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I am interested in understanding what factors govern the lower limit on the number of turns in the Secondary coil of a step down transformer. I have read that a Primary coil with too few turns will induce a very weak magnetic field in the core, but what happens if the secondary coil has a very low number of turns?

If the secondary coil uses a shorter wire, the resistance is lower and so voltage induced in the secondary coil produces a greater current, as per Ohm’s law. This greater current may be higher than the wire's maximum current capacity and cause dangerous or undesirable levels of heat in the wire, but is this the only factor governing the minimum number of turns in the secondary coil? If the secondary coil has too few turns, can the efficiency of the transformer be reduced in some other way?

If I want to have a step down transformer with as few turns as possible in the secondary coil, what problems do I need to be aware of, and how can I calculate what the minimum number of turns in the secondary coil could be?

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  • \$\begingroup\$ You set the number of turns based on your desired transformation ratio. You set the wire size based on current. \$\endgroup\$ – R Drast Aug 31 '17 at 15:16
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    \$\begingroup\$ The need to establish a complete circuit prevents the number being lower than 1. \$\endgroup\$ – Brian Drummond Aug 31 '17 at 15:23
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I am interested in understanding what factors govern the lower limit on the number of turns in the Secondary coil of a step down transformer.

Good.

I have read that a Primary coil with too few turns will induce a very weak magnetic field in the core,

You have remembered wrongly, or misread.

You need at least a minimum number of turns on the primary in order to keep the core field below its maximum.

Core materials, iron or ferrite, have a maximum magnetic field. With iron, the limit tends to be saturation, with ferrite, heating from losses tends to bite before saturation. The maximum fields are typically in the order of 1.5T for mains iron transformers, and in the range 0.1T to 0.3T for ferrite depending on frequency and grade.

The core has an area \$A\$, and the core material has a maximum field \$B_{max}\$. The transformer has an operating frequency \$f\$. We can calculate the maximum voltage induced in one turn via the maximum rate of change of flux, which is \$V_{max} = 2\pi f B_{max} A\$ for a sinusoidal input, with all terms in SI units. This gives us the peak voltage. Divide by sqrt(2) to give the RMS voltage. We need enough turns to equal our input voltage.

To give an example, I have an iron toroid with cross section 13x25mm, operating at 50Hz, and I am going to assume a \$B_{max}\$ of 1.5T.

\$V_{max} = 2 \pi \times 50 \times 0.013 \times 0.025 \times1.5 = 0.153V \$

0.153V peak is about 0.108V rms. At 240v, we would need 2222 turns on the primary, minimum. A few more wouldn't hurt.

but what happens if the secondary coil has a very low number of turns?

then you get a very low secondary voltage produced.

If the secondary coil uses a shorter wire, the resistance is lower and so voltage induced in the secondary coil produces a greater current, as per Ohm’s law.

well, yes, if you shorted the secondary. But we don't do that. The resistance of the load tends to exceed that of the winding, and so it's the load that determines the current.

You do several things to design a transformer. Note that a full design could get much more detailed, but this gives you a good start.

1) Start with a big enough core for your power. Go to a transformer catalogue, and find a transformer with a similar power to what you want to use. Halve its weight, the copper is about the same weight as the iron. Start with a core this weight.

2) Calculate the minimum number of turns needed on the primary, as above

3) Caluclate the number of turns needed on the secondary, to give your design voltage, using the same volts per turn figure.

4) Choose a the maximum thickness of wire that will half-fill the winding window. 50% is a good fill factor, you will be doing well to get a better factor.

5) Check that the current density doesn't exceed about 3A/mm2 (a rule of thumb valid for small to medium transformers) at your load power.

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  • \$\begingroup\$ See here is where the dark art comes in. The equation is a simplification and relies on the turn fully enclosing the core area. Any gap, and subsequent layers of turns have a different equation. Also shape of core, round rectangular, laminated, etc all make a difference. Also material quality makes a difference, as does wire size and winding density. It's a starting point, and if you de-rate sufficiently, probably good enough. But these are the wishy washy bits I mentioned. But +1 from me anyway. \$\endgroup\$ – Trevor_G Aug 31 '17 at 17:37
  • \$\begingroup\$ @Trevor With iron and most ferrites having ur > 1000, the difference between different layers is negligible. Shape of core has a similar relevance. Wire size needs to be enough for current, and winding density is irrelevant. Given that the resistance of Cu changes by 20% with a 50C temperature change, and Bmax can be from 1.5T to 1.9T depending on the iron, it's a good start. I've added a caution to the design process. \$\endgroup\$ – Neil_UK Aug 31 '17 at 17:58
  • \$\begingroup\$ Somehow I can't see a winding on an outer layer enclosing a circular diameter twice that of the core diameter (aka 4 times the area) having the same V per turn as the innermost turn. Granted that would be a horrible coil setup.. but you get my point. \$\endgroup\$ – Trevor_G Aug 31 '17 at 18:04
  • \$\begingroup\$ @Trevor it's not the same V/turn, but it's very, very close. To put it in perspective, it would be if the core \$\mu_r\$ was infinite. The error is an order of magnitude less than assuming infinite beta when we bias up a transistor. If we need to (rarely) the correction for finite beta is small. Compared to not being able to even estimate whether you need more or fewer primary turns as the OP couldn't, my route will get you within a few percent of where you need to be. Now if you want a dark art, it's explaining how a transistor works, Early voltage, non-constant beta ... \$\endgroup\$ – Neil_UK Aug 31 '17 at 18:12
  • \$\begingroup\$ LOL.. yes you are right about that :) and a lot of other stuff we suspend belief on.... Anyhow, I was just pointing out there are a few "don't look too close" issues with the basic formulas. None the less, they are good enough most of the time within limits. The "Rules of thumb" you mentioned is the art of it. \$\endgroup\$ – Trevor_G Aug 31 '17 at 18:18
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For an Inductor we have the following equation and we can think of transformers as coupled inductors.

$$V = L \cdot \dfrac{\text{d}I}{\text{d}t} = N \cdot A_e \cdot \dfrac{\text{d}B}{\text{d}t}$$

Where \$V\$ is the voltage across the winding, \$L\$ is the inductance of the winding, \$I\$ is winding current, \$N\$ is winding turns, \$A_e\$ is the effective area of the core and \$t\$ is time.

