question about winding a coil to correct inductance

[Ben]

Hi there,

further to my previous question I have now wound a coil to act as an
antenna for my RFID application. I don't really understand the physics
behind this, hence I don't really understand the following result, but
I'm really hoping someone can explain it to me!

Now according to the instructions of the circuit board which drives my
antenna, the antenna needs to be resonate at 125kHz, which apparently
will happen if the antenna has an inductance of 1.62mH. So, to find out
how many turns the coil needs, according to these instructions, I should
use the formula:

N=sqrt( L*(3d+9l+10e)/(0.08*d^2) )

Where N is number of coils, L is desired inductance, d is coil diameter,
l is coil width, and e is coil thickness.

I want a coil of width 10cm, diameter 1.9cm, width around 0.3cm. This
results in me needing 837 turns....

So, 1000 turns later (just to be on the safe side) I end up with a coil
which gives me a frequency of more like 75kHz when I plug it into the
board. I put in too many turns because you can easily remove turns but
you can't put them back in once you've cut the wire. But, from what I
thought I understood, more turns would equal more inductance would equal
a greater frequency? And in any case, 20% more turns wouldn't make a 50%
smaller frequency would it? So, can any tell me what I have done wrong
or which bit I don't understand?

Thanks very much,

Ben Kenward
Department of Zoology
Oxford University

 

[Ian Stirling]

Hi there,

further to my previous question I have now wound a coil to act as an
antenna for my RFID application. I don't really understand the physics
behind this, hence I don't really understand the following result, but
I'm really hoping someone can explain it to me!

thought I understood, more turns would equal more inductance would equal
a greater frequency? And in any case, 20% more turns wouldn't make a 50%
smaller frequency would it? So, can any tell me what I have done wrong
or which bit I don't understand?

Look at the resonant frequency equation.
More turns = more inductance = lower reactance = lower freq.

 

[John Fields]

Hi there,

further to my previous question I have now wound a coil to act as an
antenna for my RFID application. I don't really understand the physics
behind this, hence I don't really understand the following result, but
I'm really hoping someone can explain it to me!

Now according to the instructions of the circuit board which drives my
antenna, the antenna needs to be resonate at 125kHz, which apparently
will happen if the antenna has an inductance of 1.62mH. So, to find out
how many turns the coil needs, according to these instructions, I should
use the formula:

N=sqrt( L*(3d+9l+10e)/(0.08*d^2) )

Where N is number of coils, L is desired inductance, d is coil diameter,
l is coil width, and e is coil thickness.

I want a coil of width 10cm, diameter 1.9cm, width around 0.3cm. This
results in me needing 837 turns....

So, 1000 turns later (just to be on the safe side) I end up with a coil
which gives me a frequency of more like 75kHz when I plug it into the
board. I put in too many turns because you can easily remove turns but
you can't put them back in once you've cut the wire. But, from what I
thought I understood, more turns would equal more inductance would equal
a greater frequency? And in any case, 20% more turns wouldn't make a 50%
smaller frequency would it? So, can any tell me what I have done wrong
or which bit I don't understand?

---
Increasing the number of turns increases the inductance, and since

1
f = ------------
2pi sqrt(LC)

you can see that if L increases and everything else stays the same, f
will also increase.


Also, if L doubles for a given C, f won't double; it'll only increase by
the difference in the _square roots_ of the different LC products.

 

[John Fields]

Look at the resonant frequency equation.
More turns = more inductance = lower reactance = lower freq.

 

[Ian Stirling]

More turns = more inductance = _higher_ reactance = lower freq.

I'd like to say that you'r an idiot, and should not correct completely
correct statements.
Unfortunately, you are of course completely right.
I have absolutely no idea what I was thinking.
Thanks.
Plus, for flat coils, the simple formula is at best approximate, and
using it as a first guess then measuring is probably the easiest route.

 

[Joe Legris]

Hi there,

further to my previous question I have now wound a coil to act as an
antenna for my RFID application. I don't really understand the physics
behind this, hence I don't really understand the following result, but
I'm really hoping someone can explain it to me!

Now according to the instructions of the circuit board which drives my
antenna, the antenna needs to be resonate at 125kHz, which apparently
will happen if the antenna has an inductance of 1.62mH. So, to find out
how many turns the coil needs, according to these instructions, I should
use the formula:

N=sqrt( L*(3d+9l+10e)/(0.08*d^2) )

Where N is number of coils, L is desired inductance, d is coil diameter,
l is coil width, and e is coil thickness.

I want a coil of width 10cm, diameter 1.9cm, width around 0.3cm. This
results in me needing 837 turns....

So, 1000 turns later (just to be on the safe side) I end up with a coil
which gives me a frequency of more like 75kHz when I plug it into the
board. I put in too many turns because you can easily remove turns but
you can't put them back in once you've cut the wire. But, from what I
thought I understood, more turns would equal more inductance would equal
a greater frequency? And in any case, 20% more turns wouldn't make a 50%
smaller frequency would it? So, can any tell me what I have done wrong
or which bit I don't understand?

Thanks very much,

Ben Kenward
Department of Zoology
Oxford University

Congratulations! At least your oscillator oscillates. However, 20% more
turns should give 41% more inductance resulting in a frequency 20%
lower, not higher.

This is because the resonant frequency of a coil and capacitor is
1/2pi*SQRT(L*C) but the inductance is proportional to the square of the
number of turns on the coil. Therefore the frequency is inversely
proportional to the number of turns.

There is apparent confusion about the coil "width" and "thickness" in
that you call them both "width". Which is which?

