Back-EMF from series and parallel inductors?

[Martin G7MRV]

Bit of an odd one. OK so I know how inductors interact in series and parallel in terms of inductance value etc, but this is of course with 'normal' signals and forward currents,

What I'd like to know, is how series/parallel connection affects the back-EMF generated when a DC pulse is applied?

My thought process is that series connected, since that is the higher overall inductance, would also have the greater back-EMF? And that if two inductances are present on the same ferrite/iron dust core, the maximum back-EMF would occur if the two windings were in phase?

In practical terms - I have a small 'transformer' from an old switching supply, with two windings, 20mH and 25mH. Im using this in a simple HV generator that relies on the back-EMF through a voltage doubler. I'd like to get the highest voltage I can out of it, which means as high a back-EMF as I can get.


Cheers
Martin

[kalee20]

Back-EMF on an inductor is just inductance x rate-of-change-of-current.

In practice, inductors always have some self-capacitance, so if you pass current through an inductor and abruptly stop it (like turning off a MOSFET in series. Mechanical switches usually spark-over as contacts are separating), the highest voltage spike you see is determined by the self-capacitance as much as the inductance.

It's not possible to determine this, therefore, without knowing all the factors!

Your 'transformer' from a switching power supply - if you have two windings, one 20mH and one 25mH, because they are on the same core you also need to consider coupling between them. So connecting in series you could get much more than 45mH, the limit is actually 89.7mH. But you could also get 0.28mH if connected out of phase.

While higher inductance means more energy is stored, it also takes longer to establish a current in a large-valued inductance. And, of course, you need to ensure the thing doesn't saturate, else you're wasting current for nothing!

(Inductors in series: Ltot = L1 + L2 +2M where M = mutual inductance)

If connected in parallel, the effective inductance would be near zero, and losses would be high.

(Inductors in parallel: Ltot = (L1 x L2 - M²)/(L1 + L2 - 2M) ).

[Radio Wrangler]

Easier than that. If you have a group of inductors in series/parallel connections of any complexity, then combine them as usual (Series adds as inductance, parallel add as 1/inductances) so you get a single equivalent inductance value between two terminals.

Now look at the current between those terminals and the voltage = -L.dI/dt

The calculated equivalent inductance is just what it says.

The above assumes no mutual field coupling.

David

Crossed with Kalee20 = I must type faster!

[kalee20]

Just as an addition to the above - the formula for inductors in parallel:

Inductors in parallel: Ltot = (L1 x L2 - M²)/(L1 + L2 - 2M)

If the coupling between the two inductors is perfect (wound in close proximity on the same core), then M² = L1 x L2 so you get zero inductance, whatever the values of L1 and L2. Except in a special case:

If the inductors L1 and L2 are identical (thus wound with the same number of turns) the formula gives Ltot = 0/0 which is meaningless. A bit of mathematical trickery shows that in this special case, Ltot = L1 = L2, ie the effective inductance is the same as either considered separately. You just get the benefit of current sharing between the two windings, so it doesn't get as hot!

[Martin G7MRV]

Much as I thought then, likely best in series and in phase, and up to a point!

I expect that I wont be able to drive it up too far, as its not intended for these voltages and so the insulation is likely to break down, but i'll see just how far I can push it.

Thanks
Martin

[Martin G7MRV]

It also makes more sense to me now how the variants of these circuits using small 1:1 audio transformers are operating.

It makes sense in this case to use the transformer as a transformer, rather than consider them as a pair of inductors wired series or parallel.

The ultimate goal is 400V DC from a 1.2V AA NiMH cell, to drive a Russian SI-19BG G-M tube.

[TrevorG3VLF]

A lot of power supplies use switch mode with a high frequency amall transformer to get low voltages.
Have you cosidered using one of these cores in reverse to generate the high voltage? The insulation should be adequate for over 200V.

