In article , Patrick Turner
wrote:

John Byrns wrote:

In article , Patrick Turner
wrote:

Damping reduces Q, and increases BW.
But it also reduces Z at Fo, thus reducing gain in an amp
which must be a current source, like a pentode or j-fet,
to realise the best selectivity for the LC circuit.

This is a half truth, what matters is that the filter is correctly
terminated, not that pentode, triode or whatever drives it.

Well most IFTs made for tube radios would perform abysmally is driven
with triode amps with Ra = say 10k.
This would over damp the LC circuit in most cases.

That is true of "IFTs made for tube radios" when they are being used as
originally intended, but what happens when we hobbyists modify them for
High-Fidelity use by increasing "k" and decreasing the circuit "Q" by
adding resistors? In this case since we need external termination
resistors anyway, all we need do is connect the resistor between the anode
of the triode and the input of the IFT and all will be well, there is no
need for a current source to drive the filter, ideal would be a source
with just the required termination resistance. There is no reason why the
required termination resistance can't be connected between a low impedance
source and the input of the filter, it does not have to have one end
earthed.

Try damping an ordinary radio's IFTs with 100K, then 47k and finally 22k
for each of the 4 IF coils in a set.
Tell me what you find.

Every filter, be it an IFT or something more complex, is designed to be
terminated in specified impedances, which may be a specified resistance,
an open circuit, or even a short circuit, what matters is that the
termination is correct, not that the filter is driven by a pentode.

As far as
stage gain goes, increasing the frequency from 455 kHz to 2.0 MHz is
likely to decrease the gain by a similar amount to widening the 455 kHz
filter to the same bandwidth as the 2.0 MHz filter.

Use 3 x IFTs, and an extra stage of IF amplification.
I still reckon the 2MHz will work, and when I have time,
I'll try the idea, and tell everyone about it.

But to know any earlier, try it out for yourself.

I never said 2.0 MHz wouldn't work, in fact I specifically stated at least
once that I thought 2.0 MHz would work. If 2.0 MHz is what floats your
boat then that's what you should use, although I notice that you choose to
use the traditional 455 kHz in your radio design. What I said was simply
that 455 kHz would also work in a wideband High-Fidelity AM radio.

What I said was what I said.
You are confused.

Maybe, in what way are you suggesting I am confused? I would suggest to
you that you don't understand how to design an IF filter, and don't
understand what can be done at 455 kHz.

I know enough about IFT design, after having built my own radio.

That isn't clear at all, you seem to be obsessed with "Q", and hardly if
ever mention "k", and how it relates to "Q" in determining the
characteristics of an IFT. You occasionally mention "critical" coupling
but haven't tied that concept in with the "Q" and "k" of an IFT, nor have
you mentioned the related concept of "transitional" coupling. I would
expect to hear more mention of these concepts from someone who knows
"enough about IFT design".

I don't need to use k to confuse everyone.

An IFT is a simple RF transformer operating at a fixed F.
The magnetic lines of force from one coil react to transfer
some power from a primary LC to a secondary LC.

The coupling and insertion loss is whatever you are gonna get.
The looser the coupling, ie, the further apart the coils, the
sharper is the nose shape of the two circuits.
Let's assume you have a current source, ie, high impedance
signal source, or generator for the primary.
Assume the output from the sec goes to a high impedance load, like the grid
of a pentode tube, with little miller capacitance.
The load of the sec LC is transfered to the pri, depending on the closeness
of coupling.

Far apart gives a large insertion loss, and lowest RL for the pri signal
source,
but the response shows the attenuation
is twice that of a single LC as you move away from the centre F.
Then as you bring the coils together, the insertion loss and load value
reduces,
and the
response suddenly becomes flat topped, but the attenuation out of the
pass band is
still
twice that of a single circuit.
Then with coils even closer, the insertion loss is low, but there are
two peaks in
the response,
but outside the two peaks the response remains twice that of a single LC
circuit.

k isn't needed to be considered since we are dealing practically with
what you get
when you
use LC circuits arranged as they are in IFTs.
We simply wanna know what happens.

If you simply "wanna know what happens" why do you even need to consider
"Q"? Your narrative description above sure makes it sound like "k" is
important, you just haven't tied "Q" and "k" to the response shapes you
describe.


Regards,

John Byrns


Surf my web pages at, http://users.rcn.com/jbyrns/