# Transformer ratios - 1:2 vs 50:100

I am only the equivalent of a high school student, so please, if possible, don't answer this question with anything too complex or really advanced university level. I am very happy to research new concepts anyone mentions, but can you please keep it reasonably simple.

With transformer windings, primary vs secondary, is there any difference between or advantage to using a 50:100 ratio rather than a 1:2 ratio? With a 1:2 ratio it would seem easier to make the wires thicker and allow for larger currents, whereas with a 50:100 ratio to give the same cross sectional area of wire, the coil would have to be much longer, larger, and involve more metal in production.

So there are obvious disadvantages to using 50:100, but are there any advantages? Thank you very much.

• Consider the inductance as well as the coupling efficiency in each design. Commented Aug 18, 2014 at 11:31
• @igael The current answers look more than satisfactory to me. Is there something specific you feel is still missing? Commented Feb 2, 2016 at 21:29
• @EmilioPisanty is this against some bounty rule ?
– user46925
Commented Feb 2, 2016 at 22:24
• @igael Not at all - it's just that it's going to be very hard to fill in the gaps left by the current (rather comprehensive) answers, as I see it. If you want to reward an existing answer that's perfectly cool, but if you want new content to be posted then I would say that a clearer indication of what you're looking for would make it easier for others to provide that content. Commented Feb 2, 2016 at 23:57
• @EmilioPisanty : I try nothing, I didn't read the question
– user46925
Commented Feb 3, 2016 at 11:42

Transformers have a number of technically relevant properties that make each transformer design fairly unique. Often these parts have to be engineered for a particular application. Let's discuss a few of the most important aspects of transformer design.

From an engineering perspective a transformer has to work at a certain frequency (or a range of frequencies), it has to be able to transfer a certain amount of power and it has to have a desired voltage ratio between any two pairs of its winding.

The amount of power a transformer can transfer depends primarily on the magnetic properties and the volume of its core. This power is frequency dependent and it scales essentially with $P\propto V\times f$. Transformer design therefor has to start with the choice of a core of sufficient size that can transfer the required amount of power at the design frequency.

For a given core size and core material (which also depends on the application) the number of turns then determines the inductance and the capacitance of the primary and secondary coils. This, in turn, determines the useful impedance and frequency range of the transformer.

To make life easy for the designer, the first important figure of merit of a transformer core is called the $A_L$ value, which is usually specified with units of $\mathrm{nH}/n^2$. The inductance of a winding with $n$ turns on a core with $A_L$ is given by the formula $L=n^2A_L$. $A_L$ basically assumes that the magnetization of the core material is linear and it depends on the size and shape of the core, already, so that we don't have to calculate the effective magnetic flux area and effective magnetic path length for a given core size. As a rule of thumb, larger cores (of the same shape) made of the same material have larger $A_L$ values (because the area scales quadratically with size while the magnetic flux length only increases linearly).

Here are three examples of how this works out in practice:

Example 1: RF transformer

A primary winding on a core with $A_L=100\:\mathrm{nH}/n^2$ and $1$ turn will have an inductance of $100\:\mathrm{nH}$. This single turn transformer will only be useful for high frequencies in the $>100\:\mathrm{MHz}$ range. At the lower end of its useful design frequency (for use in typical $50\Omega$ systems) the effective impedance of the primary winding will be $$Z=2\pi\times f\times L = 2\pi \times 100\:\mathrm{MHz}\times 100\:\mathrm{nH}=62.8\:\mathrm \Omega.$$ This is suitable as an RF transformer in systems with $50\:\mathrm\Omega$ impedance but can't be used at much lower frequencies.

Windings have both an inductance and a self-capacitance. In case of this single turn RF transformer, we may have a typical winding capacitance of approx. $1\:\mathrm{pF}$. If we insert this into the formula for an LC resonance circuit, we find a self-resonance frequency of

$f_{Self-Resonance}={1\over{2\pi\sqrt{LC}}} = {1\over{2\pi\sqrt{10^{-12}\mathrm F\times 10^{-7}\mathrm H}}}\approx 500\:\mathrm{MHz}$.

