# Turning off (collapsing the field) of large inductors

In a totally unrelated search to something I was looking for, I came across an article from the EPA that describes looking at soil properties through a large inductor for geophysics.

https://archive.epa.gov/esd/archive-geophysics/web/html/time-domain_electromagnetic_methods.html

What struck me is that the premise of the project is that you impart a field into the earth with a large inductive loop, and then you read the results; however, it completely hinges on the fact that you can collapse a field instantaneously (Figure 4 in the link above, now attached below).

How would one actually collapse a large field quickly in practice? I am looking for the nuances of collapsing an inductive load that is large which potentially 1kV+ of kickback.

I found some vague reference to 10 turns in a square of 10m x 10m for 6 gauge wire for 100V at 100A for a system that runs at 120Hz. If I put resistor across the inductor to collapse it quickly, it would basically put a million volts across the resistor. There has to be something more elegant to remove the field in an inductor quickly due to this voltage spike; however, it might be as simple as a resistor.

update: I mentioned "instantaneously" as that is what the figure showed. I am aware that this is unrealistic.

update 2: As I've found more system information, I'm not even clear that the coil could get to steady state (what I mean by this is that the flux is already in the earth and thereby the coil has reached 100V@100A). I'm beginning to think that the system specs are more theoretical than pragmatic.

• "Elegant" implies solicitation of an opinion. My "clunky" may be your "elegant". – TimWescott Jun 18 '19 at 17:18
• To know the voltage, you would also need to know either the current or magnetic field that was running through the inductor before you shut it off. – Voltage Spike Jun 18 '19 at 17:22
• @laptop2d 100V at 100A! – b degnan Jun 18 '19 at 17:27
• @HarrySvensson I've updated the question to be more clear – b degnan Jun 19 '19 at 1:45
• Steady state is meaningless for an application such as this. To create a viable EM wave front you want the rapid turnoff and would 'listen' for results from that propagation. – Jack Creasey Jun 19 '19 at 4:14

How would one actually collapse a large field quickly in practice? I am looking for the nuances of collapsing an inductive load that is large which potentially 1kV+ of kickback.

As you acknowledged in your first update to your question, it is unrealistic to expect the instantaneous collapse of the magnetic field generated by the "exciting" inductor. However, taking into account some unavoidable non idealities of the circuit, you can control the fall time in the inductor within reasonable to quite low values, also avoiding the inductive kickback (or perhaps turn it into an advantage) and excessive power dissipation.

Some time ago I faced the same problem when driving some solenoid valves: we needed to make the electric part of the turn-off delay negligible respect to the mechanical part (the time needed by the "spring" to close the nozzle). This implies the need of quickly transferring the energy $$\E\$$ stored in the inductor $$\L\$$ of the solenoid (+ some mechanical energy due to the elastic potential of the "spring", but this is not relevant in the current context so I cite it only here just for the sake completeness and no more in the text below), i.e. $$E=\frac{1}{2}LI_o^2\label{1}\tag{1}$$ where $$\I_o\$$ is the on state inductor current, and this in turn implies the necessity of disposing of large powers.

The core of the circuit we developed is shown here, with obvious meaning of the inductance $$\L\$$ and of its stray series resistance $$\R_s\$$: simulate this circuit – Schematic created using CircuitLab

It is an half bridge driver, similar in spirit to the full bridge circuits proposed by Jack Creasey and laptop2d, with one significant difference: the freewheeling diodes $$\D_1\$$ and $$\D_2\$$ are not connected to the same power supply which feeds the MOSFETs, but with another one at a higher voltage level, i.e. $$\V_{DD}\ll V_{sto}\$$. This gives rise to a turn-off behavior which is analytically predictable (at least to the extent to we can consider the anode voltage of a diode nearly constant) and controlled by the ratio $$\V_{sto}/V_{DD}\$$: let's see how.

