# Electromagnetic shorting

ritecliq
Hi Guys,
I'm hoping that someone here can help me.

I have an electromagnet that when pulsed creates a short across the 30v supply. If I increase the resistance in series with the electromagnet to stop the short, will this reduce the strength of the electromagnet? Is there a way to store the energy in a capacitor then isolate the electromagnet from the ground to discharge the capacitor through the electromagnet? The electromagnet has a resistance of 5 ohms. Has 600 turns of 18 AWG wire and the iron core is 200mm long.

Regards,
Dale.

## Answers and Replies

Gold Member
You probably have an insulation breakdown. Did you use transformer wire?

Also your current is 30 V/ 5 Ohms = 6 Amps ... but 18 AWG is only rated for 2.3 Amps.

Homework Helper
I have an electromagnet that when pulsed creates a short across the 30v supply.
DC supply?
You mean that the coil trips the PSU circuit breaker?

The coil is a 5Ohm resistor - what is the current it will try to draw at 30V?
What is the current the PSU is limited to?
You can, temporarily, get high currents by discharging a capacitor through the coil - yep - or use a tougher power supply. You can increase the strength of the magnetic field by using more turns on the coil too.
Do you know how the magnetic field depends on the current and the number of turns?
For a fixed DC voltage, the current also depends on the number of turns.

Or do you mean that it was 5Ohms, but it is now closer to zero (that would be a "sort circuit")... in which case, see post #2 above.

ritecliq
Thanks for your help guys,

So I need to increase the number of turns to increase the resistance to have less amps across the wire? Will this reduce the magnetic field generated by the electromagnet?

Yes I have used brand new enamelled copper wire bought in a roll and not reused transformer wire.

I measured the 5 Ohm resistance of the coil with a multimeter. Is this wrong? The idea is to generate a strong magnetic field without shorting the dc psu.

Using the bottom formula on: http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/solenoid.html I should be generating a large magnetic field but doesn't work out that way. Am I using the wrong formula?

Gold Member
Your question is a little confusing. If you apply 30V across a 5 ohm coil the current will quickly ramp to 6 amps. Are you saying that appears as a short to your 30V supply?

Or, are you saying that it ramps to 6A and then becomes shorted?

Does your 30V supply still have 30V across it with the coil connected?

Jon B
A high current pulse applied to 600 turns of 18 gage wire will produce a hefty magnetic pulse if the power supply can handle the high current pulse. Many more turns of 18 gage magnet wire would most likely increase the strength of the field and provide a better match between it and the power source. The capacitor bank idea would be like the capacitive discharge ignitions used on older cars. Good idea.

ritecliq
Your question is a little confusing. If you apply 30V across a 5 ohm coil the current will quickly ramp to 6 amps. Are you saying that appears as a short to your 30V supply?

Or, are you saying that it ramps to 6A and then becomes shorted?

Does your 30V supply still have 30V across it with the coil connected?

Sorry to be confusing. When the electromagnet has 30V applied, it does short to my 30V supply.

Thanks,
Dale.

Gold Member
Sorry, I want to get to the bottom of the "short". It requires over 238 meters of 18G wire to produce 5 ohms. What does your meter read when the meter leads are shorted? 238/600 = 0.4 average meters of wire per turn, which implies a pretty thick core.

What are the specifications for your power supply? Can it really supply >6A at 30V (180 watts). If not, I assume that is what you mean by "becomes shorted". Does it blow a fuse? Current limit? Shut down?

Doubling the turns will increase the resistance, reducing the current. But, the increased turns will compensate. Limiting the current with a resistor is an option, but it would need to be a big power resistor.

What are you using for a core? At what point will it saturate? Do you know about magnetic saturation? http://en.wikipedia.org/wiki/Saturation_(magnetic)

As for a capacitor bank, charging and discharging such a bank is non-trivial also. It will also short your supply, so needs to be current limited as it is charged. Needs a pretty hefty switch also. How long of a pulse do you need? Does it need to be automatic?

ritecliq
Electromagnets and voltage

Hi Guys,
I have a short DC Pulse of 30V 1A and I am trying to feed this through a electromagnet to get the biggest magnetic field possible. I have been told that electromagnets require voltage and not current. Is this correct? Using different circuits, I have boosted the voltage to 200V DC. My question is: Will boosting the voltage give me a stronger magnetic field? As now I have 200V through the 5 ohm electromagnet, therefore my current across the electromagnet is 20 Amps. Am I right in assuming this?

