gwhite wrote:
Kevin Aylward wrote:

Don Pearce wrote:
On Mon, 01 Sep 2003 14:16:51 +0100, Paul Burridge
wrote:


Is a tuned load (tank circuit) a viable load for an RF amplifier
operating in class A? Or is this type of load only really suitable
for class C?

Of course you can use a tuned load with class A. But the nice thing
about a tuned load is that you don't *have* to use class A to
achieve a clean output. Of course, if you are using an amplitude
modulated signal, then you will need class A.

That's what receivers would use since power consumption doesn't matter
much. It would be true for low level TX'er stages too, again because
of power consumption. But for things like "Ham linears" it is not
necessarily true.

That is, if the input signal is already AM, you need a class A or
linear amplifier.

Let's be specific regarding the your word "linear" since "or" implies
something else other than class-A may be possible.

Yes.

Single-ended
"linear" narrow band amps may be biased class-B if the tank has a high
enough Q. The missing half cycle is restored by the so-called
"flywheel effect." In practice deep AB is also used.



If the AM is to be done at that stage itself, then its the
class c non-linear bit that makes the multiplication modulation work.

No.

No. Its Yes. Non-linear action generates the multiplication products. I
was not drawing any real distinction between class c and b in this
context. The practical difference is minimal. They both do not amplifier
the waveform in a linear manner. I was not meaning to infer that it was
an "only" c. I was referring to the fact that you need at least some
method that generates non linarity.

It can be done with class-A or class-B since the assumption is
that the amp will be driven (or nearly so) to the rails by the carrier
alone.

I agree, this is another method of generating x-product multiplication
terms. However, arguable, a class A amplifier is not really a class A
amplifier if it is driven to saturation. Its a really a switching amp
or, a pulse amplitude modulator if its rails are varying.

Kevin Aylward

http://www.anasoft.co.uk
SuperSpice, a very affordable Mixed-Mode
Windows Simulator with Schematic Capture,
Waveform Display, FFT's and Filter Design.

Kevin Aylward wrote:

gwhite wrote:
Kevin Aylward wrote:

Don Pearce wrote:
On Mon, 01 Sep 2003 14:16:51 +0100, Paul Burridge
wrote:


Is a tuned load (tank circuit) a viable load for an RF amplifier
operating in class A? Or is this type of load only really suitable
for class C?

Of course you can use a tuned load with class A. But the nice thing
about a tuned load is that you don't *have* to use class A to
achieve a clean output. Of course, if you are using an amplitude
modulated signal, then you will need class A.

That's what receivers would use since power consumption doesn't matter
much. It would be true for low level TX'er stages too, again because
of power consumption. But for things like "Ham linears" it is not
necessarily true.

That is, if the input signal is already AM, you need a class A or
linear amplifier.

Let's be specific regarding the your word "linear" since "or" implies
something else other than class-A may be possible.

Yes.

Single-ended
"linear" narrow band amps may be biased class-B if the tank has a high
enough Q. The missing half cycle is restored by the so-called
"flywheel effect." In practice deep AB is also used.


If the AM is to be done at that stage itself, then its the
class c non-linear bit that makes the multiplication modulation work.

No.

No. Its Yes. Non-linear action generates the multiplication products. I
was not drawing any real distinction between class c and b in this
context. The practical difference is minimal. They both do not amplifier
the waveform in a linear manner. I was not meaning to infer that it was
an "only" c. I was referring to the fact that you need at least some
method that generates non linarity.

It can be done with class-A or class-B since the assumption is
that the amp will be driven (or nearly so) to the rails by the carrier
alone.

I agree, this is another method of generating x-product multiplication
terms. However, arguable, a class A amplifier is not really a class A
amplifier if it is driven to saturation. Its a really a switching amp
or, a pulse amplitude modulator if its rails are varying.


Here's a "class-A" amplifier that can be amplitude modulated but yet not
saturated (assumes a constant load R):

V+
|
LC Tank
| AM RF_out
+--||--O
Carrier |
RF_in c
O--||---b Class A biased (no base bias details)
e
|
|
RF_Choke
Audio |
in c
O--||-- b Class A modulator (no base bias details)
e
|
GND


This is a single ended amplitude modulator. The top transistor could be
driven to the switch mode by the carrier, but this is not necessary to
produce AM. Practically, it will be driven to the switch mode for
efficiency reasons.

