Calculating wire lenght for an air core inductor

In summary: I have a device that can turn them on and off, will this design still work?Yes, the design will still work with the cells hooked up in series.
  • #1
Robin07
139
0
I would like to build an air coil that will create a flux field, which will in turn induce a closed loop coil, that results in an equal and opposite force. The power source is an array of PV cells hooked up in series or parallel which ever will give the air coil the greatest/strongest magnetic field. If I understand this correctly series will increase the amperage and parallel will increase voltage. The closed loop coil may or may not be constructed with a soft iron core.

To make one of these arrays I'm starting off with 6 PV cells that are rated at 1/2V @ 720ma each. How do I calculate the optimum length of wire for the air coil and the closed loop coil?

This question may be too easy to answer for you EEs, so I would rather understand the calculations involved than be given the answer, since the design may alter in the prototyping process it would be better for me to understand what I'm doing.

Thanks
Robin07
 
Engineering news on Phys.org
  • #2
I'm not quite understanding what it is you're trying to do; you may have left out some important details. PV cells produce DC, which will certainly produce a B field around your inductor, but it will not couple to a secondary winding because a static field can not transfer energy.
 
  • #3
If you put solar cells in series their voltages will add, just like batteries.

So, your cells
6 PV cells that are rated at 1/2V @ 720ma each would give 3 volts at 720 mA if they were fully lit by the sun and placed in series.

If you put them in parallel, the result will be the same voltage as anyone of them, but the combined current of all of them. Again, just like batteries.
 
  • #4
PV cells in series increases voltage. You will optimally transfer power to a resistive load if it is about 3V/0.72A = 4.16 ohms. You have a choice of several wire gauges for your coil. I list a few and their resistances:

18 Ga 6.5 ohms per 1000 ft
20 Ga 10.4 ohms per 1000 ft
22 Ga 16.5 ohms per 1000 ft (probably best choice- so about 250 feet)
24 ga 26.2 ohms per 1000 ft
26 ga Too thin for 720 mA

Use Formvar coated wire if possible. A dc current in a coil will NOT induce currents in any other coil.
 
  • #5
Thanks vk6kro, I stand corrected...series adds voltage and parallel adds amps.

Thanks for pointing that out negitron and Bob S the B field that is to be used will be switching on and off repeatedly so that the B field will be expanding and collapsing at a fairly high rate. The rate of switching is determined by the strength of the induced B field, which hasn't been determined as of yet. So, will I need to have the PV cells hooked up in series or parallel to produce the strongest B field possible, I would imagine that maximum amps will do what I'm looking for.

There must be a method to determine optimum wire length...?
 
Last edited:
  • #6
Sorry Bob S, I posted the above while you were posting your response. From your post I calculated 3V x 0.72A = 2.16 not 4.16 ohms. Can you clarify how you came to 4.16 ohms?
I understand how you determined approx. 250' of 22Ga Formvar wire to carry 4.16 ohms, its simple division, thanks.
 
  • #7
Robin07 said:
Sorry Bob S, I posted the above while you were posting your response. From your post I calculated 3V x 0.72A = 2.16 not 4.16 ohms. Can you clarify how you came to 4.16 ohms?
I understand how you determined approx. 250' of 22Ga Formvar wire to carry 4.16 ohms, its simple division, thanks.
6PV cells in seriies @ 0.5 volts each is 3 volts. 3 volts divided by 0.72 amps = 4.16 ohms.
 
  • #8
If you are planning to switch the output of the solar cell array, putting the cells in series would probably be the best way.

Putting them in parallel would give you more current, but the device you use for switching would use up some of your voltage and with only 0.5 volts, you wouldn't have any to spare.

Even in series, 3 volts isn't much and any forces you can generate will be quite small.
 
  • #9
OK, fair enough, voltage may be the better choice here. Correct me but 6 PVs hooked up in a combination of series and parallel could give me 2.5V and 1.44A, this would leave me .5V for the switching device and the maximum voltage for induction. Understand that the switching device is not a mechanism at all but a physical turning away from the source, mainly sunlight, does this switching. In consideration of that, would I still need to take into account the .5V for the switching?

I guess I should be more clear of this design. If you take a rod of let's say 100mm long
x 4mm in diameter, take a skinned softball leaving only the inner foam core. Drill and insert the rod through its' center so that the assemble is balanced. The foam core will have three channels machined into it, each 120 degrees of off center relative to each other, which will house the 250' 22Ga Formvar coated wire. This wire is and forms a closed loop c\w the PV Cells. When the cells and coil are activated, which it is part of, will setup an oriented magnetic field. And If I understand it correctly, as in a homo polar motor. When current moves from point A to point B the force that the wire experiences is tangent to the wire and this will cause the assembly to turn or in this case rotate, at which time the second set of the three PV arrays will be exposed to the light and so on... I'm tickled pink to find that I will need 250 of wire per coil because this lends to weight on the outer extremities of the ball which will likely lend it self to some sort of fly wheel principal adding source input.

