I am trying to wind my own transformer for a power supply. How many turns do I need on my primary winding?

I know that the turn ratio determines the voltage ratio, but how do I determine the actual number of turns? I can imagine that for an air-core transformer, having many windings can help keep the flux from leaking. However, the more wire I use the more materially expensive and electrically inefficient it becomes.

If I'm using a toroidal core, can I get away with just one loop for the primary?


  • 1
    \$\begingroup\$ Why do you prefer to wind your own transformer rather than buy a stock transformer? What power rating are you typing about anyway, what frequency? A one turn secondary works just fine ... but unless you are trying to melt aluminium probably not too practical. \$\endgroup\$
    – jippie
    Commented Mar 2, 2015 at 19:01
  • \$\begingroup\$ I have seen lots of programs and javascript tools etc. out there that allow you to input a few parameter and then get the turns \$\endgroup\$
    – PlasmaHH
    Commented Mar 2, 2015 at 19:54
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    \$\begingroup\$ I want to wind my own coil because I want to learn and I learn best by doing. \$\endgroup\$
    – freefood89
    Commented Mar 2, 2015 at 20:20
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    \$\begingroup\$ Thanks for the comments, but I'd like to clarify that I'm not a random kid plugging things into the wall. I studied analog integrated circuit design at a respected university and have nearly a decade of experience tinkering. I will take the necessary precautions to be safe. I just want to be more comfortable with magnetic devices because in IC design (especially in college) one does not use many inductive devices and as I've mentioned earlier I learn best by tinkering. \$\endgroup\$
    – freefood89
    Commented Mar 2, 2015 at 22:30
  • 2
    \$\begingroup\$ This URL give the detail to construct your own Transformer youtube.com/watch?v=lyZ7nM6Fo94 \$\endgroup\$
    – khaled
    Commented Dec 20, 2016 at 2:50

5 Answers 5


Here (from Wikipedia) is a fairly complete linear model of a transformer:

enter image description here

Ideally, it says: "An ideal transformer is a theoretical linear transformer that is lossless and perfectly coupled. Perfect coupling implies infinitely high core magnetic permeability and winding inductance and zero net magnetomotive force (i.e. ipnp - isns = 0)."

Thus, I_0 current is zero in the ideal transformer and you can have only one turn in the primary. However, due to leakage in the magnetic flux, you will have I_0 current. If the primary winding is too low then your leakage magnetic flux will be too high resulting in the high I_0 current, which will burn your transformer.

For the real case scenario, the following is what you get:

Note the magnetizing inductance Xm across the primary. If that inductance is too low, you'll get excessive current flowing even with no load on the secondary.

While a single-turn primary is certainly possible, with sensibly-sized cores it implies either a very low voltage (for example, a current transformer, which is typically toroidal) or a very high frequency (or both).

The inductance is proportional to the number of turns squared, and a small 120/240V 50/60Hz mains transformer primary might be some hundreds of turns, so you can see how far off a single turn is. At a fraction of a volt, or higher frequencies at relatively low voltage, a single-turn primary might make some sense.

  • \$\begingroup\$ Thanks for giving me a starting point. I will go back to my college books to figure out how to actually determine X_m. I'll be back if I don't get hundreds of turns as I expect. \$\endgroup\$
    – freefood89
    Commented Mar 2, 2015 at 20:44
  • \$\begingroup\$ it might be interesting to note that in this context, a "single turn" is actually just a wire that goes through the opening of the transformer, and the wire alone doesn't actually form a closed ring around the transformer - rather, the "single turn" is the entire rest of the circuit including the wire. If you took the wire and wrapped it once around the core, that would technically be two turns \$\endgroup\$
    – BeB00
    Commented Jun 22, 2022 at 0:07

The minimum number of turns required for a 120 V, 60 Hz primary is a closely guarded secret. :-) Not really, but everyone always talks about turns ratios, but forgets to mention the minimum number of turns for the primary. Well, the iron/silicon/metal core of the transformer can accept only so much magnetic flux before it saturates and can't take more. If you go beyond this, the inductance drops a lot and you end up drawing a lot more current off the powerline and it will get really hot - not good.

A rule of thumb, for transformer laminations you may salvage from a junked 60 Hz transformer: Number of turns needed for the 120 V, 60 Hz primary = 800/(area of the core in square inches). You measure the height of the pile of laminations, and the width of the center leg of the E lamination. This width is measured along a line that would go vertical when looking at the E as the letter E appears here. In other words, imagine a single turn of wire tight on the center core, the area of the loop this single turn forms is the area. Do not include the outer legs of the E, or the I. A bigger area will make for a lower number of turns.

