Q:
I'd like to hear more about Step 5. What are the advantages of
the balanced vs. single-ended output configurations for the NE612? Whichever
sort of output is used, is there an optimal termination for the mixer?

A: Using both outputs has the advantage of supplying double the signal for the next stage. The output impedance of the mixer is 3k ohms (from the data sheet).

Q: This is an exchange regarding the bandpass filter after the transmit mixer. The original questions begin with >>

>>I'll start my questions on the Tx bandpass filter by quoting a section from
>>the introduction in the rig's manual:

>>"Changed the Tx bandpass filter to use IF transformers. This configuration
>>makes use of the differential outputs of the NE612. Filter bandwidth is
>>increased considerably over the original configuration."

>>Is using the mixer in the differential mode any better? Why?

The outputs of pins 4 and 5 of the mixer are out of phase with respect to one
another. By using them both, as opposed to a single-ended configuration,
there's twice as much signal to work with.

>>I understand that in this filter we want a relatively broad bandwidth so any
>>signal in the 40m band passes through unattenuated. I checked out the
>>schematic for the original version (as seen in _QRP Power_), and he at first
>>used an LC filter design. What was wrong with its bandwidth that it needed
>>improved?

The original design was a little too sharp- it couldn't be modified much
beyond the original bandspread with falloff becoming noticeable. I've simply
broadened it a bit to make that adjustment a little less critical. Yes, a
resistor could have been added to swamp the response. On the new version, the
TX mixer source impedances (1500 ohms nominal at pins 4 and 5) as well as the
emitter follower stage, establish loading on that filter. The value of
coupling capacitor C31 also affects the filter response.

>>I checked the Mouser catalog and see that one of these IF transformers sell
>>for $0.65 in hundreds. So Dave probably saved a lot of real estate and maybe
>> even cut the rig's cost by going to the transformers. But why are two
>> transformers necessary; wouldn't one be sufficient? Don't two stages in the
>> filter sharpen its frequency response, whereas the original idea is to
broaden
>>it?

Here were the key design considerations:

- Real estate
- Cost
- Easier asembly

It's possible to use a single tuned circuit and still meet the FCC
requirements for spectral purity. Doing so, though, requires some pretty high
impedances, which should avoided as a general rule with transmitter design-
there's an increased susceptibility to unwanted feedback due to ground-loop
problems. I didn't have the luxury of unlimited ground plane, although I was
cognizant of ground paths, and elected for 'safe' over 'sorry'. Keeping the
impedances reasonable and still retaining a good bandpass characteristic (good
rejection away from the passband) calls for more parts.

>>Also, very puzzling to me: the IF transformers are resonant at 10.7 MHz. But
>>without knowing the values of the internal coil and cap, what's the process
>>for determining that a 47 pF cap in parallel brings resonance down to 7 MHz?

The internal characteristics are easily inferred with a signal generator and
scope. Once the resonance peak is found, extra capacitance is added in
parallel and the new peak is located.
The resonance formula is

F(mhz)**2= 25,330 / LC

where L is in microhenries and C is in pF.

Doubling the capacitance (adding a parallel external capacitance replicating
the internal one) moves the resonance to .707 of the original frequency.

For these transformers, it's approximately 5 uH and 50 pF.

Q: > Does the circuit following the mixer provide something like a
> 3K termination for the mixer? If so, how can we see this?
>

First, C33(0.01uf) has a very low impedance, so that T3 is coupled
very tightly to Q4 base. Whatever Q4's input Z is, that's what
T3's going to see. Ditto with C34(0.01uf)

Next, take a look at Q4's biasing resistors, R23(22K) and R22(10K).
Their parallel combination is 6800 ohms. So T3 isn't going to see a
load any higher than this.

Now comes the trickier part: determining the actual base input
impedance of Q4. Whatever it is, it'll be in parallel with the
bias resistors above, lowering the 6.8K load Z we have so far.
Q4 is an "emitter follower". The impedance that we'll
see (looking into its base terminal) is hie + hfe * Re
where Re is the impedance from emitter to ground.
and hie is the base-to-emitter impedance of the transistor
and hfe is the transistor's current gain (at 7 MHz.).

I'm going to make a guess at hie of about 300 - 1000 ohms.
And I'm going to guess that hfe is about 25.

If you look up the 2N4401 data sheet, you'll think I'm way off.
They list hie=700 ohms (at 6ma. Ic).
But they list hfe=150

The difference arises because these small-signal h-parameters were
measured at 1KHz, not at 7MHz. At high frequencies, current gain
drops off. At about 200MHz, it has no current gain at all.