There are a few other principals that are involved in the total design but those are on another forum(s).

I don't want to miss lead anyone here but the assemble I described has been around for a while now, I'm simply modifying to get more electricity out.

I must give credit to Xez and Berkeman for the insights. Thanks again.
 
Last edited:
  • #10
Correct me but 6 PVs hooked up in a combination of series and parallel could give me 2.5V and 1.44A, this would leave me .5V for the switching device and the maximum voltage for induction. Understand that the switching device is not a mechanism at all but a physical turning away from the source, mainly sunlight, does this switching. In consideration of that, would I still need to take into account the .5V for the switching?

You would need 5 PE cells to get 2.5 volts and this would still only be able to deliver 0.72 Amps if that is what the individual cells could deliver. If the cells were not being switched you would not lose the any voltage in a switching device.

If you used strong Neodymium magnets and good bearings it should work well. You will probably need more solar panels though. Three banks of 6 would be a start.
 
  • #11
Ok, I'm confused now, 5 PV cells hooked up in series will give me 2.5V, we agree there. Now if I hook up the 6th PV cell in parallel to the 5 PV cell array this should give me 2.5V @ 1.44 amps...No? why not, do we just negate the 6th cell? what happened to its out put? Or, let's say we were going for the most amperage. I would then hook all 6 of them in parallel and the out put should read 1/2V @ 4.32 amps. If I should need .5V for the switching device. I would then hook up 5 PV cells in parallel and hook the 6th to that in series, giving me 1/2V @ 3.2 amps. Is this true?

Where and how is it that you would use NIBs in this assembly?
 
  • #12
Robin07 said:
Ok, I'm confused now, 5 PV cells hooked up in series will give me 2.5V, we agree there. Now if I hook up the 6th PV cell in parallel to the 5 PV cell array this should give me 2.5V @ 1.44 amps...No? why not, do we just negate the 6th cell? what happened to its out put? Or, let's say we were going for the most amperage. I would then hook all 6 of them in parallel and the out put should read 1/2V @ 4.32 amps. If I should need .5V for the switching device. I would then hook up 5 PV cells in parallel and hook the 6th to that in series, giving me 1/2V @ 3.2 amps. Is this true?

Where and how is it that you would use NIBs in this assembly?

A bit of basic theory here. If you have 5 cells and put them in series and then put a 6th one across them, the 5 will try to discharge into the single cell. This is because they have a combined voltage of 2.5 volts in this case and the single cell only has 0.5 volts. The output would only be 0.5 volts if you did that.

So, you could put them in series/parallel, but you need equal numbers of cells in each branch.

You could have all 6 in parallel.
You could have two sets of 3 in series and put them in parallel.
You could have 3 sets of 2 in series and then put these in parallel.
Or you could just put them all in series.
 
  • #13
vk6kro said:
If you used strong Neodymium magnets and good bearings it should work well. You will probably need more solar panels though. Three banks of 6 would be a start.

Thanks, in this design the nibs are the bearing, reducing friction to a minimum. Steel roller bearings, no matter the quality, I have found no successes in.
 

Related to Calculating wire lenght for an air core inductor

1. How do I calculate the length of wire needed for an air core inductor?

The formula for calculating the length of wire for an air core inductor is: L = (μ0 * N^2 * A) / l, where L is the length of wire, μ0 is the permeability of free space (4π x 10^-7 H/m), N is the number of turns in the inductor, A is the cross-sectional area of the coil, and l is the length of the coil.

2. What is the purpose of calculating wire length for an air core inductor?

Calculating the wire length for an air core inductor is important because it helps determine the overall size and efficiency of the inductor. It also ensures that the inductor will have the desired inductance and can handle the expected current without overheating.

3. How do I determine the number of turns for an air core inductor?

The number of turns in an air core inductor can be determined by the desired inductance and the formula: N = √(L * l) / (μ0 * A), where N is the number of turns, L is the desired inductance, l is the length of the coil, μ0 is the permeability of free space, and A is the cross-sectional area of the coil.

4. Can I use any type of wire for an air core inductor?

The choice of wire for an air core inductor can impact its performance. Generally, thicker wire with a lower resistance is preferred for higher power applications, while thinner wire can be used for lower power applications. It is important to choose a wire that can handle the expected current without overheating.

5. How do I calculate the cross-sectional area of a coil for an air core inductor?

The cross-sectional area of a coil for an air core inductor can be calculated using the formula: A = π * (r^2), where A is the cross-sectional area, and r is the radius of the coil. The radius can be determined by dividing the diameter of the coil by 2.

Similar threads

Replies
4
Views
556
Replies
76
Views
7K
  • DIY Projects
Replies
2
Views
2K
  • Electrical Engineering
Replies
4
Views
1K
  • Electrical Engineering
Replies
2
Views
3K
Replies
10
Views
4K
  • Electrical Engineering
Replies
6
Views
2K
Replies
4
Views
2K
  • Electrical Engineering
Replies
4
Views
2K
  • Electromagnetism
Replies
6
Views
2K
Back
Top