Q: Mike, I am probably wrong but aren't D2 and C105 really part of the 2nd
mixer circuit and not the VFO? Are they essential for the VFO to work or
were they included in your list because you were following Dave's order
of building?

A: D2 and C105 are part of the DC supply for the mixer chips and the VFO. Current flows from the DC power source through D2. It is filtered by C105 and supplied to the other parts from there. If you don't load these parts your VFO will get no DC power.

As I understand it, D2 drops 8v to 7.4v. Also adds a little RF isolation for U1, U3.

Q: After stuffing the VFO parts, there seems to be an extra hole on my
SW30+ pcb. Extra hole is connected to ground. (It is also on the SW-40+)

A: This is provided so you can solder a ground lead across the top of the three crystals in the IF filter (to be installed later). Grounding the filter cases prevents "blow by" which is strong external signals effecting the IF filter section.

Q: I would like to know the effect of using the 50K variable resistor in
place of the 100K one that is used in the VFO section. As I recall the
Radio Shack part number given for this was the number for the 50K pot.
The schematic calls for a 100K pot. I have the 50K linear taper pot and
the only 100K pot I could find at the local RS was an audio taper one.

Does using the 50K decrease, increase, or have no effect on the tuning
range? And would one notice an appreciable difference in using the 100K
pot that is an audio taper vs. one that has a linear taper?

A: Question came up about the tuning pot. The only function that the
tuning pot, listed as 100K, has is to act as a voltage divider and
provide the voltage to set the bias voltage on the MV1662 varactor
diode. This in turn sets the VFO frequency.

With 8V applied to the pot, 50K will give you 0.16mA and 100K will
give you 0.08mA of current, so in the big scheme of things it doesn't
make much difference. I used the 50K myself. BTW this is the current
through the tuning pot only and just adds to the total current required
for the rig.

Q: Now, many years ago, I learned all about Colpitts oscillators, but now I have forgotten.The function of C5 has me stumped. QRP Notebook says it is a feedback cap, but I don't see how that can be. Can either of you shed some light on this?

A: Let's first examine the 3MHz. resonant components. L1 is the
resonant inductance, but resonant capacitance is due to a number of
-C6 in series with C4 in series with C5
-C8 in series with varicap capacitance (d1)
-C9 in series with C10
-C3 in series with C2.

These strings of capacitance all add up to resonate with L1 at 3MHz.
A lot of the resonant current flows down the C6/C4/C5 string, since this
string of capacitances represent the lowest reactance path of the
bunch listed above.
Now you could say that Q2 senses the resonant voltage at its base
and injects pulses of current from its emitter into the resonant circuit
in order to keep the oscillation amplitude up.
So the junction of C4 and C5 is where Q2's emitter "pumps-up" the
oscillations. That's the feedback point. Of course, Q2's emitter pulses
are timed to re-inforce the sinewave shape that the resonant circuit naturally

Q: Can somebody please help me understand what's going on by perhaps redrawing the circuit so all the series/parallel caps are more obvious, and then work through all the calculations that eventually lead to the value of the resonant circuit (near 3 MHz)?

A: OK, you've got the inductance right. Here's the string of capacitances
redrawn. They're still kinda messy because there's so many. I'm assuming the
varicap diode is 50pf. And I'm assuming that C7 (which is selected to get the
VFO on the right frequency) is 68pf. Use courier font (or something else that's
not proportionally spaced) to see the ascii circuit:
_____________________||_____________________ (_ | | ||C6 | | | (_L1 | _|_ 3300 _|_ _|_ _|_ (_2.5uH _|_ C9___ C4___ C8___ C3___ (_ ___ 10| 2700| 82| 10| (_ C7 | | | | | | 68p| C10_|_270 _|_ _|_ _|_ | | ___ C5___ D1___ C2___ | | | 2700 | 50 | 47 | gnd gnd gnd gnd gnd gnd

