On Fri, 21 Nov 2008, exray wrote:

Date: Fri, 21 Nov 2008 22:30:24 -0400
From: exray
Newsgroups: rec.radio.amateur.homebrew
Subject: Doubling

This is a really dumb question but it dawned on me that I did not know the
correct answer.

In terms of old transmitters from the 20s/30s...In a crystal oscillator I
understand the concept of setting the oscillator output tank to favor the
harmonic from the crystal. (Stop me if I'm wrong already...)

I think this is correct, but the books say that tuning the output of the
oscillator can "pull" the frequency of the oscillating crystal. I have
sometimes seen this.

But in a doubling amplifier stage am I counting on having enough harmonic
content at the input or am I creating the harmonic with the non-linearity of
the amplifier?

Despite what at least one other person responding to this said, I can rest
assure you that if you run a doubler/multiplier stage even in a linear
mode, AND if you tune the output of that stage to the multiple harmonic,
you will definitely get output at that harmonic frequency which is
stronger than the input drive voltage. In the last few years I have built
many tube stages and observed the harmonic voltage output on a wideband
oscilloscope. As a matter of fact if you ever get a wideband scope and
look at the locked output waveform as you tune through the both the
fundamental and the harmonic frequency you will be very surprised at what
you will see. All of the descriptions in all of the handbooks I have read
(a few) explain this from a theoretical perspective but don't bother to
actually show, with photographs of actual scope traces, how this works.
It would just take an extra page or two and would make people think about
what they are doing.

All amplifiers have some non-linearity, the question is what effect this
has on you meeting "purity" of emissions requirements. The more important
question is whether you are getting the gain/drive that you want from a
given stage of amplification. Reducing unwanted spurious emissions might
require more tuned circuits or measurement using a receive with an S-meter
and operated many wavelengths from your antenna. Most "appliance
operators" just buy a commercial rig and don't worry about anything;
homebrewers might not worry either if their signals go through a tuned
circuit, an antenna tuner, and an antenna for a narrow frequency range.

If you really want to blow your mind, then hook an oscilloscope to the
output of a mixer with two low harmonic content input sine waves to be
mixed. The raw output will look like hell on a scope. The only way to see
the mixed (say, difference) frequency will be to go through at least a
couple of tuned circuits that are tuned for the wanted sine wave
frequency.

I've done this stuff. There are a couple of other minor matters that are
not quite correct in our ham handbooks, too.


TIA
-Bill WX4A

On Nov 22, 8:43*pm, Stray Dog wrote:
Despite what at least one other person responding to this said, I can rest
assure you that if you run a doubler/multiplier stage even in a linear
mode, AND if you tune the output of that stage to the multiple harmonic,
you will definitely get output at that harmonic frequency which is
stronger than the input drive voltage.

Huh? No way... you MUST have non-linearities to make a doubler.

On Sun, 14 Dec 2008, Telstar Electronics wrote:

Date: Sun, 14 Dec 2008 08:20:56 -0800 (PST)
From: Telstar Electronics
Newsgroups: rec.radio.amateur.homebrew
Subject: Doubling

On Nov 22, 8:43*pm, Stray Dog wrote:
Despite what at least one other person responding to this said, I can rest
assure you that if you run a doubler/multiplier stage even in a linear
mode, AND if you tune the output of that stage to the multiple harmonic,
you will definitely get output at that harmonic frequency which is
stronger than the input drive voltage.

Huh? No way... you MUST have non-linearities to make a doubler.

All tubes (and transistors, etc) have non-linearities (if the transfer
characteristics are non-straight lines) if that is what you are talking
about.

However, I have observed output on a scope of second harmonics (and, yes,
the time base was set right and auto-self triggering) and the
amplifier was running no higher than Class B. You should actually try this
yourself and see for yourself. Tune the output to the second harmonic and
you will see grow out of the vally new "peaks" corresponding to that
second harmonic.

I don't know what the solid state gear is doing, but from many schematics
of the vintage tube gear I'm familiar with show, and measure, biasing for
linear operation, even in stages meant to multiply frequency.

