George, electrical fields add by the law of superposition;
they do not multiply. You suggest that the combined time varying
voltage resulting from superposition at the input gate nodes
produce a variation
over time in the source drain current which results as the
current passes through a resistor
in the time varying product of the voltages on these input
gates.
And that the filters unlike the diagrams in my text do not use
inductors where a flat response over a specific range of
frequencies
is desired as in this radio astronomy case. Of that sounds
reasonable.
But that muliplication occurs physically and that the filters
pick out
the terms of an equivalent sum is not reasonable. You simply
state this
in mathematical notation and links that dont explain anything.
I would accept your explanation if you could explain why the
law
of superposition does not apply here and how other filters
at the output can produce each of the two input gate frequencies
and
how electric fields multiply.
Ralph


"George Dishman" wrote in message
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"ralph sansbury" wrote in message
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"George Dishman" wrote in message
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"ralph sansbury" wrote in message
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I still dont understand what you mean by the
multiplication
of voltages arriving at the two gates of dual gate
transistor:

You produce the sum and difference frequencies by
making use of this identity:

sin(a) * cos(b) = [ sin(a+b) + sin(a-b) ] / 2

The right hand side contains the sum and difference
so is the output from the circuit. The function needed
on the right hand side is multiplication.

You mean the left hand side.

Doh! Yes, the left hand side is the multiplication.

Yes the summation is equal to
the product. And it is the summation that is produced by
adding
the voltages at the dual gate transistor along with the input
frequencies.

No. Here is a conceptual block diagram:

multiplier filter(s)
+---+
a(t) ----| | +---+
| * |---------| ~ |-- d(t)
b(t) ----| | c(t) +---+
+---+

The time-varying voltages a(t) and b(t) are multiplied
together to produce voltage c(t) = 2 * a(t) * b(t).
(The factor of 2 makes the text easier to read later.)

If a and b are sine waves with angular frequencies w_a and w_b:

a(t) = sin(w_a * t)
b(t) = cos(w_b * t)

then c(t) can also be expressed as

c(t) = sin((w_a+w_b) * t) + sin((w_a-w_b) * t)

The filter rejects the sin((w_a+w_b) * t) term so you are left
with only the difference frequency sin((w_a-w_b) * t)

Now in the real system, one signal is a pure sine wave while
the other is the continuum of frequencies 110MHz wide but
the same analysis applies to each component so you get a
continuum out of the filter but shifted down the spectrum
by the reference frequency.

The same
method works when using a wideband signal where each
component is shifted in frequency by the same amount.


My sense of this is that the sum of the voltages produces
a pattern
which contains a frequency which is the difference
frequency plus
the sum frequency plus the two input frequencies and that
the
filters in the special circuits you refer to produce
these
separate components???.

If you replace "the sum of the voltages" in the first
line by "the product of the voltages", the paragraph
is perfectly correct. The filters are conceptually
separate from the mixer but often merged in practice.

I think you are overlooking the basic physics here. You
have no
reason
for the filters at the output of the dual gate transistor
etc, and
which are an integral 'part of the mixer but conceptually
separate',
unless it
is to take the oscillation of voltage which results from the
addition of
the
separate oscillating voltages and to produce the different
component
frequencies as
different outputs. The difference frequency of oscillating
voltage is then
taken as the desired output.

Again, that is all correct except "which results from the
addition
of the" should read "which results from the multiplication of
the"

Of course the mathematical equivalence between the
specified product
and the specified sum makes it mathematically correct to say
that the
black box
circuit produces a product of the two input frequencies as
well
as the two input frequencies.

Wrong way round. You have to actually multiply in order to
create something equivalent the sum and difference.

This is adequate for engineers once the circuit has been
designed. But the first designers of the
circuit had to know the physics

I am a designer and have an honours degree in physics,
how about you?

and to put in the inductors and
capacitors in a resonant configuration




Yet again: it cannot be a resonant configuration if you want
all the band to get through without distortion, it must have
a flat passband which is easiest to envisage conceptually as
separate high and low pass filters.

etc to get the desired output ie.
to design around the problem that voltages at the input can
only
add.
Thus it is not quite physically correct to say that the
transistor multiplies the
input voltages

Sorry Ralph, it is physically accurate. Read your text
book on how a field effect transistor works and find
out the equation that relates "Id", the current from
drain to source to "Vgs", the voltage between gate and
source, in the saturation region. If it doesn't cover it
there are pages below.

and more correct to say that the transistor adds the incoming
voltages
such that
the resulting pattern can be written as the sum of the input
frequencies
and the
difference frequency and the sum frequency and that the
latter two
frequencies
are mathematically equivalent to the product of the
frequencies
etc.

Nope, you need to do some homework. You may be able
to find better than these pages but they will do for
a start. FETs are normally run in saturation which
is to the right of the dotted line on the graph:

http://ece-www.colorado.edu/~bart/bo...ter7/ch7_2.htm

with a more detailed analysis on the next page:


http://ece-www.colorado.edu/~bart/bo...7/ch7_3.htm#7_
3_2

Here's another, sorry it is in word document format but
it gives a brief look at the physics:

http://www.eng.abdn.ac.uk/~eng188/EG1567/JW-08-FETs.doc

The key equation is in the "Characteristic Equations" on page
5:

ID ~ (Idss/Vt^2)/(Vgs - Vt)^2 = Idss[(Vgs/Vt)-1]^2

George