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r-JSV
r.J.5v
Ql
lbMA
Q3, eTC.
~
+---~~-
Jfo
MA
lZOMA
The current through Ql is determined by the difference between
two current sources, one involving Q3 and the other involving Q2.
The most convenient way to find the current from each source is
to measure the voltage across each emitter resistor. (A 1 kQ
resistor should be used in series with the voltmeter probe tip to
prevent the circuit from oscillating and giving an erroneous
reading.) For Q3 and Q6, the current through R58 and R60 must
be included. If results are inconsistent, the emitter resistor should
be checked also. Check to see that the BW5 line is at the voltage
specified
(~±0.3V)
in the table on the schematic.
In the LC mode (the four wider bandwidths), the BW5 line goes to
14.8V and turns off the current source Q3. The current supplied
by Q3 in the crystal mode is now supplied through CRI and Rl3
from the BW5 line. In the LC mode, the current through Ql can
be found by subtracting the current through R 13 from the current
through R8.
Unity Gain Buffer Amplifier.
0
The Unity Gain Buffer Amplifier is the same as the 10 dB Input
Buffer Amplifier except that it has a FET input (Q5) and is con-
nected for unity gain. The input is selected by the BW5 line from
CR9 in the LC mode or from CR8 in the crystal mode.
In the crystal mode, the current through Q5 is determined by the
difference between the current sourced by Q6 and that sunk by
Q7: about 4 rnA. A significant deviation from this current should
be reflected by the gate-to-gate source voltage of Q5. The source
should be at least 0.2V more positive than the gate, but not more
than 1.5V more positive. If the difference is less than 0.2V, the
FET current is too high; if the difference is greater than l.5V, the
FET current is too low. In either case the FET could also be defec-
tive. To determine precisely the current through Q5, the dif-
ference between the current through R38 and that through R60
should be subtracted from the current through R30. If the results
are inconsistent, check the above-mentioned resistors.
In LC mode of operation, current is supplied through R37 and
CR19 from the BW5 line instead of through Q6. The difference
between the current through R37 and that through R30 yields the
FET current.
Output Buffer Amplifier
0
The output Buffer Amplifier is a complementary pair of tran-
sistors in which Q9 acts as a source follower boosted by Q 10. The
current through FET Q9 is set by R53:
I
_ Vbe (Ql 0)
o:=
FET-
196S1
-·- 7 -
o:=
3 rnA
196S1
The total current through Q9 and QlO is set by R54. The input is
selected by the BW5 line from either CR16 in the LC mode or
CR15 in the crystal mode.
Crystal Filtering Circuits
0 4D
The bandwidths 1 kHz, 3 kHz, 10 kHz, and 30 kHz are obtained
by crystal filtering. The crystals are used in series resonant mode
and can be modeled as a series resonant circuit with a parallel
capacitance:
Co
\ r - - - '
~lrTIL...---l)
Rs
L
L_
n:
The parallel capacitance (C
0 )
and series resistance (R
5 )
are not
desired and are compensated for in the circuit, resulting in this
simplified schematic of a single pole of crystal filtering:
/"IN DiODE
The PIN diode CR4 functions as a variable resistor at 21.4 MHz.
As the resistance is lowered by increasing the current in the BW6
line, the bandshape becomes narrower. The bandwidth of one
pole widens to approximately 70 kHz when the PIN is turned off
completely at the 30 kHz BW setting. (For a four-pole filter, the
bandwidth of each pole is about 2.3 times the bandwidth of all
four poles taken together. The bandwidth of two poles is about
1.5 times the bandwidth of all four poles taken together).
A simplified schematic of a crystal pole, including compensation
for R
5
and C
0
in the crystal and input capacitance of the buffer
amplifier, is shown in Figure 8-60.
CIS.Qii>
/ t
A6J'JSTABL£
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R25>1:
!/Y UNITY
(?tJ./N
(6£N£RA7£1J
BY
eUri'"£R
03 + POSITIVE.
L7
AMPLIFIER
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Figure 8-60. Crystal Pole, Simplified Schematic
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