Q: >>So what is accomplished by
clamping the negative portion of the drive
>>signal? It would seem that we're wasting valuable signal by shunting it
It's not really shunted to ground. The combination of C35 and D6 is a 'clamp'
circuit. When the voltage at the base of Q6 goes very far negative, D6 begins
conducting, and as such, it's charging up the .01 coupling capacitor. This
negative side of the voltage swing normally contributes nothing to driving the
class C stage, so there's no signal loss per se. On the positive side of the
input signal excursion, the presence of this stored charge on C35 actually
drives the base of Q6 a bit harder. The improvement is about 2 dB in terms of
making the final easier to drive. You can verify this by setting the output
power around 1.5 watts and removing the diode- you'll see output power drop.
The signal at the base of Q6 (without the diode) is about a volt positive
(and noticeably "squashed") on the positive half-cycle of the input waveform.
This is because conduction on the PA starts when the base is about one diode-
drop above ground and doesn't vary much in voltage as the drive is further
increased. Without the diode, the negative signal swing can be quite large-
it's not unusual to see a few volts of negative signal swing. With the clamp
diode in place, the negative swing is constrained to near ground, the DC
average is pushed upward, and the base of Q6 is driven harder. For the
curious, the presence of this added diode appeared to have no adverse effect
on spectral purity.
Paul (AA1MI) and I had a recent
private discussion of the
SW40+ circuitry involving Q5, and its collector load thru
T4, the ferrite 8:1 transformer. We both agree that it may
be useful to others....(its l-o-n-g)
[[start of log]]
Original posting from Paul (AA1MI) to QRP-L:
I've been looking over and playing with the latest installment (RF
buffer/driver) and have a few questions for our online sages...
1. The design consists of a 3-transistor output stage (Q4, Q5 and Q6). Why is
it necessary to have three stages? Wouldn't just two suffice -- one to buffer
the mixer output/filter, in turn driving the PA transistor (here, Q6)?
2. I think I understand the purpose of T4 (improving gain on a Class A amp?),
but I sure would appreciate it if somebody could walk me through the process
of determining how many turns on what kind of core are needed -- and *why*.
Response from Glen (VE3DNL)
>1. The design consists of a 3-transistor output stage (Q4, Q5 and Q6). Why is
>it necessary to have three stages? Wouldn't just two suffice -- one to buffer
>the mixer output/filter, in turn driving the PA transistor (here, Q6)?
Good question. Dave has designed the RF amp stages very conservatively.
Good approach. Consider that these transistors are CHEAP!!! like about ten
cents each. If adding one extra makes your design more stable or re-
then its money well spent. You COULD do it with two, but you'd have to push
'em to the limit.
>2. I think I understand the purpose of T4 (improving gain on a Class A amp?),
>but I sure would appreciate it if somebody could walk me through the process
>of determining how many turns on what kind of core are needed -- and *why*.
I went thru the design of this stage and found that Dave pushed the
collector EXACTLY right...beautiful job.
Its a class-A amp. To get max power out of it, you want the collector current
to swing from its quiescent value (I recall its something like 20 - 30 ma)
down close to zero ma., and up to twice quiescent.
At the same time, collector voltage should swing from +12v down close to
the emitter voltage, and up to +24 v on the overshoot.
For both current and voltage swing to reach these limits simultaneously
requires a specific collector load Z. That's where the turns-ratio of T4
If quiescent Ic = 25mA, and collector voltage swing is 10v (some is taken
up by the emitter resistors), then Rc for best efficiency is 10v/.025
ohms, or 400 ohms.
With a turns ratio of 8:1, the impedance ratio is 64:1
This means that the base circuit must be around 400/64 ohms (about 6 ohms).
This load is mighty non-linear and can't be measured easily.
(and remember that these numbers are from memory, and might not be exact).
To choose the turns ratio requires you to know what impedance
Q6's base circuit presents to the 1-turn winding of T4.
So you've gotta choose a core that gives an inductive impedance of much
greater than 6 ohms per turn^2.
A rule of thumb is to design for between 5 to 10 times the 6-ohm load.
So at 7MHz you want about 50 ohms minimum inductance.
That's an inductance of 1uh per turn.
Choose a core using the "AL" winding value (units of henries-per-turn-squared)
sometimes they spec AL in mH/t^2 or nH/t^2. I'd expect that the FT37-43 core
has an AL value exceeding 1uH/t^2 (more is OK, less isn't).
