I have read and learnt a lot of things about transformers in various places (school, internet, books) but there is something that I consider fundamental that is yet not clear to me.
We connect an inductor at mains. This is the primary. So, current flows through that inductor and establishes a magnetic field inside the inductor. I suppose this field contains energy, according to the 0.5LI^2 common equation. Label it E1.
This field is characterised by a certain flux that is confined inside the magnetic material of the core. Thus, it is linked to the secondary winding, another inductor, and according to Faraday, it will induce an EMF across it.
Now the tricky part. Connect a load on the secondary winding. An EMF is present, so current will flow across the load. So this current will also create a flux that will be linked to the primary this time. Consider this field as carrying a total energy of E2.
I know that the primary will counteract and induce an "opposite current" according to turns ratio to extinguish that new field.
So, lets take a transformer is rated at VA voltamperes. So it can ideally contain (this is my understanding) a maximum total of VA energy. Label it E.
What is the relationship of these three energies? We are concerned about short circuits, so is E2 the most important? Because if ideally the primary cancels out the effect of that field, and no additional flux is produced, how can the transformer suffer? If this is the case, then why can't we connect a huge load of 1000VA perhaps across a 10VA transformer? Is the winding resistance the problem? The heat source or whatever?
Also, magnetising current is commonly specified as 1/10 to 1/100 of nominal current. Thus, I cannot expect E1 to be roughly equal with E. So it is only used to create the voltage on the secondary winding? That's all?
So, I have come to realise that the flux induced by the magnetising current (so secondary has no load) is used only to create the voltage on the secondary winding, and that the amount of E2 is limited by E, so that E2 must always be smaller or equal (ideally) to E.
I speak of energies since I read that the flux that can be contained in the core can be expressed in terms of VA. But then again, flux is dependant on the primary voltage. Does this mean that by construction, we are already close to E even at no load? This confuses me - we are concerned about a short circuit, but we also design our transformers not to come into saturation given the mains voltage.
These hysteresis curves can be viewed as V-I characteristics. But that I, is it the magnetising current? The total primary current? Or what else?
Excuse me for not being clear, but I tried to explain myself. But in a nutshell, I have not yet understood the flux/energy limitations of a transformer, or its mechanism all together, I don't know. Anyone who is willing to explain to me anything of the above, is welcome!
We connect an inductor at mains. This is the primary. So, current flows through that inductor and establishes a magnetic field inside the inductor. I suppose this field contains energy, according to the 0.5LI^2 common equation. Label it E1.
This field is characterised by a certain flux that is confined inside the magnetic material of the core. Thus, it is linked to the secondary winding, another inductor, and according to Faraday, it will induce an EMF across it.
Now the tricky part. Connect a load on the secondary winding. An EMF is present, so current will flow across the load. So this current will also create a flux that will be linked to the primary this time. Consider this field as carrying a total energy of E2.
I know that the primary will counteract and induce an "opposite current" according to turns ratio to extinguish that new field.
So, lets take a transformer is rated at VA voltamperes. So it can ideally contain (this is my understanding) a maximum total of VA energy. Label it E.
What is the relationship of these three energies? We are concerned about short circuits, so is E2 the most important? Because if ideally the primary cancels out the effect of that field, and no additional flux is produced, how can the transformer suffer? If this is the case, then why can't we connect a huge load of 1000VA perhaps across a 10VA transformer? Is the winding resistance the problem? The heat source or whatever?
Also, magnetising current is commonly specified as 1/10 to 1/100 of nominal current. Thus, I cannot expect E1 to be roughly equal with E. So it is only used to create the voltage on the secondary winding? That's all?
So, I have come to realise that the flux induced by the magnetising current (so secondary has no load) is used only to create the voltage on the secondary winding, and that the amount of E2 is limited by E, so that E2 must always be smaller or equal (ideally) to E.
I speak of energies since I read that the flux that can be contained in the core can be expressed in terms of VA. But then again, flux is dependant on the primary voltage. Does this mean that by construction, we are already close to E even at no load? This confuses me - we are concerned about a short circuit, but we also design our transformers not to come into saturation given the mains voltage.
