(?/4) = Z (?/4 / )/Z0 the normalized input impedance • Z0 Z (0) =1/ zL = y L the normalized load admittance. Then (4.102) can be written in the compact form
We see here that the normalized input impedance equals the
inverse of the normalized load impedance. This is the definition
of an impedance inverter device.
Such a TL impedance inverter would be impractically long for
our uses. Instead, we can make an impedance inverter using
discrete L’s and C’s. From
Notice that the magnitudes of the inductive and capacitive
reactances are equal in this circuit. This strictly can occur only at
a single frequency. So this impedance inverter will be a narrowband device, which is ok for us since the IF Filter will have a very narrow passband. Let’s verify the operation of the impedance inverter in The input impedance is
Therefore
where zi and zL are the normalized input and load impedances,
respectively. That is, the normalized input impedance equals the inverse of the normalized load impedance. All quantities have been normalized to the inverter reactance, X. Cascade An Impedance Inverter to a Series Resonator
Now, let’s examine what happens when an impedance inverter is
placed in front of a series resonant circuit
The impedance inverter, according to (5.41), provides a
normalized input admittance yi of
yi = zL = jxL ? jxc + r
What does the RHS of (5.42) represent for yi? It is a parallel
resonant circuit! To see this, consider
By inspection we see that i L c yi = ? jbL + jbc + g
Consequently, this circuit is equivalent to a series RLC with an
impedance inverter provided [comparing (5.42) and (5.43)]
jxL = jbc , ? jxc = ? jbL and r = g
or
bc = xL , bL = x and gc = r
Hence, we conclude that a series RLC circuit connected through
an impedance inverter appears to the input terminals of the
inverter exactly equivalent to a parallel RLC circuit. Cool! Cohn Filter
Your text provides a wonderful description of how the IF Filter
works in the NorCal 40A. This filter is a fourth-order Cohn filter
built from four quartz crystals and five identical capacitors.
The four element Cohn filter that forms the Intermediate
Frequency (IF) Filter in the NorCal 40A is:
To understand the operation of this filter, the text first adds
fictitious L and C elements. The unsigned reactances of L and C
are equal so their series impedance is zero at f0.
The next step is to recognize the presence of impedance
inverters positioned between each quartz crystal.
Replace these tee networks with impedance inverter circuits
What remains are two capacitors in series with the crystals X2
and X3. We can now see that the purpose for C9 and C13 is to
ensure that X1 and X4 also see the same capacitances.
Now, substitute the equivalent series LC network for each of the
quartz crystals (and the two series C’s):
Our qualitative analysis will begin at the far right of this
equivalent circuit:
1. The series LC network 4 connected to impedance inverter C
appears as a parallel LC network at the left. This is connected
to series LC network 3.
2. The series network 3 – and now “parallel” network 4 –
appears through impedance inverter B as “parallel” network 3
and series 4.
3. Finally, the series network 2, “parallel” 3 and series 4 appear
to 1 through impedance inverter A as series 1 connected to
“parallel” 2, series 3 and “parallel” 4, as shown below.
We can recognize this equivalent circuit as a fourth-order,
bandpass LC ladder filter! Consequently, this fourth-order Cohn
filter using quartz crystals is effectively a fourth-order, bandpass
LC ladder filter.
Lastly, as mentioned earlier, C9-C13 must be chosen properly if
they are to facilitate the impedance inverter operation. In
particular, their reactance at 4.91 MHz (the IF) must closely
match L (which includes the L of the crystal and the “fictitious”
L of the impedance inverter).
While not quantitative in nature, the discussion here at least
illustrates how the Cohn filter achieves its bandpass nature Loaded and Unloaded Q
In Prob. 14, you will measure the loaded Q of the crystal defined
As
In other words, stating that a Q value is a “loaded Q” implies
that losses from the crystal and the circuit it’s connected to are
both included. You’ll likely measure Qloaded ?O (10,000 ) The unloaded Q of the quartz crystal
is typically much larger since it includes only the losses in the
crystal. As we’ve mentioned before, Qcrystal ?O (150,000 ).
