Maker Pro
Maker Pro

Dumb newbie has another question...

A

Andrew Holme

Jan 1, 1970
0
Dave said:
How can I increase the Q of a tank circuit? ANY ideas would be
appreciated...

Thanks,

Dave
[email protected]

Reduce the load on it:
1. increase load impedance;
2. change the tap point (inductive or capacitive tap).

Use higher quality (Q) components :)

Optimise the L/C ratio
 
B

Bob Eldred

Jan 1, 1970
0
Dave said:
How can I increase the Q of a tank circuit? ANY ideas would be
appreciated...

Thanks,

Dave
[email protected]

You can drive the tank with negative resistance to increase the Q. You do
this with an amplifier appropriately connected to provide positive feedback
sensing the current in the tank. A bridge circuit can be used to do this.
Care must be taken as the circuit can easily burst into oscillation if too
much positive feed back is applied. Needless to say, the bandwidth of the
amplifier must be great enough to cover the frequency of interest. Key
words: positive feedback, negative resistance, negative impedance
converters, NIC, gyrator, variable damping.
Bob
 
A

Andrew Holme

Jan 1, 1970
0
Bob said:
You can drive the tank with negative resistance to increase the Q.
You do this with an amplifier appropriately connected to provide
positive feedback sensing the current in the tank. A bridge circuit
can be used to do this. Care must be taken as the circuit can easily
burst into oscillation if too much positive feed back is applied.
Needless to say, the bandwidth of the amplifier must be great enough
to cover the frequency of interest. Key words: positive feedback,
negative resistance, negative impedance converters, NIC, gyrator,
variable damping.
Bob

Is that a "Q-multiplier" ?
 
D

Dave

Jan 1, 1970
0
no_one said:

In other words, higher quality components? Or are you thinking something
else. Really *am* new to RF...

Dave
db5151
 
D

Dave

Jan 1, 1970
0
Andrew Holme said:
Reduce the load on it:
1. increase load impedance;

Okay, I *think* I am working on that...

2. change the tap point (inductive or capacitive tap).

Use higher quality (Q) components :)

Optimise the L/C ratio
How would I optimise the LC ratio? Seems like I saw something about this in
one of my books, but now can't find it. What should I look up?


Thanks,

Dave
[email protected]
 
D

Dave

Jan 1, 1970
0
Bob Eldred said:
You can drive the tank with negative resistance to increase the Q. You do
this with an amplifier appropriately connected to provide positive feedback
sensing the current in the tank. A bridge circuit can be used to do this.
Care must be taken as the circuit can easily burst into oscillation if too
much positive feed back is applied. Needless to say, the bandwidth of the
amplifier must be great enough to cover the frequency of interest. Key
words: positive feedback, negative resistance, negative impedance
converters, NIC, gyrator, variable damping.
Bob

Negative resistance, like a tunnel diode? Maybe with it's own tank circuit?
(I don't *think* that would be hard to implement with my setup.) I am
working on a simple RF amplification stage that provides positive feedback,
could that help? Was going to use a Darlington pair of 2N3906's until I saw
if/how it worked, then rebuild it with a pair of NTE 10 low-noise VHF/UHF
amplifiers. Starting with 3906's cause that's what I have handy. Does that
sound like a decent idea? If so, should I just cut to the chase and go with
the NTE 10's?

Will have to look negative impedance converters. What is a gyrator? And how
could I use variable damping?

Sorry for all the questions. Total newbie. Thanks for the help.

Dave
[email protected]
 
A

Andrew Holme

Jan 1, 1970
0
Dave said:
How would I optimise the LC ratio? Seems like I saw something about
this in one of my books, but now can't find it. What should I look up?

There is stuff about this in the RSGB and ARRL handbooks.

Loaded Q = X / R = wL / R = 1 / wCR

Where w = omega (2*pi*f)

You can increase Q by increasing L and reducing C - but only to a point;
you're also increasing the losses in L.
 
K

Ken Smith

Jan 1, 1970
0
[QUOTE="Dave said:
You can drive the tank with negative resistance to increase the Q. You do
[...]
Negative resistance, like a tunnel diode?[/QUOTE]

Just to help explain:

At low frequencies, you can do this:


R1
----/\/\/\---------
! !
From -------+---------!+\ !
Tank ! >------+
--!-/ !
! !
+--/\/\/-----
! R2
\
/
\ R3
/
!
GND

The current in R1 is:

Vin * (R2/R3) / R1

If you assume +1V coming from the tank circuit and follow through the
circuit, you should see the basic idea of how an amplifier makes a
negitive resistance when it feeds back into the tuned circuit.