These hysteresis curves can be viewed as V-I characteristics. But that I, is it the magnetising current? The total primary current? Or what else?
Excuse me for not being clear, but I tried to explain myself. But in a nutshell, I have not yet understood the flux/energy limitations of a transformer, or its mechanism all together, I don't know. Anyone who is willing to explain to me anything of the above, is welcome!
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Transformer loading is dominated by heating in the winding resistance.
Off-load, a transformer is limited by flux saturation in the core. This gets easier as a load is applied. Some cheap transformers approach saturation when off-load so they can't cope with a rise in primary voltage.
Off-load, a transformer is limited by flux saturation in the core. This gets easier as a load is applied. Some cheap transformers approach saturation when off-load so they can't cope with a rise in primary voltage.
Not a useful way to think about it. The transformer does not 'contain' energy equal to the VA rating. A correctly operated transformer contains very little energy as the primary and secondary currents set up opposing fluxes so the next flux is low - hence stored magnetic energy is low.
Yes, I know it is not a good way. Just used to illustrate that I have read that flux is dependant on VA. And from your answer, I deduce that the hysteresis curves refer to the magnetising current alone, am I right?
So, assume I have a 10VA transformer core, and I manage to wind it using heavy copper, so that winding resistance is very low. Can I safely load the secondary to 50VA? I mean, if that is the case, why wouldn't we just do that and forget about big irons?
A good designed 10VA transformer as well as a 1kVA will not approach saturation off-load. If the had the same voltages and winding resistance, could they cope with the same load?
So, assume I have a 10VA transformer core, and I manage to wind it using heavy copper, so that winding resistance is very low. Can I safely load the secondary to 50VA? I mean, if that is the case, why wouldn't we just do that and forget about big irons?
A good designed 10VA transformer as well as a 1kVA will not approach saturation off-load. If the had the same voltages and winding resistance, could they cope with the same load?
A 10VA core does not have enough physical room for a 100VA winding. For best transformer efficiency you want copper and iron losses to be roughly equal. If you want a smaller transformer then you have to increase the frequency, and use an appropriate core material. The flux is related to the inefficiency of the transformer - the degree to which primary and secondary currents fail to exactly cancel.
There is a deep mystery in transformers, as a perfect transformer would have exact cancellation and so zero flux! The answer lies in EM theory and the vector potential (A) which can be non-zero even when the fields are zero - IIRC. I think Feynman talks a little about this -or maybe I saw it in another textbook.
There is a deep mystery in transformers, as a perfect transformer would have exact cancellation and so zero flux! The answer lies in EM theory and the vector potential (A) which can be non-zero even when the fields are zero - IIRC. I think Feynman talks a little about this -or maybe I saw it in another textbook.
Yes
A properly designed 10VA (or any VA for that matter) will use all of the winding area for the primary and secondary(ies).
What counts is the copper area, the way the copper is distributed has no influence. Often the characteristics of a core are given in terms of unit-turn for that reason.
If you had an enormously more effective conductor than copper, say superconductive, your 10VA could become a 1KVA
With a 1KW load, the output of a 10VA would collapse completely because of the internal resistanceA good designed 10VA transformer as well as a 1kVA will not approach saturation off-load. If the had the same voltages and winding resistance, could they cope with the same load?
Thanks for your answers! So I assume it goes down to the fact that we are not able to wind heavy copper around a small core.
Still, can't bigger cores carry more VA? I have read this in engineering books. Saying more VA, I am referring to the V-A product of the primary. I assume it is also the VA at the primary off-load.
Elvee, I said "assuming they have same winding resistances".
DF96: You mean the flux caused by load current demands?
Still, can't bigger cores carry more VA? I have read this in engineering books. Saying more VA, I am referring to the V-A product of the primary. I assume it is also the VA at the primary off-load.
Elvee, I said "assuming they have same winding resistances".
DF96: You mean the flux caused by load current demands?
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The resistances can be equal in more than one way. There is an infinite number of couples meeting that condition in fact, but anyway it is not very interesting because it is completely hypotheticalElvee, I said "assuming they have same winding resistances".
