**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

capacitances:

-C6 in series with C4 in series with C5

-C8 in series with varicap capacitance (d1)

-C7

-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

follows.

**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

current.

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 =

10*10^-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

shortly.

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

-(1)---C6-----(2)--

| | | | |

| 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:

Ct=(C9||C10)+C6||((C4||C5)+(C8||D1)+(C2||C3))

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