Any feedback would be appreciated
-Dale

2021 Award
ritecliq said:
I have been told that electromagnets require voltage and not current. Is this correct?
No. The voltage is needed across the coil to get the current flowing through it, but current is then limited by the resistance.

ritecliq said:
As now I have 200V through the 5 ohm electromagnet, therefore my current across the electromagnet is 20 Amps. Am I right in assuming this?
No. Ohms law says I = V/R. Therefore I = 200/5 = 40 amp maximum current.

How short is a short DC pulse? One word has not been mentioned; Inductance.

The magnetic field as measured in “ampere*turns” is proportional to the number of turns but there are two problems. 1. Resistance is proportional to the number of turns. That limits current. 2. Inductance is proportional to the square of the number of turns. Inductance limits the rise time of the current.

The voltage across the coil due to inductance will be; V = inductance * di/dt
The voltage across the coil due to resistance will be; V = I * R
The total voltage is the sum of those two.

I suspect the negative inductive voltage spike that must occur when you stop the current has broken through the insulation of your coil. You now have a partial short circuit in your coil.

You need to have some device such as a reverse biassed power diode across the coil to catch that spike and so prevent damage to the coil.

For a, four times faster rising pulse, you need to use half as many turns with twice the thickness wire. That needs half the voltage and twice the current.

ritecliq
No. The voltage is needed across the coil to get the current flowing through it, but current is then limited by the resistance.

No. Ohms law says I = V/R. Therefore I = 200/5 = 40 amp maximum current.

How short is a short DC pulse? One word has not been mentioned; Inductance.

The magnetic field as measured in “ampere*turns” is proportional to the number of turns but there are two problems. 1. Resistance is proportional to the number of turns. That limits current. 2. Inductance is proportional to the square of the number of turns. Inductance limits the rise time of the current.

The voltage across the coil due to inductance will be; V = inductance * di/dt
The voltage across the coil due to resistance will be; V = I * R
The total voltage is the sum of those two.

I suspect the negative inductive voltage spike that must occur when you stop the current has broken through the insulation of your coil. You now have a partial short circuit in your coil.

You need to have some device such as a reverse biassed power diode across the coil to catch that spike and so prevent damage to the coil.

For a, four times faster rising pulse, you need to use half as many turns with twice the thickness wire. That needs half the voltage and twice the current.

Thanks for the help. Sorry I can't seem to grasp this. I meant to say 40 Amps across the coil. The coil is 600 turns (3 layers of 200) of 1mm enamelled copper wire. My pulse is rather short, just a discharge from a 4700uF Cap. I'm trying to make the biggest magnetic field I can with the 30V 1Amp stored in the cap. I generally do have a reversing diode across the coil.

I will try your idea of twice as thick wire with half the turns. Do you know anyone who could tutor me on this?

Regards,
Dale.

2021 Award
What is the inside diameter of the coil you have now?

You must charge the capacitor from your supply through a resistor of about 30 ohms to limit the supply current into the capacitor. That will prevent the supply fuse blowing.

You must have a reverse biassed power diode, rated at about 30 amps, across your coil. That will protect the coil from a spike, and the capacitor and supply from the negative 30V expected after the pulse.

You should disconnect the charging supply from the capacitor before connecting the capacitor to the coil.

Your 4700uF capacitor will have some resistance in it's structure. That will also tend to limit the maximum current.

To understand what is happening with your circuit you need to develop an understanding of voltage, current and resistance = Ohm's Law. Then charge, power and energy. Then capacitance and inductance.

Tutor? You are probably better asking questions here on PF in the short term. That way you will get the help you need from those who can give it.