Is the basic answer to your question. All magnetic cores will saturate it you allow \$B\$, the flux density to become too large. Knowing what the voltage across a winding should be and how long it's there we can come up with this related equation.

$$N = \dfrac{1}{B_{max} \cdot A_e}\int V \text{d}t$$

\$ B_{max} \$ is the maximum allowable flux density and this will depend on the frequency of operation and the material properties of the core.

I usually start by designing the lowest voltage winding first and then increase turns if required to get the turns ratio I want.

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  • \$\begingroup\$ :) as I mentioned.. a dark art... \$\endgroup\$ – Trevor_G Aug 31 '17 at 16:00
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If I want to have a step down transformer with as few turns as possible in the secondary coil, what problems do I need to be aware of, and how can I calculate what the minimum number of turns in the secondary coil could be?

It can be as low as one turn on the secondary (but read further down because if you want to step down 240 V AC to 80 VAC, having 3 turns on the primary will be a disaster).

If the secondary coil uses a shorter wire, the resistance is lower and so voltage induced in the secondary coil produces a greater current, as per Ohm’s law.

Secondary current is only due to the secondary voltage being applied to a load - the secondary doesn't inherently circulate current without a load so it's ohms law, the output voltage and the load that governs secondary current.

This greater current may be higher than the wire's maximum current capacity and cause dangerous or undesirable levels of heat in the wire

You can overload a transformer by trying to take too much current from it and it will overheat for sure.

is this the only factor governing the minimum number of turns in the secondary coil?

Your secondary coil has to be designed from thick enough wire to sufficiently power the anticipated load at full current without over-heating.

If the secondary coil has too few turns, can the efficiency of the transformer be reduced in some other way?

Now we're getting to the important stuff. If you design your secondary with too few turns then, for the required step down ratio from the primary voltage, there will be too few turns on the primary AND too few turns on the primary usually means tears and disappointment.

You have to forget the secondary when designing the primary (despite other claims). You have to consider how much flux is produced in the core when the primary connects to the AC power line. Too few a number of turns will mean too much magnetization current in the primary and the core will saturate and you may, in worst case scenarios, get melt down or a fire.

The primary is an inductor after all and if the inductance is too low it draws too much current to magnetize the core irrespective of what the secondary is doing so, you ignore the secondary when designing the primary.

Once you have calculated the primary turns (in order to avoid excessive core saturation) you simply wind the secondary turns based on the voltage step down ratio required.

There are of course a few more subtleties involved but the above is the main issue.

I have read that a Primary coil with too few turns will induce a very weak magnetic field in the core

You may have read it but at best it's misleading.

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The number of turns in the secondary is governed by whatever step-down ratio you require.

That is: Secondary_Turns = Primary_Turns x Vin / Vout.

But I think you know that...

Your real question ought to be, how do I calculate the number of turns required on the primary. That's defined by a much more complex equation and is dictated by the geometry, materials, and design of the transformer core along with the applied primary voltage and frequency and power levels you need to transfer.

The latter is all a bit of a dark art.. and way beyond my experience level to quote for you.

As far as currents are concerned, that is a matter of using the appropriate gauge (diameter) of wires on both sides rather than the actual length of wire. That and having enough magnetics to support those currents.

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  • \$\begingroup\$ it's not that dark, Trevor, you just have to be prepared to keep your eyes open long enough \$\endgroup\$ – Neil_UK Aug 31 '17 at 16:51
  • \$\begingroup\$ :) lol @Neil_UK . Ya I mean but there are so many wishy-washy magnetic factors involved you really need quite a bit of experience to do transformer design well. Even then, I'm pretty sure most off the shelf transformers are "lets build it first" then write the spec sheet after we test it and try to sell that. \$\endgroup\$ – Trevor_G Aug 31 '17 at 16:54
  • \$\begingroup\$ read my answers on calculating transformer turns, I'm not sure I'll go into full detail on this one, but there are another two with good details. Very little is wishy washy, it just seems that way if you're not familiar with it. \$\endgroup\$ – Neil_UK Aug 31 '17 at 16:58
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    \$\begingroup\$ old answers are not easy to find. After I've posted this answer, I may have a look. Remebering a key word used in the text is about the best way to do it, which only the author is likely to have the luxury of. \$\endgroup\$ – Neil_UK Aug 31 '17 at 17:07
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    \$\begingroup\$ If the core goes through and around the outside of that mile wide turn of wire then back to the beginning then yes, it will induce the same voltage. However, leakage inductance will be massive (henries no doubt) and so as soon as you drew some current from it you'd be scuppered. Think about a half turn winding where the remainder of the turn is the circuit it connects to if I were to be more practical. \$\endgroup\$ – Andy aka Aug 31 '17 at 19:12

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