Help on reducing/removing inductance

[zexclo]

Say I have no choice, but have to do multi-coils, is there any way which I can use capacitance to reduce the impedance in the circuit?
As I have to design this circuit where the coil is designed to be Helmholtz coil, and I am trying to ensure the field strength of the coil is the strongest at the centre, which I conducting some studies on the effect of the magnetic waves.
What I was told is the impedance is causing the field to be not in the centre, hence I am thinking if adding the capacitor will help to "reposition" the strongest field back at centre.

 

[BobW]

The problem that I previously mentioned is that the total length of wire in your coils must be much shorter than the free space half-wavelength of the frequency at which you are operating. The half-wavelength of 50 MHz is 3 metres. Your total length of wire must be much less than this. By "much less" I would estimate no more than 1/10, or 0.3 meters. If your wire length is more than this, you will have uneven current distribution in the coils, and hence an uneven magnetic field, and there is NOTHING that you can add externally, whether it be capacitance or magic pixie dust, to correct this problem.

Meanwhile, I just went back and re-read the posts to make sure that I haven't missed any important specs, and found several confusing statements.

First of all:

Hi, currently I working on a helmholtz coil, to generate a radio-frequency at the centre of the coil, a field strength of 1A to 2A is put into the coil, and the frequency ranges from 10Mhz to 50Mhz.

"field strength of 1A to 2A" makes no sense. The unit A (Amperes) is current, not magnetic field strength. Field strength is measured in Teslas.

Other things that you've posted that are contradictory or confusing:

"Currently, a side of the coil has 74 turns, diameter of 20.5, total length is 48.82m."

"The length of my coil currently is 97.64m, which is adjustable (buy more or cut)"

"Currently, this is my coil, the distance between the two coils are 10.5cm (half the Diameter of the coil)."

"the length of the coil is about 12 cm"

So let's get some basic information:
What is the diameter of your coils?
What is the axial spacing between your coils?
What field strength, in Teslas, do you want to achieve?

And one more confusing item:
"I am trying to make a magnetic coil, to generate a field strength where the maximum strength is at the centre of the coil."
Again, this is confusing, because a Helmholtz coil is intended to create a uniform field. So, do you want a uniform field, or something where the field increases towards the centre?

 

[zexclo]

Firstly, sorry for the confusion.

This is the situation,I am constructing a Helmholtz coil, the input current to the current is 2A, and frequency is about 50Mhz.

Information about the Helmholtz coil.
The diameter of the coil is 20.5cm.
Axial Spacing between the coils is 10.5cm,
inductance is 4.87mh
resistance is 1.43ohm

What I want to achieve,
1. the helmholtz coil able to generate a uniform field.
2. reduce or remove/neutralise the impedance

Thanks so much for the patience and for assistance, so sorry for the confusion

 

[BobW]

One question still unanswered:
What field strength, in Teslas, do you want to achieve?

 

[zexclo]

whats is the greatest field strength that can be achieved from this setup? I am tasked to achieve the highest field strength from this as available

 

[ronsimpson]

I have not done the math. 50mhz, 2A into 5mH will take serious voltage. Do the math! What is the impedance of 5mH at 50mhz? Now what voltage will it take to get 2A? (is it mH or uH? I can't remember) How will you generate this?

Helmholtz coil

I question this word "Helmholtz".

Field strength has to do with amp X turns. 74T x 2A. But we think you can only use 1 turn at 50mhz. So you field strength will be much smaller. I think you will need 1 turn and 140A. And the big question is "How will you generate this power?"

 

[zexclo]

I have not done the math. 50mhz, 2A into 5mH will take serious voltage. Do the math! What is the impedance of 5mH at 50mhz? Now what voltage will it take to get 2A? (is it mH or uH? I can't remember) How will you generate this?

Yup, agree, now that you mentioned. I am really unsure, and reading wiki and stuff just confuses me further.

From the helpful people around, I gathered that, its quite impossible to get any current passing through the wire at 50 Mhz from the designs of 1 and 2,therefore unable to generate any significant magnetic field.

So, maybe I am thinking lets ignore all this information about the 50 Mhz etc, and allow me to hopefully simplify the problem.

I have a power source (power unknown) and a Helmholtz coil.
There is strong impedance/inductance from the helmholtz coil, which according to my friends (and i measured, its 4.87mh)
1. are affecting the "location" of the wave and
2. supposedly, when the input of the voltage/current into the Helmholtz changes, the field strength of the helmholtz should change "significantly". However, currently probably due to the strong impedance/inductance, its not exactly that strong.

