# Transformer literature

Discussion in 'Electronics Resources' started by mentaaal, Apr 29, 2008.

1. ### mentaaal Thread Starter Senior Member

Oct 17, 2005
451
0
Hey guys,

I am finding that transformer theory is not very well explained (perhaps because it is a difficult subject but anyway.) i am trying to find some book or reference that explains the working of a transformer in great detail. I have real AAC's own info on transformers, and while being extremely useful i am looking for something a bit more detailed so i can get a better understanding.

Could anyone point me in the right direction?

Cheers

2. ### mik3 Senior Member

Feb 4, 2008
4,846
70
Its better to search the internet for some information. Some authors explain something in a different way than others do. Some people understand better the one way and some other the other ways. The best think is to read many resources as take many explanations .

3. ### Caveman Senior Member

Apr 15, 2008
471
1
I went through this at one point. What exactly are you looking to understand better? Is it a particular application or circuit?

4. ### mentaaal Thread Starter Senior Member

Oct 17, 2005
451
0
Ok caveman, thanks for offering your help. The following text is just a copy of an email i have sent to my lecturer detailing my problem.

I am trying to understand flux in the core of a transformer. At present i understand the formula for voltage and flux.

This is the way i see it for a potential transformer:
The current in the primary multiplied by the number of turns causes a magnetomotive force in the core. The flux in the core is determined by the value of the reluctance by: mmf = reluctance*flux. I am deducing myself now that if the secondary is open circuited, the secondary coil appears as a large reluctance in the core inhibiting the flux to very low levels. This could explain why a potential transformer utilises very little power when unloaded as the flux is low and consequently the input current is low.

This could also explain why a current transformer's secondary coil should never be open circuited with the primary powered as the large reluctance introduced by the open circuit secondary will impede the flow of flux. As the formula for flux is the amount of current through the conductor multiplied by the number of turns, if the flux is impeded, then so too will the primary current. This drops the supply voltage across the current transformer as the current transformer will then appear to be a large impedance which is hazardous for obvious reasons.

The reason why i am asking this is because some people have told me that the flux in a transformer doesnt change and others have told me that it is dependant on how loaded the transformer is. Is my line of thinking above correct?

5. ### Caveman Senior Member

Apr 15, 2008
471
1
Going back to basic principles (Ampere's law), the flux in the core is simply related to the current flowing in the windings around the core. But note that this is all of the current flowing through the core, primary and secondary. And each is weighted by the number of turns.

I actually just realized that this means that an ideal transformer that is loaded has no stored energy, ie. no flux. Never thought about it like that. Seems weird.

If the secondary is unloaded, it doesn't come into play in a theoretical transformer. (In a real transformer, the capacitance between the secondary's windings will still load it a bit.) This is why a transformer utilizes very little power when unloaded. Basically in this case it is an inductor, so the (self)-inductance is the only load to the primary circuit.

I don't think the reluctance theoretically changes.

A current transformer has, for example, one turn around the primary and 4000 turns around the secondary. If you were to provide a 40A AC current through the primary then it wants to put 10mA on the secondary. So a 1k resistor across it will produce 10V on the secondary. This is 100mW taken from the primary side. So the voltage drop on the 40A side is 2.5mV.

However, as you increase the resistance (or impedance) on the secondary, the amount of voltage output increases. If you open circuit it, it will generate a theoretically infinite amount of voltage. It does this because it wants to maintain the current. This is why you should always load a current transformer. The voltages generated can be very dangerous.

6. ### mentaaal Thread Starter Senior Member

Oct 17, 2005
451
0

Also further to your point about the unloaded transformer acting as an inductor, when you say inductor, do you mean an inductor with a closed core or a solenoid with an iron bar in it? because in the latter case, the flux would be very low in comparison to the first case due to the reluctance of the air.....

7. ### mik3 Senior Member

Feb 4, 2008
4,846
70
I actually just realized that this means that an ideal transformer that is loaded has no stored energy, ie. no flux. Never thought about it like that. Seems weird.
[/QUOTE]

I think you are false because if there is no flux then there will be no voltage induced on the secondary.