"Telstar Electronics" wrote in message ...
On Nov 22, 8:43 pm, Stray Dog wrote:
? Despite what at least one other person responding to this said, I can rest
assure you that if you run a doubler/multiplier stage even in a linear
mode, AND if you tune the output of that stage to the multiple harmonic,
you will definitely get output at that harmonic frequency which is
stronger than the input drive voltage.

Huh? No way... you MUST have non-linearities to make a doubler.

Actually you do not need any nonlinearity to make a doubler (quadrupler, etc.).

Assume you have two Class B (or AB) stages that are driven in push-pull. The outputs are connected in parallel. And to make things even more linear, let each stage have a resistive load. Each stage will produce a linearly amplified (but inverted) version of the input signal FOR THE POSITIVE HALF of the driving waveform only. Being driven 180 degrees out of phase with the input signal, the second stage will produce a linearly amplified but (again inverted) version of the input signal FOR THE NEGATIVE HALF of the driving waveform. Both outputs will have a DC offset of the plate (collector, drain) voltage.

The resultant waveform with the outputs in parallel will look like much like a full wave rectified version of the input signal subtracted from the plate voltage. To express this mathematically, let the input signal be expressed as:

Vin = A sin(wt)

Now let the voltage gain of each stage be "-k" and the plate voltage be "B". The resultant waveform of the two stages connected in parallel will be:

Vout = B - abs[A*k sin(wt)] where "abs" is the absolute value

Vout = B - A*k sin(wt) for 0 wt Pi and
= B + A*k sin(wt) for Pi wt 2Pi or alternately for -Pi wt 0

We can then calculate the Fourier series of this waveform to determine its spectrum. I will not present the calculations here as it is too difficult to show the integration over defined integrals using only plain text (and I doubt many readers will have math fonts anyway). If you wish to see the math for the Fourier series for a number of functions, read:
http://www.maths.qmul.ac.uk/~agp/calc3/notes2.pdf or
http://www.physics.hku.hk/~phys2325/notes/chap7.doc.

Vout = B - 2*A*K/Pi * [1 - SUMMATION {2*cos(nwt)/(n*n - 1)] for n=2, 4, 6, 8...

Note that the original frequency has been eliminated and that only even order harmonics are present, and that the amplitudes drop off quite rapidly. For example, the fourth harmonic will be one fifth of the second harmonic.

For those that need a simplified explanation of Fourier series, Don Lancaster wrote a good article that can be found at:
http://www.tinaja.com/glib/muse90.pdf. I always thought Don had a ham license but I could not find one.

In a real implementation of this multiplier, a tuned circuit would be used as the plate load. The Q of this tuned circuit will assure that only the second harmonic is present in the output. The two stages would need to be well balanced if cancellation of odd harmonics and the fundamental is required.

73, Dr. Barry L. Ornitz WA4VZQ


POSTSCRIPT:

Now let me describe how it is possible to produce ONLY the second harmonic. Instead of using two Class B or AB stages, it is possible to use triodes operating where their plate current is proportional to the square of the grid voltage. Driving the two such stages in push-pull with the outputs in parallel with a resistive load, the output waveform will be:

Vout = B - A*A*k sin(wt)*sin(wt)

Using a trigonometric identity {see:
http://en.wikipedia.org/wiki/List_of_trigonometric_identities},

sin(x)*sin(x) = sin(x)^2 = 0.5[1-cos(2x)]

thus Vout = B - A*A*K/2 + A*A*k/2 cos(2wt)

This shows that only the second harmonic is found at the output.

NoSPAM wrote:
"Telstar Electronics"
wrote in message
...
On Nov 22, 8:43 pm, Stray Dog
wrote:
? Despite what at least one other person responding to this said, I
can rest
assure you that if you run a doubler/multiplier stage even in a linear
mode, AND if you tune the output of that stage to the multiple
harmonic,
you will definitely get output at that harmonic frequency which is
stronger than the input drive voltage.

Huh? No way... you MUST have non-linearities to make a doubler.