>From Paul to Glen:
A few follow-up questions:
Why do you say the load is highly nonlinear? And how does somebody go about
estimating it in order to size T4?
>So you've gotta choose a core that gives an inductive impedance of much
>greater than 6 ohms per turn.
Why? I only need 6 ohms, right, so one turn does the trick?
> A rule of thumb is to design for between 5 to 10 times the 6-ohm load.
Rule of thumb for what? I'm not sure what you're referring to here...
> So at 7MHz you want about 50 ohms minimum inductance.
> That's an inductance of 1uh per turn.
How did you arrive at that figure (1 uH/turn)?
>From Glen to Paul
>Why do you say the load is highly nonlinear? And how does somebody go about
>estimating it in order to size T4?
Q5 is the last linear stage: all the active devices beyond are "ON"
during part of the (7MHz) cycle, and "OFF" during other parts of the cycle.
This includes D6 and Q6, D7-D10. (D12 should never conduct in normal
When "ON", the base of Q6 is VERY low impedance. When "OFF" it is very high
impedance. Trying to figure out the average impedance isn't easy. Same with
I suppose you could say that Q5 has an input Z that varies with
instantaneous input signal, but it varies only slightly. When we say that Q5
is linear, we assume that input Z and output Z DON'T vary with input signal,
and its a pretty fair approximation.
We can't make the same approximation with D6 and Q6 because their
input Z and output Z vary WILDLY with instantaneous input signal level.
>>So you've gotta choose a core that gives an inductive impedance of much
>>greater than 6 ohms per turn.
>Why? I only need 6 ohms, right, so one turn does the trick?
>> A rule of thumb is to design for between 5 to 10 times the 6-ohm load.
>Rule of thumb for what? I'm not sure what you're referring to here...
Well, we don't want the inductive reactance of the transformer to
load down Q5's collector. Q5 should be seeing mostly a resistive load,
consisting of the (transformed) impedance of Q6's base circuit. Since
T4's inductive reactance is in parallel with Q5's collector, and in parallel
with (transformed) load of Q6, it should be much larger than the resistive
load that Q5 sees. This will assure that most of Q5's collector current
goes into the load, and not into T4's inductive reactance.
This is an important thing to see. Ask more questions if you're
not completely sure about what's going on here.
>> So at 7MHz you want about 50 ohms minimum inductance.
>> That's an inductance of 1uh per turn.
>How did you arrive at that figure (1 uH/turn)?
What inductance gives 50 ohms at 7 MHz?
It is 50/(2*PI*7e6). That works out to a little more than 1e-6 henry (1uH).
Since T4's secondary winding is 1 turn, then AL is 1uH/turn^2
>From Paul to Glen:
First, I noticed that many example circuits use a tuned tank at the output,
putting a cap across the transformer. Why not here?
You say, "since T4's inductive reactance is in parallel with Q5's
Hold on!! The transformer is in series with Q5's collector, not in parallel,
right? Even looking at the Handbook's equivalent circuit for a transformer,
all the elements are series (except for stray capacitances).
You then again say that T4's reactance is also "...in parallel with
(transformed) load of Q6..." Again, I don't see any parallel circuit here; the
secondary of T4 is in series with the base of Q6, which presents the load.
>From Glen to Paul:
>I noticed that many example circuits use a tuned tank at the
>output, putting a cap across the transformer. Why not here?
Yes, a tuned collector load could work. It'd be pretty low Q because
of the low impedances involved. But still, the tuned circuit might require
a trim-capacitor to resonate at 7MHz. Costs more, and its another tuning
stage that builders can mis-align. The broad-band transformer that Dave
uses is cheap, requires no tuning, and has about the same performance.
>You say, "since T4's inductive reactance is in parallel with Q5's
>Hold on!! The transformer is in series with Q5's collector, not in parallel,
>right? Even looking at the Handbook's equivalent circuit for a transformer,
>all the elements are series (except for stray capacitances).
Didn't mean to add confusion.
I've pulled out the only Handbook available,(1978) and am
referring to their audio transformer modelling discussion...
The Handbook talks about a few different loss mechanisms in their
transformer: due to magnetizing current, leakage current, and winding
resistance losses, and core losses. In the '78 book, they show the transformer
model you've described:
It shows Rc (core losses) in parallel with the primary
It shows Rp (primary wire resistance) in series
It shows Xp (leakage inductance loss) in series
Then it shows an ideal transformer labelled "PRI" and "SEC"
On the secondary side:
It shows Xs (leakage inductance loss) in series
It shows Rs (winding resistance loss) in series
Then the transformer model's output Es.