[Argus25]

Originally Posted by Martin G7MRV

In practical terms - I have a small 'transformer' from an old switching supply, with two windings, 20mH and 25mH. Im using this in a simple HV generator that relies on the back-EMF through a voltage doubler. I'd like to get the highest voltage I can out of it, which means as high a back-EMF as I can get.


Cheers
Martin

I'll answer your question in a different way, because regardless of whether you use one winding, or the two in series aiding to create a larger inductance, you will need to know how to calculate the peak voltage you will get 1/4 of a cycle into the oscillation, after your switching device (presumably an NPN transistor) switches off, its all about the equivalence of electric and magnetic field energy.

Firstly, when you drive an inductor like this in a single ended manner, do not forget the energy recovery diode (a BY228 works fine for most applications) from the transistor's collector to emitter, it stops the collector going negative with respect to the emitter which occurs at/after 1/2 a cycle into the oscillation, and it also returns magnetic energy stored in the core, back to the power supply.

When your driver circuit (maybe a 555 or similar) switches the transistor on, the supply voltage V is applied across the inductor, inductance L. Initially the current climbs at a rate of V/L linearly. After a while the rate tapers off, in the usual inverted exponential form and would settle on V/R where R is the resistance of the inductor & wiring. As noted you don't want to go near that region. keep in the linear zone ideally.

At the end of the time you allow the current to build up for, there will be some current I in the inductor and the magnetic energy field will have a value of (I^2)L/2. Then, when your transistor switches off, and the field attempts to collapse, this energy starts to exchange with the electric field of the capacitance.

But what is the self capacitance of your inductor ? (You can actually estimate it easily by putting it on a generator with two high value series resistors like 4.7M , so as not to damp it and check is self resonance with a generator with the scope across one resistor) . The transistor also adds some capacitance at its collector.

At 1/4 cycle into the oscillation all the stored magnetic field energy is transferred to the electric field of the capacitance (ignoring losses in resistances, core material, dielectrics etc), so to calculate the peak voltage Vp you make (I^2)L/2 =(Vp^2)C/2 and solve for Vp.

In practice it is better in most cases to lower the peak voltage achieved by adding lumped capacitance across the coil, (it can be added across the diode connections). This way the oscillation frequency is lower, the 1/2 cycle of oscillation is broader, making it easier to peak rectify without damping the peak, it also swamps out the coil and transistor's self capacitance, making it dead easy to calculate the voltage you get. You could start with 1000 to 2000pF. Then you also don't have to know exactly the self capacitance properties of your inductor or transistor either. Also, it is often necessary to have the lumped capacitance to ensure that the max Vce voltage of your transistor is not exceeded or you will send it quickly to la-la land.

One design trick is to start with a large capacitance, say 0.1uF at least, and progressively lower it in steps, until your peak voltage at the transistor's collector is acceptable for the transistor and you have achieved the value you want. It saves destroying transistors.

Obviously, what I have just described to you is the H output stage of a TV set. And even generating high voltages in this manner, if you need really high voltages an overwind is required. However, for example, a typical small TV lopty, running from 12V can produce a peak collector voltage of 300v, and that is with it tuned down in frequency with added lumped capacitance.

[Argus25]

PS: just one thing on the terminology, the notion of "back emf" is applicable when you are powering the coil from the supply, the inductance opposes the change in current by producing a terminal voltage as noted by David in post #3 as
V = - L.di/dt.

note the minus sign.

But this is not what happens when the coil is de-powered and you get your voltage peak, due to the collapsing magnetic field, as described by Lenz's law of induced current and the voltage you get is in opposite polarity to what was applied.

Often the term back emf is incorrectly used. For example a person might say that the back emf from the coil burnt their relay contacts (the relay contacts switching a coil can arc over when the contacts open), but in fact this is incorrect, it just sounds plausible because you appear to be getting voltage back from the inductor. What really happened is that the stored magnetic field rapidly collapsed inducing a high voltage on the coil terminals causing the contacts to arc over.

[Martin G7MRV]

So what is the 'correct' terminology in this case?

