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Old March 16th 04, 01:22 AM
John Larkin
 
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On Mon, 15 Mar 2004 18:40:07 +0000 (UTC), Joe McElvenney
wrote:

Hi,

http://www.aade.com/lcmeter.htm


Or, as you have a counter, you could save yourself the money by
copying his oscillator and use a calculator to find the value of L
or C from the change in frequency.


Cheers - Joe


To save $100, I'm not going to copy *anybody's* circuits.

John

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Old March 16th 04, 11:10 AM
Joe McElvenney
 
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Hi,

To save $100, I'm not going to copy *anybody's* circuits.


An unfortunate choice of words on my part perhaps; possibly 'copy
his technique' would have been better. A technique, I should add, that
has been in the public domain for a very long time. The basic idea
being -


1. Take an oscillator tank circuit with fixed C1 or L1.

2. Measure frequency F1.

3. Connect unknown Cx across C1 or Lx in series with L1.

4. Measure new frequency F2.

5. Calculate Cx (Lx) using a scientific calculator, remembering
that F2/F1 = sqrt(C1 + Cx) or sqrt(L1 + Lx) then we have,

Cx = C1([F1/F2]^2 - 1) or Lx = L1([F2/F1]^2 - 1).

For best accuracy you would need to know the value of the strays
which may easily be worked out using the same technique in reverse, or
swamped, as the fancy takes you.

On the subject of copying, I would guess that 95% of homebrew
projects rely on pre-published information; there must have been
hundreds of 'chinese copies' of commercial rigs produced by hams over
the years. And it goes without saying that homebrewing not buying is
what we are supposed to be talking about in this group.


Cheers - Joe


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Old March 16th 04, 11:30 AM
Joe McElvenney
 
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OOPS!!

5. should have read -

Calculate Cx (Lx) using a scientific calculator,
remembering that F2/F1 = sqrt([C1 + Cx]/C1)

or sqrt([L1 + Lx]/L1) thus we have,

Cx = C1([F1/F2]^2 - 1) or Lx = L1([F2/F1]^2 - 1).


Cheers - Joe


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Old March 16th 04, 11:30 AM
Joe McElvenney
 
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OOPS!!

5. should have read -

Calculate Cx (Lx) using a scientific calculator,
remembering that F2/F1 = sqrt([C1 + Cx]/C1)

or sqrt([L1 + Lx]/L1) thus we have,

Cx = C1([F1/F2]^2 - 1) or Lx = L1([F2/F1]^2 - 1).


Cheers - Joe


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Old March 16th 04, 11:10 AM
Joe McElvenney
 
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Hi,

To save $100, I'm not going to copy *anybody's* circuits.


An unfortunate choice of words on my part perhaps; possibly 'copy
his technique' would have been better. A technique, I should add, that
has been in the public domain for a very long time. The basic idea
being -


1. Take an oscillator tank circuit with fixed C1 or L1.

2. Measure frequency F1.

3. Connect unknown Cx across C1 or Lx in series with L1.

4. Measure new frequency F2.

5. Calculate Cx (Lx) using a scientific calculator, remembering
that F2/F1 = sqrt(C1 + Cx) or sqrt(L1 + Lx) then we have,

Cx = C1([F1/F2]^2 - 1) or Lx = L1([F2/F1]^2 - 1).

For best accuracy you would need to know the value of the strays
which may easily be worked out using the same technique in reverse, or
swamped, as the fancy takes you.

On the subject of copying, I would guess that 95% of homebrew
projects rely on pre-published information; there must have been
hundreds of 'chinese copies' of commercial rigs produced by hams over
the years. And it goes without saying that homebrewing not buying is
what we are supposed to be talking about in this group.


Cheers - Joe




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