I looked for a more RF specific model that included some stray capacitances
but didn't find any. In our application (T4) this model is OK because the
coupling coefficient between primary and secondary is almost 1.0
However, the model shown doesn't discuss the path for magnetizing
current. Since coupling is tight, Xp and Xs are small. And Rp and Rs are
small too, since we have so few turns.
Consider what happens in the extreme case where we disconnect the
load. Secondary current goes to zero. Its a pretty good approximation to
assume the whole secondary part of the transformer disappears.
So you're left with Rc, Rp, Xp and the primary winding "PRI".
What isn't discussed in the '78 Handbook clearly is that in this model,
some current still flows thru the primary side in this case. It flows
around thru PRI, and is mostly inductive in nature (meaning that the
primary voltage leads primary current by close-to-90-degrees).
This is the magnetizing current.
And we'd like this current to be small, compared to the primary current
that flows when the load IS connected.
To accomplish this, we need to have the inductance of PRI to be much
larger than the transformed load resistance. This is where the "rule-of-thumb"
of *5 to *10 comes in, to make magnetizing current 5 to 10 times lower
than transformed load current.
OK, now back to the T4 being in parallel with the collector....
I consider that the collector output voltage is measured with-respect-to
For RF purposes, the top end of T4 is at ground too, because of the
bypassing effect of C111 and C110. So in my mind, T4 is actually in
parallel with Q5's collector. Most times, the supply voltage and ground
are the same for AC signals. Make SURE you see this...its REALLY important.
>You then again say that T4's reactance is also "...in parallel with
>(transformed) load of Q6..." Again, I don't see any parallel circuit here;
>the secondary of T4 is in series with the base of Q6, which presents the load.
The secondary winding of T4 is in PARALLEL with its load.
I think we'd both agree that D6 is in parallel with R29. Can you see that
the base of Q6 is in parallel as well? If so, then you'll have to agree
that the secondary winding of T4 is in parallel as well. I'm mentally
shorting out C35, because its a very low impedance for RF.
Q: >My interest has been
"peaked" while reading the first chapter of "The
>Art of Electronics," a text recommended on QRP-L. I have been reading a
>section on Thevin's equivalent circuit which the author's use as a basis
>for understanding impedance matching between circuits.s
Thevenin and Norton equivalents are VERY powerful analytical
tools. Used extensively to simplifiy linear circuits. Learn 'em well.
>1. What are the input-output impedances associated with the circuitry
>associated with Q4 and Q5? I would appreciate the logic supporting the
>answer. I am comfortable with only resistance and DC circuits at this
>time, but I am not quite certain how to incorporate active devices and
>reactance components in the computations.
You need to learn a little more about active circuits before you
can work out Q4's input Z. But you can find an upper limit by calculating
the equivalent resistance of Q4's bias resistors, R23(22K) and R24(10K).
You should be able to see that Q4's base will be in parallel with these
resistors, and can only reduce the result of the above calculation.
Since we're dealing with AC signals, consider the top of R23 to
be at AC ground potential. So for AC, R23 and R24 are in parallel.
>2. C114 by-passes only a part of the Q5 emitter resistance. Why? I
>"thought" good practice dictated a by-passing of the total emitter
>resistance was necessary to assure a "stable" circuit, that is, to avoid
>oscillation. Obviously, I have missed something in my past
Actually, the circuit is more stable with some of the emitter
resistance un-bypassed. The gain is lower as a result. And Q5's input
impedance is higher too. Q5 becomes a more linear amplifier as well.
A good, conservative design.
>3. The directions for peaking T2 and T3 do not exclude the use of a
>metallic tool. I used both a metal small screwdriver and a homebrewed
>plastic "screwdriver" with no discernible difference in adjustment.
>However, I have some other kit building experiences with the adjustment
>of coil/capacitor/ferrite which required the use of a plastic tuning
>tool. What gives?
T2 and T3 are cup-cores. The screwdriver slot is mostly out of
the flux path. This design is mostly self-shielding.
And T2, T3 tune fairly broadly too, so you don't notice much de-tuning.
I'd still use a plastic tool though.
A slug-tuned threaded core with a hex-hole is a different story.
Here, the flux-path is terribly distorted by a metal tool.