How can I prevent electromagnetic shorting in my electromagnet?

[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.

 

[Baluncore]

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