The turn-off behavior of the circuit
When the MOSFETs $$\M_1\$$ and $$\M_2\$$ are turned off, the above circuit is equivalent to the following one: simulate this circuit

The initial current $$\i(0_+)\$$ is equal to $$\I_o=V_{DD}/R_s\$$, since the inductor tends to keep constant the magnetic energy stored within it: therefore the two diodes $$\D_1\$$ and $$\D_2\$$ start to conduct and can be seen as two voltage generators with $$\V_{A_{D_1}} = V_{A_{D_2}} = V_\gamma \$$. Writing down the mesh equation we have $$\begin{split} V_{sto}+V_{A_{D_1}} + V_{A_{D_2}} + R_s i(t) + L\frac{\mathrm{d}i(t)}{\mathrm{d}t} &=0 \\ \Updownarrow \qquad\qquad\qquad &\\ V_{sto}+2V_\gamma + R_s i(t) + L\frac{\mathrm{d}i(t)}{\mathrm{d}t} &=0\\ \end{split}$$ Applying the Laplace transform to the equation we get $$\frac{V_{sto}+2V_\gamma}{p} + R_s I(p) + pL I(p)-Li(0_+)=0$$ and after solving for $$\I(p)\$$ and applying the inverse Laplace transform we have $$i(t)=\frac{V_{DD}}{R_s}e^{-\frac{R_s}{L}t}-\frac{V_{sto}+2V_\gamma}{R_s}\left[1-e^{-\frac{R_s}{L}t}\right]\label{2}\tag{2}$$ This equation is (nearly) correct only for values of $$\t\$$ for which $$\i(t)\ge 0\$$: on the other hand, we want to know the collapse time $$\t_c\$$, i.e. the time needed for $$\i(t)\$$ to go from $$\I_o\$$ to $$\0\$$, since this is exactly the time needed for the magnetic field to collapse, i.e. to reach the zero magnetic energy condition in \eqref{1}: \begin{align} i(t_c)=0\iff &\frac{V_{DD}+V_{sto}+2V_\gamma}{R_s}e^{-\frac{R_s}{L}t_c}-\frac{V_{sto}+2V_\gamma}{R_s}=0\\ \\ \iff & e^{-\frac{R_s}{L}t_c}=\frac{V_{sto}+2V_\gamma}{V_{DD}+V_{sto}+2V_\gamma} = \left[1 + \frac{V_{DD}}{V_{sto}+2V_\gamma}\right]^{-1} \\ \\ \iff & \color{blue}{t_c = \frac{L}{ R_s } \ln \left(1 + \frac{V_{DD}}{V_{sto}+2V_\gamma}\right)} \label{3}\tag{3} \end{align} As stated above, the larger the ratio $$\V_{sto}/V_{DD}\$$ the smaller the collapse time $$\t_c\$$, ideally going as close to zero as needed. Note also that, apart from the dissipation on $$\D_1\$$, $$\D_2\$$ and $$\R_s\$$ the largest part of the energy \eqref{1} is feed back to the power supply $$\V_{sto}\$$, increasing enormously the efficiency $$\\eta\$$ of the circuit: thus the inductive kick is used to feed back into a power supply the inductor stored energy (note: the power supply must be able to sink the current), avoiding both the excessive dissipation of a snubber circuit and the breakdown of the MOSFET $$\M_2\$$.

Notes

• Some notes:
• the $$\D_1\$$ and $$\D_2\$$ must be fast $$\I_o\$$ rated diodes. In this case $$\I_o\simeq 100\mathrm{A}\$$: however very high current Schottky diodes, in Si or SiC technology, are readily available and thus this is not a problem.
• The MOSFET should be able to withstand very large drain source voltages, i.e. $$\V_{DS}\simeq V_{sto}\$$: even this is not a problem since we can use totem pole arrangements and SiC devices in order to realize the required performance. New edit: as remarked by Jack Creasey in his comments, in applications where field inversion is needed (as it appears this is the case) you can apply the same concept to a full-brigde driver, provided you avoid the effects of the body-drain diode of the upper MOSFETs, i.e. the short circuit of the terminals of the loop inductor to $$\V_{DD}\$$ during the inductive kick. This basically can be accomplished in two ways
1. Put a diode with the same characteristics of $$\D_1\$$ in series to each upper MOSFET, and put a freewheeling diode in parallel to this arrangement: this however rises the power dissipation of the circuit.
2. Put two MOSFETs in antiseries in place of each upper MOSFET and again put a freewheeling diode in parallel to this arrangement: this possibly increase the complexity of driver circuits. simulate this circuit