Amplitude modulation can be "made" via linear methods. "Multipliers"
cannot be generally stated to be either linear or non-linear. A system
which includes a multiplier must be put through the linearity test to
see if the configuration is linear or non-linear. IOW, it can be
either.

+------+
x(t) O---| h(t) |---O y(t)
+------+


linearity:

a·x1(t) = a·y1(t)
b·x2(t) = b·y2(t)

if x(t) = a·x1(t) + b·x2(t)

then

y(t) = a·y1(t) + b·y2(t)

If that is true, then the system is linear. This can be true for
systems with multipliers.


This system is linear and has a multiplier (it is not time invariant):

The System
+---------------+
| |
in | /¯¯¯\ | out
x(t) O--------( X )---------O y(t)
| \___/ |
| | |
| | |
| O |
| cos(w_c·t) |
+---------------+


It produces a DSB signal (y(t)). w_c·t could be "added in later"
(linearly) to y(t) in the proper amplitude and phase and the resultant
signal would for all practical purposes be indistiguishable from
standard AM. No non-linear circuit was used but yet AM was produced.
Not convenient, but it does dispel the "non-linearity is required"
myth. Also, a multiplier can be viewed as a MISO system.

gwhite wrote:
Kevin Aylward wrote:

gwhite wrote:
Kevin Aylward wrote:


It can be done with class-A or class-B since the assumption is
that the amp will be driven (or nearly so) to the rails by the
carrier alone.

I agree, this is another method of generating x-product
multiplication terms. However, arguable, a class A amplifier is not
really a class A amplifier if it is driven to saturation. Its a
really a switching amp or, a pulse amplitude modulator if its rails
are varying.


Here's a "class-A" amplifier that can be amplitude modulated but yet
not saturated (assumes a constant load R):

Ho hummm...


V+
|
LC Tank
| AM RF_out
+--||--O
Carrier |
RF_in c
O--||---b Class A biased (no base bias details)
e
|
|
RF_Choke
Audio |
in c
O--||-- b Class A modulator (no base bias details)
e
|
GND


This is a single ended amplitude modulator. The top transistor could
be driven to the switch mode by the carrier, but this is not
necessary to produce AM. Practically, it will be driven to the
switch mode for efficiency reasons.

And you think that this is me to me? I suppose you aint read many of my
10,000+ posts.

Err..you've missed the cap from the top transistor emitter to ground. If
the bottom transistor circuit was a true current source at rf, the top
transistor could not effect the output current at all.

Oh, and its not very linear with required to audio input signal without
an emitter resister anyway. The distortion of an basic tranister amp for
2nd Harmonic = Vi(mv)%, that is 10 mv will get you 10% distortion.


Amplitude modulation can be "made" via linear methods.

Nope. Not a chance.

"Multipliers"
cannot be generally stated to be either linear or non-linear.

If one input of a multiplier is held constant, the other input has a
linear response. If the other input is a function of time, the response
to the first input is non-linear. That is, it dose *not* satisfy a(f(t))
= f(at).

The above circuit relies on the fact that the "re" of the top
transistor, is a function of its emitter current set by the bottom
transistor, via the equation re=1/40.Ic. e.g. see
http://www.anasoft.co.uk/EE/index.html

A system which includes a multiplier must be put through the linearity
test to see if the configuration is linear or non-linear. IOW, it
can be either.

Ho humm again. You confuse where the term linearity is to be applied.


+------+
x(t) O---| h(t) |---O y(t)
+------+


linearity:

a·x1(t) = a·y1(t)
b·x2(t) = b·y2(t)

if x(t) = a·x1(t) + b·x2(t)

then

y(t) = a·y1(t) + b·y2(t)

Linearity can more easily be expressed as:

a(f(t)) = f(at)


If that is true, then the system is linear. This can be true for
systems with multipliers.

Nope. A signal being acted on by a multiplier is a non-linear system if
the second input is non constant with time. Your way of base on this
one.



This system is linear and has a multiplier (it is not time invariant):


Nope, its not.

The System
+---------------+
| |
in | /¯¯¯\ | out
x(t) O--------( X )---------O y(t)
| \___/ |
| | |
| | |
| O |
| cos(w_c·t) |
+---------------+


It produces a DSB signal (y(t)). w_c·t could be "added in later"
(linearly) to y(t) in the proper amplitude and phase and the resultant
signal would for all practical purposes be indistiguishable from
standard AM.