So let's do the math now. Start at the end, working out the series combination
of C2 and C3: 10*47/(10+47) = 8.25pf.....we'll call this "Ca"

Do the same with C8 and D1: 82*50/(82+50) = 31.06pf......call this "Cb"
Do the same with C4 and C5: 2700*2700/(2700+2700)=1350pf...call this "Cc"
Now we'll add Ca, Cb, and Cc together since they're all in parallel:
8.25 + 31.06 + 1350 = 1389.31pf
Now this capacitance is in series with C6, so let's combine the two:
1389.31*3300/(1389.31+3300) = 977.70pf.....call this "Cd"
Let's work out the series combination of C9 and C10:
10*270/(10+270) = 9.64pf....call this "Ce"

To finish off, we've got C7, Ce, and Cd all in parallel, so we add 'em:
68 + 9.64 + 977.7 = 1055.34pf
This is the capacitance that resonantes with L1, so we can work out
the resonant frequency:
F = 1/(2*PI*SQRT(2.5e-6 * 1055.34e-12) = 3.0985 MHz. Pretty close.

>3. How would I go about calculating the efficiency (Pout / Pin) for the
>oscillator circuit?

May I suggest that this isn't the kind of circuit where efficiency is
important. Its not a power oscillator. The two loads (U1 mixer and U5 mixer)
are both pretty high impedance and don't draw significant power from
this VFO.
And the DC power drawn by the VFO is a very small fraction
of total transmitter power.
According to Dave's DC voltage chart, emitter voltage is 2.5v DC. That's
across the emitter resistor (R17) of 2.2k. So the VFO draws 1.14mA DC
current from the supply. That's still a fraction of the 16mA receiver
So you might think, "why can't we draw more power from the VFO,
so that we don't need as much gain in the transmitter chain?". There's a
good reason: U5 (the transmit mixer) can't handle big signals. It sets the
limit on how much 7MHz. energy is available.

Q: More Math questions answered.

lets see if we can walk back through this....

first lets agree on a couple of conventions...

a) lets represent the series capacitance operation, 1/(1/c+1/c+...) as the
operator ( || ), note that this is the same as resistors in parallel. Also
two forms of exponential notation will help in ascii, 10 pf = 10E-12 =

b) since my ascii art is miserable to say the least, lets redraw the
schematic in words the same way it would be done in a spice or ARD
simulation, that is by series and parallel connections to numbered nodes.
You'll probably want to do this w/ pencil and paper... To do this we'll
agree that node 1 is the first node and we'll connect the inductor L1
there. A node is simply a point in the circuit where things are connected
together. Node 0 is usually ground. the other numbers will fall out

Since we're redrawing this to see the VFO LC pieces we'll ignore all others
for just a minute....

L1 is connected from node 1 to node 0, or "IND 1 0 <value>; L1"
C9 is in series with C10, connected from node 1 to 0
C6 is in series from node 1 to node 2
C4 is in series with C5, connected from node 2 to 0
C8 is in series with D1, connected from node 2 to 0
C2 is in series with C3, connected from node 2 to 0

| | | | |
| C9 C4 C8 C2
L1 | | | |
| C10 C5 D1 C3
| | | | |

we could also designate nodes between each series cap element (and would
need to for any simulation input. Also, the note that the series
combination of C9-C10, C4-C5, and C2-C3 are used to match impeadance by a
technique called a capacitive split. A capacitive split works analagous to
a tapped transformer. more on that later...

Now, the total capacitance can be written (remember the || notation) as:


Using circuit values for the SW30+, and assuming D1=120pf

Ct(pf)=(9.64)+3300||(1650+60+8.25) = 1140 pf, or 1140E-12

F=1/(2*pi*f)(L*C)^1/2 and L= 1/ (2*pi*f)^2*C

So for 10.12 Mhz w/ IF = 7.68 Mhz, LO = 10.12 - 7.68 = 2.44 Mhz

L = 1/ ((2*3.14*2.44E6)^2*1140E-12) = 3.736E-6 or 3.736 uh