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Old March 15th 04, 11:02 PM
Paul Burridge
 
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On Mon, 15 Mar 2004 08:10:02 -0800, John Larkin
wrote:

On Mon, 15 Mar 2004 07:40:36 -0800, "Tim Wescott"
wrote:


Google for "LCR" meter; see what you come up with (mostly high $
instruments, probably).


This one is great:

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


Crumbs! Thanks, John!
--

The BBC: Licensed at public expense to spread lies.
  #22   Report Post  
Old March 16th 04, 12:01 AM
Active8
 
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On Mon, 15 Mar 2004 15:12:24 +0000, Paul Burridge wrote:

On Mon, 15 Mar 2004 12:48:43 -0000, "Leon Heller"
wrote:


"Paul Burridge" wrote in message
. ..
Hi all,

I've got quite a huge stash of inductors in my parts bin. The colour
codes don't always seem to relate to the values I've been able to
measure, with my multi-function DVM, however, and I can't accurately
measure any inductor about 10uH. Is there a circuit anywhere that
would enable me to get a reasonably accurate idea of the values I've
got down to say 100nH or thereabouts?


If you have a signal generator you can connect them to a suitable capacitor
and measure the resonant frequency.


Hi Leon,

That's what I have been doing, in fact. It's just I'd prefer to have a
more convenient, portable method to get instant read-outs of coil
values...


So why beat yourself to death building a sweep gen to test filters
when you got that there sig gen?
--
Best Regards,
Mike
  #23   Report Post  
Old March 16th 04, 12:01 AM
Active8
 
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On Mon, 15 Mar 2004 15:12:24 +0000, Paul Burridge wrote:

On Mon, 15 Mar 2004 12:48:43 -0000, "Leon Heller"
wrote:


"Paul Burridge" wrote in message
. ..
Hi all,

I've got quite a huge stash of inductors in my parts bin. The colour
codes don't always seem to relate to the values I've been able to
measure, with my multi-function DVM, however, and I can't accurately
measure any inductor about 10uH. Is there a circuit anywhere that
would enable me to get a reasonably accurate idea of the values I've
got down to say 100nH or thereabouts?


If you have a signal generator you can connect them to a suitable capacitor
and measure the resonant frequency.


Hi Leon,

That's what I have been doing, in fact. It's just I'd prefer to have a
more convenient, portable method to get instant read-outs of coil
values...


So why beat yourself to death building a sweep gen to test filters
when you got that there sig gen?
--
Best Regards,
Mike
  #24   Report Post  
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

  #25   Report Post  
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



  #26   Report Post  
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


  #27   Report Post  
Old March 16th 04, 11:10 AM
Joe McElvenney
 
Posts: n/a
Default

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


  #28   Report Post  
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


  #29   Report Post  
Old March 16th 04, 11:30 AM
Joe McElvenney
 
Posts: n/a
Default

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


  #30   Report Post  
Old March 16th 04, 01:02 PM
Paul Burridge
 
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On Mon, 15 Mar 2004 19:01:01 -0500, Active8
wrote:

So why beat yourself to death building a sweep gen to test filters
when you got that there sig gen?


Because it has to be small and light enough to be mounted on our robot
for field checks. Try doing that with a 120lb Marconi. :-)

--

The BBC: Licensed at public expense to spread lies.
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