Author Options:

protecting from EMF? Answered

so i am working on some high power geophysical transmitters which are being damaged by back emf.
can anyone help me with what would be the best way to protect the instrument from this.
we are mainly blowing the switching IGBT's.
the transmitter is supplied DC from a 30kw powersupply, feeding a set of IGBT's which are switched at a selected frequency usually 0.5 to 2hz at 50% duty cycle into a loop of wire ranging in resistance but usually around 1 ohm. we run this at up to 150A at 150V and our loads can be very inductive, usually 300mx300m 25sqmm copper wire.
im thinking diodes or caps for a bit of protection.
im open to any ideas



1 year ago

Iceng and I have kicked this over off-list.

The diodes on the transistor models are called intrinsice diodes, and aren't real device you can play with. You would really need to add external, and FAST high current REAL external diodes.

Can we not do something active to clamp the load as well ?

If we have the bridge arranged with transistor

1 2

3 4

It needs properly modelling in a SPICE tool though.

Referring to Jack's image, modified by me.

Lets start with everything OFF.

1.) Switch on 1 and 4.

2.) switch OFF 4, then very quickly switch ON 2

3.) Switch OFF 1 then switch ON 3

4.) Switch OFF 2, then very quickly switch on 4

5.) Switch off 3

Repeat. 1..5 as needed

Cycle 1, Current flows left to right in the load.

2. Current flows is turned off, by switching off right of bridge, but
VERY quickly, the top of the bridge is turned on, energy in the loop
decays through load resistance (and RC snubber) There HAS TO BE A DELAY
before you switch on 2 after 4 is turned off, you MUST NOT let current
punch through the bridge from top to bottom.

Cycle 3 Current in loop is Zero, so we can start the other cycle.

Cycle 4, as 2 above.


2 years ago

Ok, so my electronics knowledge is limited, no formal training, only on job experience so please dont expect me to understand everything straight away.

not many people know much about these sort of instruments which is why i am working with them and not a trained professional.

IGBT's are used as we do need to switch off as fast as possible, this is one of the most important things with geophysics as its what we are actually looking at because the turnoff is when the sensor/magnetometer actually starts recording. the switching is controlled via gps timing as it needs to be precise, hence why mechanical switching is not used. The switching creates a square wave, pos on, off, neg on, off. the unit switches like this constantly at a frequency set be us. its during the turnoff when our receiver samples data.

the igbt we use, which im quite sure does not have a diode already.


so the big issue is the flyback voltages during each switch off which is usually twice per second or more, and im assuming these are getting quite high.

No the load is not overly inductive, just for us it is as we usually use smaller cable etc.

What is a snubber and where does it go in the curcuit?

Jack A Lopezbenmerlingeo

Answer 2 years ago

It kind of looks like you have 4 IGBTs wired as an H-bridge,


There is no way to tell from this picture, which pair of red and black wires is the voltage source, and which is the load, your big long loop.

I am sort of guessing the DC supply is connected to those jumper bar things, of which the red bar one is connected to the positive side of the supply and terminals C1, on both IGBT modules, and the black bar is connected to the negative side of the supplyterminals E1, on both IGBT modules.

And the load, is connected between terminal C2E1 on one IGBT module, and C2E1 of the other IGBT module.

I will attach a picture, a hand drawn circuit diagram, of the way I am imagining this.

In response to your question, "What is a snubber and where does it go in the circuit?"

In the section from H and H, that I upped a picture of yesterday, we can see (in Figures 1.95 and 1.96) that the place where it goes, is typically, wired in parallel with the load, which in both those figures is just an unlabeled inductor.

In Figure 1.95 the snubber is a diode. In Figure 1.96 the snubber is an RC, a resistor in series with a capacitor.

Using a diode as a snubber will not work for your load, because you are doing this thing where you alternately change the polarity. Besides it would have to be a really beefy diode, one that could handle the same, like 150 A, current as your load loop.

Regarding a RC snubber, I think that is just going to give a RLC that rings, unless you pick pick R and C big enough to make the response overdamped,


so you don't get that ringing noise. However if you make R too big, it is kind of the same as not putting any snubber there at all.

Actually, for something like your application, I am thinking perhaps a snubber that is just a resistor, particularly one that is larger than the resistance of the loop, by a factor of about 10 or 100.

This is why, in the circuit I drew, I put a 100 ohm resistor wired in parallel with the load, which I was imagining to be a 1 ohm resistor plus 300uH inductor.

