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Transformers have hundreds of turns on both the secondary and primary winding, and as a result use very thin copper wires for each. But, why do they not just use fewer turns on each winding and get the same voltage ratio?

More importantly, why not use fewer turns of a thicker wire for an increased VA? (instead of 1000:100 turns of 22 awg wire, why not 100:10 turns of 16 awg wire if this would increase VA)

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    \$\begingroup\$ Are you basically asking, "Why would a transformer designer needing a transformer taking 120 VAC input and putting out 12.6 VAC, and therefore needing say a 10:1 turns ratio, use 1000 turns on the primary and 100 turns on the secondary instead of 600 turns on the primary and 60 turns on the secondary? What factor makes that choice?" Is that your question? \$\endgroup\$ – jonk Feb 4 '17 at 18:00
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    \$\begingroup\$ "Transformers have hundreds of turns on both the secondary and primary winding". No, they don't, at least not always. A great example is a soldering gun. Those usually have a single-turn secondary. \$\endgroup\$ – Olin Lathrop Feb 4 '17 at 18:12
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    \$\begingroup\$ Transformers often end up using 10% of the rated power current just for core magnetization to improve coupling factor closer to 1. So even a soldering gun has a thousand turns on the primary to accomplish this 100mA or so current V/(2pifL) then use >1 A at 120V for 125W. The number of turns dictates the value of primary L, not the wire diameter. The single turn secondary permits the high current boost ratio. So the smaller the transformer, the more turns needed to raise the no load impedance and reduce the no load current to <=10% \$\endgroup\$ – Sunnyskyguy EE75 Feb 4 '17 at 20:17
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    \$\begingroup\$ If it helps to understand this more intuitively, fewer turns makes for a horrible magnet. Also, NO turns makes it functionally into a dead short, which is super useful for a current sensing transformer, but silly and dangerous for a potential transformer, since dead shorts across meaningful amounts of voltage tend to explode. \$\endgroup\$ – Sean Boddy Feb 5 '17 at 2:44
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    \$\begingroup\$ Read the question as: why do transformers use so many turns to transform... \$\endgroup\$ – zx8754 Feb 5 '17 at 15:32
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When you apply voltage to the primary winding of a power transformer, some current will flow, even when the secondary is open circuit. The amount of this current is determined by the inductance of the primary coil. The primary must have a high enough inductance to keep that current reasonable. For 50 or 60 Hz power transformers, this inductance is pretty high, and you typically cannot get there with a small number of turns in the winding.

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    \$\begingroup\$ Correct, and also mention the real-world core permeability and core dimensions. For example, if the mu of iron happened to be 1000x higher, then 1-turn primaries would work just fine. Or, wind our 1-turn primaries on meter-wide multi-ton iron cores. (Heh, or ditch the 60Hz and use a 30KHz power grid, as done in aerospace.) \$\endgroup\$ – wbeaty Feb 6 '17 at 20:22
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    \$\begingroup\$ @wbeaty No, permeability does not affect saturation. If you want a 1 turn primary in that size of core, you'd need iron that saturated at 2000T instead of 2T. Metre-wide cores would work! \$\endgroup\$ – Neil_UK Feb 7 '17 at 10:10
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    \$\begingroup\$ @mkeith While the inductance keeps the current low, the inductance is dependent on the core permeability, which will collapse if the core flux rises above saturation. We must design enough turns to keep the core field low enough. If we could double the iron permeability, while that would halve the magnetising current drawn, it would not halve the number of turns we could use. \$\endgroup\$ – Neil_UK Feb 7 '17 at 10:12
  • \$\begingroup\$ Note that the frequency is important - a similarly VA rated transformer running on an aircraft at 440Hz will be much smaller and require fewer turns (and thus less copper, less weight, etc). \$\endgroup\$ – Adam Davis Feb 7 '17 at 14:46
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If you had only 1 turn on an iron core it might have an inductance of (say) 1 uH. When you apply two turns inductance doesn't double, it quadruples. So two turns means 4 uH. "So what?" you may say!

Well, for a given AC voltage applied, the current taken by that two turn winding is one-quarter of the current for a single turn winding. Take note because this is fundamental in understanding core saturation.

What causes core saturation (something that is to be largely avoided)? The answer is the current and the number of turns. It's called magneto motive force and it has dimensions of ampere turns.

So, with two turns and one quarter the current, the ampere turns (magneto motive force) is half that of a single turn winding. So, immediately we can observe that if two turns took the core to the "edge" of saturation, a single turn coil would significantly saturate and be a big problem.

This is the fundamental reason why transformers use many primary turns. If a certain transformer has 800 turns and is at the point of saturation, significantly reducing the turns will saturate the core.

