Transformer winding

[anishkgt]

as per the link in this post

This site is handy:
http://ludens.cl/Electron/trafos/trafos.html

its 697 turns. So would that have any affect on the secondary current when i use a 2 AWG flexible copper wire ? On a normal MOT i'd get 1100amps at the secondary. would i get comparatively more or less with the difference in the Gauge at the primary.

[MrAl]

Hi,

Probably dont want to waste your time with aluminum.

The other thing to consider next is the window area.  The idea is to fill up the window as much as possible so that you get the full area used up.  The reason is that the power rating will be based on the window being full with copper which is 1 turn with copper that fills up the whole window.  This is because of the resistance of copper vs the cross section of the copper.  That's of course if you want to make full use of the core.  IF not, then you can go with almost anything.  Note that 22 gauge wire will only result in maybe three layers, which is barely using any of the core, but if the secondary takes up the rest of the room then your optimized.  If not, then you can increase the wire size of the primary and secondary.

[anishkgt]

Thanks AI.


Will not be using Aluminium for sure. Now about the size of the wire for the primary. What is the difference with the gauge of 22 and 16. isn't 16 more space taking as that is more thicker ? Moreover i will be filling the remaining window with the secondary with 2 or 3 Gauge flexible copper wire whichever is available here. probably 3 or 4 turns if possible to get the max current out of it.

Question is if am drawing that much current at the secondary would that heat up the primary with a thin wire like 22 AWG wire ? or would the sweet spot be 16AWG ?

[james_s]

#16 is thicker and will take up more of the window, it will also result in lower resistive losses and less sag under load. How much is something you will have to calculate, nobody can give you a definitive answer without measuring the core and making all these calculations themselves.

[anishkgt]

Thanks James Will stick to the 16AWG then. the calculations is something i can't figure out like the flux and stuffs. Well i will just go with the excel sheet. So would the flux density be 1 ? how is that calculated ?

[MagicSmoker]

Question is if am drawing that much current at the secondary would that heat up the primary with a thin wire like 22 AWG wire ? or would the sweet spot be 16AWG ?

In a typical transformer the usable window area is split evenly between the primary and secondary windings (note that only 40-50% of the actual window area can be used).

The design process to follow:

1. Calculate the number of primary turns required given the applied voltage, effective core area (A_e) and peak flux density swing. For typical silicon steel mains frequency transformers a flux density swing of 1-1.2 Tesla (or 10000-12000 Gauss) is commonly used. The formula for calculating turns, assuming sine wave excitation at 50Hz, cm² for area and a flux density swing of 10000G is:

N = V_RMS / 0.0222 * A_e

Earlier on you measured the center area and it came out to around 25.4cm², so that means a 230VAC/50Hz primary on this transformer will need 408 turns.

This seems reasonable, as a typical 50/60Hz transformer of this size will need around 2 turns per volt.

[anishkgt]

Thanks magicSmoker. Appreciate it.


So that would be 480 turns of 16AWG and the transformer is switched on for about 500ms max and mostly 200 - 300ms. So would 25 AWG be ok or i stick with 16AWG like James had suggested.

What does Ae stand for ? how is the peak density flux calculated ?

[MagicSmoker]

So that would be 480 turns of 16AWG and the transformer is switched on for about 500ms max and mostly 200 - 300ms. So would 25 AWG be ok or i stick with 16AWG like James had suggested.

You keep asking this question but we can't answer it because we don't know the window area of the transformer! A good rule of thumb is that 40-60% of the total window area can be occupied by copper and this area is split evenly between the primary and secondary windings (with perhaps a bit more allotted to the primary, since it also has to make up for any losses). So to determine what gauge wire to use you need to first measure the window area, then assign 20-25% of that area to the primary and another 20-25% to the secondary. Here is a handy wire chart I found with my google machine: http://diyaudioprojects.com/Technical/American-Wire-Gauge/

#16 AWG has an area of 1.31mm² so 408 turns would require 535mm² (or 5.35cm²). You need a total window area of at least 21.4cm² to use #16 for the primary, then.

[MrAl]

Thanks AI.


Will not be using Aluminium for sure. Now about the size of the wire for the primary. What is the difference with the gauge of 22 and 16. isn't 16 more space taking as that is more thicker ? Moreover i will be filling the remaining window with the secondary with 2 or 3 Gauge flexible copper wire whichever is available here. probably 3 or 4 turns if possible to get the max current out of it.

Question is if am drawing that much current at the secondary would that heat up the primary with a thin wire like 22 AWG wire ? or would the sweet spot be 16AWG ?

Hi,

Yes #16 will take up more room than #22, but if that was the only consideration then why not use #34 wire as that takes up less room than #22.

The whole thing is based on how much room the primary takes plus the secondary, and because there is a relationship between the primary turns and secondary turns  for both voltage and current, this creates a situation for optimization of primary and secondary wire sizes.  Once you pick the primary wire size you know the secondary wire size because you know the primary voltage, secondary voltage, turns ratio, and thus you know the secondary current.
For example, if you use a wire size for the primary that has cross sectional area 1 unit and the current is 1 amp and voltage 10 volts, then if the turns ratio is such that the secondary is 1 volt then the second current is 10 amps.  This means you know the wire size for the secondary as it must have a cross sectional area of 10 times that of the primary, which would then be 10 units.
What this means is that we know the primary size and secondary wire size, and we then proceed to see how it fits on the core in the window area.  If it fits perfectly then we've optimized the design for the given input voltage.  That gives us the maximum current output on the secondary.  If it is too large  we have no choice but to lower the primary wire size which then makes the secondary wire size smaller too.  If it is too small then we know we might be able to increase the primary wire size (and thus the secondary wire size also) but there is a catch and that is if we get too close to the max going up even one wire size may go over the limit of the window area so we have to back up and call the first try the optimum.

