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Old March 27th 04, 02:22 PM
Paul Burridge
 
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Default What's the value of this inductor?

Hi guys,

I have some surplus axial inductors marked as follows:

uH27 -K
GEL

I would have thought this implied 0.27uH or 270nH? GEL is presumably
the maker. However, there's no resonance at 17.6Mhz with 300pF in
parallel as might be expected of an inductor of that value. There *is*
however, a resonance at 50.6Mhz., which implies that the inductors are
actually 33nH. I'm confused. Anyone have any idea as to how to
determine the value absolutely? Are makers' marking to be trusted at
all?
--

The BBC: Licensed at public expense to spread lies.
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Old March 27th 04, 06:20 PM
Bob
 
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I would have guessed 27 microhenries with a 20% tolerance.

Bob


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Old March 27th 04, 08:10 PM
Tim Wescott
 
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Paul Burridge wrote:

Hi guys,

I have some surplus axial inductors marked as follows:

uH27 -K
GEL

I would have thought this implied 0.27uH or 270nH? GEL is presumably
the maker. However, there's no resonance at 17.6Mhz with 300pF in
parallel as might be expected of an inductor of that value. There *is*
however, a resonance at 50.6Mhz., which implies that the inductors are
actually 33nH. I'm confused. Anyone have any idea as to how to
determine the value absolutely? Are makers' marking to be trusted at
all?


50MHz is pretty high -- are you sure it's not 27nH with stray
inductance? Did you measure more than one to check for tolerance as the
OP brings up?

---------------------------------------
Tim Wescott
Wescott Design Services
http://www.wescottdesign.com
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Old March 28th 04, 08:35 PM
Paul Burridge
 
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On Sat, 27 Mar 2004 12:10:36 -0800, Tim Wescott
wrote:

50MHz is pretty high -- are you sure it's not 27nH with stray
inductance? Did you measure more than one to check for tolerance as the
OP brings up?


I'm not sure of anything! These small inductors are notoriously
difficult to measure accurately. Even lab-grade kit can differ in
measured values by as much as 100% or more and the risk of parasitic
resonances is an additional problem. :-(
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Old March 28th 04, 08:35 PM
Paul Burridge
 
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On Sat, 27 Mar 2004 12:10:36 -0800, Tim Wescott
wrote:

50MHz is pretty high -- are you sure it's not 27nH with stray
inductance? Did you measure more than one to check for tolerance as the
OP brings up?


I'm not sure of anything! These small inductors are notoriously
difficult to measure accurately. Even lab-grade kit can differ in
measured values by as much as 100% or more and the risk of parasitic
resonances is an additional problem. :-(


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Old March 28th 04, 12:33 AM
John Jardine
 
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Paul Burridge wrote in message
...
Hi guys,

I have some surplus axial inductors marked as follows:

uH27 -K
GEL

I would have thought this implied 0.27uH or 270nH? GEL is presumably
the maker. However, there's no resonance at 17.6Mhz with 300pF in
parallel as might be expected of an inductor of that value. There *is*
however, a resonance at 50.6Mhz., which implies that the inductors are
actually 33nH. I'm confused. Anyone have any idea as to how to
determine the value absolutely? Are makers' marking to be trusted at
all?
--

The BBC: Licensed at public expense to spread lies.


0.27uH is 5ohms at 5MHz.

Set to 5MHz and wind up your 50ohm signal generator to max output (say a
100mV or so). Measure it's voltage using the scope.
Short the inductor across the sig genny terminals and measure the voltage
across the inductor.
Ignore all irrelavencies such as phase angles, stray capacitances, Q factor,
series resistance,skin effects, self resonances etc. Just treat the inductor
as if it's a straight forward potential divider resistance in series with
the 50 ohms of the genny.

Eg. Sig genny open circuit output at 5megs is 100 mV peak-peak
Inductor connected and voltage measured is now 10mV peak-peak.
Therefore 1/10th of the available voltage is across the inductor and 9/10
lost across the genny internal 50ohms.
So inductor looks like 5 ohms therefore,
5ohms = 2 x Pi x 5megs x L?
=270nH. Simple !
Using this method it's easy to get down to the odd few nH.
Exact same method as used by a nice 'inductance meter' design that turned up
in Radcom about 12 years ago

regards
john








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Old March 28th 04, 12:56 AM
John Jardine
 
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John Jardine wrote in message
...