If you want to work at higher frequencies, you can't just an op-amp
because it doesn't make gain at those frequencies. We can instead use
just a transistor to do it. I suggest you think about this circuit:

VCC
!
+---!!---GND
Bias !
C1 ! !/
-------!!-----+--+-----! Q1
( ! !\e
( C2 --- !
( L1 --- !
! ! !
GND +-------+
! !
C3 --- (
--- ( L2
! ( RF choke
GND !
GND



C1, C2 and C3 in series form the capacitor of the tuned circuit.

Chances are L2 actually as a resistor in series with it. It is to pass
the DC of Q1 but not any of the RF stuff.


If Q1 has enough gain, this circuit is an oscillator. Q1's emitter looks
like a negitive resistance hooked onto C3. You can adjust the gain of Q1
by adjusting the current through it. Just below the point of oscillation,
the circuit will have a very high Q because the negitive resistance from
Q1 will cancel most of the resistances in the tuned circuit.



What is a gyrator? And how

A gyrator is any circuit that uses an amplifier to make what appears to be
an inductor or a capacitor.

Here's a dumb gyrator:



V1
-----+--------
! !
\ R1 !
/ !
\ !
! !!- d
+-----!! Q1 MOSFET
! !!- s
--- !
--- C1 /
! \ R2
GND /
\
!
GND


Assume V1 has been 10V for a long time. C1 will have 10V on it. R2 will
have lets say 5V on it (Vgs=5V). Some current will flow in R2 and through
Q1.

Now lets assume that V1 steps up to 20V. The voltage on C1 will rise
slowly towards 20V. The voltage on R2 will rise slowly towards something
near 15V and the current in R2 will as a result rise slowly. This circuit
looks like an inductor in that the current increase lags the voltage
increase but it contains no real inductors.

With op-amps, you can make large high Q inductors this way.
 
R

Rich Grise

Jan 1, 1970
0
In other words, higher quality components? Or are you thinking something
else. Really *am* new to RF...

So, draw a graph. Overlay a graph of Xc = 1/(2*pi*f*c) and Xl = 2*pi*f*l,
and draw a phasor diagram of them, along with R, whether it's in series
with C, in series with L, in parallel with both, in parallel to series RC,
whatever.

Nothing to it! ;-P

Cheers!
Rich
 
R

Rich Grise

Jan 1, 1970
0
Okay, I *think* I am working on that...

How would I optimise the LC ratio? Seems like I saw something about this in
one of my books, but now can't find it. What should I look up?

That's the crux of the whole thing.

This one item is the thing that you were supposed to have learned in the
last three or so months in this class.

If you haven't got it after three months of classroom instruction, how do
you expect to get it from a freaking newsgroup post?

Good Luck!
Rich
 
R

Rich Grise

Jan 1, 1970
0
Is that a "Q-multiplier" ?

Yes, and sometimes "BFO".
And sometimes, "Regenerative receiver". ;-)

Cheers!
Rich

[0] Beat Frequency Oscillator.
 
D

Don Pearce

Jan 1, 1970
0
Is that a "Q-multiplier" ?

Yes, and sometimes "BFO".
And sometimes, "Regenerative receiver". ;-)

Cheers!
Rich

[0] Beat Frequency Oscillator.

A BFO is different - it is an oscillator used to generate the tone you
hear from a CW Morse code transmission - nothing to do with
regeneration or Q multiplication.

d

Pearce Consulting
http://www.pearce.uk.com
 
T

Tam/WB2TT

Jan 1, 1970
0
Dave said:
Okay, I *think* I am working on that...


How would I optimise the LC ratio? Seems like I saw something about this
in
one of my books, but now can't find it. What should I look up?



Thanks,

Dave
[email protected]
Dave,
To a first approximation, if you have an R, an L, and a C in parallel, Then
at resonance
X(L) = X(C) and Q = R/X(L).
Now, to make Q larger, make R bigger, or the inductor smaller. There is a
limit as to how far you can go with this, because the loaded Q can not be
bigger than the unloaded Q of the inductor.
; you will then need a better inductor. For example, when you get to around
50 MHz, it is not unusual to see air wound inductors made with #10 wire. On
the other hand, below about 20 MHz, you are probably best off with an
inductor wound on a toroidal core. Amidon publishes Q curves for various
toroids for different L and frequency combinations.