Our RF transformer is therefore limited to a useful operating range of $100\:\mathrm{MHz}$ at the low end because of its inductance and $500\:\mathrm{MHz}$ at the high end because of its self-resonance frequency. With careful design techniques this range can be improved quite a bit, but RF transformers rarely have wider bandwidth ranges than $1:100$ and many work best over no more than a couple of octaves.

Example 2: Switching power supply transformer

The same core wound with $50$ turns will have an inductance of $$L=50^2\times100\:\mathrm{nH}/n^2=2500\times100\:\mathrm{nH}=250000\:\mathrm{nH}=250\mathrm{\mu H}.$$ Because the design with a $50$-turn primary winding has an inductance that is 2500 times higher, it will perform well in applications that are running at frequencies 1000 times lower than our first example and is therefor useful at frequencies of around $100\:\mathrm {kHz}$. Such a transformer will, for instance, be found in switching voltage converters, which are typically operating in the $50kHz-4MHz$ range.

Because a larger number of turns means that we have to use thinner wires with thinner insulation, the medium frequency transformer with its $50$ turns on the primary has a much higher winding capacitance (in the range of $10\:\mathrm{pF}$ to hundreds of $\mathrm{pF}$, depending on how much care is put into the winding scheme), which means it has a much lower self-resonance frequency. Technically useful designs will have self-resonance frequencies around the $10\:\mathrm{MHz}$ range.

Example 3: Audio transformer

If we want to build transformers for much lower frequencies, then we need cores with much higher $A_L$ values (e.g. $5\:\mathrm{\mu H}/n^2$) and we will need to add hundreds or thousands of turns. A 1000-turn transformer on a $A_L=5\:\mathrm{\mu H}/n^2$ core will have an inductance of $L=10^6\times 5\:\mathrm{\mu H}=5\:\mathrm{H}$. This transformer will have an impedance of $Z\approx 600\:\mathrm \Omega$ at $20\:\mathrm{Hz}$ and will typically be used in audio amplifiers.

The $1000$ turn audio transformer, on the other hand, will perform over a range of approx. $15\:\mathrm{Hz}$-$25\:\mathrm{kHz}$ if wound really well, but there is some art and manufacturing know-how to making these wide-band transformers. A poorly calculated, poorly wound device will not work well at all in audio applications.

There are additional considerations that limit the performance of a transformer. For applications which transfer significant amounts of power one also has to take the resistance of the winding and the skin effect into account, both of which lead to $I^2R$ losses and heating. Additional losses are caused by the hysteresis of the core material's magnetization curve and the eddy currents that can be induced in electrically conductive core materials like transformer steel and some ferrites. These core losses have to be carefully considered during material selection and they are also core shape dependent, which leads to a great number of core geometries for different applications.

Designing a high quality transformer requires that the design engineer picks the right core (shape, material and size) and that the correct wire diameter and number of turns for each winding are used. These devices also have to be tested for their performance before being used in a circuit. For some transformers (like $50/60\:\mathrm{Hz}$ power transformers) this is comparatively easy, but for signal and especially wide-band RF transformers often a significant amount of iterative optimization is needed.

• Great (+1). This answer really adds something! Commented Jan 24, 2016 at 19:35
• You make excellent points. I think the answer would be stronger if it was divided into a section that just described the principles, and another that did some calculations. Mixing them, in my mind, makes it less powerful than it would otherwise be. Commented Feb 2, 2016 at 15:36
• @Floris: You are right... it's just a heap of examples. It started out as a comment and then just began mushrooming, I am afraid. I'll edit it some later. Commented Feb 2, 2016 at 20:05
• @CuriousOne - there is a lot of good material there, but I know how that goes. You look well on your way to get a 500 point bounty for this. Make it count! Commented Feb 2, 2016 at 20:07

There appears to be some confusion. The ratio is the number of windings of the primary coil divided by that of the secondary coil. Thus 1:2 and 50:100 is the same thing. However, the number of windings itself is far from a trivial matter. To learn more have a look at: http://en.wikipedia.org/wiki/Transformer

Edit: In view of Rennie's remark the following. If a lower number of thicker windings is used more current runs through the coil. This can cause saturation of the iron core resulting in more losses and (inacceptable) heating (actually this phenomenon was/is used in transformers for battery chargers for cars where it is used for limiting the maximal current).