8. ### Caveman Senior Member

Apr 15, 2008
471
1
They are both inductors. One is an air core, the other is an iron core.

All a transformer is is an inductor with at least one other winding. That other winding provides what is called "mutual inductance" as opposed to the normal inductance which is technically called "self inductance". The ideal transformer equation is:

V1 = L1*dI1/dt + M*dI2/dt
V2 = L2*dI2/dt + M*dI1/dt

So you can see that if the secondary is not loaded, there will be no dI/dt for the secondary current, and therefore it reduces to a simple inductor.

9. ### Caveman Senior Member

Apr 15, 2008
471
1
Flux is proportional to current, not voltage. This is one of those things where everything theoretically is simultaneous. Of course we need flux in the core, but at every instance in the ideal case, the energy that is stored in flux is transferred out immediately. So no flux is in the core.

There will still be voltage on the secondary, because if there is not, there will not be the cancelling secondary current.

Of course this is just the ideal situation, so it can never really happen. There is always some amount of self inductance that will store energy in a magnetic field, ie. in flux.

10. ### mentaaal Thread Starter Senior Member

Oct 17, 2005
451
0
I understand what you are saying caveman, and even though my lecturer forgot to do mutual inductance with us last semester i can follow it and it makes sense. But, when you say that the transformer is acting like an inductor when the secondary is unloaded, is it behaving like regular inductor with an iron bar through it or an inductor with a closed path of iron - like a transformer with no secondary coil installed. The reason why i am asking this is because surely the second model - with the closed path would have more flux. If its behavour is acting like the first inductor, then it would kind of fit my wildly speculative and unfounded theory as stated earlier....

11. ### Caveman Senior Member

Apr 15, 2008
471
1
They will act differently, ie. they will have different amounts of flux. This is simply due to their different permeability values.

12. ### triggernum5 Senior Member

May 4, 2008
216
1
Caveman is correct for an ideal transformer.. Simplest way to look at it, is if power_in = power_out 100% of the time, then there is no net power in the core, there is no net flux..

13. ### thingmaker3 Retired Moderator

May 16, 2005
5,073
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So... does anyone have any good book titles?

14. ### Wendy Moderator

Mar 24, 2008
21,838
3,047
I came across one in my library, but that was many 10's of orbits ago. Something that jumped out at me though, the emf from the unloaded secondary still induces a back emf in the primary, it isn't just an inductor. This is the main reason an unloaded transformer draws no current, if it were simply an inductor it would still have a reactance. Or am I misreading what you are saying?

Next time I go to the library I'll see if I can find the title again, but books aren't generally on the internet, and they are usually available at the library. I was studing up on the subject way back when, was thinking of winding a do it yourselfer for some serious power. Nowdays I think I would go the switching route.

15. ### triggernum5 Senior Member

May 4, 2008
216
1
That goes toward pulling the current through the primary each cycle.. If the primary coil was to spend all of its energy on inducing the secondary, then no current would flow through it without the back emf..

16. ### Wendy Moderator

Mar 24, 2008
21,838
3,047
Huh? Could you try restating that please?

17. ### triggernum5 Senior Member

May 4, 2008
216
1
Inductors resist changes in current increases, and in current decreases.. In a perfect standalone inductor, you essentially get a push/pull on each cycle (pushing to magnetic field, pulling back to current) that accounts for 100% of the energy in.. Since the secondary steals the magnetic energy to induce its current, it must return the energy needed for actual current, not just the theoretical perfect inductor induced voltage..
Without the back emf, you'd have reaction without action..

18. ### Caveman Senior Member

Apr 15, 2008
471
1
An unloaded transformer does draw current. An inductor is really just a simplified transformer. The ideal transformer equation is:

Vprimary = Lprimary*dIp/dt + Mp2s*dIs/dt
Vsecondary = Lsecondary*dIs/dt + Mp2s*dIp/dt.

Where Mp2s is the mutual inductance. You can see that in a perfect transformer that has the secondary open circuited, it will look like a simple inductor because the dIs/dt term is 0.