Hence, I am thinking, if I can simply look at this circuit at a LC circuit, where the Helmholtz coil is the inductor, with the addition of the capacitor can help resolve the issue?

and also, for Helmholtz coil,I believe that there is always the issue of impedance, how do you "eliminate" the problem and allow the field strength to be controlled by the power source.

Correct me if I am wrong, impedance and inductance is the response from the circuit when one changes the current/voltage into the circuit, and this should be more troublesome for Helmholtz coil?

 

[Tony Stewart]

mh is incorrect and sometimes is misread as millihenry
use uH or better use omega icon and click 3rd letter then ALWAYS capital letters for names abbreviated in units of measure and never multipliers unless it is ambiguuous like mV and MV

So names like Ampere, Volt Watt and Henry are always capitalized when abbreviated and NOT when in full like millihenries.

You may want to reconsider your impedance, Q interwinding errors and determine that you length to diameter ratio is too long for the error tolerance on the diameter and spacing.
( read about HELICAL antenna) I've never seen many more than 10 turns for high gain but used many quad helix 10 turn antenna in my day.

But you want high Q then X(f)/R is the deciding factor then interwinding capacitance and Self resonant frequncy (SRF) and cut the length to diameter ratio to 10:1 , use 1/8" copper tubing. and a Mylar or Polyurethane, cap ultralow ESR cap for tuning the coil. ESR ought to be 10mΩ and copper tubing under 1mΩ/m

Then design it to be self-resonant with drain or collector on coil to Vcc and feedback in colpitts style to emitter and base with small values.

 

[BobW]

Hence, I am thinking, if I can simply look at this circuit at a LC circuit, where the Helmholtz coil is the inductor, with the addition of the capacitor can help resolve the issue?

Yes, you can do this as long as your operating frequency is much less than the self-resonant frequency of the coil. As you approach the self-resonant frequency, the current distribution in the coil will become very uneven. Above self-resonance there will be current reversals within the coil that you cannot compensate for. So, that will be the ultimate limitation on frequency. Given that your coils each contain 48.82 metres of wire, the self-resonant frequency will be about 3 MHz. (I'm assuming that you are connecting them in parallel rather than in series, otherwise the self-resonant frequency will be 1.5 MHz.) That figure assumes a velocity factor of 1.0, and ignores self-capacitance of the coil. In reality, the self-resonance will be considerably lower than 3 MHz. Let's be generous, and say 2.5 MHz. As I said earlier, you'll want to operate well below the self-resonant frequency, probably by a factor of 10. So, your upper frequency limit will be about 250 kHz. So, if you want to limit your experiments to less than 250 kHz then you can use the LC resonance formula to determine the correct capacitance value to cancel out the inductive reactance of the coil:
[latex]C=\frac{1}{4 \pi^2 f^2 L}[/latex]

 

[zexclo]

mh is incorrect and sometimes is misread as millihenry
use uH or better use omega icon and click 3rd letter then ALWAYS capital letters for names abbreviated in units of measure and never multipliers unless it is ambiguuous like mV and MV

Will do so from now on, thanks!

The second part am taking some time to digest...
Am looking into the possibility of designing it to be the self-resonant with drain or collector on coil. Will look into this area, after trying out BobW suggestion on reducing the self-resonant frequency of the coil. Going to try to do it later today.

May I know how you derive the self-resonant frequency of the 48.82m to be around 3Mhz? Is there a formula, as I am looking into the possibility of reducing the length for making the coil...

 

[BobW]

May I know how you derive the self-resonant frequency of the 48.82m to be around 3Mhz? Is there a formula, as I am looking into the possibility of reducing the length for making the coil...

It's the speed of light divided by the length of wire in the coil. In actual fact electromagnetic waves will propagate somewhat more slowly than the speed of light when in a coil (or transmission line). That's where the velocity factor comes in. I used a velocity factor of 1.0 for the initial estimate, although it will most likely be no higher than 0.8.

 

[zexclo]

It's the speed of light divided by the length of wire in the coil. In actual fact electromagnetic waves will propagate somewhat more slowly than the speed of light when in a coil (or transmission line). That's where the velocity factor comes in. I used a velocity factor of 1.0 for the initial estimate, although it will most likely be no higher than 0.8.

From these, can I say that if I am using shorter wire, say 20m,
the self-resonance frequency will be

taking speed of sound to be 300,000,000m/s
length of each wire to be coil, 20m
no. of coils, 2

Assuming,velocity factor of 1
For self-resonant frequency,it will be 300,000,000 divide by (2x20m) = 7.5Mhz?

 

[BobW]

Speed of light, not speed of sound. But your number 300,000,000 m/s is correct.

And yes your calculation is correct. But remember that this is an estimate only, because the velocity factor will always be less than 1.0, and the self-capacitance of the coil will also reduce the self-resonant frequency.