So to recap, once you choose the primary wire you know the secondary wire size, and then you can try to figure out if it all fits in the window, and this will result in three possible cases:
1. It fits perfectly.
2.  It goes over the window area limit.
3.  It does not use up the whole window area.

Depending on the case you may or may not try to increase or decrease the primary (and thus secondary) wire size.

Because of skin depth you also dont want to use too big of a secondary wire size.  1/4 inch diameter is probably a good limit.  If you need heavier wire, use two turns wound bifilar.

Quick example:
primary 100v, secondary 10v, which means the turns ratio is 1/10.
If we pick the area of the wire size for the primary of 1 unit the secondary wire size has area 10 units.
If the required number of turns on the primary is 200, then the required secondary turns is 20.  We see if that fits in the window area and if not, increase both wire sizes or decrease both wire sizes and see if it is a good fit after that.
This can all be put into a formula but i am not sure if you would be comfortable with that.

[james_s]

You don't always have to increase both the primary and secondary wire size. If you have space to increase the size of either one, you will reduce the resistive losses in that winding which will improve the overall efficiency of the transformer and reduce the temperature rise.

[MrAl]

You don't always have to increase both the primary and secondary wire size. If you have space to increase the size of either one, you will reduce the resistive losses in that winding which will improve the overall efficiency of the transformer and reduce the temperature rise.

Hi,

In the raw theory where we are working with a continuously variable wire size the optimum is to change both wire sizes.  This is of course before the actual construction starts.  Under this kind of theory increasing one wire size reduces resistance but that's all.  Increasing both reduces the resistance more and also increases the current.
Of course because in practice we have to work with integer wire sizes there may be times when if we did increase one the other would not increase that much and so we would not have to change that wire size, but in general they both change to get to the optimum, and why not get to the optimum.  If we dont want the optimum then anything is possible.  The optimum wire sizes leads to the best possible design for a given transformer window area.

Interesting is if we have wire size with area A on the primary and wire size with area B on the secondary, since the optimum wire size is based on the area A it will be N*A where N is the turns ratio.  Since the number of secondary turns is p/N where p is the number of primary turns and N*A the wire area, the total secondary area is p/N*N*A=p*A, which is the same as the primary total area.  This of course means that the secondary winding should take up the same area in the window as the primary winding, in a perfect world.  Since the winding goes on layer by layer, the optimum size wire for the primary winding is the size that leads to a winding build height of 1/2 of the window width.  That's for the ideal case.

[MagicSmoker]

...
What does Ae stand for ? how is the peak density flux calculated ?

Forgot to answer this - Ae is the accepted/standard terminology for the area of the center pole (or the cross-sectional area of a toroid) in a transformer or inductor (Ae means "Area, effective"). Note that sometimes the center pole area - usually abbreviated as Ac - will also be specified, particularly for E-shape cores, and Ac will likely be slightly higher in value than Ae; in such cases you use Ae to determine the number of turns, but Ac to determine core losses (if core volume, V or Ve or Vc, is not given).

The usual procedure for designing a transformer is to set the peak flux density ahead of time based on either the saturation flux density or the loss factor (usually in mW/mm³). If the operating frequency is low enough for a given core material then the design will be saturation limited, whereas at high frequencies* core losses start to dominate and flux density will need to be reduced to keep those losses in check.

Most grades of silicon steel appropriate for transformers will have a saturation flux density around 1.5 Tesla (15000 Gauss), so if you aren't core loss limited then 1.2T is a relatively safe value. Note that the number of turns is inversely proportional to peak flux density and core area, and directly proportional to applied voltage, so for a given core area and applied voltage, increasing the turns decreases the flux density which in turn decreases the core losses (but more turns means higher winding losses). Similarly, a larger core area will need less turns per applied volt for a given flux density, and it is for this reason alone that larger cores can handle more power (this often comes as a surprise to those new to transformer design).

Normally you want to split the primary into two halves of equal number of turns and sandwich the secondary between them to minimize proximity effect losses and leakage inductance, but in this particular application the proximity losses are irrelevant and more leakage inductance is a good thing (it will help limit short circuit current), so wind the primary first, then wrap it with yellow mylar (polyester) tape (e.g. - 3M #56) and wind the secondary on top.


* - what is "high" frequency very much depends on the core material and lamination thickness, if applicable. 400Hz would be considered a high frequency for 0.25mm thick silicon steel lamination, while 25kHz would still be considered a low frequency for most power ferrites (e.g. - Ferroxcube 3F3, Magnetics Inc. R, etc.).

[MrAl]

Hi again,

I forgot to mention something that is very useful also.

When dealing with wire sizes like 22 gauge AWG wire, it helps to remember that if you go DOWN in wire gauge number three times then the current capability is doubled.  So for example if we were using 22 gauge at 1 amp and we wanted to know the wire size for 2 amps, just go DOWN three wire sizes, which brings us to number 19 gauge.
Looking at a wire size table, we see that #22 is good for about 1 amp with 600 circular mils per amp current density, and #19 is good for twice that, 2 amps at 600 cm/amp.
Note that the diameter does not go up by 2 because the area changes in two dimensions, so the diameter goes up by the square root of 2 which is about 1.4142 .
Also remember that you can only do this so many times before it becomes inaccurate, so it's not a good idea to start with wire gauge 36 and try to come up with a wire size that works at 30 amps.