From: Edward Pacyna (email@example.com)
The subject circuit (QQ, 6/92) was lifted from Solid State Design, Chapter on
Test Equipment and Measurements, figure 91. The authors didn't say much on how
on how it works, so here's the scoop.
The receive signal path is through the two series connected diodes. On receive
these diodes are turned on by forward bias. The path is established because the
transistor has been turned on because base current flows through the 10K base
resistor. At DC, you now have a path from the 13.5V supply through the 2.7K
resistors, diodes, inductor and the CE junction to ground. At AC (i.e. RF) you
have two paths. The 50 ohm path (from the collector of the power amp to the
receiver input) and a 500 ohm shunt path (parallel combination of the two 2.7K
resistors and the reactance of the inductor which is about 1K ohm). This is one
reason there is some loss of signal (the other is due to the forward voltage
drop across the diode junctions).
When transmitting, some RF is picked off the final amp and half wave rectified.
This resultant voltage is than passed through a R/C network (which hopefully
filters and smooths to DC) to establish a base current that turns on a second
transistor (it's CE junction goes to ground). This now turns off the first
transistor (because it's base current is removed) and opens up the DC path
discussed above. Now the two two series diodes have no forward bias and the
50 ohm signal path opens up (high impedance).
So what's wrong with this picture?
The two transistors are in the common emitter configuration. Therefore they
can act as a switch (saturated mode) or amplifier (active mode), or maybe both.
If you put a scope on the half wave rectified signal discussed above you'll
see that thats its not DC. The bottom half of the 14Mhz waveform gets clipped
and you have a train of periodic positive waveforms (i.e. 14Mhz raw DC). The
470pf capacitor helps filter and smooth it out, so don't remove it). It passes
through one transistor (perhaps even being amplified) on the second. Keep in
mind that even if a switch is 100% saturated, its not an ideal switch and it
will have a Vce(sat) drop. The next transistor passes (or amplifies) the wave-
formwave on the the diode circuit bias circuit and back to the power amp! The
purpose of the .01 uF cap on the collector is to provide an AC ground for the
inductor and transistor collector. Don't leave home without it. Without alot
of analysis the simplest way to correct this cicuit would be to drive it with
a pure DC signal (i.e. from keying transistor vs RF amp output sample). You
may need to rebias the transistor (make sure its saturated w/o exceeding the
Also, why such a complicated circuit. Why not use the simple series L C with
2 back to back clamping diodes? Only problem is that the X(c) and X(l) each
need to be about 500 ohms which makes it suitable for monoband rigs only (but
there are ways around this).
Time for lunch, so just a fast note on the MC3340 peaking when varying the
voltage from 0 to 6V. I looked at the data sheet last night. The attenuation
vs voltage curve starts at 3.5v (0 attenuation) to 6.0v (80 dB attenuation).
Device behavior at less than 3.5v is not specified and I suspect that there
is attenuation (thus your peaking). So just control it with a voltage starting
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