• Initially we tried this circuit with $$\V_{DD}=24\mathrm{V}\$$ and using various Zener diodes: we reached the sought for $$\t_c\$$ value at the price of a large power dissipation. After that, we used a power supply and solved the problem.
• The fact that we are able to get lower and lower values of $$\t_c\$$ does not imply an unphysical behavior of the circuit: after all, we are dealing with electromagnetic fields, which admits a "shock" behavior i.e. extremely fast changes in their intensities. What makes unrealistic instantaneous variation of the fields are the real characteristics of the means used to generate them, not limited to but including circuits.
• Very nicely explained. A couple of notes 1) you don't need a separate power supply, you can use a pre-charged capacitor (large enough to hold the inductive energy) This allows you to dissipate the energy (discharge the cap) over a much longer time. 2) you can do this with an H-bridge too, you simply need a series diode from the drive supply and series diode to the cap storage. – Jack Creasey Jun 24 '19 at 15:17
• @JackCreasey thank you for your observations. I was aware of the possibility of using a polarized capacitor, similar to the possibility suggested by DirkBruere: I did not chose it (in my former application) since we have higher voltage power supplies in the same system and we wanted to use all the energy we recover. On the other hand I did not investigated the use a full bridge circuit. – Daniele Tampieri Jun 24 '19 at 15:31
• If your 24V supply was via a regulator, I'd suggest you were simply using its output capacitor as a reservoir. The largest complexity is that the clamp supply you use has to be able to sink the current peak without going out of tolerance (if its a regulator). The H-bridge only becomes necessary is you want to be able to reverse the projected field. If a unipolar field is sufficient, then your circuit would work fine. – Jack Creasey Jun 24 '19 at 15:38
• @SunnyskyguyEE75 No, I don't. The information provided in the paper clearly shows an inductive loop (not connected to the ground in any way). I have no idea what you are suggesting in your answer. – Jack Creasey Jun 25 '19 at 15:27
• @SunnyskyguyEE75 You are trying to solve another problem ….nothing to do with the question asked. The question asked is about projecting a magnetic field ….not measuring the resistivity of the ground using direct probing. Whether one method is better than another is not part of the question (or the paper). While I have some problems with the content of the paper, the question was specific and related to creating a magnetic wave front. – Jack Creasey Jun 25 '19 at 18:26

I want to point out one thing that the other answers are ignoring, or at least glossing over.

In this measurement, we stop the flow of current in our inductor, and this causes current to flow in the earth below it. That means we're effectively using the earth itself as the secondary coil of a flyback transformer.

In a flyback transformer, the primary current can indeed be shut off very quickly, precisely because the magnetic flux is able to couple to the secondary and produce a current there, and we don't need to absorb the stored magnetic energy in the primary side circuit to quench the current.

The typical waveforms for a typical flyback transformer (with a copper wire secondary rather than soil) look like To the extent that the experiment works, and we do generate a current in the earth, we won't need any voltage limiting device on the primary to ensure we can quench the current. The resistance of the earth will eventually absorb the energy that was stored in the magnetic field.

It may be a good idea to include a spark gap or other voltage limiting device, to protect the primary side circuit from damage in case the earth resistance is very high, or the device is accidentally operated in air instead of near the ground. But if this voltage limiter actually operates, it will likely disturb the measurement so that we don't get a result usable for determining the properites of the earth being studied.

You should also notice this paragraph in the page you referred to:

In figure 4 and in this discussion, it was assumed that the transmitter current is turned off instantaneously. To actually accomplish this with a large loop of transmitter wire is impossible, and modern transmitters shut the current down using a very fast linear ramp. The duration of this ramp is maintained as short as possible (it can be shown to have an effect similar to that of broadening the measurement gate widths) particularly for shallow sounding where the transient decays very rapidly at early times. The duration of the transmitter turn-off ramp (which can also be included in modern inversion programs) is usually controlled by transmitter loop size and/or loop current.

