Dumb newbie has another question...

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

Reply to
Bob Eldred
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How can I increase the Q of a tank circuit? ANY ideas would be appreciated...

Thanks,

Dave snipped-for-privacy@hotmail.com

Reply to
Dave

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

Dave db5151

Reply to
Dave

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 snipped-for-privacy@hotmail.com

Reply to
Dave

feedback

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 snipped-for-privacy@hotmail.com

Reply to
Dave

Reply to
no_one

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

Reply to
Andrew Holme

Is that a "Q-multiplier" ?

Reply to
Andrew Holme

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.

Reply to
Andrew Holme

In article , Dave wrote: [...]

[...]

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.

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.

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kensmith@rahul.net   forging knowledge
Reply to
Ken Smith

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

Reply to
Tam/WB2TT

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

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Reply to
Don Pearce

Gold plating is not used in electronic assemblies because of gold's *superior* conductivity characteristics. Both copper and silver beat it here--at a lower cost.

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Gold's resistance to oxidation is the reason it is used.

Reply to
JeffM

this

Then

around

On

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 snipped-for-privacy@hotmail.com

Reply to
Dave

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 snipped-for-privacy@hotmail.com

Reply to
Dave

feedback

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 snipped-for-privacy@hotmail.com

Reply to
Dave

If you are at frequency F, X(L)= 2 x Pi x F x L, where F is in Hertz, and L is in Henrys. So, at 8 MHz, for example, a 10 uH inductor has a reactance of

2 * 3.14 * 8E6 * 10E-6 = 50.24 Ohms. If you want the Q to be at least 100, R has to be 50.24 * 100 = 5024 OR GREATER,

RF chokes are NOT OK for tuned circuits. They are specifically made to be fairly low Q. I don't remember what your load was, but the gate input of a FET is about as good (high impedance) as you can get. What is your frequency, what Q do you want, and what is the load impedance. What are you driving it from? that has an effect also.

Tam

Reply to
Tam/WB2TT

a

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you

Damn. Now I understand a little better why I am seeing what I am seeing. The amazing thing is that my project works as well as it does. And now I understand why the schematic for the end product (that I hope to end up with someday) uses the components it does (air-wound coils/FET's etc.) Thank you so much for the illuminating information. I went to an electronics trade school 24 years ago and learned enough to fix computers, but RF is so totally different as to be mind-boggling. That mind-bogglingness is attractive to me though, and as a hobby I don't think anything else could quite match it for obsessive interest. I am currently playing with shortwave receiving rigs, but people keep telling me I ought to get into HAM radio, and I may someday. Once I understand a few things.

Many thanks. More questions later, I am sure.

Dave snipped-for-privacy@hotmail.com

Reply to
Dave

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.

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"What is now proved was once only imagin\'d." - William Blake, 1793.
Reply to
Paul Burridge

============================

If R is in series with L and both are in parallel with C then, at resonance, all is fully explained by -

Q = 1 / 2 / Pi / F / C / R

or

Q = 2 * Pi * F * L / R

and that's all there is to it.

People who missed out arithmetic at school have a hard time understanding radio. Using arithmetic it is so precise and simple to describe.

--
Reg.
Reply to
Reg Edwards

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