This is not a linear circuit. You need to understand what linear means.
A linear system, cannot produce frequencies that are not in the input,
essentially, by definition. With all due respect, I would guess you
don't have an EE B.S. degree. This is all pretty basic stuff really.

No non-linear circuit was used but yet AM was produced.

Nonsense. Your pretty misguided on this. You can not achieve
multiplication without a non-linear circuit. For example, Gilbert
multipliers use the fact that Id=Is.exp(vd/Vt). That is it logs, adds
and antilog. Balanced switching mixers use switches. Fet mixers use
their square law response.

Not convenient, but it does dispel the "non-linearity is required"
myth.

Its not a myth. I know of no way whatsoever to generate an analogue
multiplication x product terms without having a device satisfying the
property of a.f(t) != f(at), i.e. a non-linear device. Please feel free
to suggest one, but file your patent first.

Oh, by the way...your trying to teach your granny to such eggs son.

Kevin Aylward

http://www.anasoft.co.uk
SuperSpice, a very affordable Mixed-Mode
Windows Simulator with Schematic Capture,
Waveform Display, FFT's and Filter Design.

Kevin Aylward wrote:


Oh, by the way...your trying to teach your granny to such eggs son.

Should have been:

Oh, by the way...your trying to teach your granny to suck eggs son.

Kevin Aylward

http://www.anasoft.co.uk
SuperSpice, a very affordable Mixed-Mode
Windows Simulator with Schematic Capture,
Waveform Display, FFT's and Filter Design.

If you're really interested in getting it right, it should have been:

Oh, by the way...you're trying to teach your granny to suck eggs son.

"You're" is a contraction for "you are".
"Your" is an adjective, meaning "of or relating to you or yourself".

And a comma would have been appropriate between "eggs" and "son", making
it better yet if it had read:

Oh, by the way...you're trying to teach your granny to suck eggs, son.

Roy Lewallen, W7EL

Kevin Aylward wrote:
Kevin Aylward wrote:


Oh, by the way...your trying to teach your granny to such eggs son.


Should have been:

Oh, by the way...your trying to teach your granny to suck eggs son.

Kevin Aylward

http://www.anasoft.co.uk
SuperSpice, a very affordable Mixed-Mode
Windows Simulator with Schematic Capture,
Waveform Display, FFT's and Filter Design.

Roy Lewallen wrote:
If you're really interested in getting it right, it should have been:

Oh, by the way...you're trying to teach your granny to suck eggs
son.

"You're" is a contraction for "you are".
"Your" is an adjective, meaning "of or relating to you or yourself".

And a comma would have been appropriate between "eggs" and "son",
making it better yet if it had read:

Oh, by the way...you're trying to teach your granny to suck eggs,
son.

Roy Lewallen, W7EL

I stand corrected, but maybe "sonny boy" may have made the point
stronger?

Kevin Aylward

http://www.anasoft.co.uk
SuperSpice, a very affordable Mixed-Mode
Windows Simulator with Schematic Capture,
Waveform Display, FFT's and Filter Design.

Roy Lewallen wrote:
If you're really interested in getting it right, it should have been:

Oh, by the way...you're trying to teach your granny to suck eggs
son.

"You're" is a contraction for "you are".
"Your" is an adjective, meaning "of or relating to you or yourself".

And a comma would have been appropriate between "eggs" and "son",
making it better yet if it had read:

Oh, by the way...you're trying to teach your granny to suck eggs,
son.

Roy Lewallen, W7EL

I stand corrected, but maybe "sonny boy" may have made the point
stronger?

Kevin Aylward

http://www.anasoft.co.uk
SuperSpice, a very affordable Mixed-Mode
Windows Simulator with Schematic Capture,
Waveform Display, FFT's and Filter Design.

If you're really interested in getting it right, it should have been:

Oh, by the way...you're trying to teach your granny to suck eggs son.

"You're" is a contraction for "you are".
"Your" is an adjective, meaning "of or relating to you or yourself".

And a comma would have been appropriate between "eggs" and "son", making
it better yet if it had read:

Oh, by the way...you're trying to teach your granny to suck eggs, son.

Roy Lewallen, W7EL

Kevin Aylward wrote:
Kevin Aylward wrote:


Oh, by the way...your trying to teach your granny to such eggs son.


Should have been:

Oh, by the way...your trying to teach your granny to suck eggs son.