Actually you do not need any nonlinearity to make a doubler
(quadrupler, etc.).

Assume you have two Class B (or AB) stages that are driven in
push-pull. The outputs are connected in parallel. And to make things
even more linear, let each stage have a resistive load. Each stage
will produce a linearly amplified (but inverted) version of the input
signal FOR THE POSITIVE HALF of the driving waveform only. Being
driven 180 degrees out of phase with the input signal, the second
stage will produce a linearly amplified but (again inverted) version
of the input signal FOR THE NEGATIVE HALF of the driving waveform.
Both outputs will have a DC offset of the plate (collector, drain)
voltage.
Class B or even Class AB in the circuit you described are non-linear.
Try that circuit
with Class A biasing.

Bill K7NOM


The resultant waveform with the outputs in parallel will look like
much like a full wave rectified version of the input signal subtracted
from the plate voltage. To express this mathematically, let the input
signal be expressed as:

Vin = A sin(wt)

Now let the voltage gain of each stage be "-k" and the plate voltage
be "B". The resultant waveform of the two stages connected in
parallel will be:

Vout = B - abs[A*k sin(wt)] where "abs" is the absolute value

Vout = B - A*k sin(wt) for 0 wt Pi and
= B + A*k sin(wt) for Pi wt 2Pi or
alternately for -Pi wt 0

We can then calculate the Fourier series of this waveform to determine
its spectrum. I will not present the calculations here as it is too
difficult to show the integration over defined integrals using only
plain text (and I doubt many readers will have math fonts anyway). If
you wish to see the math for the Fourier series for a number of
functions, read:
http://www.maths.qmul.ac.uk/~agp/calc3/notes2.pdf
http://www.maths.qmul.ac.uk/%7Eagp/calc3/notes2.pdfhttp://www.physics.hku.hk/%7Ephys2325/notes/chap7.doc or
_http://www.physics.hku.hk/~phys2325/notes/chap7.doc
http://www.physics.hku.hk/%7Ephys2325/notes/chap7.doc._
Vout = B - 2*A*K/Pi * [1 - SUMMATION {2*cos(nwt)/(n*n - 1)] for n=2,
4, 6, 8...

Note that the original frequency has been eliminated and that only
even order harmonics are present, and that the amplitudes drop off
quite rapidly. For example, the fourth harmonic will be one fifth of
the second harmonic.

For those that need a simplified explanation of Fourier series, Don
Lancaster wrote a good article that can be found at:
http://www.tinaja.com/glib/muse90.pdf. I always thought Don had a ham
license but I could not find one.

In a real implementation of this multiplier, a tuned circuit would be
used as the plate load. The Q of this tuned circuit will assure that
only the second harmonic is present in the output. The two stages
would need to be well balanced if cancellation of odd harmonics and
the fundamental is required.

73, Dr. Barry L. Ornitz WA4VZQ


POSTSCRIPT:

Now let me describe how it is possible to produce ONLY the second
harmonic. Instead of using two Class B or AB stages, it is possible
to use triodes operating where their plate current is proportional to
the square of the grid voltage. Driving the two such stages in
push-pull with the outputs in parallel with a resistive load, the
output waveform will be:

Vout = B - A*A*k sin(wt)*sin(wt)

Using a trigonometric identity {see:
http://en.wikipedia.org/wiki/List_of_trigonometric_identities},

sin(x)*sin(x) = sin(x)^2 = 0.5[1-cos(2x)]

thus Vout = B - A*A*K/2 + A*A*k/2 cos(2wt)

This shows that only the second harmonic is found at the output.

"Bill Janssen" wrote in message
...
NoSPAM wrote:
"Telstar Electronics"
wrote in message
...
On Nov 22, 8:43 pm, Stray Dog
wrote:
? Despite what at least one other person responding to this said, I can
rest
assure you that if you run a doubler/multiplier stage even in a
linear
mode, AND if you tune the output of that stage to the multiple
harmonic,
you will definitely get output at that harmonic frequency which is
stronger than the input drive voltage.

