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So the specifications of the electromagnet :
24v 0.2Amp
Holding force 500lb
The input is 24v, .  once the voltage is disconnected the Back EMF will occur, at that instant there still is a magnetic field which collapses very fast, in collapsing it 'cuts' the coil at high speed and the higher the 'Q' coil the more voltage is 'generated'. I imagine that my electromagnet that can lift 500lb with 24v will generate a huge CEMF.
So my question is, approximately how much Cemf should I expect and could I use it to recharge my batteries? Let's say I switch it on for 1sec than switch it off.
There are techniques for back emf suppression,
http://www.progeny.co.uk/Back-EMF-Suppression.aspx
I would like to redirect it, transform it to make it usable and recharge my batteries with it. If I manage to do it, will I get at least as much power from the cemf in order to be able to recharge my batteries (or capacitator)

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## 17 Replies

Basically, forget about doing calculations to figure this one out. There are too many unknowns and practical considerations, and limitations. Even if you could, it would basically require a high level of calculus and if nothing else, a VERY deep understanding of how to look at a schematic and figure out what formulas to use. If you have the time and patience to learn and figure this out, go ahead!

In general, if the inductance is high enough, the EMF spike will be about 10 times higher than the voltage going through an inductor. I learned that when I was working on my singing arc project. (at the time, I did not even understand what inductance was, and why switching the coil on and off would cause nasty voltages of the primary of a flyback, or why the current wasn't extremely high while the MOSFET I was using was on for the short pulses and durations.)

steveastrouk (author)2015-01-17

that x10 'rule of thumb' is no way accurate, it depends very much on the volume of the magnetic circuit. It's common to see spikes >>1000x

-max- (author)2015-01-17

I never said it was. It can in many cases give some idea of what to expect in my experience. Spikes of 100x the voltage with what conditions? 1000V IGBTs and high inductance low ESR inductors?

steveastrouk (author)2015-01-18

'in general the .......the emf spike will be about 10x'. Even a mechanical switch can break the current fast enough to be nasty. Relay coils are the killer parts, rarely inductors, except in power supplies. I've seen pcb's with tracks blown of by them.

-max- (author)2015-01-17

1000x *

Maxc13 (author)2015-01-18

Best Regards

iceng (author)2015-01-16

You are sort of assuming the OP's coil will survive the thousand vicious kick back transient voltage spikes without wire to wire shorting damage :)

iceng (author)2015-01-16

When you apply power to your 24v 200ma coil you will establish a magnetic field.

After a second or two your magnetic flux field is fully established and the only activity is the resistive power loss 24x0.2=4.8Watts.

When you interrupt the current to the coil, it 'the coil' wants to maintain the current flow and generates a reverse voltage as the flux field collapses. If you do not provide some circuitry to allow the current to dissipate it 'the coil' can increase the voltage high enough to damage wire to wire insulation and short some turns ! This kick-back action is used to good measure in inverter power supplies.

The energy stored in a coil is P = L I x I / 2 that's Inductance_in_Henrys times the Current_in_Amperes Squared divided by Two..I'm fairly sure your unit is less then a 1000mH = 1Henry which is only .02J =20mJouls or 20milliwatts in a second. That is a tiny energy very easily dissipated by a 1N4002 inverse diode. Remember you cannot get any more energy out then you put in EVER!!

A quick look at some electromagnets you can buy R-3030-24

http://www.magnetechcorp.com/round.html

shows a 400lb lift 24volt 1amp 3"dia 3"high unit for \$145.

Maxc13 (author)2015-01-18

Thank you your replies are always excellent

Regards

steveastrouk (author)2015-01-16

You need to study more circuit theory. The voltage is GENERATED by "Faraday's voltage law", axiomatically. The collapsing magnetic field creates the back EMF.

I respectfully disagree with my colleague that the time to establish the field is as much as two seconds.

iceng (author)2015-01-16

I did not say, it took as much as one second let alone two !.

I merely stated that after a second or two the field is steady state and established. That is a fair and honestly true statement.

Also think back to the magnetic motor starters that we used and the large coils with the copper rings we could add to increase the coil delay by half seconds allowing the machine to accelerate before giving it full power...

steveastrouk (author)2015-01-16

He he.

Anyway, we can only guess at the inductance. I agree with your napkin-sketch maths.

iceng (author)2015-01-16

Yes, it was a bit of a brain teaser to word it so, to give that impression :)

Like most of us, I had to guess at the inductance to make a case for lack of high energy....

There have been several interesting astronomer TV specials on Dark matter and energy, ..... we may have to work on linking to the dark for the over_unity community here at ibles ! #@##@## ;-)

-max- (author)2015-01-16

Actually, what you describe is a boost converter! Instead of a inefficient spark gep, how about a diode instead so that way after power is applied, the capacitor reaches the battery voltage very quickly, and then the back EMF will instantaneously rise until it exceeds the voltage across the capacitor + voltage drop across the diode, then it can rise much more and instead charges the capacitor a bit. Do this over and over many, many times a second and the capacitor will quickly charge up and up and up.

-max- (author)2015-01-16

The EMF kickback from what I understand is related to the instantaneous change in current through the coil over time. Mathematically, this is di/dt. When this crazy di/dt thing is a really big (positive or negative) number, it means that the current flow changing really fast, like what happens when a switch is flipped from off to on, or vice versa. You may recognize this as a derivative from calculus if you have taken any calc course. (It is a simple concept to me, it is a lot like slopes of lines, but for everything!)

Since the di/dt through an inductor is basicly directly proportional to voltage across it, the faster the current changes, the bigger the voltage across it is. In other words, if we had a perfect inductor connected across, say, 12V, then you can say that 12 (divided by L) will tell you how fast the current is changing. In the case of a 0.01H inductor, thats 12/0.01 = 1200A/second!!!!!! Resistance and ohms law in the real world will be in effect and will limit that though :(

Anyway, my point is, when you remove power, all of the sudden the current when from BIG to nothing almost instantly. That means the "di" part is a very VERY big thing, while the dt, or change in time is still almost zero! SOOO that means di/dt is almost infinite, and the voltage would be as close to infinite as possible!!!

In the real world, there is resistance, ESR, arcing inside the switch which will prevent instantaneous current change because of course the faster current changes = more volts, so basicly the calculations become VERY difficult if you want to get a accurate figure for the voltage transients. (it is hard to predict the imperfections of the switch, and how fast it can turn ON to OFF, and how much voltage it can handle between the contacts as it turns off.) Basically, it is far easier to built it up and see if it all works. In my experience, these voltage transients can be several hundred volts. However, if you use these voltage transient and feed them through a diode and into a capacitor, you will end up with a crude boost converter that thothe voltage spikes can be harnessed to charge up a capacitor, and ultimatly a battery.

steveastrouk (author)2015-01-16

The ability of the magnet to lift is also severely affected by the surfaces doing the lifting.

steveastrouk (author)2015-01-16

TANSTAAFL.

There ain't no such thing as a free lunch. You COULD get some of your energy back, but a lot has to disappear in Ohmic heating of the coil, and some radiates away as RF. Every 1/2 second pulse WILL burn 2.4 joules of energy, whatever you do.