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.