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I don't think it's cause he's a brit Space Varmint =)
 
I have been using Google to find a hint to the reactance desired for the coil in the FET oscillators under discussion. Not found yet.
Here is a page with a lot of depressing math.
**broken link removed**
Download the pdf for the full article and other articles included. It's a small file.
EDN PDF
 
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I have been using Google to find a hint to the reactance desired for the coil in the FET oscillators under discussion. Not found yet.
Here is a page with a lot of depressing math.
**broken link removed**
Download the pdf for the full article and other articles included. It's a small file.
EDN PDF

Take a look at chap 2 in this paper. Good theory.

**broken link removed**
 
I have been using Google to find a hint to the reactance desired for the coil in the FET oscillators under discussion. Not found yet.
Here is a page with a lot of depressing math.
**broken link removed**
Download the pdf for the full article and other articles included. It's a small file.
EDN PDF

I think your putting too much into the wrong emphasis unless your one of those analytical engineers. I'm in applications myself. You know, make it work and move on.

But if you must know, you use the two formulas I gave you. These will give you the plug in values for the grand daddy:


f=1/ (2pi * square root of LC)

more simply put, when the inductive reactance is equal to the capacitive reactance, this will be your frequency because they are 180 degrees out of phase thereby cancelling the reactive impedance to that frequency. In effect, you have zero resistance at one frequency.

Now, what I was saying about application. You got a variable capacitor, this will vary the capacitive reactance. So just get the oscillator to start up at power on. You have won 95% of the battle right there. Now check your range by varying the capacitor. If you need to go higher in frequency, just remove a turn or two on the coil.

Next, start worrying about the mixer and crystal filter. Here is the real work. Oh, one more thing on the oscillator. Try to make it clean as possible. You want good symmetry in your sinusoidal wave form. Unless you have allot of expensive equipment,
you won't be able to measure things like phase noise and so forth. With a scope, make sure you got no humps in your sign wave. These are harmonic energy and will cause you to pick up out of band signals. Keep it clean!
 
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SV, I know the generalities. The only specific thing I'm asking is a good ac resistance to shoot for for the coil. I could look at a lot of circuits and work backwards to see if it exists.

I'll check the pdf Mike suggests. Perhaps it is in there.
It will be different for fet vs bi-polar and I would think it will be different if the coil in the tuned circuit is in the emitter/source or base/gate or collector/drain sub-circuit. It just seems a good idea to me to know what reactance to design for.
 
Let me see if I can clarify for you. Look at attached image. Circuit is a 1MHz colpits osc. Notice the tuned tank consist of 100uh inductor and split caps at 500pf each which means cap value is 250pf.

**broken link removed**

Now if you plug these numbers into the formula Fo= 1/(2pi√LC)
You get a value of 1MHz. This result should give an indication that above formula works.

So you should start with two known values, say Frequency and capacitance.
Transpose above formula to solve for L. L= 1/(4pi²CF²)

If you don't want to work out the math, try this online resonance calculator.



This should get you close to the ballpark figure for the inductor. Most LC oscillators I have seen use a variable inductor. The kind that you adjust the core with little plastic screwdriver.

There is really a lot of ugly math involved, but above method should get you in the ballpark.
Hope that helps.

Oh, this app note also has a few good tips.
https://www.electro-tech-online.com/custompdfs/2009/04/lonsosc.pdf
 

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SV, I know the generalities. The only specific thing I'm asking is a good ac resistance to shoot for for the coil. I could look at a lot of circuits and work backwards to see if it exists.

I'll check the pdf Mike suggests. Perhaps it is in there.
It will be different for fet vs bi-polar and I would think it will be different if the coil in the tuned circuit is in the emitter/source or base/gate or collector/drain sub-circuit. It just seems a good idea to me to know what reactance to design for.

If you think application, than what you are stuck with is a variable capacitor as one half of your frequency determining components. Sure the impedance is a factor but to a lesser degree than the actual X of L and X of C values. And so, with most of the variable capacitors I've seen, they are always a very low value. You want to work with that because if you start adding parallel capacitance to the tank resonant circuit, you will significantly reduce the oscillator bandwidth. Meaning, try not to pad the oscillator. Use only the variable cap if possible. We are talking VFO (variable frequency oscillator).