Huh? No way... you MUST have non-linearities to make a doubler.
Actually you do not need any nonlinearity to make a doubler
(quadrupler, etc.).
Assume you have two Class B (or AB) stages that are driven in
push-pull. The outputs are connected in parallel. And to make things
even more linear, let each stage have a resistive load. Each stage will
produce a linearly amplified (but inverted) version of the input signal
FOR THE POSITIVE HALF of the driving waveform only. Being driven 180
degrees out of phase with the input signal, the second stage will
produce a linearly amplified but (again inverted) version of the input
signal FOR THE NEGATIVE HALF of the driving waveform. Both outputs will
have a DC offset of the plate (collector, drain) voltage.
Class B or even Class AB in the circuit you described are non-linear. Try
that circuit
with Class A biasing.

Bill K7NOM


All that is really required is that the active devices have a different
gain with positive input signals than with negative input signals. This is
easily achieved with Class B and Class AB stages. As long as both stages
are identical the fundamental and odd order harmonics will cancel. You are
correct that with two Class A stages where the gain is identical for either
polarity of input, the output signal will perfectly cancel. To make the
method work here, you could synchronously switch the input signal between
two perfectly linear stages. My point was that a full-wave rectified
signal contains only even order harmonics.

In the real world, as Stray Dog pointed out, ALL amplifier stages are
nonlinear to some degree. The reason that Class AB and B amplifiers are
considered linear RF amplifiers is that the tuned circuit on the output
supplies supplies the "missing half" of the waveform. Without the tuned
circuit, harmonics of the 2nd, 4th, 6th, etc. order as well as the
fundamental are present. Odd order harmonics are only found if the gain is
nonlinear for positive input signals. The tuned output stage passes the
fundamental and suppresses the harmonics.

Thanks for pointing this out, Bill.

73, Barry WA4VZQ

On Dec 14, 10:27*pm, "NoSPAM" wrote:
*Actually you do not need any nonlinearity to make a doubler
(quadrupler, etc.).

You mean to tell me that you take a clean sine wave... pass it
through... say a single-ended class A amp... and you can put a tank on
the output of that amplifier... and tune for a harmonic? You will get
nothing.

"Telstar Electronics" wrote in message
...
You mean to tell me that you take a clean sine wave... pass it
through... say a single-ended class A amp... and you can put a tank on
the output of that amplifier... and tune for a harmonic? You will get
nothing.

Class A means that plate current is flowing throughout the entire cycle of
the input wave with the tube operated between cutoff and saturation. It
says nothing about the linearity of the tube's transconductance (plate
current as a function of grid voltage). With real devices, the
transconductance curve is ALWAYS nonlinear to some degree, producing
distortion (and harmonics). As you decrease the drive to a single-ended
Class A amplifier, you are working on a smaller and smaller portion portion
of the transconductance curve which decreases distortion. In the limit
where only an infinitesimal part of the transconductance curve is used, you
will get no distortion and no harmonics. Of course, in this situation the
tube produces NO output.while drawing current from the power supply.

The scheme that I was talking about, known as a push-push doubler,
generally uses the tubes operated in Class B although AB operation will
work too, but it produces less harmonics. The real advantage of a
push-push doubler is that odd order harmonics and the fundamental cancel
out, making the resultant waveform easier to filter.

73, Barry WA4VZQ

On Mon, 15 Dec 2008, Telstar Electronics wrote:

Date: Mon, 15 Dec 2008 06:16:29 -0800 (PST)
From: Telstar Electronics
Newsgroups: rec.radio.amateur.homebrew
Subject: Doubling

On Dec 14, 10:27*pm, "NoSPAM" wrote:
*Actually you do not need any nonlinearity to make a doubler
(quadrupler, etc.).

You mean to tell me that you take a clean sine wave...

You might want to consider qualifying your thinking on this by setting a
specification for harmonic distortion (in other words, you might need to
consider how much of that "clean sine wave" signal has other components
in it, including non-harmonic componentes)

pass it
through... say a single-ended class A amp...