There is a measure called magnetic reluctance which has to be considered when selecting the cross-section of a core from a given material. I think typically this goes up with the VA rating.
In other words, just as the cross section of a wire goes up with the rated current, so the cross section of a core goes up with the rated VA. It may not be the best analogy, but a convenient way of understanding why the cores have to get bigger and heavier, not just larger in diameter.
In a normal good designed transformer, no significative energy is stored in it.
As I know, energy stored in the core is very small or null, a small amounts of it in the leakage inductance.
A special case in where energy is stored deliberately, in a special form on fly "back transformers", in SMPS's and horizontal deflections in CRT monitors and TV's, although this kind of transformers aren't true transformers, better called are mutually coupled inductors.
In this cases, primary and secondary currents don't flow simultaneously, which some special performance are obtained and a well defined gap is inserted in the core to obtain a specific inductance, for example a 1:1 "fly back transformer" can give voltage ratios of 1:10, 1:4, 1:1, 2:1, or even 10:1 etc. This cases, Li²/2 * duty ratio is stored in the gap of the transformer (note that the reluctance of the ferrite /iron is so small compared which one of the non magnetic gap (usually air), so when it energy is released in the secondary (and primary windings) is recovered as a different voltages, although same power input as output(s) is maintained.
As I know, energy stored in the core is very small or null, a small amounts of it in the leakage inductance.
A special case in where energy is stored deliberately, in a special form on fly "back transformers", in SMPS's and horizontal deflections in CRT monitors and TV's, although this kind of transformers aren't true transformers, better called are mutually coupled inductors.
In this cases, primary and secondary currents don't flow simultaneously, which some special performance are obtained and a well defined gap is inserted in the core to obtain a specific inductance, for example a 1:1 "fly back transformer" can give voltage ratios of 1:10, 1:4, 1:1, 2:1, or even 10:1 etc. This cases, Li²/2 * duty ratio is stored in the gap of the transformer (note that the reluctance of the ferrite /iron is so small compared which one of the non magnetic gap (usually air), so when it energy is released in the secondary (and primary windings) is recovered as a different voltages, although same power input as output(s) is maintained.
PMI: This is actually what I am asking. VA used for what?
As I understand from the above, there is no additional flux, ideally, due to load current. Because the one that is created by the secondary current is cancelled by an opposite one coming from the primary current that is generated to supply the secondary current.
So, in what way could the VA used by a 10VA transformer for the establishement of the mutual flux differ from the VA needed by a 1kVA one? If they have same primaries (please ignore primary copper losses to get my point), how would they use different VA seen by the mains?
And as answered above, I realise that even then you can't use a universal small core, since for high currents you need heavy copper, and all this cannot fit around a small core. If I am not mistaken.
Posts 4 and 5 state this fact, and seem quite well written to me.
As I understand from the above, there is no additional flux, ideally, due to load current. Because the one that is created by the secondary current is cancelled by an opposite one coming from the primary current that is generated to supply the secondary current.
So, in what way could the VA used by a 10VA transformer for the establishement of the mutual flux differ from the VA needed by a 1kVA one? If they have same primaries (please ignore primary copper losses to get my point), how would they use different VA seen by the mains?
And as answered above, I realise that even then you can't use a universal small core, since for high currents you need heavy copper, and all this cannot fit around a small core. If I am not mistaken.
Posts 4 and 5 state this fact, and seem quite well written to me.
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I never said that in my post, or at least did not try to imply that. Just because energy input equals energy output plus energy lost as heat, does not mean that you can reduce the medium which conducts that energy to nothing, and still have an output!
Energy supplied to a lamp equals the energy produced as light plus heat, but will the lamp work without a cord?
Alternating current through the primary produces magnetic field in the core.
A magnetic field in the core produces an alternating current in the secondary.
No magnetic field -> No output current.
The difference between VA input and VA output is due to copper and core losses, and determines the efficiency of the transformer.
To make a higher secondary current, or a higher secondary voltage with the same current, you need a core which can conduct more magnetic flux.
Assuming the same meterial is used, this means that higher VA requires a core with a bigger cross-section.