Therefore your inductor value will be contingent upon the capacitor more than anything else. So your usual formulas will apply.

f = 1/ (2pi * square root of LC)

So start with Xc for a given mid frequency. So your cap is static within it's assigned range. Take a mid-range value of approximately 150pf. Then use the formula Xc= 1/ (2pi * fC)

example:

frequency = 15 MHz
Capacitance = 150pf

then XC = 70.7 ohms

Next you want to match the inductive reactance of the coil because:

f= [2pi*fL = 1 / (2pi*fC)]

When the inductive and capacitive reactance are equal and using the formula directly above, it can be broken down to:

f = 1 / (2pi * square root of LC)

So:

XL = 2pi*fL

solving for L we get:

L = XL / 2pi*f

70.7 ohms / 6.28 * 15 times 10 to the 6th = .75 uH

That should put you in the ball park except for one thing. Your mid rage frequency for your local oscillator will be offset by the frequency of the crystal filter depending on if you use high side or low side injection. The formula will apply directly to a signal generator for testing the hf band.
 
It just seems a good idea to me to know what reactance to design for.

You think wrong - no need (or use) to calculate the reactance, you just need to calculate the inductance of the coil from the resonant frequency required with the specific capacitance you're using.

The formula has been posted here multiple times, but I'm sure you already knew it.
 
ok, thanks. Everybody is saying that I should not think about a desired reactance to shoot for but just match the tuning cap that I can find somewhere. That will be the practical thing to do.
Since I will have to order any tuning caps I need I'll think about tuning range required and go from there.

At resonance reactance is ideally infinite or zero so I'm barking up the wrong tree anyway :) arf arf
 
If you think application, than what you are stuck with is a variable capacitor as one half of your frequency determining components. Sure the impedance is a factor but to a lesser degree than the actual X of L and X of C values. And so, with most of the variable capacitors I've seen, they are always a very low value. You want to work with that because if you start adding parallel capacitance to the tank resonant circuit, you will significantly reduce the oscillator bandwidth. Meaning, try not to pad the oscillator. Use only the variable cap if possible. We are talking VFO (variable frequency oscillator).

Therefore your inductor value will be contingent upon the capacitor more than anything else. So your usual formulas will apply.

f = 1/ (2pi * square root of LC)

So start with Xc for a given mid frequency. So your cap is static within it's assigned range. Take a mid-range value of approximately 150pf. Then use the formula Xc= 1/ (2pi * fC)

example:

frequency = 15 MHz
Capacitance = 150pf

then XC = 70.7 ohms

Next you want to match the inductive reactance of the coil because:

f= [2pi*fL = 1 / (2pi*fC)]

When the inductive and capacitive reactance are equal and using the formula directly above, it can be broken down to:

f = 1 / (2pi * square root of LC)

So:

XL = 2pi*fL

solving for L we get:

L = XL / 2pi*f

70.7 ohms / 6.28 * 15 times 10 to the 6th = .75 uH

That should put you in the ball park except for one thing. Your mid rage frequency for your local oscillator will be offset by the frequency of the crystal filter depending on if you use high side or low side injection. The formula will apply directly to a signal generator for testing the hf band.

Instead of using all those calculations, you could just follow the steps I suggested. The result is the same, but less work.
 
Hi, Mike
I just tried the Resonant Frequency Calculator and inputing any cap value between 2 and 250 pf and frequency 8 mhz I only get NaN for an inductance. (Firefox v3 java enabled).
 
L = 1 / (C*[2pif]squared)

what's in the brackets is squared...kapeesh?

Anyway, I was just showing the relationships. It's absurd to say that reactance doesn't matter. In a Hartley circuit we know it should be a high Z. So it really goes back to what I've been trying to say from the beggining. If one of them doesn't matter in practical application, then none of them do. The best advise I can give is the physical construction of the coil. After you get it running, you can go back into your mad scientist closet or what have you, and calculate away if you want. Me, I'd much rather just get the thing running and play with it until it looks and acts right, then start licking my chops at building the next stage, and cross my fingers and hope that it doesn't load the piss out of my oscillator to where I'm practically back to the drawing board (figuratively speaking). This is what my original instructions should help him avoid.
 
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L = 1 / (C*[2pif]squared)

what's in the brackets is squared...kapeesh?

It is spelled capisci if your Italian... :)

If you pay attention you will note.
L= 1/(4pi²CF²)

2 squared is 4 and C x f squared is same thing as you wrote above. Capisci?

If you prefer, use attached formula, same thing. :rolleyes:
 

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It is spelled capisci if your Italian... :)

If you pay attention you will note.
L= 1/(4pi²CF²)

2 squared is 4 and C x f squared is same thing as you wrote above. Capisci?

If you prefer, use attached formula, same thing. :rolleyes:

Hey Mike? How do make that little 2 ? The squared sign.
 
I am sorry, I don't quite understand your question...
 
Hey Mike? How do make that little 2 ? The squared sign.

Okay, I think I get your question now. Use the advanced editor option. It allows use of symbols and such like Ω²√'

Look for advanced option, was that your question?
 
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