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Gold Member
It is certainly difficult to piece together exactly what you are doing from all the words. A drawing would certainly help. You didn't mention what you are using for a switch and how long it is closed. Does it stay closed for full discharge?

When you connect the 4700uF capacitor to the inductor you have an RLC circuit which will behave differently based on the C,L and R terms. Eventually all the energy will be dissipated. But how it behaves while dissipating is value dependent (you should read about series RLC circuits).

Saying you have 30V 1A stored in the Capacitor is inaccurate. You have 30V stored on the capacitor (a capacitor charged to 30V). It can supply huge currents (but the inductor will "resist" and the current will ramp up).

When you apply the capacitor to the Inductor the 30V appears across the coil instantly. The coil current starts at 0 and ramps up, building a field. At the same time the voltage across the capacitor starts to drop since you are removing charge. What happens when the capacitor voltage gets to zero will depend on several factors.

You can read about RLC circuits here http://en.wikipedia.org/wiki/RLC_circuit.

ritecliq
Hi Guys,
Attached is the simple diagram.

When the DPST switch closes, the capacitor charges up. Then the DPST is open to isolate the electromagnet from the supply. When the push button switch closes, the cap discharges through the 20 ohm resistor into the electromagnet. The 20ohm resistor is to protect from switch bounce and the inrush of current to the electromagnet. Hope this diagram helps in explaining things.

The size of the iron core coil is 40mm x 10mm x 200mm wrapped with 600 turns of 1mm enamelled copper wire.

Yes the push button switch stays closed for the full discharge of the cap to the electromagnet.

-Dale.

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2021 Award
1. The 20 ohm resistor is quite unnecessary. Initial current will be limited by the coil's inductance, then final current by the coil's resistance.

2. The diode across the coil protects the winding's insulation and the switch from an inductive spike and arc. But the diode will keep the current flowing in the coil, which will lengthen the tail of the magnetic pulse time.

3. The iron core will need to be laminated or of an iron powder or ferrite material. (Transformers are wound on laminated cores that have laminations thin enough to become magnetised in about 5 milliseconds). If it is a solid iron core as suggested then it will take one second or more to build up the magnetic field. If you want the fastest rise in magnetic field you cannot use a solid mild steel core, you must avoid magnetic materials and use air. How fast do you want the field to rise?

ritecliq
1. The 20 ohm resistor is quite unnecessary. Initial current will be limited by the coil's inductance, then final current by the coil's resistance.

2. The diode across the coil protects the winding's insulation and the switch from an inductive spike and arc. But the diode will keep the current flowing in the coil, which will lengthen the tail of the magnetic pulse time.

3. The iron core will need to be laminated or of an iron powder or ferrite material. (Transformers are wound on laminated cores that have laminations thin enough to become magnetised in about 5 milliseconds). If it is a solid iron core as suggested then it will take one second or more to build up the magnetic field. If you want the fastest rise in magnetic field you cannot use a solid mild steel core, you must avoid magnetic materials and use air. How fast do you want the field to rise?

Hi,
The core is a laminated core, exactly the same as a transformer core. Sorry I have another electromagnet that is a solid core. Got them mixed up. But the one I am using has a laminated core. 5 milliseconds I'm sure is enough. I have been looking into RLC circuits.

The aim is to create the biggest and sustained magnetic field I can create with the voltage stored in the capacitor. With the info and the diagram I provided, do you think that I need to make another laminated core electromagnet with 400 turns of 2mm wire?

Also I can take out the de-bounce resistor. From my understanding, this should give me more current across the coil. The formula's is what tricks me up. Slowly it is sinking in. Besides a PHD, Is there anything else I can do to achieve my goal?

Thanks,
Dale.

Gold Member
1. You definitely want a protection diode in case of switch bounce and pre-mature release.
2. Not sure you need a DPDT to charge the capacitor. A SPST might suffice.

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/indsol.html says your inductance with an iron core and 600 turns is over 18mH. (I said 200 relative perm, 0.36cm area, 20cm long)

If the circuit is underdamped then the polarity on the capacitor will reverse, which you don't want.