You might also want to consider, here, too, how much harmonic distortion
THAT class A amplifier also causes which makes a contribution to the
output.

and you can put a tank on
the output of that amplifier... and tune for a harmonic? You will get
nothing.

You might even more also want to consider that any tuned circuit will pass
energy not at the resonance of that tuned circuit.

You would probably contribute to your own enlightenment if you actually
did some real experiments on this. It does not take long to do.

Back when I was an undergraduate student with major in physics (BS, 1966),
I worked in a Mossbauer Effect spectrometer lab and we built most of our
equipment (dual delay line pulse amplifiers, regulated DC power supplies,
repairing survey meters, etc) my boss had me build a waveform converter
that used a network of resistors and diodss to convert a sawtooth waveform
to sine wave and he was doing this because the book he got the circuit
from said that there would be less than 1% harmonic distortion and he was
interested in that specification for the spectrometer drives and all of
our commercial high quality signal generators were worse in that
specification, particulary at the very low frequencies we ran the drives
at (less than one cycle per second).

So, you have to define what you mean by "clean sine wave." But, I'll also
say that, no, you will not get nothing if you tune to the second harmonic
and have a linear amplifier (unless, maybe, you have a _perfect_ sine wave
and a _perfect_ linear amplifier [the rest of you guys might want to comment
on this yeah, I know about Fourier analysis, too]).

"Telstar Electronics" wrote in message
...

You mean to tell me that you take a clean sine wave... pass it
through... say a single-ended class A amp... and you can put a tank on
the output of that amplifier... and tune for a harmonic? You will get
nothing.

Of course you will. No active device is perfect.

I decided to illustrate the fact that a single ended triode operated in
Class A can produce harmonics. For a tube, I used a 6C4 (1/2 of a 12AU7)
operated with 300 volts on the plate, a grid bias voltage of -7 volts,
driven with a pure sine wave of 14 volts peak-to-peak. The high driving
voltage was chosen to illustrate my earlier points, but the stage _IS_
operated Class A with the plate current between cutoff and saturation.

Since the "rec"groups are not supposed to have binaries in them, I placed
the graphics as PDF attachments to a post entitled "Harmonics generated by
a Class A stage" in the "alt.binaries.ham-radio" newsgroup. If anyone
wishes to see these curves and their newsgroup provider does not provide
this group, I apologize. I believe Google Groups may not provide binaries,
so I suggest getting a real newsreader and a good newsfeed.

The first graph is entitled "Transconductance.pdf" and it shows the plate
current as a function of the grid voltage. This data was obtained directly
from the General Electric datasheet, ET-T1604 dated March, 1960. Since
Excel stinks when plotting and doing calculations with data that is not
best expressed in a bar chart, I used an evaluation copy of PSIPlot from
Poly Software International (http://www.polysoftware,com) to generate the
plots. {Real scientists and engineers never use a bar chart except when
making presentations to brain challenged management!} :-)

The driving waveform and the resultant plate current waveform are shown in
the graph entitled "Waveforms.pdf". The obvious flattening is due to
cutoff being approached at the crest of the driving waveform. After all,
the transconductance curve is not perfectly a straight line.

Finally, the spectrum of current waveform is plotted in the graph called
"Spectrum.pdf". The spectrum has been normalized with respect to the DC
output. The scale of the X-axis is slightly off but it was not worth my
time correcting it. The fundamental is about 60 to 70 percent of the DC
output, and the second harmonic is about 40 percent of the DC output. All
higher harmonic are less than one percent of the DC output except the
fifth. Higher harmonics are still greater than one tenth of a percent of
the DC up to the _13th_ harmonic. Harmonics beyond the 14th are still
readily measured.

In conclusion, even single ended Class A amplifiers generate harmonics.
If a lower driving voltage were used, the amplitudes of the harmonics would
be reduced, but the fundamental would also be reduced. Please follow-up to
the "rec.radio.amateur.homebrew" newsgroup. Golden-eared audiophools will
be ignored.

73, Dr. Barry L. Ornitz WA4VZQ