There is a better explanation complete with the basic formulas and a picture on Wikipedia, here:
Transformer - Wikipedia, the free encyclopedia
Consider that the cost of a large and heavy core is a big part of the transformer cost. If the cross-section did not matter, would we have such heavy cores?
Read the previous posts and you will realise why I don't understand that.
I realise that secondary voltage is dependent on secondary windings, off-load. As mentioned earlier, when a load is connected to the secondary, current flows and produces a flux, which is then cancelled out by means of the primary. So I don't see why additional flux is needed to up the VA in the secondary.
Believe me, this is my thinking too: I couldn't realise that it is possible to make the core tiny and get kVA's of load! But the previous posters underline that the limitations that prevent the use of a tiny core is that you need heavy copper for high loads, which can't be wound on a tiny core.
But for some reason, your approach seems intuitive and correct to me. This is why I am so confused.
I realise that secondary voltage is dependent on secondary windings, off-load. As mentioned earlier, when a load is connected to the secondary, current flows and produces a flux, which is then cancelled out by means of the primary. So I don't see why additional flux is needed to up the VA in the secondary.
Believe me, this is my thinking too: I couldn't realise that it is possible to make the core tiny and get kVA's of load! But the previous posters underline that the limitations that prevent the use of a tiny core is that you need heavy copper for high loads, which can't be wound on a tiny core.
But for some reason, your approach seems intuitive and correct to me. This is why I am so confused.
It does seem that those posts are saying that, but sometimes what you type does not come out quite right.
It does not mean there is no flux in the core, it is just a mathematical construct, like two sides of an equation, or debits and credits in a business account. At any time, the accounts balance, but that does not mean no money is flowing through them.
I think that the "cancelling" is like two sides of an equation being equal. It is just terminology. You can say that you get a salary, and that you use that salary to buy things. You could say that your expenses cancel out your salary. It does not mean that there is no money going from one place to another.
But I never said there is no flux. I said there is no flux due to load current. Of course there is mutual flux, this is the way of inducing an EMF into the secondary using the turns ratio. This is why I asked you about the VA - since the load has no contribution to the total flux (according to previous posts), then I don't see why higher flux is needed for higher loads.
That may not be exactly what those posts meant. Core size will increase with VA rating, not just wire gauge.
Hello AJT,
no, I don't make my own transformers! I just want to become familiar with their operation.
Can you explain the "winding window"? You mean what is previously stated, that is the fact that you can't wind heavy copper on a small core?
PMI, I think the posts meant that. Of course VA rating for the core will increase with size, but if bigger size is needed to wind, how can we deduce that it it needed for anything else?
I own Fitzgerald's "Electric Machinery", where the following is stated:
"the rms voltamperes required to excite the core to a specified flux density is equal to.." , giving an expression that shows that this VA required to excite the core are proportional to the flux required and the volume of the core. So I understand that you need higher VA to excite the bigger core, not in order to manage a higher load, but just because it is bigger.
no, I don't make my own transformers! I just want to become familiar with their operation.
Can you explain the "winding window"? You mean what is previously stated, that is the fact that you can't wind heavy copper on a small core?
PMI, I think the posts meant that. Of course VA rating for the core will increase with size, but if bigger size is needed to wind, how can we deduce that it it needed for anything else?
I own Fitzgerald's "Electric Machinery", where the following is stated:
"the rms voltamperes required to excite the core to a specified flux density is equal to.." , giving an expression that shows that this VA required to excite the core are proportional to the flux required and the volume of the core. So I understand that you need higher VA to excite the bigger core, not in order to manage a higher load, but just because it is bigger.
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winding window....the open space between the center leg....
yes you can, to the extent that it fits the window....
An externally hosted image should be here but it was not working when we last tested it.
yes you can, to the extent that it fits the window....
ideally this should be zero at no load, in the real world this VA is very small, for the same core area, it will decrease with increase in primary turns, and vice versa...
loading the secondary, the primary winding takes current from the source according to its turns ratio...
primary flux increases until a point is reached that flux can no longer increase, this is called saturation under load...another way to do this without load is to increase the primary voltage until huge currents flow thru the primary winding..(see my first line)
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