I think I picked 100 ohms, because that would give only a moderate amount of wasted power, which is (150 V)^2/(100 ohm) = 225 W

which is of course tiny, compared to the power wasted (as ohmic losses) by the loop itself

(150 V)^2/(1 ohm) = (150V)*(150A) = 22500 W = 22.5 KW

Of course, the dissipated power from this resistor-snubber would make some heat, i.e. around 225 W of heat, so it will have be the kind of kind of resistor that comes with a big heat sink.

I don't know if there are any inexpensive, essentially resistive, mains powered electrical consumer goods, like incandescent light bulbs, or an old toaster or sandwich maker, that are compatible with 150 VDC, but it would be convenient if that would work, since that kind of stuff is already designed to dissipate the heat.


2 years ago

You should model the behaviour of the load with its real resistance and inductance.

What does "very" inductive mean ? I make the inductance around 300mH, which isn't THAT much

Have you looked at the actual load voltage and current waveforms ?

This sounds like you need snubbers.

Jack A Lopezsteveastrouk

Answer 2 years ago

Hey, I see what you did there. You multiplied 300 m times 1e-6 H/m, the approximate value of mu zero.


You know, sometimes I wish these people would up some pictures of the weird stuff they ask questions about, because this must be a pretty spectacular current loop, just because it is huge, assuming one of its dimensions is like 300 m.

Also, I guess it might be hard to photograph, unless there is a way to get above it somehow, like a nearby hill, or an aircraft flying over it, or something like that.

benmerlingeoJack A Lopez

Answer 2 years ago

its just a square loop of wire laying on the ground 300m x300m, for this setup we use 25sqmm double insulated wire which gives a resistance of 1 ohm, so we use 150V to produce 150A. This is connected to the transmitter which is connected to a 30kw dc 3 phase powersupply (sorenson sga series) running off a large diesel genset.

i would probably have some pics somewhere if u were interested

Jack A Lopezbenmerlingeo

Answer 2 years ago

I think I can picture it. It is a square, so the total length is 4*300 = 1200 m.

The bulk resistivity of copper is about 17e-9 ohm*m, and a cross-section of of 25 mm^2 is 25e-6 m^2, so the resistance of the whole 1200 m length is

R= rho*L/A = (17e-9)*(1200)/(25e-6) = 0.81 ohm

In terms of AWG


Copper wire with cross-section of 25 mm^2 is, I guess, a size in between 3 and 4 AWG, from looking at the table there.

Also, regarding pictures, I was looking at the Wikipedia article for "Transient electromagnetics"


and that article has a picture of something similar. It looks like a regular 12-gon being towed by a helicopter, although from that picture it is kind of hard to guess at the scale of it.

That article linked to this page as one of its references,


and wow! It would take a while to read all of those, but it kind of looks like as good a place as any to get an overview of how this stuff works.

BTW, I do appreciate that picture of your IGBT switches, and I will comment on that, as a reply to that post.


2 years ago

so it seems i need a diode of some sort

can anyone point me in the right direction as to what sort and size i would need to get and exactly where in the curcuit it should go?


2 years ago

Remember component intrinsic inverse parallel ( Freewheeling ) diodes are mainly made for speed !

You need to add an external inverse parallel diode to pass ALL of the inductive energy and protect the intrinsic diode !!

The procedure, when you turn the IGBT OFF, is the freewheeling diode starts passing reactive current in nanoseconds and also starts heating up but your external high current diode begins to take take the major portion of this current in microseconds later protecting the intrinsic diode and damage it would cause to your power IGBT...


Answer 2 years ago

Snubbers are for narrow high dv/dt voltage spikes that can be caused by power line noise, other devices in your circuit or reactive kick-backs across your IGBT..

Bipolar devices can act as a capacitor for a fast (high dv/dt) rising voltage transients and get turned on at the worst time for a circuit...

BTW don't forget the snubber must also be discharged without causing excessive di/dt damage in the form of IxT^2 fusing when you later turn ON your IGBT again !


Answer 2 years ago

Right.. Intrinsic means built_inside the device... And very fast but not able to carry the same current as the switching device for more time then the microseconds the external diode needs to begin to pass current in parallel to the built in fasty that you should parallel. You should keep the wires as short as possible and very close to the component, otherwise longer wires add inductance which slows response time...

Keep in mind that your slow switching time can still generate a kilocycle = kilohertz oscillation referred to as ringing which could easily damage your IGBT...

Jack A Lopez

2 years ago

I am guessing what you want is a big freewheeling, flyback diode, for to give big current in an inductor a place to go, at that moment when a switch opens. There is a Wikipedia page for this,


I don't know exactly what your circuit looks like, but maybe the Wiki article I have linked to will give you some ideas.