Kevin Aylward

http://www.anasoft.co.uk
SuperSpice, a very affordable Mixed-Mode
Windows Simulator with Schematic Capture,
Waveform Display, FFT's and Filter Design.

Kevin Aylward wrote:


Oh, by the way...your trying to teach your granny to such eggs son.

Should have been:

Oh, by the way...your trying to teach your granny to suck eggs son.

Kevin Aylward

http://www.anasoft.co.uk
SuperSpice, a very affordable Mixed-Mode
Windows Simulator with Schematic Capture,
Waveform Display, FFT's and Filter Design.

Kevin Aylward wrote:

gwhite wrote:
Kevin Aylward wrote:

gwhite wrote:
Kevin Aylward wrote:


It can be done with class-A or class-B since the assumption is
that the amp will be driven (or nearly so) to the rails by the
carrier alone.

I agree, this is another method of generating x-product
multiplication terms. However, arguable, a class A amplifier is not
really a class A amplifier if it is driven to saturation. Its a
really a switching amp or, a pulse amplitude modulator if its rails
are varying.


Here's a "class-A" amplifier that can be amplitude modulated but yet
not saturated (assumes a constant load R):

Ho hummm...


V+
|
LC Tank
| AM RF_out
+--||--O
Carrier |
RF_in c
O--||---b Class A biased (no base bias details)
e
|
|
RF_Choke
Audio |
in c
O--||-- b Class A modulator (no base bias details)
e
|
GND


This is a single ended amplitude modulator. The top transistor could
be driven to the switch mode by the carrier, but this is not
necessary to produce AM. Practically, it will be driven to the
switch mode for efficiency reasons.

And you think that this is me to me? I suppose you aint read many of my
10,000+ posts.

I'm pretty sure I won't bother at this point.

Err..you've missed the cap from the top transistor emitter to ground. If
the bottom transistor circuit was a true current source at rf, the top
transistor could not effect the output current at all.

True. It should be there.

Oh, and its not very linear with required to audio input signal without
an emitter resister anyway. The distortion of an basic tranister amp for
2nd Harmonic = Vi(mv)%, that is 10 mv will get you 10% distortion.

The point wasn't to design a perfect amp. The point was to illustrate
you're wrong.

Amplitude modulation can be "made" via linear methods.

Nope. Not a chance.

Oh, but it can.

"Multipliers"
cannot be generally stated to be either linear or non-linear.

If one input of a multiplier is held constant, the other input has a
linear response. If the other input is a function of time, the response
to the first input is non-linear. That is, it dose *not* satisfy a(f(t))
= f(at).

No, false, or whatever negation pleases you best. You were already
given the answer. You confuse time-invariance with linearity. You need
not take my word for it. Consult any Signals and Sytems text, any
Linear Systems text, or any Communications text. IIRC, the following
was a homework problem in Stremler's text:

Determine linearity
Determine time-invariance

The System
+---------------+
| |
in | /¯¯¯\ | out
x(t) O--------( X )---------O y(t)
| \___/ |
| | |
| | |
| O |
| cos(w_c·t) |
+---------------+

http://www.amazon.com/exec/obidos/tg...glance&s=books

I suppose if you want to make up your own definition of linearity, you
can get whatever anwswer you wish.


A system which includes a multiplier must be put through the linearity
test to see if the configuration is linear or non-linear. IOW, it
can be either.

Ho humm again. You confuse where the term linearity is to be applied.

I defined the system quite clearly. I mean, I don't think it can be
made simpler.

You can stick with the formal definition given in pretty much every
Signals and Sytems text, any Linear Systems text, or any Communications
text, or you can make up your own and get the answer that pleases you.


+------+
x(t) O---| h(t) |---O y(t)
+------+


linearity:

a·x1(t) = a·y1(t)
b·x2(t) = b·y2(t)

if x(t) = a·x1(t) + b·x2(t)

then

y(t) = a·y1(t) + b·y2(t)

Linearity can more easily be expressed as:

a(f(t)) = f(at)

Except that isn't "the" definition (hey, but it is true if a = 1). It
doesn't even meet you own description:
"A linear system, cannot produce frequencies that are not in the input,
essentially, by definition." -- Kevin Aylward

I would like to see you apply this "definition."

If that is true, then the system is linear. This can be true for
systems with multipliers.

Nope. A signal being acted on by a multiplier is a non-linear system if
the second input is non constant with time. Your way of base on this
one.