Once you have the number of primary turns, then you can do the turns ratio to get the number of turns for the secondary. Add a few more turns to make up for resistance of the wire voltage losses. If your powerline frequency is 50 Hz, you need 60/50 times the above result for your primary for 120 V, and twice that for 240 V.


A transformer has a maximum volts per turn. You need to have sufficient number of turns on the primary so that the primary voltage you apply, divided by the number of turns, does not exceed this volts per turn.

The maximum voltage per turn for a transformer is controlled by three things.

  • the operating frequency
  • the core cross section
  • the maximum core flux you can use

Taking the last item first, the limitation on core flux depends on the core material. If it's iron for a mains transformer, then the maximum is given by saturation, 1.5 T to approaching 2 T, depending on cost and quality of the iron. If it's a high frequency ferrite transformer, then core heating will probably limit you to less than the DC saturation level of typically 400 mT, operating figures of 50 mT to 200 mT are common. If it's air core, then obviously neither saturation nor core heating play any role, and the flux will be limited by primary inductance, primary I2R heating, or the primary current you are able to deliver.

Assuming a sinusoidal waveform for both voltage and flux (other waveforms can be used with a change in details, square voltage and triangular flux is the other common one), then as the flux swings through zero, its rate of change is maximum, and it generates the maximum voltage in the primary (and any other winding sharing the core).

The rate of change will be affected by the frequency. You can draw a single cycle and put slopes on it, or just accept the formula that swinging between fields of B and -B at a frequency f,

max rate of change = 2πfB

The magnitude of the voltage per turn is just the rate of change of flux in Webers/s. One Weber is one Tesla operating over an area of 1 m2. Now we know the maximum voltage per turn when we have a core area of A

max volts per turn = 2πfBA

To give a concrete example, let's say we have an iron core, 12 mm by 25 mm cross section, operating at peak 1.5 T, at 50 Hz.

The maximum volts per turn = 2π x 50 x 1.5 x 0.012 x 0.025 = 0.141 V

If we are going to connect this to standard 240 V mains, then we first need to calculate the maximum voltage that will produce. 'Standard' 240 V mains can exceptionally go up to 265 V. It's also an RMS figure, so we need to multiply by 1.414 to get to peak voltage, which is 375 V. Finally we can compute the primary turns as 375/0.141 = 2657 turns.


The calculations for the transformer are complex; however, since you want a toroid where windings are always on top of each other you can't just make something, measure and adjust - you want to know before laying out the coils so I suggest to just bite the bullet and start understanding the formulae.

If your power source is 120V and you want to get 12V then the smallest secondary is one turn and your primary can't have less than an integer multiple of 10 turns. This is only close to real life for high frequencies, for 50/60 Hz frequencies of typical household mains the number of turns in the primary will be in the thousands and the number of turns in the secondary must reflect that.

A workable shortcut will be to grab a ready-made toroid transformer that has its secondary on top, remove it, figure out the turns ratio by winding and measuring test coils, then wind the desired secondary - this can be done without calculations.

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    \$\begingroup\$ Thanks. I am actually not afraid of math. I'm trying to get over my uneasiness with magnetic devices. \$\endgroup\$
    – freefood89
    Commented Mar 2, 2015 at 20:21
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    \$\begingroup\$ the issue with magnetics is that fundamentals are quite trivial but you're restricted by tons of secondary factors. For example, when winding a transformer you will be concerned with how much wire you can actually pack in the gap of your core of choice. Smaller cores require more turns for given power and frequency, to keep the core from saturating (saturated core will "disappear" and your coils will act like they are in free air). Again, if you want to experiment, find a transformer in which the secondary can be easily removed, then remove it and wind your own. You'll learn much by doing it. \$\endgroup\$ Commented Mar 2, 2015 at 20:32
  • \$\begingroup\$ hi, thanks again for the additional input. I went through my textbook again and realized that I totally forgot about phenomena like the skin effect and the proximity effect. I may need to follow advice from you and others and start from modifying a commercial transformer. \$\endgroup\$
    – freefood89
    Commented Apr 19, 2015 at 1:54

https://en.wikipedia.org/wiki/Transformer gives the universal EMF equation as: E(rms) = 4.44fNaB(peak), where: E(rms) is the rms voltage that you are applying to the primary f = the operating frequency N = the number of turns a = the core cross sectional area in square meters B(peak) = the peak magnetic flux. This is a property of the core material, typically 1.2 - 2 for silicon steel and 0.7 for soft steel.


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