1A current at 11 MHz into a toroidal circuit

In summary, the conversation revolves around the calculation of voltage needed to generate a 1A current at 11 MHz into a toroid coil with specific dimensions. The inductance parameter, A_L, and the number of turns are important factors in determining the voltage needed, and the formula V= 2 \pi fIuN^2r^2 / D can be used to calculate it. However, the correct values for mu, the radius (r) and diameter (D) of the toroid must be used for accurate calculations. The use of the term "solenoid" in this conversation refers to the shape of the core, and not necessarily its function as an actuator. Overall, the purpose of the project
  • #1
jumpjack
222
3
I need to generate a 1A current at 11 MHz into a solenoid around a toroid like this:
http://cgi.ebay.it/TOROIDE-T200-2-ROSSO-1-30-MHZ-NUOVO-/280683303561?pt=Ham_HF_VHF_UHF_SHF&hash=item415a075289

Out Diameter= 5cm
In Diameter = 3 cm
wire diameter: 1mm
spires number: 100

Which device can I use/build to generate such a current inside this circuit?
 
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  • #2
That very much depends on the number of turns you intend to use.

The coil specifies an inductance of 12 nH per Turn squared, so for example with just one turn you'd only need about one volt @11 MHz to get you a one amp current. At 35 turns however you'd need nearly 1000 volts @ 11MHz.

Why don't you tell us what you're trying to build.

BTW. A "solenoid around a toroid" is just called a toroid. The term "solenoid" is used to describe a straight cylinder shaped coil and the term "toroid" is used to described the coil that is wound on a circle like a doughnut shape.
 
  • #3
uart said:
That very much depends on the number of turns you intend to use.

The coil specifies an inductance of 12 nH per Turn squared, so for example with just one turn you'd only need about one volt @11 MHz to get you a one amp current. At 35 turns however you'd need nearly 1000 volts @ 11MHz.

Why don't you tell us what you're trying to build.

BTW. A "a solenoid around a toroid" is just called a toroid. The term "solenoid" is used to describe a straight cylinder shaped coil and the term "toroid" is used to described the coil that is wound on a circle like a doughnut shape.

Thanks for clarifications.
But then how do you name the core of a toroid?!? Doughnut does not look very "scientific" :tongue:

Back to physics... how do you calculate voltage from spires number?
I'd need 100 spires! Does it mean 3 kV?
 
  • #4
jumpjack said:
Thanks for clarifications.
But then how do you name the core of a toroid?!? Doughnut does not look very "scientific"
That's why we give it the name toroid.

Back to physics... how do you calculate voltage from spires number?
I'd need 100 spires! Does it mean 3 kV?

No it's proportional to the number of turns squared, so it more like about 8 kV

Once again. To avoid potentially wasting peoples time could you please tell us what you're trying to achieve with this exercise. :)
 
  • #5
uart said:
No it's proportional to the number of turns squared, so it more like about 8 kV

Once again. To avoid potentially wasting peoples time could you please tell us what you're trying to achieve with this exercise. :)

Sorry but I can't.
But which is the formula you used?
 
  • #6
jumpjack said:
Sorry but I can't.
Yeah it's "free energy" related isn't it? Usually when people are have little idea what they're doing and are asking "not quite right" questions, then it's "free energy" related. Having little idea of what you're doing is actually a prerequisite for working on "free energy" you know. :uhh:[tex]L = A_L N^2[/tex]
 
  • #7
uart said:
Yeah it's "free energy" related isn't it?
Wrong.
Better luck next time. :wink:

uart said:
[tex]L = A_L N^2[/tex]
I can't see V in this formula, so how do I get V from it and my data?!?
 
  • #8
jumpjack said:
Wrong.
Better luck next time. :wink:
Well you'll have to forgive me. "Free energy" (as with other banned topics) really is one of the most common reasons why people are reluctant to divulge what they are trying to achieve.
I can't see V in this formula, so how do I get V from it and my data?!?

[tex]V= 2 \pi f L I[/tex]
 
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  • #9
uart said:
Well you'll have to forgive me. Free energy (as with other banned topics) is one of the most common reasons why people are reluctant to divulge what they are trying to achieve.
Fuuny, forum engine automatically links "free energy" to REAL free energy rather than the one you're talking about! :wink:



uart said:
[tex]V= 2 \pi f L I[/tex]
thanks.

So let me see if I understood:
[tex]V= 2 \pi f A_L N^2 I[/tex]

N= spires number
What is Al?
 
  • #10
[itex]A_L[/itex] is a parameter of the core and is given on the data sheet as 12 nH/N^2. (One nH is 10^(-9) H).

It's only valid if the core is not saturated, meaning that the flux density is less than about 250 to 300 mT for most ferrite cores and less then about 1000 to 1500 mT for most powdered iron cores.

The flux is equal to (L I / N) and the flux density is equal to the flux divided by the cross sectional area.
 
