Transformer Design

[MrAl]

Hi,

Yes there are formulas that provide for the selection of a core based approximately on the power that needs to be handled and the window area. I'd have to look them up now. But you still have to satisfy that equation with the number of turns for a given core area. The more turns the less saturation. In this case though you are working at 1000Hz which is a little higher than typical for steel that is usually used at 50 to 400Hz. I think you might use steel that is made for 400Hz and lower the flux density.
You really should get some data on the type of steel you intend to use. When you find the steel you find the manufacturer and when you find the manufacturer you find the data and design examples. One such manu is called (or was called) Magnetics, Inc. They have a site on the web. You can learn a lot from their literature.

 

[maheshyedla]

Hi all,

How do we calculate the current density of the transformer at 1000HZ frequency.

 

[MrAl]

Hello again,


Calculate the diameter of the wire (or look it up) in inches from the AWG wire number N:
D=92^((36-N)/39)/200

Calculate the area of the wire (or look it up) in square inches from the diameter D:
Aw=0.25*pi*D^2

Calculate the current density knowing the wire area Aw and the current A:
J=Aw/A

J is in units of square inches per ampere.

You can also calculate the circular mils of the wire from the diameter D:
Acm=D^2*1e6

A typical straight run wire circular mils would be something like 350 circular mils per amp, but for a transformer you would want much less.

For example, a AWG wire number 10 wire has diameter 0.1019 inches and area in square inches of 0.0081548, and area in circular mils is 10383, and at 30 amps that would be about 346 circular mils/amp.

 

[maheshyedla]

Hi,

Once again thanks for the feedback.. I didn't understood the Diameter formula which you have given. I mean the numbers like 92,36,39 and 200.

And also I have seen some papers that they have given like J = I/Aw ... So which one is correct.

SO could you please explain me.

Hello again,


Calculate the diameter of the wire (or look it up) in inches from the AWG wire number N:
D=92^((36-N)/39)/200

Calculate the area of the wire (or look it up) in square inches from the diameter D:
Aw=0.25*pi*D^2

Calculate the current density knowing the wire area Aw and the current A:
J=Aw/A

J is in units of square inches per ampere.

You can also calculate the circular mils of the wire from the diameter D:
Acm=D^2*1e6

A typical straight run wire circular mils would be something like 350 circular mils per amp, but for a transformer you would want much less.

For example, a AWG wire number 10 wire has diameter 0.1019 inches and area in square inches of 0.0081548, and area in circular mils is 10383, and at 30 amps that would be about 346 circular mils/amp.

 

[MrAl]

Hi,

Once again thanks for the feedback.. I didn't understood the Diameter formula which you have given. I mean the numbers like 92,36,39 and 200.

And also I have seen some papers that they have given like J = I/Aw ... So which one is correct.

SO could you please explain me.

Hi again,


Actually they are both correct as long as you observe the units being used. However, Amps/SquareInch is the more common form of the actual current density, but often you have a target goal of a certain area per amp, which would be square_inches per amp or circular_mils per amp. For example you may want a goal of obtaining 300 circular mils per amp. Note that for the formula i gave i also provided the units.

But if you'd like to use amperes per square inch then use:
J=I/Area

Next, the formula:
D=92^((36-N)/39)/200

provides the diameter D of the wire in inches knowing the American Wire Gauge number, such as #10, #8, #32, etc. When you go to buy wire usually it is given in the gauge number like that. To get the diameter from that gauge number just plug it into the formula for D above where the gauge number is N.
For example, to calculate the diameter of a number 10 gauge wire (AWG only) we would plug in the number 10 for N like so:
D=92^((36-10)/39)/200

and all you do next is the math required for that formula, which is first subtract that 10 from 36, then divide that by 39, then take 92 up to that power, then divide by 200. For a #10 wire the diameter D comes out to a little over 0.1 inch. You can test yourself using that result.

 

[johansen]

maheshyedla how much money do you think this transformer is going to cost you?
the reason i ask is because its going to take you about 4 iterations to build this thing...

you might find my spreadsheet to be useful **broken link removed**
note that at 10 Khz and high currents most of the copper loss is due to proximity effect, not resistance, and my model doesn't do any of that.

if you are using sheet copper then just change the maximum fill percentage to get a more realistic number for the amount of copper you think you can actually stuff in the transformer.

pay attention to the thermal time constant calculated for the core and for the copper (they are separate) to get an idea about the cooling required.

 

[maheshyedla]

Hi Johan,


I wan unable to open the spreadsheet.. which you have provided in the above post. could you provided me with some pdf or word file if possible.

Thanking you,

maheshyedla how much money do you think this transformer is going to cost you?
the reason i ask is because its going to take you about 4 iterations to build this thing...

you might find my spreadsheet to be useful **broken link removed**
note that at 10 Khz and high currents most of the copper loss is due to proximity effect, not resistance, and my model doesn't do any of that.

if you are using sheet copper then just change the maximum fill percentage to get a more realistic number for the amount of copper you think you can actually stuff in the transformer.

pay attention to the thermal time constant calculated for the core and for the copper (they are separate) to get an idea about the cooling required.

 

[johansen]

its an open office file. i could save it in excel but i was having problems getting the formulas to work

 

[maheshyedla]

Hi all,


Could I please know the AL and effective permeability(Ue) values of the Iron cores. I tried to find it but unfortunatelly I couldn't find it in the datasheets. So, please can any help me with these two values.

Thanking you.

 

[johansen]

you do not need to know the AL and the permaibility of the iron cores.

for two reasons. first is the permeability is mostly dependent on the air gap and not the iron.
you cannot control the air gap beyond a +/-50% estimation.

instead iron is classified under watts per kilogram of iron loss and magnetizing VA per kilogram of reactive power drawn when the flux level is known. (say, 1.7T field strength at 60Hz would typically be 1-2 watts per kilogram and 5-15 VA/kilogram)
note that the VA does not include the copper loss under no load.

if you are familliar with microwave oven transformers, they draw about 4 amps no load current, but 2/3rds of the real power loss under no load condition is copper loss.
in a modern high efficiency transformer the copper loss under no load is actually negligible.

the second reason you do not need to know AL is because it is a function of the following: permiability, air gap, and the physical dimentions.