Paul Burridge wrote in message
...
Hi guys,

I have some surplus axial inductors marked as follows:

uH27 -K
GEL


[clip]
Therefore 1/10th of the available voltage is across the inductor and 9/10
lost across the genny internal 50ohms.
So inductor looks like 5 ohms therefore,

[clip]

********.
Screwed myself by rushing for easy numbers

0.27uH is actually 8.5ohms at 5MHz

For the 10mv example L would actually be (10mV / (100mv-10mv)/50) =5.55
ohms, equiv' to 176nH.

Sorry!


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Old March 28th 04, 08:35 PM
Paul Burridge
 
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On Sun, 28 Mar 2004 00:56:05 -0000, "John Jardine"
wrote:


John Jardine wrote in message
...

Paul Burridge wrote in message
...
Hi guys,

I have some surplus axial inductors marked as follows:

uH27 -K
GEL


[clip]
Therefore 1/10th of the available voltage is across the inductor and 9/10
lost across the genny internal 50ohms.
So inductor looks like 5 ohms therefore,

[clip]

********.
Screwed myself by rushing for easy numbers

0.27uH is actually 8.5ohms at 5MHz

For the 10mv example L would actually be (10mV / (100mv-10mv)/50) =5.55
ohms, equiv' to 176nH.

Sorry!


Okay, thanks John (and others). I'll give this ingenious solution a
bash and see what turns out. Been struck by a virus on my main
computer so can't get net access as frequently as I would wish so may
have to report back 2morrow....
  #9   Report Post  
Old March 28th 04, 08:35 PM
Paul Burridge
 
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Default

On Sun, 28 Mar 2004 00:56:05 -0000, "John Jardine"
wrote:


John Jardine wrote in message
...

Paul Burridge wrote in message
...
Hi guys,

I have some surplus axial inductors marked as follows:

uH27 -K
GEL


[clip]
Therefore 1/10th of the available voltage is across the inductor and 9/10
lost across the genny internal 50ohms.
So inductor looks like 5 ohms therefore,

[clip]

********.
Screwed myself by rushing for easy numbers

0.27uH is actually 8.5ohms at 5MHz

For the 10mv example L would actually be (10mV / (100mv-10mv)/50) =5.55
ohms, equiv' to 176nH.

Sorry!


Okay, thanks John (and others). I'll give this ingenious solution a
bash and see what turns out. Been struck by a virus on my main
computer so can't get net access as frequently as I would wish so may
have to report back 2morrow....
  #10   Report Post  
Old March 29th 04, 02:25 AM
Nick Kennedy
 
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OK, but remember that the voltage drop across the inductor plus that across
the source resistance don't add directly to the source voltage. With your
example of 10 mV across L and 100 mV open circuit voltage, the voltage
across the 50 ohm source resistance is 99.5 mV. That's because they're 90
degrees out of phase and add in a right triangle fashion.

Maybe you considered that, and that's why you chose a small value of drop
across the inductor. When that's the case, the hypotenuse and x-side of the
triangle are pretty close to equal. But users of this method need to
understand this simplification and the implied restrictions.

I've used something similar to measure the inductance of chokes too large
for my meter, such as transformer windings and power supply chokes. You can
put a known resistor in series with the choke, put 60 Hz AC across it, and
measure the two drops. Then calculate the inductance using a method similar
to what you've described. But keep the sum of squares relationship in mind.
(Supply voltage squared equals drop across resistance squared plus drop
across inductance squared.) If the inductor has significant resistance, it
should be added to the value of the series resistor.

Regards,

Nick, WA5BDU
in Arkansas

"John Jardine" wrote in message
...


0.27uH is 5ohms at 5MHz.

Set to 5MHz and wind up your 50ohm signal generator to max output (say a
100mV or so). Measure it's voltage using the scope.
Short the inductor across the sig genny terminals and measure the voltage
across the inductor.
Ignore all irrelavencies such as phase angles, stray capacitances, Q

factor,
series resistance,skin effects, self resonances etc. Just treat the

inductor
as if it's a straight forward potential divider resistance in series with
the 50 ohms of the genny.

Eg. Sig genny open circuit output at 5megs is 100 mV peak-peak
Inductor connected and voltage measured is now 10mV peak-peak.
Therefore 1/10th of the available voltage is across the inductor and 9/10
lost across the genny internal 50ohms.
So inductor looks like 5 ohms therefore,
5ohms = 2 x Pi x 5megs x L?
=270nH. Simple !
Using this method it's easy to get down to the odd few nH.
Exact same method as used by a nice 'inductance meter' design that turned

up
in Radcom about 12 years ago

regards
john





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