There are, of course, other pages out there that describe the same thing, but confusingly, they call it different names; i.e.

"snubber diode, commutating diode, freewheeling diode, suppressor diode, suppression diode, clamp diode, or catch diode"

I have also hear the phrase, "inductive kick", used to describe the surge of energy the flyback diode helps to fix.

icengJack A Lopez

Answer 2 years ago

And "EMF" stands for ( Electro Magnetic Force ) a term used with non static driven electric motors usually meaning the repulsive and attractive magnetic forces that concentrate in the air-gap of rotary and linear machines... Perhaps a better terminology would be to call the damaging reactive effect to collapsing magnetic fields... Or in my layman's wording the propensity of a coil always attempting to maintain a current status Quo with any dirty trick possible...

In my fast MOSFET inverter days a straight six inch length of Litz wire

https://en.wikipedia.org/wiki/Litz_wire had enough inductance to generate a hundred volt spike when current was interrupted very rapidly... Ergo the need for a Very very fast inverse intrinsic diode to catch the resulting voltage spike and turn it into a reactive current that dissipates in the coil resistance...

Jack A Lopeziceng

Answer 2 years ago

Yeah, there's energy in them thar magnetic fields!

BTW, as part of a top level answer to this question I upped a picture of a page from a well known textbook, with a circuit diagram of an opening switch in series with an inductor, with a drawing of tiny explosion, plus the word, "BOOM!"

Jack A Lopez

2 years ago

Hey, I remembered where I first came across that phrase, "inductive kick", and it was in an old textbook on my shelf.

I re-read that section, and it had the usual suggestion, which is a diode placed in anti-parallel with the load.

It also has a suggestion for a RC snubber, made from a resistor plus capacitor.

Also it has a cute little illustration (Fig 1.94) of an opening, exploding, switch accompanied with the word, "BOOM!"

So, I decided it would be worthwhile, to take a picture of this section, "1.31 Inductive loads and diode protection", from pp 52-53 of Horowitz and Hill's The Art of Electronics. 2nd Ed., and uh, share this with the rest of the class.


2 years ago

I might be shooting in the wrong direction here but here are my thoughts:
With such low frequencies, I would try to avoid all electronics.
But unlike what Jack suggest in terms of a mechanic switch I was thinking more in the direction of car distributer - the one for the ignition on old cars...
Only thing really needed from it is the idea though ;)
Depending on the rotation speed and amount of contacts you would not need much hardware to create a powerful switch that works the same way.
Only big difference is that instead of a high voltage with a few mA you need to switch just a few hundred volts but massive Amps.
A insulated disk of suitable size with brass rods as the contacts and a rotor with a suitable brass or hard copper contacts would be well capable of switching the load.
Contact size, number of contacts, disk size and rotation speed give the switching frequency.
To make this work with such high currents you need to prevent arching and burning out the contacts.
If I am not mistaking submerging the whole thing in isolatin oil, like transformer oil, would prevent the worst of the arching and also provide easy means of cooling, including filtering the oil if required.
To avoid stress on the power supply I suggest to use a good capacitor bank to take the main load of the impulse.

A totally different approach to the problem could go like this:
Use a high voltage power supply to drive a voltage multiplier.
The output can be DC depending on the configuration, the input is always AC.
If I understand the math behind it correctly it should be fairly simple to create enough stages to get the mains frequency down to what you need.
This would be of course a very high voltage that is building up in the last stage!
If you feed that voltage through a spark gap into an impulse transformer you get massive AMPS at a low enough voltage.
My fellow professionals here will correct me if I am wrong but I think if spark gap size and distance is correct and preferably in a vacuum enviroment then only a spark gap and maybe two diodes would be required on the output of the transformer to produce a stable enough signla in your loop coil.

Jack A Lopez

2 years ago

Hey, I was just thinking. You are switching some rather large currents, and large power.

I was looking at the Wikipedia article for IGBT,


and I noticed that some of the hardware in their example pictures, particularly those with current ratings in the 100s of amperes, that these things tend to be built as modules, with the freewheeling diodes built in.

So I guess what I am thinking is that my previous answer might have been kind of naive suggestion, if freewheeling diodes are something you already know about, particularly if they are already built into your thing.

The data sheet for your IGBT, of course, should have details about this.

Also, the final thing I was going to mention is that since your switching speed is so incredibly slow, a mechanical switch would probably work well too, if you could find one big enough.

This idea of a mechanical switch came to me, when I was trying to think of the last time I switched a 150 A current, and it might have been the last time I started my car. I think starter motors want current in approximately that range, around 100 A.