• In a closely coupled metal or ferrite transformer, what you say is correct. However, this is a loosely coupled air core transformer and the primary voltage will rise to a very high voltage and flashover somewhere if you try to turn off the driving device. – Jack Creasey Jun 19 '19 at 16:22
• @JackCreasey, notice the quote I included from the source material. They don't actually try to shut off the current instantaneously. And I deliberately said "very quickly" rather than "instantaneously" in my description of how the flyback transformer operates. – The Photon Jun 19 '19 at 16:25
• Nothing operates instantaneously, that's hardly worth pointing out. It's still not like a transformer in any real sense. As I showed in the answer I gave you need to control the back EMF, the ground coupling will NOT be enough to prevent ultra high voltages developing so this is NOT like a flyback transformer IMO. – Jack Creasey Jun 19 '19 at 17:05
• @JackCreasey, if it isn't a transformer then it doesn't operate like the source material says it does. From OP's linked page: "The process of abruptly reducing the transmitter current to zero induces, in accord with Faraday's law, a short-duration voltage pulse in the ground, which causes a loop of current to flow in the immediate vicinity of the transmitter wire, as shown in figure 3. " – The Photon Jun 19 '19 at 17:11
• I never said it didn't act like a transformer. I said it was an air core inductor and acts very differently to a transformer with metal or ferrite core material between primary and secondary. The coupling is very low in this case so the secondary load has very little effect on the primary. – Jack Creasey Jun 19 '19 at 17:15

Can you collapse an inductors magnetic field instantaneously?

NO ….you can neither create nor collapse a magnetic field instantaneously since there is always some L and R time constant involved.
You CANNOT reverse the voltage on an inductor either (at least not in any practical circuit) and get lower discharge times.

Perhaps a simple example may help: simulate this circuit – Schematic created using CircuitLab

In the circuit above the inductor is a very low value and may be typical of what you might have in a large air cored loop, though the value is not really important.

Here I've shown the inductor as having about 0.5 Ohm resistance, and fed from 12V by a 0.5 Ohm resistor. There is an L and R time constant to build up the magnetic field, and when you turn OFF M1 drive I've shown a Zener Gate feedback drive to dissipate the energy as quickly as possible.
In the schematic and values above it takes about 500us to increase L1 current to 10A, and about 20us to decrease it to zero. The practical limit here is the voltage you allow to develop across M1. Increasing the voltage decreases time taken to discharge L1.

Update: If you want to delve into the math involved in characterizing and selecting components, supply voltage etc, this ST application note may be well worth reading.

It is worth noting:

1. Using a spark gap of any form is an unlikely method of discharging an inductor since it has a trigger voltage and a maintain voltage (which would actually extend the collapse time). Even with a professional vacuum/gas spark gap they take time to operate, usually in the 1us range for small units. However, once triggered they maintain at a much lower voltage (typically 70-100V) so don't provide the capabilities needed (maintain a high voltage).

2. Propagation of an EM field is always less than the speed of light and medium dependent (go study Maxwells equations). A constant field is not subject to propagation delay, only a changing field, so you actually need the current to be changing continuously to create EM wave fronts.

Update: Since the comment about being able to reverse the voltage across an inductor caused some commotion, let me add this:

You can of course use for example an H-bridge to feed an inductor and this will reverse the voltage, but it DOES NOT provide a shorter discharge time in any practical circuit, and the point of the OPs question was in seeking to collapse the field quickly.

Here is a schematic for an H-bridge using 100V, but only 10A (because of the FETs available to be simulated): simulate this circuit

The waveforms look like this:

Notice that the rise and fall times of the inductor current are the same, and it's not possible to change them without changing the voltage applied.
In my original circuit the falltime was 0.1 of the risetime and could be dropped even further by altering the clamp voltage.

The only way to shorten the time taken to reach max current is by increasing the clamp voltage or decreasing the resistance (of the inductor). If you switch the voltage polarity on L1 then the rise and fall times are always equal.
It is only by clamping the back EMF of the collapsing field to a much larger voltage than the supply that you can reduce the falltime.

• Thank you for your answer; however, I was looking for something more in line with the 100V, 100A (that I might have added later). This is a bit of a rabbit hole question as it's outside of my immediate knowledge in the application space. – b degnan Jun 19 '19 at 13:56
• @bdegnan Of course, and you did add the parameters later. However, now you know what is required you can scale to any voltage or current you want. – Jack Creasey Jun 19 '19 at 14:36
• As the geophysicists think you can collapse it quickly, even if you say 1 second to collapse the field, that is a lot of energy that you'll have to dissipate. I think that your gas spark gap might be what is done commercially. – b degnan Jun 19 '19 at 14:41
• You cannot reverse the voltage on an inductor? Sure you can. class D amplifiers, BLDC motor drivers, etc do this all the time. You can't reverse the current, but you can reverse the applied voltage, or, indeed, apply any voltage you are capable of generating at any time. You just can't get around V = Ldi/dt. – mkeith Jun 19 '19 at 16:07
• @mkeith You CANNOT reverse the voltage on an inductor such as this in any practical circuit because there will be a point in the crossover where neither drive device (such as an Hbridge) is ON and the voltage will clamp with the diodes in the H-bridge devices. This will distort your wave front, which you don't want to do. Don't conflate the simplicity of a Class D amp to what is being attempted at 100V/100A. I carefully spelled out "in any practical circuit". If you can come up with a practical way of doing this, by all means post it ….I'll happily upvote it if it looks like it can work. – Jack Creasey Jun 19 '19 at 16:16