What happens when the core saturates you might ask. Inductance starts to drop and more current is taken and this saturates the core more and well, you should see where this is going.

Note that this answer has not considered anything other than the primary winding; in effect we are just talking about the primary magnetisation inductance - it is this and this alone that can saturate the core. Secondary load currents have no part to play in core saturation.

Also note that transformers used in high speed switching power supplies have relatively few turns; 10 henry at 50 Hz has an impedance of 3142 ohms and 1 mH at 500 kHz has exactly the same impedance. For a core that naturally produces 10 uH for a single turn, to wind 1 mH requires ten turns (remember it's turns squared in the formula for inductance). For the same core at 50 Hz (impractical of course), 10 henry requires 1000 turns.

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  • \$\begingroup\$ Comments are not for extended discussion; this conversation has been moved to chat. \$\endgroup\$ – Dave Tweed Feb 6 '17 at 17:57
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    \$\begingroup\$ @DaveTweed I disagree with such an early removing of comments that point out a serious technical flaw of an answer. \$\endgroup\$ – Massimo Ortolano Feb 6 '17 at 20:41
  • \$\begingroup\$ And I disagree with @MassimoOrtolano when he asserts that core saturation is not caused by current. Bio Savart informs that magnetic flux is directly proportional to current. And it matters nothing at all if the thing is a transformer or a loop antenna. I've heard the arguments and accept that you can use volt seconds but why does massimo deny the link between current and flux. Now that is what I call a technical flaw. Why is massimo not giving the same protracted treatment to other answers that say the same? \$\endgroup\$ – Andy aka Feb 6 '17 at 20:52
  • \$\begingroup\$ @MassimoOrtolano: The comments have not been removed, they've just been moved to a chat room. Follow the link provided above. And please continue the discussion there. If you reach any conclusions, post them here. \$\endgroup\$ – Dave Tweed Feb 6 '17 at 21:25
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If you have an iron core for a transformer, one of its specs is "how many turns one winding must have per one volt when the frequency is given". One can't bypass this spec and have fewer turns without having the following consequences

  • reduced efficiency
  • more unwanted transversal current that only causes losses, but do not anything useful for the voltage transformation process

The transversal current can be made lower by increasing the inductance of the primary winding.

The turns/volt spec is a consequence of the following list of facts which all have a tendency to make the coil inductances smaller:

  • the iron material has limited magnetic permeability
  • the iron core can't be made of full iron. It's divided to thin insulated layers to keep the eddy currents small enough in the core. The insulation takes its space and that's off from the iron
  • the magnetic flux of one winding partially bypasses the iron and the other windings
  • too much transversal current causes magnetic saturation in the iron. The saturation radically reduces the magnetic permeability

How one can fight against these by adding more turns? It's because the inductance grows as the square of the number of the turns. One can arque: But the magnetization (=turns x current) grows too! True, but it grows only linearly, so enough turns, then finally the inductance is high enough to outfight the drawbacks.

Exactly, not all drawbacks. The space is limited. Thus more turns means that the wire must be thinner. This increases the resistance and the resistive losses (=heating).

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Transformers work by transferring energy via magnetic flux from one side to the other.

Both sides are made up by inductors, the primary inductor creates a magnetic field,which is induced into the secondary inductor.

The inductance determines the ability to create magnetic flux (\$ \Phi \$) from a current and is proportional:

$$ L = { d\Phi \over di } \text{ and } d\Phi = L * di $$

An inductor's inductance is determined by the number of turns (beside the area, or size):

$$ N = { µ N² A \over l } \text{ (simplified, reduced winding-area-length relation) } $$

See Wikipedia on Inductance

A small transformer is usually desirable, so more turns is better than bigger size (simply put).

The inductance has to match the mains frequency. Otherwise the primary winding would either now allow enough electrical and thus magnetical current to flow (for higher frequencies) or is more like a short circuit (for lower frequencies). Both is not desirable.

Lower frequencies require higher inductance (=more turns or bigger cores). This is the reason why switching power supplies, utilizing higher frequencies in the hundrets of kHz - MHz range, use so small transformers while being able to transfer much more power compared to conventional transformers.

A quote from the Wikipedia article on transformers:

The EMF of a transformer at a given flux density increases with frequency.[16] By operating at higher frequencies, transformers can be physically more compact because a given core is able to transfer more power without reaching saturation and fewer turns are needed to achieve the same impedance.

(Emphasis mine.)

See Wikipedia on Effect of frequency on transformers

So,

  • the power the transformer needs to transfer is determined by the current flowing through its coils
  • the current the wire has to conduct determines the wire thickness (which plays into the size)
  • the size of the coil and the number of turns determine the inductance
  • the inductance at a certain frequency determines the ability to transfer energy

Conclusion: you'd need to make the transformer physically bigger to reduce the number of windings. When reducing the number of windings you lower the efficiency and increase the losses. And this is usually not desirable.