What is Re? Maximum value is 500ohms due to R24. But in
parallel with R24 is Q5 and its bias resistors. Bias resistors
R25(2.2K) and R26(470) work out to 387 ohms.
I'd estimate that Q5 has about the same Z as its bias resistors.
So in parallel with R24 is 387 ohms, in parallel with another
387 ohms.
So total Re is 500||387||387 = 140 ohms

So in summary, for Q4 we have:
hie = approx 700 ohms
hfe = 25
Re = 140 ohms

And its input Z is 700 + 25 * 140 = 4200 ohms.

This is in parallel with the bias resistors of 6.8K

So the total Z that T3 sees is 6800||4200 = 2600 ohms

Now this is a minimum estimate, because I've assumed that the wiper
of R24 is right up at the top. If its down lower, the total Z figure
above will be higher, because Re will be larger.
Dave has chosen bias resistors on the low side. This provides
temperature stability, and also means that the load that T3 sees
doesn't change wildly as you turn the R24 trimpot.

Q: Summary of transmit mixer problem.

I am building the SW30+ and have run into a problem. When I completed Lesson
4, I observed that Q3 was switching properly and that there was a complex
waveform on pins 4 and 5 of U5. Feeling confident that all was well, I moved
on to Lesson 5 and installed T2 and T3 and related components. The complex
waveform observed on pins 4 and 5 of U5 are also present on the input of T2 but
the output of T2 is flat. After checking all component values, orientation,
solder joints etc. I finally concluded that T2 was probably working as
designed, ie the input signal was outside of the bandpass of T2.

With that in mind I started backtracking and taking a closer look at the
waveforms on U5 and made the following observations.

1) LO as measured at the base of Q2 is about 2.439 mhz and 2.6V p-p.
2) LO as measured at pin 2 of U5 is about 0.16 V p-p.
3) Pin 7 of U5 is a beautiful modulated sinewave (NOT WHAT I EXPECTED). P-P
voltage is about .05 V and it appears to be a signal of about 24-25 mhz moduled
by at about a 2.4-2.5 mhz rate. I was expecting to see a 7.68 mhz sinewave
here. I have rechecked Y5, RFC2, C28, and C29. All values are proper,
connections look good, etc.

What should I be seeing on Pin 7 of U5? Assuming that what I am observing is
not right, any ideas of what might be wrong?

TURNED OUT THAT THE CRYSTAL (Y5) WAS BAD. UP AND RUNNING NOW.......

Q: >When I tweak T3 and T2 (while observing waveform on scope) I find the setting
for T3
>that gives me the largest amplitude to be with the slug turned all the
>way CCW (to the stop). T2's peak is roughly mid-way between the stops. T3
>never really shows a peak - just gets larger as I go CCW. Is this okay?
>

Glen, VE3DNL, suggested that perhaps my scope probe was adding enough
capacitance to the circuit to cause this problem. Following his suggestion, I
went ahead and installed the Q4 buffer (and related resistors) and redid the
alignment with my scope probe at the emitter of Q4 rather than the base. This
effectively isolated the probe from the bandpass filters. The results proved
to be quite interesting. There is now a definate peak about mid-range in T3.
What is kind of weird is that as soon as I go slightly past the peak (CCW) the
signal degenerates into a real mess.

Dave Benson suggested a different solution which was to add 10-22pf capacitance
across T3 (ie C32). I will try that this evening and report the results. I'm
betting that the two solutions togther solve the problem.

More on this problem

The procedural problem (measuring the waveform at Q4's base rather than at the
emitter) did not result from any ambiquity in your manual. The Elmer 101 Part
5 had us stop prior to installing Q4 or any of its biasing resistors and then
instructed us to measure the waveform specifically at the base of Q4 (no signal
at emitter until buffer components are added in subsequent lesson).

As I said, I went ahead and put the buffer section in and while that helped it
did not totally solve the problem. I was then seeing a peak more toward the
middle of T3's range but when I hit the peak (going CCW) rather than seeing a
reduction in amplitude what I saw was a degeneration of the waveform which
required I unkey and rekey the transmit section.

I took your advice and put 22pf in C32 and found that threw T3 all the way to
the CW stop. Reduced it to 10pf and found a nice solid peak near mid-range.
That, however then caused T2 to hit its stop. I then put 10pf in for C30 and
am pleased to say that I now have both transformers showing nice solid peaks
near mid-range.

Even more....

>>Dave Benson suggested a different solution which was to add 10-22pf
capacitance
>>across T3 (ie C32). I will try that this evening and report the results.
I'm
>>betting that the two solutions togther solve the problem.

I assumed that you were looking at Q4's emitter, thus isolating the probe
capacitance from the tuned circuit. As I look back, though, I didn't
expressly say that anywhere in the manual- an oversight on my part. Scope
probe capacitance will indeed upset the operation of that bandpass filter, and
what you reported makes sense in light of the presence of probe capacitance.
I'd recommend moving the measurement point to Q4's emitter and repeaking
T2/T3- the extra capacitance I suggested earlier shouldn't be necessary.

73, Dave- NN1G