> Hi,
>
> I am using the following magnetic core to make a solenoid of inductance 12uH (A11)
>
> http://www.ferroxcube.com/prod/assets/3c94.pdf
>
> Part Number: ROD 6/ 20 /3C 90. Its 6mm in diameter, 20.3 mm in length and ui = 2300.
>
> I used the 20 AWG gauge solid wire.
>
> You can find the schematic of the circuit at the following link
>
> http://img217.imageshack.us/img217/3215/inductorcore.jpg
>
> The input to the inductor is 40 V peak to peak sine wave with a frequency of 100KHz.
>
> Problems:
>
> 1. The solenoid gets hot after 40 minutes when I put it in the magnetic field generated by some other coil. I do not know how much magnetic field is present. But the solenoid (A11) is getting hot.
>
> The magnetic core is rated for 100KHz and its Hysteresis curve is thin (page 121, Figure 3). I do not know what is going on here. Can anyone give any suggestions. How can I keep the temperature constant like 20 degrees above ambient temperature and do not let it increase with time?
>
>
> 2. Calculation of the self inductance of this solenoid (A11)
>
> I found the following formulas on the internet
>
> a) L = (u0 x N x N x A ) / l
>
> u0 = 4x pi x10^-7
> A = cross sectional area pi x r x r
> N = Number of turns
> l = length of the wire
>
> b) L = (u0 x ur x N x N x A ) / l
>
> Which one is correct? And if formula b is correct than what will be the value of ur for the core that I am using?
>
> jess
>

I'm estimating that you have about 90 or so turns on the rod to get the
12uH. From that I estimate that your resistance is about 1 ohm at
100kHz. The reactance of your specified inductance (12uH) is about 7.5
ohms, so the impedance magnitude is about 7.6 ohms. Your 14V RMS (40V
p-p) will result in about 1.86A. Put that through about 1 ohm and you
have about 3.5W dissipation in a device that is roughly the size of a 2W
through-hole resistor.
This does not include the load.
My conclusion is that you need more inductance.
John

Reply by Uwe Hercksen●June 20, 20122012-06-20

jsscshaw88@gmail.com schrieb:

> Problems:
>
> 1. The solenoid gets hot after 40 minutes when I put it in the magnetic field generated by some other coil. I do not know how much magnetic field is present. But the solenoid (A11) is getting hot.
>
> The magnetic core is rated for 100KHz and its Hysteresis curve is thin (page 121, Figure 3). I do not know what is going on here. Can anyone give any suggestions. How can I keep the temperature constant like 20 degrees above ambient temperature and do not let it increase with time?

Hello,
try to increase the distance between both coils. Also try to rotate one
coil by 90 �.
Bye

Reply by Bill Sloman●June 20, 20122012-06-20

On 6/19/2012 11:29 PM, jsscshaw88@gmail.com wrote:

> Hi,
>
> I am using the following magnetic core to make a solenoid of inductance 12uH (A11)
>
> http://www.ferroxcube.com/prod/assets/3c94.pdf
>
> Part Number: ROD 6/ 20 /3C 90. Its 6mm in diameter, 20.3 mm in length and ui = 2300.
>
> I used the 20 AWG gauge solid wire.
>
> You can find the schematic of the circuit at the following link
>
> http://img217.imageshack.us/img217/3215/inductorcore.jpg
>
> The input to the inductor is 40 V peak to peak sine wave with a frequency of 100KHz.
>
> Problems:
>
> 1. The solenoid gets hot after 40 minutes when I put it in the magnetic field generated by some other coil. I do not know how much magnetic field is present. But the solenoid (A11) is getting hot.
>
> The magnetic core is rated for 100KHz and its Hysteresis curve is thin (page 121, Figure 3). I do not know what is going on here. Can anyone give any suggestions. How can I keep the temperature constant like 20 degrees above ambient temperature and do not let it increase with time?
>
>
> 2. Calculation of the self inductance of this solenoid (A11)
>
> I found the following formulas on the internet
>
> a) L = (u0 x N x N x A ) / l
>
> u0 = 4x pi x10^-7
> A = cross sectional area pi x r x r
> N = Number of turns
> l = length of the wire
>
> b) L = (u0 x ur x N x N x A ) / l
>
> Which one is correct? And if formula b is correct than what will be the value of ur for the core that I am using?

Both formula's are correct, but the effective permeability - u - of the
ferrite rod will be hard to work out.
A quick google search on "inductance of a solenoid" found this
http://en.wikipedia.org/wiki/Solenoid
amongst others.
http://hyperphysics.phy-astr.gsu.edu/hbase/electric/indsol.html
offers you a calculator. This tells you that the effective permeability
of iron is about 200 in this context. Your ferrite rod will have a lower
effective permeability, but probably not all that much lower.
This isn't a clever way to make an inductor.
--
Bill Sloman, Nijmegen

Reply by ●June 19, 20122012-06-19

Hi,
I am using the following magnetic core to make a solenoid of inductance 12u=
H (A11)
http://www.ferroxcube.com/prod/assets/3c94.pdf
Part Number: ROD 6/ 20 /3C 90. Its 6mm in diameter, 20.3 mm in length and u=
i =3D 2300.
I used the 20 AWG gauge solid wire.=20
You can find the schematic of the circuit at the following link=20
=20
http://img217.imageshack.us/img217/3215/inductorcore.jpg
The input to the inductor is 40 V peak to peak sine wave with a frequency o=
f 100KHz.=20
Problems:
1. The solenoid gets hot after 40 minutes when I put it in the magnetic fie=
ld generated by some other coil. I do not know how much magnetic field is p=
resent. But the solenoid (A11) is getting hot.=20
The magnetic core is rated for 100KHz and its Hysteresis curve is thin (pag=
e 121, Figure 3). I do not know what is going on here. Can anyone give any =
suggestions. How can I keep the temperature constant like 20 degrees above =
ambient temperature and do not let it increase with time?=20
2. Calculation of the self inductance of this solenoid (A11)=20
I found the following formulas on the internet
a) L =3D (u0 x N x N x A ) / l=20
u0 =3D 4x pi x10^-7
A =3D cross sectional area pi x r x r=20
N =3D Number of turns
l =3D length of the wire
b) L =3D (u0 x ur x N x N x A ) / l=20
Which one is correct? And if formula b is correct than what will be the val=
ue of ur for the core that I am using?=20
jess