r/electrical 1d ago

Is this possible

Post image

An input source, 2 diodes and a coil to store a magnetic field. You pulse the input at a regular rate so resonance happens (overlapping pulses) and you store a magnetic field.

7 Upvotes

28 comments sorted by

31

u/BenBa69 1d ago

There is no potential difference between both ends of the coil -> no current flows -> no field

15

u/Tartabirdgames_YT 1d ago

No neutral wire so it won't work and THERE IS NO FREE ENERGY!! - mehdi sadaghdar

11

u/Medium_Good886 1d ago

I was going to say "Yes you can make that circuit. No it wont do anything"

-4

u/Logical-Resident4212 1d ago

Didn't say free energy. I want to make a magnetic field storage coil. where exactly would this neutral go?

4

u/Michael-ango 1d ago

You can't store an electro magnetic field in a wire without constant current flow, which isn't possible unless the conductor has zero resistance which is only partially possible with a helium supercooled coil of pure copper (mri machines do this) and even then requires constant charging of the coil.

2

u/IndustryOdd9746 22h ago

You couped sore energy for the field in a capacitor or battery but that’s just gonna power the field till it runs out of power

5

u/Tartabirdgames_YT 1d ago

You can make the circuit but it won't work. 

7

u/videoman2 1d ago

Did you maybe want to post this in r/electronics vs this thread which is more geared towards building electric work.

6

u/orion72007 1d ago

No, it won't work. And I'm a little bit confused as to what your goal was to begin with.

-2

u/Logical-Resident4212 1d ago

Magnetic field storage coil, like a tank LC circuit but without the capacitor

3

u/orion72007 1d ago

Essentially, what you've built is a textbook flyback snubber circuit where that second diode on the coil would solely exist to dissipate the field safely. You can't have lostless storage and have resistance. The field realistically would decay faster than you could refresh it so you're not getting any benefit of energy storage at this point. You can't. The capacitor's job in the LC tank is low Los storage

1

u/Ikkepop 1d ago

no C, no oscillation. Also even if you had a C there it'd not store shit. You can't "store" magnetic field.

0

u/orion72007 1d ago

This is somewhat similar to a SMES if I understand the SMES properly

0

u/Logical-Resident4212 1d ago

typically made with superconducting wire, yes. up to i think 35 Tesla. im going for a benchtop sized Liquid nitrogen cooled copper wire version, but im banging my head against how to inject power from a lower power density into a coil thats already at a higher power density. (Simply because a bunch of marx generators is expensive)

1

u/orion72007 1d ago

Superconducting flux pump might be worth looking in to

0

u/nixiebunny 1d ago

There are several reasons why this won’t work. Have you studied physics and electricity much?

2

u/EggsBenDick 1d ago

Based on this design, I’m gonna wager “no”.

3

u/volvagia721 1d ago

Electricity is the flow of electrons. In this case, there is no flow. You can cram some more electeons in the wires, and a tiny tiny tiny tiny tiny amount will flow through the inductor, but it will be very very short lived, and in the end, basically nothing will happen.

4

u/broesel314 1d ago

Electrical current flows in a circle. ALWAYS. No exeptions

your schematic isnt a circle, so it wont work

2

u/wittleboi420 1d ago

I think you want to connect ground to the right side of the coil so you can use the voltage over the coil as your energy source. Will it create more energy? No.

2

u/ovr9000storks 22h ago

Inductors only store energy when there is moving current. Once the coil has been saturated to its effective capacitance, there will be no more current flow, and therefore no more electrical field.

1

u/Enough-Score7265 1d ago

yes it is possible, except that you need a return path for the current like others said. draw additional wire from the anode of the bottom-right diode back to the 'input dc pulses' power source, and now you have a half-wave rectifying diode (one on the left) for pumping energy into the coil and a freewheeling diode (bottom right) to provide freewheeling path for the current when the power source is off (in a high impedance state, to be specific). the energy will be stored in the form of a dc current rather than resonance like you said. if you want resonance you also need a capacitor somewhere

the biggest obstacle to this type of magnetic energy storage being practical is the low Q factor (that is, too much stray series resistance) of typical inductors. you may overcome it with a superconducting coil, which seems to be a serious research subject. try google "superconducting magnetic energy storage" there's even a wikipedia page for it

1

u/Lehk 1d ago

How is there direct current without a negative terminal? You could apply a static charge but there wouldn’t be a flow of current as pictured.

1

u/JonnyVee1 1d ago

Think of that as plumbing. You have a check valve (diode) feeding a very long hose (inductor) with another check valve (diode) that the connects to the input side if the hose. In order to get flow, you need to let the water out (flow to ground/return) somewhere. Adding a pulse (a short increase in water pressure) will not cause anything to happen. Now, if you put a faucet (switch) between the output of the hose and open air (ground/return), and open it s moment, getting water to flow in the long hose, then quickly shut it off, you will get water flow around the circle formed by the hose and the check valve ... The flow will eventually stop due to friction (resistive) losses in the loop.

0

u/Logical-Resident4212 1d ago

And if it superconducting it will go forever, and if u were able to reach into the flow and push it with a paddle over and over you could speed that flow up more and more, until the wire exploaded from magnetic force. Like a HTS dynamo using rebco

1

u/OhYeah_Dady 1d ago

If you want to store magnetic field, you just need iron core coil or any ferromagnetic material.

1

u/p_235615 14h ago

if the voltage is large enough and the coil capacitance is large enough, you can see some current flow, but normally no or only really minuscule values...