Damping factor less than 0.707 (critical damping) will have ringing.

[STRIKE]damping = (R/2) sqrt (L/C)[/STRIKE] <------ EDIT: THIS FORMULA IS WRONG

Assuming R is 5 ohms and C=4700uF, there will be adequate damping if sqrt (L/C) > 0.2828
L/C > 0.53
L > 2.4mH

You need to consider losses in the core and ESR of the capacitor as part of the total effective R.

Check my numbers because I'm notoriously bad at cranking out numbers.

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2021 Award
@meBigGuy. There is a diode across the coil in the diagram.

The capacitor is therefore prevented from becoming significantly reverse charged, there can be no oscillation.

Once the capacitor is discharged, the inductor current circulates through the coil and diode only. The diode keeps the voltage across the inductor low, which will increases the current and field decay time, since V=L*di/dt.

Gold Member
You are right, of course.

ritecliq
1. You definitely want a protection diode in case of switch bounce and pre-mature release.
2. Not sure you need a DPDT to charge the capacitor. A SPST might suffice.

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/indsol.html says your inductance with an iron core and 600 turns is over 18mH. (I said 200 relative perm, 0.36cm area, 20cm long)

If the circuit is underdamped then the polarity on the capacitor will reverse, which you don't want.

Damping factor less than 0.707 (critical damping) will have ringing.

damping = (R/2) sqrt (L/C)
Assuming R is 5 ohms and C=4700uF, there will be adequate damping if sqrt (L/C) > 0.2828
L/C > 0.53
L > 2.4mH

You need to consider losses in the core and ESR of the capacitor as part of the total effective R.

Check my numbers because I'm notoriously bad at cranking out numbers.

@meBigGuy
Forgive my ignorance but how did you work out the coil radius?

-Dale.

Gold Member
I guessed. And made a mistake (or 2). I just said that 10x40 was 400mm^2 which is not equal to 0.4cm^2, but rather to .04cm2. I typed in a radius that gave a close number (should have just entered the area) And I also ignored the wire thickness. using .04cm^2 gives an inductance of ~2mH. Using twice that gives 3.6mH.

ritecliq
I guessed. And made a mistake (or 2). I just said that 10x40 was 400mm^2 which is not equal to 0.4cm^2, but rather to .04cm2. I typed in a radius that gave a close number (should have just entered the area) And I also ignored the wire thickness. using .04cm^2 gives an inductance of ~2mH. Using twice that gives 3.6mH.

Hi meBigGuy,
Ok so the radius 400mm^2 expressed in meters = .04m^2 or 40cm^2. A coil radius (r) of 3.57 gives an area (A) of 40cm^2. Is this correct? So the coil has an inductance of 1811mH?

Thanks,
Dale.

Gold Member
doh ---- what the heck am I doing. 10mm (1cm) x 40mm (4cm) is 4cm^2 which is 180mH.

I just played with a simulator at http://www.falstad.com/circuit/ at it showed ringing with 5 ohms, 180mh and 4700uF, which does not match what I calculated earlier.

I mistyped the damping equation. It should be Damping = (R/2) sqrt (C/L). (I wrote L/C and believed it)

So, for 5 ohms, 4700uF, 180mH the damping is 0.4 which means it will ring.

The simulator showed a peak current of 2.9 amps in about 40ms. The inductance slows the current rise and the capacitor voltage is dropping so you don't get the current you would expect.

Gold Member
Actually, I am the wrong person for this. My practical magnetics knowledge sucks. Baluncore is the guy. I can't help you with saturation, etc.

ritecliq
Actually, I am the wrong person for this. My practical magnetics knowledge sucks. Baluncore is the guy. I can't help you with saturation, etc.

Hi Guys,
I am looking at having a new electromagnet created using 2mm wire and 200 turns at 200mm. I want to use sheets of mu-metal or metglas (higher permeability) as the core. Is this recommended and will this still give me magnetisation in 5 milliseconds like the iron laminated core?

Thanks,
Dale.

2021 Award
@ meBigGuy. I'm no magnetician.

@ ritecliq: Well, does it rise in 5 msec ?
Have you measured the time profile of the current in the coil ?