If you tap down on the inductor, say connect the load to the center tap of
the inductor, the effective load the inductor sees is 4x the R value, but
you will get 1/2 the voltage into the load.

Lastly, use decent capacitors. Mica is best. The El Cheapo capacitors sold
for bypassing will not work in a tuned circuit.

Tam
 
P

Paul Burridge

Jan 1, 1970
0
How can I increase the Q of a tank circuit? ANY ideas would be
appreciated...

Use a decent quality capacitor with low internal losses and wind your
own coil. Make it from silver/silver plated/gold plated wire and wind
it with nice, open, wide diameter, well-spaced turns and no core
material.
 
D

Dave

Jan 1, 1970
0
Rich Grise said:
That's the crux of the whole thing.

This one item is the thing that you were supposed to have learned in the
last three or so months in this class.

If you haven't got it after three months of classroom instruction, how do
you expect to get it from a freaking newsgroup post?

Good Luck!
Rich

What class? I haven't been in a class on this topic in over 20 years. Since
then I spent 12 years swapping boards and 11 years sleeping (currently on
disability for a sleep disorder.) Consequently, I have forgotten more than
I ever knew. That's why I am asking, and am not up on some of the current
technology. Also why I appreciate the help, and patience...

Dave
[email protected]
 
D

Dave

Jan 1, 1970
0
Tam/WB2TT said:
Dave,
To a first approximation, if you have an R, an L, and a C in parallel, Then
at resonance
X(L) = X(C) and Q = R/X(L).
Now, to make Q larger, make R bigger, or the inductor smaller. There is a
limit as to how far you can go with this, because the loaded Q can not be
bigger than the unloaded Q of the inductor.
; you will then need a better inductor. For example, when you get to around
50 MHz, it is not unusual to see air wound inductors made with #10 wire. On
the other hand, below about 20 MHz, you are probably best off with an
inductor wound on a toroidal core. Amidon publishes Q curves for various
toroids for different L and frequency combinations.

If you tap down on the inductor, say connect the load to the center tap of
the inductor, the effective load the inductor sees is 4x the R value, but
you will get 1/2 the voltage into the load.

Lastly, use decent capacitors. Mica is best. The El Cheapo capacitors sold
for bypassing will not work in a tuned circuit.

Tam

Uh HUH. Okay. I do appreciate the help. I am currently working with a
big, clunky variable capacitor and some RF chokes, connected by wire and a
switch (to choose which inductor is in the circuit). Unless I am mistaken,
my R is already quite low because of this. So I should ADD R to the circuit?
Would it be best to have a 1:1 relationship between R and X(L)? Or should I
go for 2:1, 10:1, or 100:1? My old textbook doesn't seem to cover anything
like this. :( One more thing, are RF chokes okay for tank circuits, or
should I try something else?

Thanks much for the help...

Dave
[email protected]
 
D

Dave

Jan 1, 1970
0
Paul Burridge said:
Use a decent quality capacitor with low internal losses and wind your
own coil. Make it from silver/silver plated/gold plated wire and wind
it with nice, open, wide diameter, well-spaced turns and no core
material.

Thank you. I think this answers something I just asked Tam, about the
practicallity of using RF chokes for my inductors. I *think* I have
something on winding inductors. Maybe in the ARRL handbook I bought late
last year (couldn't wait, too impatient.)

Much appreciation.

Dave
[email protected]
 
D

Dave

Jan 1, 1970
0
Bob Eldred said:
You can drive the tank with negative resistance to increase the Q. You do
this with an amplifier appropriately connected to provide positive feedback
sensing the current in the tank. A bridge circuit can be used to do this.
Care must be taken as the circuit can easily burst into oscillation if too
much positive feed back is applied. Needless to say, the bandwidth of the
amplifier must be great enough to cover the frequency of interest. Key
words: positive feedback, negative resistance, negative impedance
converters, NIC, gyrator, variable damping.
Bob

DAMN Bob. I looked all this up on Google, and it sounds like it is just
what I need. Do you *teach* somewhere?

Many thanks.

Dave
[email protected]
 
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