Another way to collapse the field - connect it to a capacitor. Wait until the capacitor voltage rises to a peak and then disconnect it. The voltage can be calculated from the energy of the inductor e.g. $$LI^2 = CV^2.$$ The fall time is determined to be half of one sine wave whose frequency is $$f=\frac{1}{2\pi\sqrt{LC}}$$ You also get to save most of the energy for the next cycle.

• Sine waves don't work well for system identification unless you sweep them aka chrip. Square waves have plenty of frequency content. – Voltage Spike Jun 25 '19 at 17:52
• @laptop2d Except you are never going to get a square wave, just edges that have a high frequency sine wave component. If you want a higher frequency, use a smaller capacitor at a higher voltage. – Dirk Bruere Jun 26 '19 at 8:28
• you need multiple frequencies, not just one, to do a system identification. This means you would have to sweep your capacitance values, which may prove difficult – Voltage Spike Jun 26 '19 at 14:03
1. They ain't collapsing no field instantaneously. At best it's limited by the speed of light; in reality they're generating some high voltage discharge, or the energy of the field is being dissipated into the soil.
2. In open air, you'd discharge the coil into a spark gap. "Instantaneous" is literally wrong, but figuratively, if you get the coil strength down to zero in less time than you need to excite any interesting modes in the thing you're testing, that's "instantaneous enough".
• You could reverse the polarity of the drive signal to the coil and use the reverse current to charge a battery. – scorpdaddy Jun 18 '19 at 17:25
• I am well aware of this, which was why I thought I should ask. – b degnan Jun 18 '19 at 17:29
• It seems the point of this instrument is to use the earth itself as a (very poorly conductive) secondary of a flyback transformer. So you really want the primary energy to be transferred into the earth, not into a spark gap or other device. – The Photon Jun 18 '19 at 17:31
• @ThePhoton First, if the soil is conductive enough (what's "enough"? I dunno) then there won't be a spark. Second, even if there is a spark, the coil will still be interacting with the surrounding soil. I assume that more details could be found by chasing the relevant papers. – TimWescott Jun 18 '19 at 17:38

Here is a schematic from others that use TDEM's to do research, which is an H-bridge with TVS diodes to short out the high voltages that may come across the inductor when it is disconnected: This would be a starting point for a circuit, but for the high currents and high voltages of a 100V 100A design there would need to be some modification.

These are some recommendations (I've never designed such a high voltage circuit, 60V is my max with h-bridges but the principles are the same):

• The first thing is the gate drive, nmos are best for matching and the most common mosfets, the gate needs to be higher voltage than the source of the mosfet, or it will not 'switch' on.

• Switching time is important, mosfets can go from a few mΩ's of resistance to more than 10^9Ω's. The problem is in between these two values, the fet's resistance can dissipate significant amounts of heat, so you want the fet fully off or fully on, but not in between. Gate drivers can help this problem.

• Make sure the drain to source voltage is higher than 100V, in your case, you'd want it significantly higher if the voltage of the inductor is spiking.

• Select FET's that can handle current higher than 100A, or parallel them (if you have smaller ones, like 50A, then you could use two or three of them to give you 100A or 150A. I'd probably go with a higher current rating than use need with parallel fets, the 100A fets are expensive.

• The switches in the h-bridge (vertical pair) must never be turned on at the same time, so use drivers that have logic built in or in the controller build in logic to ensure this happens. This includes system power up and power down (sometimes the logic can turn gates on as it's powering down or up). Use pull up or pull downs to make sure this doesn't happen.

If you've never built an h-bridge before, this may become a challenge. These are the basics of h-bridge design. There are devices that switch higher currents and higher voltages with h-bridges, like welders and inverters.

• The diodes are used to store energy in a capacitor to limit the high voltage. They don't need to be TVS and the diodes are not clamping the voltage. – Jack Creasey Jun 21 '19 at 18:37