We don't know what you do with the field, or how you close the magnetic path outside the core.
Nor do we know why you are doing this, so it is difficult to advise you.

Why a more expensive core? That will increase the inductance and so slow the rise-time of the field current.

The field can be estimated in terms of ampere*turns. So you need lots of amps and turns... But;
1. You must use few turns to keep the inductance down, so current will rise fast. Unfortunately, inductance is proportional to turns squared.
2. You must select wire that is fat enough to keep the resistance down, so current limit is higher.
There is no point in a very fast rise if the current is then limited by resistance, as you would be better increasing the number of turns.

Once the current is circulating in the coil and diode, the field decay is proportional to the coil resistance.

The absolute limitation on coil current is the quality or internal resistance of the 4700uF capacitor.
Maybe you should use several low esr capacitors in parallel. But the best rise-time and energy transfer will be when coil resistance is similar to capacitor esr, so get the coil resistance down towards Cesr first.

Keep your core and so your coil diameter small, that makes a lower resistance coil and therefore higher current. The core material must not saturate at your maximum ampere*turns. That will set the core cross sectional area needed.

If you knew what you wanted you could juggle all those numbers and produce an optimum design.

Gold Member
You need to understand how the inductance, resistance, and capacitance conspire to cause a 180mH 5ohm coil with a 4700uF capacitor charged to 30V to only rise to 3A and take 40ms to do it. (maybe you understand already)

Think of the 5 ohm resistance as being a resistor in series with an ideal coil. The voltage across that ideal coil is what controls the rate of current increase (di/dt = V/L). But, as the current increases, the voltage across the coil decreases, so the current increases slower. The voltage across the coil decreases for 2 reasons: Drop across the resistor and capacitor discharge.

For example, once the current reaches 3A there is a 15V drop across the resistor and the capacitor has discharged, maybe by 5 volts.

When the 30V was initially applied, the current slew rate (di/dt) would be (based on V = Ldi/dt) V/L = 0.166/ms so it would (if not for resistance drop and capacitance droop) rise to 6 amps in 40ms. But that can only happen with an ideal voltage source and ideal inductor.

When the coil voltage drops to 10V, current will only slew at 0.016ma/ms.

Does that make sense? This ignores any saturation effects.

leftcliq
Hi Guys,
I am planning to use the 12AWG wire with a length of 20 meters which has a resistance of 0.02 ohms. So 30 Volts divided by the total resistance ( 0.02 ohms + resistance of the 4700uF Cap) will give me the total amps across the coil? I am trying to get my coil resistance down as far as possible.

I haven't wound the coil yet, still trying to do the math. I am planning on using a much smaller laminated core then previous. The core size is 13mm x 13mm x 75mm.

I thought about using several low ESR capacitors in parallel but wouldn't there resistance(s) add to the total resistance of the circuit? I thought I would do better with just 1 low ESR capacitor.

The idea is to use this 30V 1A to create the strongest electromagnetic field possible. I am some what flexible in my size of the core, number of turns etc.

Regards,
Dale.

Gold Member
Draw a real capacitor as an ideal capacitor with a series resistor. Draw several real capacitors in parallel.

oneamp
No. The voltage is needed across the coil to get the current flowing through it, but current is then limited by the resistance.

No. Ohms law says I = V/R. Therefore I = 200/5 = 40 amp maximum current.

How short is a short DC pulse? One word has not been mentioned; Inductance.

The magnetic field as measured in “ampere*turns” is proportional to the number of turns but there are two problems. 1. Resistance is proportional to the number of turns. That limits current. 2. Inductance is proportional to the square of the number of turns. Inductance limits the rise time of the current.

The voltage across the coil due to inductance will be; V = inductance * di/dt
The voltage across the coil due to resistance will be; V = I * R
The total voltage is the sum of those two.

I suspect the negative inductive voltage spike that must occur when you stop the current has broken through the insulation of your coil. You now have a partial short circuit in your coil.

You need to have some device such as a reverse biassed power diode across the coil to catch that spike and so prevent damage to the coil.