  • #11
I found this for a toroid:
19451e2f18b11c62eb5764f0d4136b29.png


So is this correct?
[tex]V= 2 \pi f L I [/tex]
[tex]L = (uN^2r^2)/D[/tex]

Hence:
[tex]V= 2 \pi fIuN^2r^2 / D [/tex]

So for 11 MHz and 1A:
[tex]V= 6,28 * 1.1 e^7 * u * 100^2 * (2 * 10^-2)^2 / (5 * 10^-2)[/tex]

I think I get 69 MV ?!?
 
  • #12
What value are you using for mu? Your formula looks ok but I suspect you have the wrong value for mu. The inductance parameter "A_L" is basically just an indirect way of specifying mu.

Edit : No you've also got the wrong values for r and D. You'd do better to use the area (A) and length (l) values given on the datasheet than the *approximation* r^2/D anyway.
 
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  • #13
What would a solenoid be doing with 11 MHz anyway? I'm just an inexperienced newbie, but I'm having a really hard time believing that anything mechanical could actuate 11 million times a second.
 
  • #14
Hi Jiggy. The term "solenoid" really just describes the geometry (long straight cylinder) of the core. Because long straight cylinder cores are naturally used for linear actuators, a lot of people use the term "solenoid" synonymously with "solenoid valve" or "solenoid actuator", but it doesn't have to be an actuator.

In this case the OP was actually referring to a "solenoid wound in the shape of a toroid", which doesn't make sense. He was really just referring to a toroid inductor all along. What he wants to do with it however is still anyone's guess. :)
 
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  • #15
uart said:
What value are you using for mu? Your formula looks ok but I suspect you have the wrong value for mu. The inductance parameter "A_L" is basically just an indirect way of specifying mu.

Edit : No you've also got the wrong values for r and D. You'd do better to use the area (A) and length (l) values given on the datasheet than the *approximation* r^2/D anyway.

Actually on datasheet I have internal radius, external radius and thickness.
I consider u=1.26 e-6 and ur = 800 (iron)

But doing calculations again I get around 6 MV anyway...
 
  • #16
- That won't be pure iron. Powdered Iron cores can have [itex]\mu_r[/itex] anywhere from 5 to 500.

- The term [itex]r^2/D[/itex] is just an approximation for Area/Length. Use the correct expression which is [itex]A/l[/itex] as given on the data sheet.
 
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  • #17
So I have this formula:

X(t)= k0 u0 ur N I O cos (O*t) / (4 pi^2 r R)

I'd need x=9,8 , but if I could just obtain X=0,098 could be acceptable in a first instance.

k0=4,31E-10
u0=1,26E-06
ur depends on material: 800 iron, 5000 ferrite, 100.000 permalloy
N = spires
I = current
O = 2 pi f (f=frequency)
r = core thickness
R = core radius
 
  • #18
none?
 
  • #19
Energy in an inductor is equal to the current squared times the inductance all halved. (E=1/2*L*I*I) This is based on a one second interval. With AC it gets a bit tricky. You establish the 1 Amp current flow in one direction then reduce it to zero and then make it flow in the opposite direction. So wouldn't the equation now be;
E=1/2*L*I*I*f
L is inductance in Henries
I is current in Amps
f is frequency
Where are you going to get the energy to run through this inductor?
 
  • #20
That's why I wrote:
I'd need x=9,8 , but if I could just obtain X=0,098 could be acceptable in a first instance.
 

1) What is the significance of a 1A current at 11 MHz in a toroidal circuit?

The 1A current at 11 MHz is the amount of electrical current passing through the toroidal circuit at a frequency of 11 megahertz. This current is important because it determines the strength and efficiency of the circuit's performance.

2) How does the frequency of 11 MHz affect the behavior of the toroidal circuit?

The frequency of 11 MHz is in the radio frequency range, which is commonly used for communication and transmission of signals. This frequency allows the toroidal circuit to efficiently transfer and manipulate electrical energy, making it suitable for use in various electronic devices.

3) What is a toroidal circuit and how does it differ from other types of circuits?

A toroidal circuit is a type of electrical circuit that utilizes a toroidal (doughnut-shaped) core to concentrate and control the flow of electrical energy. This design allows for a more compact and efficient circuit compared to traditional linear circuits.

4) What factors affect the performance of a toroidal circuit operating at 1A and 11 MHz?

The performance of a toroidal circuit is affected by several factors, including the quality of the core material, the number of turns in the winding, and the frequency and amplitude of the input current. Other factors such as temperature, impedance, and resistance can also impact the circuit's performance.

5) What are some practical applications of a toroidal circuit operating at 1A and 11 MHz?

Toroidal circuits are commonly used in electronic devices such as transformers, inductors, and power supplies. They are also used in radio and communication systems, as well as in medical equipment, automotive electronics, and renewable energy systems. The compact size and high efficiency of toroidal circuits make them ideal for these applications.

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