 

[zexclo]

ok, shall try to do this,
length of each wire to be coil = 6m,
no. of coil = 2
Total length = 12m

Shall post my updates on the progress here.

 

[ronsimpson]

length of each wire to be coil = 6m,
no. of coil = 2
Total length = 12m

Why not put the two coils in parallel not series?
no. of coil = 2
Total length = 6m

 

[JimB]

This is a long drawn out thread for just a few turn of wire!

When you have made a suitable coil, what are you going to use to drive 2 amps at 50 MHz through it?
What are you going to use to measure 2 amps at 50 MHz?
Why are you trying to do this? What are you really trying to do?
Are you really going to do this at 50MHz, or is that just some frequency plucked out of thin air?

So, to try and get you going in the right direction, I made two single turn coils and made some measurements.

Coil number 1. A 240mm diameter single turn coil.
Measuring at 50MHz, it had an inductance of 1uH.
It had a self resonant frequency of about 650 MHz. I think, it was a bit difficult to get sensible results.
Calculating, it should resonate at 50MHz when connected in parallel with a 10pF capacitor.
Connecting a variable capacitor set to 10.5pF, I measured a resonant frequency of 47.6MHz.

Coil number 2. A 115mm diameter single turn coil.
Measuring at 50MHz, it had an inductance of 0.358uH.
It had a self resonant frequency of about 167MHz.
Calculating, it should resonate at 50MHz when connected in parallel with a 29pF capacitor.
Connecting a variable capacitor set to 28.5pF, I measured a resonant frequency of 49.1 MHz.

So where is this leading?
If you want to do things at 50MHz, you will not do them with multiturn coils of wire.
Can I suggest that you experiment with small diameter coils, say 100 to 150mm diameter, single turn, connect the two coils in series and make the coil assembly series resonant using a capacitor in the 15pF region, the exact value will depend on the inductance of the coils.

Why make it series resonant?
This all come back to the device which you are going to use to drive this thing.
A series resonant circuit is low impedance, so you do not need a high voltage to get 2 amps to flow.
There will however be high voltages across the coil and the capacitor, so you will need good insulation and a high voltage rated capacitor.

JimB

 

[ronsimpson]

Years ago I made FM radio transmitters. I have experiance with 100mhz resonant transformers. Here is a picture of the capacitors I used to make the inductor resonate. These look like door knobs for size.

You might be able to use a smaller cap like this. Notice the wide spacing between plates. I think your voltage will be very high.
(9 to 38pF) I don't know the voltage or current rating.

 

[JimB]

Nice capacitors there Ron.

This whole enterprise is a non-trivial task.
It also occurs to me that it should be taking place in a screened room.

JimB

 

[ronsimpson]

It also occurs to me that it should be taking place in a screened room.

My guess is that radios for miles around will hear this very clearly. This will upset the FCC or what ever government agency that controls the radio waves in your country. It has been too many years but I think there are radio frequencies set aside for something like this. Example the frequencies for transmitting wireless energy to cell phone chargers. Can't remember but no radio is at these frequencies.
If I had to choose a frequency why not 41-45mHz, 10.7mHz, 455khz or 226khz or 110khz. If you know how radios work you know why these numbers.

 

[zexclo]

Why not put the two coils in parallel not series?

It just random, am just testing. It can be either in parallel or series. Although for series, the self-resonance value will be of a smaller value.

When you have made a suitable coil, what are you going to use to drive 2 amps at 50 MHz through it?
What are you going to use to measure 2 amps at 50 MHz?
Why are you trying to do this? What are you really trying to do?
Are you really going to do this at 50MHz, or is that just some frequency plucked out of thin air?

I have a micro-motor power supply that give 10/50Mhz, 2A p-p, which will be connected to the coil. (I dun really understand what it meant by 10/50Mhz micro-motor power supply, but I am assuming that this should means that the frequency is 50Mhz?）
An oscilloscope will be connected to the coil to measure the frequency, voltage across it.
I am conducting an experiment using the Helmholtz coil, but before that experiment, I want to be able to control the Helmholtz coil. By control, I meant that as the power input increases, the measurements(i.e frequency and voltage measure in the oscilloscope) should respond according and not "swing" or vary wildly.
It should be 10/50Mhz according to the power supply, or at least the power supplied is 10/50Mhz.

It had a self resonant frequency of about 650 MHz. I think, it was a bit difficult to get sensible results.

May I ask, how you find the self-resonant frequency of 650 MHz, do you use equipment for measurement or by calculations?

Yea, I am conducting an experiment in a lab. Its actually a school experiment.