In regards to the relay as mentioned - the term 'back-EMF' is what was taught on my HND Electronic & Electrical Engineering back in the '90s, and seems to be a very well known and understood term in relation to relay drive protection.

[Skywave]

Originally Posted by Argus25

What really happened is that the stored magnetic field rapidly collapsed, inducing a high voltage on the coil terminals causing the contacts to arc over.

I'm a bit puzzled by that remark. The high voltage appears across the coil - not the relay contacts. So why, therefore, should that HV cause the coil contacts to arc over, since there are no electrical connections between those contacts and the coil itself?

Al.

[ms660]

Originally Posted by Skywave

I'm a bit puzzled by that remark. The high voltage appears across the coil - not the relay contacts. So why, therefore, should that HV cause the coil contacts to arc over, since there are no electrical connections between those contacts and the coil itself?

Relay contacts are often in series with the relay coil, for example in some tape recorder transport circuits.

Lawrence.

[Argus25]

Originally Posted by Skywave

I'm a bit puzzled by that remark. The high voltage appears across the coil - not the relay contacts. So why, therefore, should that HV cause the coil contacts to arc over, since there are no electrical connections between those contacts and the coil itself?

Al.

I was referring generally to relay contacts switching an inductive load or some coil, and not referring to the relay coil or its inductance.Sorry I did not make that clear.

[Argus25]

Originally Posted by Martin G7MRV

So what is the 'correct' terminology in this case?

In regards to the relay as mentioned - the term 'back-EMF' is what was taught on my HND Electronic & Electrical Engineering back in the '90s, and seems to be a very well known and understood term in relation to relay drive protection.

I was once taught as a boy that Fish feel no pain because they have no nerves so its ok to beat them over the head with a club, then I learnt that the Dogfish has practically identical cranial nerves to a Human and makes a great basic anatomical model. Or that Lemmings committed suicide, but really they were pushed over the cliff, or accelerated & flung off a rotating carousel, for the benefit of the documentary film makers.

You cannot believe everything you read or are told. Or sloppy usage of terminology taught in courses. That includes any of my remarks too as I could be wrong, so check it out for yourself.

The correct terminology for this type of supply is "flyback supply" or "flyback supply with energy recovery" to be specific or another term that is very apt I have seen is "Impulse supply" which I quite like when the method of high voltage generation doesn't specifically refer to television use. I have seen the term "flywheel" applied as some people have called the energy recovery diode a flywheel diode, but its poor because it crosses over with flywheel diode arrangements in horizontal sync separators.

[G8HQP Dave]

My understanding is that 'back EMF' is used to refer to two situations:
1. the voltage developed across an inductance when the current is sharply reduced (often to zero) - the voltage depends on the stray capacitance too.
2. the voltage developed from the motion of an electric motor - the voltage depends on motor speed.

Hence if a voltage appears across some contacts in series with a relay coil when the contacts break then this is back EMF. You can consider it arising from LdI/dt, or you can consider it being due to kinetic energy being transformed into potential energy - the two explanations are both correct and give the same result.

[Herald1360]

The term for small flyback supplies (as used for LED drivers from a single cell, for instance) that I like is "Joule Thief".


Derives from their ability to extract the last drop of energy from the battery (within the constraints of the active device turn on voltage).

[Skywave]

Originally Posted by Argus25

What really happened is that the stored magnetic field rapidly collapsed, inducing a high voltage on the coil terminals causing the contacts to arc over.

Originally Posted by Skywave

I'm a bit puzzled by that remark. The high voltage appears across the coil - not the relay contacts. So why, therefore, should that HV cause the coil contacts to arc over, since there are no electrical connections between those contacts and the coil itself?

Originally Posted by Argus25

I was referring generally to relay contacts switching an inductive load or some coil, and not referring to the relay coil or its inductance. Sorry I did not make that clear.

Ah! Thanks: all is now clear.

