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Transformer question: Explanation for number of turns

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localbroadcast

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Hello! New here but not to electronics!

This may be a very basic concept that I should know, but for some reason I can't figure out the logic behind it!

I want to know what the reason is for choosing the number of turns to use for the primary winding of a transformer.

I understand that if I have 10 turns on the primary and 1 turn on the secondary, and apply 120v to the primary, then I will get 12v on the secondary.

The question is... If I was designing a transformer, why would I choose to use 10 turns on the primary and 1 turn on the secondary? Why wouldn't I choose 20 turns on primary and 2 on secondary? Or even 1000 turns on the primary and 100 turns on secondary?

I know there has to be more to choosing the number of turns than just the ratio! Anyone want to shed some light on this topic? Thanks!
 

SunnySkyguy

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No load excitation current will be governed by the input inductive impedance, X(f)= 2pi*f*L where Ipri=Vin (max)/X(f).

This is typically <= 10% of the rated current.
 

Orson Cart

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You need enough turns on the driving side (primary) so that with the applied volt-seconds on one half cycle you do not exceed the peak flux the core can handle, actually you need to keep it lower than this due to core losses due to the flux swing at applied frequency, Erms = Bpk /( 4.44 F N Ae) where Erms = applied volts as a sine wave, Bpk = peak tesla in the core, = 1.2 for normal transformer steel (to keep losses down to manage-able levels), F = frequency 50 or 60 Hz, N = turns, Ae = area of core in m^2, so 50 x 50mm = 0.0025 m^2. So if you know the others, and you do now, you can calc the min N for your application. The wire size is a whole other passel of working out...
 

localbroadcast

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You need enough turns on the driving side (primary) so that with the applied volt-seconds on one half cycle you do not exceed the peak flux the core can handle, actually you need to keep it lower than this due to core losses due to the flux swing at applied frequency, Erms = Bpk /( 4.44 F N Ae) where Erms = applied volts as a sine wave, Bpk = peak tesla in the core, = 1.2 for normal transformer steel (to keep losses down to manage-able levels), F = frequency 50 or 60 Hz, N = turns, Ae = area of core in m^2, so 50 x 50mm = 0.0025 m^2. So if you know the others, and you do now, you can calc the min N for your application. The wire size is a whole other passel of working out...


Ok Ok.. So by using this formula, and selecting the range you would like to operate in on the BH curve, you can select your number of turns.. If you have too many turns or too few turns, then you will be operating in an undesirable section of the BH curve.. Too far along the curve and the core will saturate, so no matter how much you increase current you won't have much increase in flux, basically, right? And not far enough along the BH curve and you are basically using a core area that is larger than needed, and you are wasting money on core that is not being utilized.

Do I have the correct idea here?
 

smxx

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Hello
There is a coupling between two Magnetic fields.for low coupling ,we have much loss of energy passed between them.
And rule of core is important too.1sf magnetic field store energy to the core (like storage) and other take energy from core.
It like a lever. For 1/2 change ratio we have many choice for diameter (hardness) of lever.
But for much force we can't use narrow lever.
 

Orson Cart

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The formulae give you the minimum turns you can use, you can use more turns to reduce core losses further...

- - - Updated - - -

The load current does not affect the core flux except for very separated winding configurations...
 
T

treez

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you always use number of turns to completely fill the bobbin.

The best way is to just do it anyhow and see what you get in terms of saturation current level, magnetising current, etc etc..then tweak it till its right.

mind your hysteresis losses...so calculate your delta B, because the flux density swinging about widly as in a smps core will give hyteresis loss of heating.

what is the application, is it smps?

Mind also your conduction losses, and your skin effect losses.......skin effect.......you should calculate your skin depth, then generally make up your turns out of wire that is around 1 skin depth deep or less.....after three skin deoths in radius the skin effect gets much worse.

there is also the dreaded proximity effect, where windings near each other strangle off the current in each other, increasing conductor impednace
 

localbroadcast

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you always use number of turns to completely fill the bobbin.

The best way is to just do it anyhow and see what you get in terms of saturation current level, magnetising current, etc etc..then tweak it till its right.

umm.. This sounds like terrible advice!!


In my experience, back when I was just out of school and I remembered all the formulas much better, the basic formulas give you theoretical values that are close enough to the actual values that trial and error just seems like a ridiculous suggestion.

Now that my memory has been jogged a bit and I remember the BH curve again, it is absolutely clear to me now why the number of turns in the coils is not always the lowest common denominator for the needed turns ratio!! the knee area of the BH curve is the most efficient operating area!!
 

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