Again, you confuse linearity and time-invariance.

This system is linear and has a multiplier (it is not time invariant):


Nope, its not.

Oh, but it is. Some rather trivial math can lead you to understand what
linear means.

The System
+---------------+
| |
in | /¯¯¯\ | out
x(t) O--------( X )---------O y(t)
| \___/ |
| | |
| | |
| O |
| cos(w_c·t) |
+---------------+


It produces a DSB signal (y(t)). w_c·t could be "added in later"
(linearly) to y(t) in the proper amplitude and phase and the resultant
signal would for all practical purposes be indistiguishable from
standard AM.

This is not a linear circuit. You need to understand what linear means.
A linear system, cannot produce frequencies that are not in the input,
essentially, by definition.

You don't know what linear means. There it is.

With all due respect, I would guess you
don't have an EE B.S. degree.

Cut the chest puffing.

This is all pretty basic stuff really.

On that much we certainly agree, Comm101 ought to do it.

No non-linear circuit was used but yet AM was produced.

Nonsense. Your pretty misguided on this.

Well I won't take you word for it, and you need not take mine. You
could make it easy just by applying the linearity test. That is,
linearity as it is defined in every Signals and Sytems text, any Linear
Systems text, or any Communications text. I'm not as "original" as you,
I simply trust the guys who wrote the books.

You can not achieve
multiplication without a non-linear circuit.

How about this one:

The System
+---------------+
| |
in | /¯¯¯\ | out
x(t) O--------( X )---------O y(t)
| \___/ |
| | |
| | |
| O |
| 2 |
+---------------+

It looks functionally to be an amp with a gain of two. Is a "gain of 2"
circuit non-linear? Isn't that a multiplier in there?

For example, Gilbert
multipliers use the fact that Id=Is.exp(vd/Vt).

I think that descibes pretty much every bipolar. So I guess transistor
amps cannot be made linear, or at least function sufficiently linear for
the purpose of electronic designers. That is an "interesting"
contention.

That is it logs, adds
and antilog. Balanced switching mixers use switches. Fet mixers use
their square law response.

Not convenient, but it does dispel the "non-linearity is required"
myth.

Its not a myth. I know of no way whatsoever to generate an analogue
multiplication x product terms without having a device satisfying the
property of a.f(t) != f(at), i.e. a non-linear device. Please feel free
to suggest one, but file your patent first.

Why not dispense with the snidery, and simply prove your contention by
applying the linearity test to this one:

The System
+---------------+
| |
in | /¯¯¯\ | out
x(t) O--------( X )---------O y(t)
| \___/ |
| | |
| | |
| O |
| cos(w_c·t) |
+---------------+

Hey, I'll even give you a hint.

Let x1(t) = a1·cos(w1·t)
x2(t) = a2·cos(w2·t)

Then y1(t) = a1·cos([wc + w1]·t)/2 + a1·cos([wc - w1]·t)/2
y2(t) = a2·cos([wc + w1]·t)/2 + a2·cos([wc - w1]·t)/2

?
If x(t) = x1(t) + x2(t) does y(t) = y1(t) + y2(t)
(This *is* the linearity test.)

?
[x1(t) + x2(t)]·cos(wc·t) = x1(t)·cos(wc·t) + x2(t)·cos(wc·t) = y(t)

The answer is yes: the system is linear, by the linearity test.

If x(t) = a·cos(w·t) is applied to the system,
and produces y(t), does x(t-to) applied to the
system result in y(t) except all the t's are
replaced by t-to? (May all the t's in
the y(t) case be replaced by t-to, and
have the system response reflect that the input
was x(t-to).) (This is the time-invariance test.)

y(t) = x(t)·cos(wc·t) = a·cos(w·t)·cos(wc·t)
= a·cos([wc + w]·t)/2 + a·cos([wc - w]·t)/2

Now put the "to" into the input and see what
comes out

What? = x(t-to)·cos(wc·t) = a·cos(w·[t-to])·cos(wc·t)
= a·cos([wc·t + w·[t-to])/2 + a·cos([wc·t - w·[t-to])/2
| |
+---------------------------+
not t-to,

So the system is *not* time-invariant.


This is Comm101 stuff, if you're ever interested enough to crack a book
open, and actually work the problems.


Oh, by the way...your trying to teach your granny to such eggs son.

You're wrong about that one too: we have no relationship. I hope that
comforts you, as I know it does me.