[Argus25]

Originally Posted by G8HQP Dave

My understanding is that 'back EMF' is used to refer to two situations:
1. the voltage developed across an inductance when the current is sharply reduced (often to zero) - the voltage depends on the stray capacitance too.
2. the voltage developed from the motion of an electric motor - the voltage depends on motor speed.

Hence if a voltage appears across some contacts in series with a relay coil when the contacts break then this is back EMF. You can consider it arising from LdI/dt, or you can consider it being due to kinetic energy being transformed into potential energy - the two explanations are both correct and give the same result.

Not exactly, I beg to differ.

Back emf applies when the voltage you apply to an inductor is immediately applied or increasing. The polarity of the voltage generated attempts to match the applied voltage, it has the same polarity as the supply, so this opposes the current. That generated voltage is correctly called "back emf".

However, when the applied voltage is disconnected (after a magnetic field is established), the the inductor attempts to maintain the current and the voltage it generates as the field collapses has an opposite polarity to the previously connected supply. This is not "back emf" and its an example of Lenz's law , the rate of change of magnetic flux generates the voltage.

Since, when the inductor looks electrically open from the DC perspective, the potential generated can only establish a current by charging the coil's self & other stray capacitances, the peak voltage of the initial 1/2 cycle reaches a very high value, especially as the capacitances are small and the system then simply has decaying oscillations.

"back emf" is correctly applied to motors, which act as generators, and generate near the same polarity and potential of the applied voltage, and less when you load them, so the current increases.

Here are some applications of Lenz's law lifted from the net:

Lenz’s Law Applications

The applications of Lenz’s law include:

Lenz’s law can be used to understand the concept of stored magnetic energy in an inductor. When a source of emf is connected across an inductor, a current starts flowing through it. The back emf will oppose this increase in current through the inductor. In order to establish the flow of current, the external source of emf has to do some work to overcome this opposition. This work can be done by the emf is stored in the inductor and it can be recovered after removing the external source of emf from the circuit
This law indicates that the induced emf and the change in flux have opposite signs which provide a physical interpretation of the choice of sign in Faraday’s law of induction.
Lenz’s law is also applied to electric generators. When a current is induced in a generator, the direction of this induced current is such that it opposes and causes rotation of generator (as in accordance to Lenz’s law) and hence the generator requires more mechanical energy. It also provides back emf in case of electric motors.
Lenz’s law is also used in electromagnetic braking and induction cooktops.

[Argus25]

To clarify a little more, the term "back emf" I think was coined to describe "counter-emf".
Counter emf was to describe something being countered, in this case the voltage or emf source applied to an inductor (or as pointed out applied to a motor).

In the case of an inductor with a previously established magnetic field, that gets disconnected from an external circuit and source of emf, nothing in this case is being "countered". The voltage generated is an emf of induction from the collapsing field, similar (but much more rapid) to what it would be if you physically pulled a magnet out of a coil.

This is why "back emf" or a back emf supply is an inappropriate term for these types of impulse power supplies that rely on a collapsing field for a high voltage pulse.

Looking at the plethora of examples on the internet where collapsing fields generating a voltage peak have been called "back emf" its not a wonder there is so much confusion about what it means.Vintage textbooks appear to have been much tighter with the use of the terminology.

[Argus25]

Also, I appear not to be the only person who has thoughts like this . I found this on a forum from somebody remarking on the "back emf misnomer":

When i say "Back emf" I mean (counter emf), "counter emf" or "back emf" being the force resisting the applied emf.

The collapsing of the magnetic field of a coil does not produce "counter emf" it produces emf, counter emf is in opposition to applied emf,
the emf produced when the magnetic field of a coil collapses does not oppose the applied emf, it emulates it or aids it.

Energy returning to the supply is not back emf either it is reactive power. That's how I see it.

Back emf or counter emf can only be the force that opposes emf while the emf is applied.

So at least I'm not entirely alone with the view.But this was posted on a "free energy" forum!