• I don't think this answers the question. My interpretation is that the OP is well aware that the ratio stays fixed at 1:2. The question is whether achieving this ratio by a few turns of thick wire is different from using many turns of thin wire. Just linking to Wikipedia does not constitute an answer to this question. Commented Jan 22, 2016 at 8:12
• @JohnRennie I added a few details. The OP should let me know if he/she has further questions. Commented Jan 22, 2016 at 10:07

I cannot possible improve on CuriousOne's excellent answer but I can try and respond to Carl Witthoft,s comment: "Consider the inductance as well as the coupling efficiency in each design."

When not loaded the primary coil of a transformer acts as a self inductance. The self inductance is proportional to the number of turns squared whereas the resistance of the wires will depend on the number of turns $N$. This is an approximating because the coils do not all have the same circumference.

For a given supply voltage the current depends on something like $1/N^2$ and if the coils have resistance the ohmic heating ($I^2R$) depends on something like $1/N^3$. Furthermore the B-field depends on NI and so something like $1/N$. So the chances of saturation are reduced.

Finally if the transformer with no load is connected to the mains supply then on average over a cycle there is no power dissipated but current still have to flow through the supply cables which have resistance. With a larger self inductance the current is smaller and so is the ohmic heating in the supply lines.

• Current flows in the supply lines = current flows in the transformer. This is because of losses in the coil / core: a real transformer gets warm even if nothing is connected to the secondary. Commented Feb 2, 2016 at 15:34
• @Floris I was thinking more of the utility supplier having to pay for the heat generated and not the customer. I did mention copper losses. Commented Feb 2, 2016 at 15:58
• I think too that it is ;) Thank you for your answer ...
– user46925
Commented Feb 2, 2016 at 16:11

There is actually a very simple answer to this question:

The absolute number of turns which each of the transformer winding should have (with fixed turns ratio) depends entirely upon which voltage and current ratings the transformer is intended to be used for.

Let me explain. For instance, measurement transformers for high voltages typically have a huge number of turns (of very thin wire) whereas the low voltage side of arc furnace transformers usually consists of only a few turns, but made of really massive copper or aluminum bars.
The voltage rating together with the number of turns determines (at a given operating frequency) the total magnetic flux through the transformer core, whereas the current rating determines the necessary cross section area (i.e., the thickness) of the wire a winding is made of.

In power applications, the number of turns in a winding is usually chosen in such a way that the magnetic flux density in the core at the rated voltage is of the same order of magnitude as the saturation flux density of the core material (about 2 Tesla for modern transformer steel). Further decreasing the number of turns would lead to saturation of the transformer core, losses would rapidly increase, and finally the transformer would stop to operate as a transformer altogether.
On the other hand, when increasing the number of turns you will not fully utilize the core anymore (or, alternatively, you can make the core slimmer), but at the same time you need more conductor material for the winding (remember that the thickness of the wire was determined by the current rating!). So in effect the transformer will become more bulky and expensive.

In other words, within a certain region of turn numbers there is a trade off between winding and core sizes. Smaller turn number means larger core but smaller winding volume, larger turn number means smaller core but larger winding volume. It's an optimization problem for the specific application you have in mind. In power applications, a close-to-optimal solution often has the propery that conductor losses in the windings and core losses are balanced (i.e., are roughly equal to each other).

A point which I didn't mention above is that in many applications one also has requirements on the so-called short circuit reactance of the transformer, which roughly varies as the square of the number of turns and therefore can influence the choice of that number.
Another point may be requirements on the self resonance frequencies of a winding (already mentioned by CuriousOne above), which are roughly inverse proportional to the the turn number.

Note that in the spirit of the original question I avoided the use of formulas here, but if desired everythig can be backed up with quantitative calculations.