What kind of capacitor are you using? Its value may off by 20% or more
and may vary significantly with temperature.

--
Joe Legris
============================================================================
Check out my Ebay listings:
Brand new electronic components in lots of 10+
http://cgi6.ebay.com/ws/eBayISAPI.dll?ViewSellersOtherItems&userid=tachogram
============================================================================

 

[Roy J. Tellason]

Increasing the number of turns increases the inductance, and since

1
f = ------------
2pi sqrt(LC)

you can see that if L increases and everything else stays the same, f
will also increase.

Where'd you learn how to do math? Try that statement out with any simple
integer in the bottom half of a fraction, and see how well it works...

 

[John Fields]

Where'd you learn how to do math? Try that statement out with any simple
integer in the bottom half of a fraction, and see how well it works...

---
Twasn't a problem with the math, it was a problem with the English.

"f will also increase" should have read "f will decrease".

Thanks.

 

[Paul Burridge]

Increasing the number of turns increases the inductance, and since

1
f = ------------
2pi sqrt(LC)

you can see that if L increases and everything else stays the same, f
will also increase.

Decrease!

 

[John Fields]

Decrease!

 

[default]

Hi there,

further to my previous question I have now wound a coil to act as an
antenna for my RFID application. I don't really understand the physics
behind this, hence I don't really understand the following result, but
I'm really hoping someone can explain it to me!

Now according to the instructions of the circuit board which drives my
antenna, the antenna needs to be resonate at 125kHz, which apparently
will happen if the antenna has an inductance of 1.62mH. So, to find out
how many turns the coil needs, according to these instructions, I should
use the formula:

N=sqrt( L*(3d+9l+10e)/(0.08*d^2) )

Where N is number of coils, L is desired inductance, d is coil diameter,
l is coil width, and e is coil thickness.

I want a coil of width 10cm, diameter 1.9cm, width around 0.3cm. This
results in me needing 837 turns....

So, 1000 turns later (just to be on the safe side) I end up with a coil
which gives me a frequency of more like 75kHz when I plug it into the
board. I put in too many turns because you can easily remove turns but
you can't put them back in once you've cut the wire. But, from what I
thought I understood, more turns would equal more inductance would equal
a greater frequency? And in any case, 20% more turns wouldn't make a 50%
smaller frequency would it? So, can any tell me what I have done wrong
or which bit I don't understand?

Thanks very much,

Ben Kenward
Department of Zoology
Oxford University

Sounds roughly correct. Inductance increases with the square of the
turns, more inductance equals lower frequency: what you have . . .

There's somewhat more going on than the formulas tell you. The
location of nearby metal, the wire diameter, length:diameter ratio,
and insulation all play a role in the actual inductance - most
inductance calculations are simplified or only work well within very
specific limits.

There's also the fact that you are concerned with the frequency, which
in turn, will be affected by other circuit parameters - like voltage
to the oscillator, capacitors across the coil, internal capacitance of
the driving circuitry, self resonance of the coil, temperature affects
on the components etc..

I take it this antenna also serves to determine the frequency?

You could have wound the correct coil and lowered the frequency with a
cap across the coil or some ferrite material for a core or a strip of
aluminum foil near/on the coil.

 

[Tony Williams]

N=sqrt( L*(3d+9l+10e)/(0.08*d^2) )
Where N is number of coils, L is desired inductance, d is coil
diameter, l is coil width, and e is coil thickness.
I want a coil of width 10cm, diameter 1.9cm, width around 0.3cm.
This results in me needing 837 turns....

I can't seem to get anywhere near 837 turns for 1.62mH.
Are the defs of d/l/e correct, and what are the units?

So, 1000 turns later (just to be on the safe side) I end up with
a coil which gives me a frequency of more like 75kHz...etc.

F-resonant is approximately proportional to the reciprocal
of the turns-count. So if 1000 turns resonated at 75KHz
then it might be closer to 600 turns for 125KHz.

 

[Ben]

Hi there,

further to my previous question I have now wound a coil to act as an
antenna for my RFID application. I don't really understand the physics
behind this, hence I don't really understand the following result, but
I'm really hoping someone can explain it to me!

Now according to the instructions of the circuit board which drives my
antenna, the antenna needs to be resonate at 125kHz, which apparently
will happen if the antenna has an inductance of 1.62mH. So, to find
out how many turns the coil needs, according to these instructions, I
should use the formula:

N=sqrt( L*(3d+9l+10e)/(0.08*d^2) )

Where N is number of coils, L is desired inductance, d is coil
diameter, l is coil width, and e is coil thickness.

I want a coil of width 10cm, diameter 1.9cm, width around 0.3cm. This
results in me needing 837 turns....

So, 1000 turns later (just to be on the safe side) I end up with a
coil which gives me a frequency of more like 75kHz when I plug it into
the board. I put in too many turns because you can easily remove turns
but you can't put them back in once you've cut the wire. But, from
what I thought I understood, more turns would equal more inductance
would equal a greater frequency? And in any case, 20% more turns
wouldn't make a 50% smaller frequency would it? So, can any tell me
what I have done wrong or which bit I don't understand?

Thanks very much,

Ben Kenward
Department of Zoology
Oxford University

Thanks a lot for all your responses, folks... It seems I got it the
wrong way round - after removing turns, I was eventually able to get
into the frequency range, and get my antenna to read an RFID tag! It
seems this is quite a trial and error thing - the eventual optimum
number of turns for maximum read range was a fair way off the number
given by the maths.

Cheers,

Ben Kenward
Department of Zoology
Oxford University