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The peak magnetic field in the core is related to the peak applied voltage per turn. The larger the area of the core, the more volts per turn can be generated.

The magnetic field in the core can't be allowed to exceed a certain saturation value, if it does then the permeability of the iron drops, and the transformer has to draw orders of magnitude more current to maintain the magnetisation. So this strictly limits how many volts per turn can be supported, and so gives you a minimum number of turns for any winding.

For a typical small (50 VA, ish?) toroidal core I have to hand, the core cross section is 25 mm by 13 mm. If I run the core with the flux peaking at ±1.8 T at 50 Hz, it will generate about 170 mV peak per turn. So a 12 Vrms winding would need 100 turns, the 240 V mains winding would need 2000. I could use more turns than this, but fewer turns would push the core into saturation.

If I used a core with the cross sectional area of a railway sleeper, 130 mm x 250 mm, I could get 12 Vrms in a single turn, but also a rather unwieldy transformer.

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  • \$\begingroup\$ It may be useful to note that a "railway sleeper" what folks in the US would refer to as a "tie"; when I first saw the term, before I read the dimensions, I thought the author was referring to a Pullman-style carriage. \$\endgroup\$ – supercat Dec 22 '17 at 21:13
  • \$\begingroup\$ Could you kindly provide some kind of numerical reference to where you get these figures from? I've looked around on the internet a bit and beyond the basic N1/N2 formula and some "magic number" formulas I'm having trouble finding a coherent answer that both describes why one cares about the number of turns, frequency, and core size of a transformer. I'd appreciate it if you also simply had a reference document with this information - with all this (mis)information spreading around I'm afraid I might just need to read a textbook to design a simple transformer. \$\endgroup\$ – hedgepig Jan 9 '18 at 12:40
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    \$\begingroup\$ @inkyvoyd 25mmx13mm is measured from my core with calipers, 1.8T for peak flux comes from data sheets for transformer iron. The transformation between flux, area, frequency and voltage comes from Faraday's Law. You can see a worked example of this in action in another of my answers here \$\endgroup\$ – Neil_UK Jan 9 '18 at 14:49
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Your basic premise is false, so the question can't really be answered.

Transformers come in many many varieties of voltage and current for their inputs and outputs. Some use many turns of thin wire (high voltage, low current). Some use few turns of thick wire (low voltage, high current).

So the answer to "Why do they not...", is "They do" (when it's appropriate).

To those that don't like this answer

I see this answer has gotten a number of downvotes, and about the same number of upvotes. Obviously it's controversial. Some see it as low quality, especially after others have speculated on the OP's true meaning in comments.

Despite what others think the OP meant, he started out with a blantantly false premise, which is that transformers have 100s of turns on both their primaries and secondaries and that "thin" copper wire is always used. It then sounds like one of those "Why doesn't everyone do it this other obvious way" rhetorical questions.

This is what I answered. It is the correct answer to the question as interpreted above. Perhaps that's not what the OP meant to ask. Perhaps it is. Note that the OP hasn't been back to provide any clarification or edit the question at all.

A much better question would have been about the tradeoffs of fewer turns of thick wire versus more turns of thin wire. That asked respectfully without first passing judgement or pre-supposing false premises would have gotten a very different answer. However, again, that's now what was actually asked, and not even what it seems the OP meant.

Even if the OP does come back and change the question, I will let this answer stand as a reminder to ask questions properly and unambiguously, and to not start out by stating wrong assumptions as facts.

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  • \$\begingroup\$ Do not flag answers as low quality unless they are spam or not an answer. If you don't like it then downvote it. \$\endgroup\$ – Voltage Spike Feb 6 '17 at 22:30
  • \$\begingroup\$ @laptop2d: Who is that directed to? \$\endgroup\$ – Dave Tweed Feb 6 '17 at 23:06
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    \$\begingroup\$ @laptop2d no, thats what the "not an answer" flag is for. this is the epitome of low quality. \$\endgroup\$ – Passerby Feb 7 '17 at 0:10
  • \$\begingroup\$ @laptop2d also, the system automatically places highly down voted answers into that queue. You really need to re-read the guidance on Low Quality before making statements like that. \$\endgroup\$ – Passerby Feb 7 '17 at 4:38
  • \$\begingroup\$ @Passe and others. See addition to answer. This answers what was asked. We might disagree on interpreting this ambiguous question, but this is a valid answer to one interpretation that can't be discounted. \$\endgroup\$ – Olin Lathrop Feb 7 '17 at 11:58

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