Square wave magnetic field - Page 1
Forum section: electronics / projects

author: ricko_uk
author_rank: Super Contributor
timestamp: May 10, 2020, 02:42:35 pm
content: Hi,if you drive a coil with a square wave you clearly don't get a true square wave current (hence magnetic field intensity) across it.What is the best way to "force" a square wave current through it "as square as possible"?For example increasing the voltage when ramping it up would increase the steepness of the front rising edge but then it would overshoot the desired amplitude. So maybe there are some clever circuits?Also, by using a ferrite core (maybe of some ferrrites with special characteristics) would it help or worsen the sharpness of the square wave current through it?Many thanks

author: Buriedcode
author_rank: Super Contributor
timestamp: May 10, 2020, 03:33:48 pm
content: Increasing voltage across the coil (inductor) or reducing the inductance of the coil. Thats the only way to "increase the steepness" as you put it.I'm sure you're aware of di/dt = V/L which makes that pretty clear. Inductors, by their nature resist changes in current.

author: duak
author_rank: Super Contributor
timestamp: May 10, 2020, 05:53:27 pm
content: BuriedCode summed it up, I think. The problem with trying to change the current rapidly is that the winding develops a counter-EMF that opposes the change in current. There is a way to reduce the inductance and that is to add something called a shorted turn that turns the inductor into a transformer. By adding a shorted turn, the counter-EMF is reduced (ie. shorted out) and the winding current can increase much more rapidly. Once the current reaches its desired value the driver reduces the applied voltage to where the requisite winding current can be maintained. If the winding current isn't changing then no current is induced in the shorted turn and it has no effect on the magnetic flux. Since the magnetic flux is determined by the winding current, the number of turns and the magnetic reluctance then when the current reaches the desired value, the flux also reaches the desired level.The shorted turn is usually a heavy wire or foil. It will take some experimentation to determine the optimum configuration.This technique is used in hard drive head actuators to improve their spreed. Look for "shorted turn actuators"

author: ricko_uk
author_rank: Super Contributor
timestamp: May 10, 2020, 10:13:43 pm
content: Thank you Duak and Buried code.Duak, that's very interesting!!

author: duak
author_rank: Super Contributor
timestamp: May 11, 2020, 12:26:32 am
content: I should have added that with a shorted turn, the self-inductance also drops. If the repetition rate and current is high, the windings could get hot.

author: jbb
author_rank: Super Contributor
timestamp: May 11, 2020, 08:43:15 am
content: Shorted turn aside, I think the method used by Magnetic Resonance Imaging (MRI) people boils down to use few turns (reduce inductance), drive with high currents (to get your Ampere-turn product up for the magnetic field), and drive with lots of voltage to get the dI / dt up.Of course, you now have a high supply rail and a high drive current, which is a recipe for high dissipation in a linear power amplifier. If you get desperate you could look at something like a Class G amplifier.Given magnetic field is a function of current, I guess you want a current control loop.

author: Siwastaja
author_rank: Super Contributor
timestamp: May 11, 2020, 10:17:17 am
content: Frequency?Required rise/fall time?Higher frequency current ripple acceptable?If the frequency is fairly small (say, tens of kHz max?) and you can accept some ripple, it would be easy to just use enough voltage to get the risetime you need (di/dt = V/L), then pulse width modulate based on current feedback to force it close to a square wave shape.Voltage limits rise time and the ratio between the square wave frequency and chopping frequency determines the amplitude of ripple.

author: donmr
author_rank: Regular Contributor
timestamp: May 11, 2020, 02:13:33 pm
content: The are no truly square electrical waves either. Some just have faster rise times than others.

author: OwO
author_rank: Super Contributor
timestamp: May 11, 2020, 03:51:05 pm
content: Having a square wave magnetic field is like having a square wave voltage waveform on a capacitor. The capacitor would require current spikes on the square edges, and the coil would require driving high voltage, narrow pulses onto the coil at the rising and falling edges of the square wave (and a small amount of voltage afterwards to maintain the current). The magnetic field in a coil is always proportional to current, so you want a square wave current source.The shorted turn approach would allow you to get a square current waveform, but not magnetic field. On short timescales the shorted turn will develop a current opposite to the driving current on the main coil, so as to cancel the magnetic field. That's why the current can be allowed to rise fast (the magnetic field can not, without a lot of energy input).A different approach to getting very sharp edge magnetic fields is to set it up as a propagating wave rather than a near field. For example any TEM waveguide terminated on one end will look like a purely resistive load looking in from the other end, and so you can drive sharp edges into it. The H field inside the waveguide at a point will have a square waveform.

author: David Hess
author_rank: Super Contributor
timestamp: May 11, 2020, 06:33:49 pm
content: This problem is the mirror of driving a capacitance, like with electrostatic deflection plates, with a square voltage; the available current limits the slew rate on each edge.When driving a current into an inductance, the maximum voltage divided by the inductance will limit the current slew rate. So what is required is a current source with high compliance voltage.

author: Benta
author_rank: Super Contributor
timestamp: May 11, 2020, 06:51:18 pm
content: Quote from: David Hess on May 11, 2020, 06:33:49 pmTo be precise, it's the mirror of driving a capacitance with a square-wavecurrent.

author: David Hess
author_rank: Super Contributor
timestamp: May 11, 2020, 07:00:25 pm
content: Quote from: Benta on May 11, 2020, 06:51:18 pmThe electrostatic field is proportional to the voltage. The magnetic field is proportional to the current. Driving a capacitance with a current yields a triangular rather than square electrostatic field.The non-ideal series resistance of the inductor limits performance when driven with a voltage but not when using a current.

author: awallin
author_rank: Frequent Contributor
timestamp: May 12, 2020, 05:22:26 am
content: maybe google for what they typically do in BEC or other atomic physics experiments where they need helmholtz coils around the experiment?https://www.physics.utoronto.ca/~astummer/Archives/2007%20Mag-O-Matic%202/Mag-O-Matic%202.html40-100A with rise-time of 100 us (!?) - but probably not at 100% duty cycle, just short bursts.