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Coupled inductors thus transformer

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Andy

Sun Mar 17 2013, 03:43AM

Registered Member #4266 Joined: Fri Dec 16 2011, 03:15AM
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Hi
What would be the difference from a coupled inductor and transformer?
If you have a CI with the primary 1mH and the secondary 1H, would the primary have a time constant of 1.001H or 1mH?

Would this work

Thanks

Dr. Slack

Sun Mar 17 2013, 07:12AM

Registered Member #72 Joined: Thu Feb 09 2006, 08:29AM
Location: UK St. Albans
Posts: 1659

Andy wrote ...

If you have a CI with the primary 1mH and the secondary 1H ...

... then the primary would have an *inductance* of 1mH.

It will also have a time constant when coupled with an external resistance, and its own intrinsic resistance.

Once fouled up with the mutual inductance to the secondary, the secondary parameters, and some inter-turn capacitance, you will have a very complicated system.

But if you're asking whether it will work as a typical flyback (which has an open circuit secondary when charging), then charging the 1mH primary will work with a time constant mostly governed by the primary parameters.

Andy

Sun Mar 17 2013, 07:36AM

Registered Member #4266 Joined: Fri Dec 16 2011, 03:15AM
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Thanks

But if you're asking whether it will work as a typical flyback (which has an open circuit secondary when charging), then charging the 1mH primary will work with a time constant mostly governed by the primary parameters.

In this case(1mH/1H) would the voltage on the secondary be 1000 times more?
What if it had a extinal inductor of 1uH parallel to the primary?

Dr. Slack

Sun Mar 17 2013, 07:52AM

Registered Member #72 Joined: Thu Feb 09 2006, 08:29AM
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Why would you parallel a 1uH inductor with a 1mH primary? It will hog practically all of the primary current.

For 100% coupling, ignoring all other parameters, for a given voltage across the 1mH winding, the 1H winding voltage will be 31.62 times higher (sqrt(1000)). But that's not how you work a flyback, that's how you work a forward converter.

What are you trying to do, flyback or forward? The advice in each case is rather different.

Andy

Sun Mar 17 2013, 08:23AM

Registered Member #4266 Joined: Fri Dec 16 2011, 03:15AM
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I'm trying to send alot of current throught a large inductor then cut the power. I thought that if you have three coils two primary and one secondary, the two primary cancel each other(opposite direction) but add to the secondary, you could send alot of power throught the primary without inductance or impedance limiting current.

Dr. Slack

Sun Mar 17 2013, 04:49PM

Registered Member #72 Joined: Thu Feb 09 2006, 08:29AM
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Posts: 1659

Andy

Mon Mar 18 2013, 11:03PM

Registered Member #4266 Joined: Fri Dec 16 2011, 03:15AM
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Thanks. Got to many thoughts bouncing around :)
With a transformer that is 1:1, what convens the voltage to the level of the primary. Some types can have 40 turns, others need 400, is it just the core width?

Dr. Slack

Mon Mar 18 2013, 11:14PM

Registered Member #72 Joined: Thu Feb 09 2006, 08:29AM
Location: UK St. Albans
Posts: 1659

I would refer to to the HVWiki, the transformers page

, but frankly I just see a wall of maths, that will mean little to you until you have a good understanding of transformers from a different source, by which time you won't need that page.

So ...

The voltage on a turn is governed by the rate at which the flux in the core can be changed. Period.

a) For normal AC transformers which operate at a fixed frequency, and tend to operate at the maximum possible flux, the volts per turn is fixed for any given core area. You cannot exceed this volts/turn without drawing large and potentially damaging currents.

b) If you don't operate at fixed frequency, but can change the flux real fast by breaking the primary current, you can get very high voltages indeed, usually high enough to damage something unless you take additional steps to control it.

Andy

Tue Mar 19 2013, 02:56AM

Registered Member #4266 Joined: Fri Dec 16 2011, 03:15AM
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One thing about the wiki, it says volts=n*(flux/t), and to workout the flux flux = v/(4.44*f*N).
Why is voltage the thing that gets divided, I thought that current and turns is what make mmf.
Is it correct that the high the voltage that you have on the primary, the more turns you will need on both to make a strong enough field.

15000/4.44*40*200p = 2.36mWb
200s*0.0023Wb/0.025(40) = 18.4volts

that's at 1:1, wouldn't it be the same?

Dr. Slack

Tue Mar 19 2013, 08:39AM

Registered Member #72 Joined: Thu Feb 09 2006, 08:29AM
Location: UK St. Albans
Posts: 1659

Andy wrote ...

and to workout the flux flux = v/(4.44*f*N).

Ugh. I hate formulae like that. The 4.44 is a constant that sweeps up various others like sqrt(2) and 2pi in a non-obvious way. There might even be some metric/imperial conversion factors in there for all I know or frankly care. For basic understanding, I much prefer to go back to first principles.

The dimensions of f are 'per second'. Play around with that formula, moving stuff from side to side, and you will get to volt.seconds = k * flux. If you are getting confused between mmf needing current, but flux being given in terms of voltage, that's probably the simplest and most comprehensive place to start.

I'm not going to spell it out in this post, but I will try to update the wiki having learnt lessons from our discussion.

For an ideal transformer with 100% coupling ...

To determine the voltage on any turn of a single winding inductor or a many-winding transformer, you need to know the rate of change of flux.

To determine the flux in the core, you need to know the total magnetising currrent. This is the signed addition of all the currents in all the turns of all the windings.

In general, the permeability of the core is so high, that a very small current, an order of magnitude or two smaller than the transformer's rated current, is sufficient to provide enough mmf to bring the core up to full flux. To a first approximation, we can use zero for the magnetising current. This means you can have any flux you like, for zero current. This means that the winding current does not determine the flux, to our first approximation.

Wait, you can't do that! Why not? The difference between an ideal transformer, and a good transformer, is only one of losses, of detailed quantative behaviour. A good transformer will have a bit of volt-drop, lose a bit of heat, not be 100% efficient, but will otherwise be indistinguishable at the few percent level from an ideal transformer.

So what determines the flux then? We know that an ideal transformer will draw no current when connected to the supply, and we know that this is sufficient to magnetise the core. If there is any voltage across the conductors of the windings, we know that a current will flow. Therefore we know that the back emf on the primary, generated in the windings by the changing flux, is equal at all times to the supply voltage. Therefore we know that the flux is the integral of the supply voltage.

So, as an exercise, take the primary voltage as sin(wt) where w=2pi * 50 or 60 Hz. Integrate this to a -k.cos(wt) flux variation in the core. Work out the constants of proportionality (use SI throughout), and see if you can derive the 4.44fN formula. Let me know how many nasty constants got swept up in the 4.44 when you're done.

I said the flux was the integral of the applied voltage. Now for the calculus sucker punch - 'ah, what about the constant of integration? got you there!' It turns out the constant of integration is very real and physical, because if you switch a big transformer on at the wrong point of the input mains waveform, you can trip breakers. Instead of swinging between -0.9 and +0.9 of max flux as it does in the long term, the transformer may initially try to swing between 0 and +1.8 of max flux, and draw huge current. In an ideal transformer, this would persist for all time, but a real transformer has losses, and the initial transient dies away, within the overload tolerance time of the breaker if the system has been designed correctly.

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