For a, four times faster rising pulse, you need to use half as many turns with twice the thickness wire. That needs half the voltage and twice the current.

How do you know the spike will happen when power is switched off, instead of when power is switched on? Aren't the charge and discharge current functions (di/dt) equal?

Thanks

RBTO
This might help:

The magnetic field strength of the coil is directly related to the number of turns and the amount of current in the coil. The amount of flux developed is then related to this and the material used for the core (its reluctance). The better that material is, the more flux, however the core must provide a complete path (donut) or air will come into play for any gap that exists and thus reduce the flux. A problem with better core material is it causes the coil to have a higher inductance (read below). Air will produce the least inductance.

The maximum current you can get in the coil will depend on the coil resistance and the voltage applied to it. If you want a large current, the resistance needs to be low and/or a high voltage needs to be applied. If a high voltage is applied, the coil insulation will need to withstand it. Example: if you have 10 turns and apply 10kV, the turn-to-turn insulation will need to withstand 1kV (assuming they're wound side-by-side). (The 10kV distributes evenly across the 10 turns.) The maximum current isn't necessarily V/Rc however (read on).

The pulse of current developed (hence the magnetic field) will ramp up to a theoretical maximum of (V/Rc) where Rc is the resistance of the coil over a period of time determined by the inductance of the coil, its resistance, and any external series resistance in the circuit (i.e., voltage source and its connections). The less inductance (fewer turns and less permeable core) and resistance, the faster this current will ramp up. If you are supplying the coil from a capacitor (the most logical choice for a high current pulse), the current will not reach the maximum predicted value (V/Rc) because the capacitor voltage drops as the current is ramping up. The larger the capacitance, the less this will be a factor, but the size of your capacitor will require some practical considerations, and larger capacitors tend to have more inductance themselves (compared to smaller valued capacitors). Leads to the coil can also have inductance (as can the capacitor) and will add to the overall inductance limiting how fast the current can rise (and what voltage the capacitor will have left when that current hits its maximum).

To develop a high magnetic pulse, you need low inductance and resistance in the coil, and you need to apply a high voltage to the coil (probably from a capacitor source as mentioned). Adding iron to the core will allow you to direct and concentrate this field, but at the expense of considerable added inductance. Paralleling smaller capacitors to obtain a larger capacitance will allow you to decrease capacitor inductance. Keep in mind the capacitor you end up with must be rated for a higher working voltage than its fully charged value.

Part of your challenge is to switch the capacitor to the inductor load using a switch which closes very fast and has low resistance itself. Otherwise, the switch will limit the current and thus the magnetic field.

If your coil shorts, an Ohmmeter may not show it, since the coil may have sparked over (between windings) when the voltage was applied to it (or a self-induced voltage caused the breakdown). Those shorts usually clear themselves once the spark terminates, and the coil will look fine otherwise. Breakdown is very unlikely at a low voltage unless the current flow in the coil is suddenly changed, and then the coil will develop its own high voltage which can break down insulation. This is not the same as suddenly applying a high voltage, and the self-induced voltage could be considerably higher than any applied voltage.

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2021 Award
oneamp said:
How do you know the spike will happen when power is switched off, instead of when power is switched on? Aren't the charge and discharge current functions (di/dt) equal?
Initially there is no current in the inductor so there is no energy stored in the inductor. Subjecting the inductor to a voltage step results in an increasing, (integrating), current flow through the inductor. The step voltage applied to the inductor when first connected to the charged capacitor is finite and well controlled.

When a current is flowing through an inductor, it's terminals can be shorted, and the current continues to flow through the short in support of the magnetic field. But if the short or switch is then opened the voltage must rise rapidly in an uncontrolled way, v = L * di / dt. If there is no other conduction path, (a flyback diode for example), the voltage spike will strike an arc as the switch is opened and the current will continue to flow through that arc.

The Kettering ignition coil used with spark plugs in petrol engines works on that principle.
http://en.wikipedia.org/wiki/Inductive_discharge_ignition#Distributor_Ignition_Systems

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