HF toroidal transformer - choosing Bpk vs primay turns

Thread Starter

macafeeje

Joined Sep 9, 2021
6
Hello,

This is my first time designing a transformer based power supply. I have wound a couple of experimental toroids so far with varying results and am still not sure exactly on how to calculate the number of primary turns.

Using a T94-2 core as an example (because I have one sitting around, datasheet link: micrometals T94-2) i'd also like to know the process to determine if a specific core is suitable or not.

Do I select Bpk based on an acceptable core loss, maybe 5-10% of rated load (method 1), or do I calculate the required primary inductance and then check that Bpk is within an acceptable range (method 2). Does the primary inductance matter?

Method 1:
Vprimary = 8.5Vrms
Iprimary = 0.470A
power ~= 4W
core loss 10% = 400mW / 2.16cm^3 = 185mW/cm^3 (not sure if i'm converting this right, using Ve from the datasheet?)
switching freq: 50kHz
Bpk ~= 1000G
Nprimary = (8.5*10^8) / (4 * 0.362 * 1000G * 50000) = 12 turns

Now, i'm not sure if the H - DC graph in the datasheet is purely for a DC analysis, or if I can plug my values above into it, which gives me 1.2Oe, well within the saturation limit of the core.

Method 2:
Vp = 8.5Vrms
Ip = 0.470A
power ~= 4W
switching freq: 50kHz
So I need an inductive reactance of 8.5 / 0.470 = 18ohm
Inductance of the coil is :. Xl / (2 * pi * f) = 18 / (2 * pi * 50000) = 57300nH
Using the value Al from the datasheet, number of turns for the required inductance is sqrt (57300 / 8.4nH/N^2) = 82 turns
Bpk = 143G, core loss of 51mW.

H - DC more saturated than method 1, but still >99%.

Building two similar cores using method 1 and method 2 as an experiment (T68-1 cores under different load conditions), the second method produced a better output. The low turns core appeared to saturate.

If it's not obvious at this point, I don't know what i'm doing! I've read countless guides on how to calculate the number of primary turns, but they always seem to pick the Bpk value out of no where. There is probably something fundamental I am missing so any guidance on determining the number of primary turns and the suitability of a core would be appreciated (not including inductive, capacitive and resistive losses also associated with a transformer, one step at a time). :)
 

Ian0

Joined Aug 7, 2020
5,140
If you are making a transformer rather than an inductor, iron powder is not a good choice. Use ferrite.
Bpeak for ferrite is about 0.2T, but the optimum (the lowest losses) may be as low as 50mT. Core losses are proportional to Bpeak raised to the power of 5/2. As Bpeak decreases you need more turns which result in higher conduction losses.
 

Thread Starter

macafeeje

Joined Sep 9, 2021
6
Hi Ian, okay - not set on a core just yet, so thanks I will look into that. Would just like to understand where the numbers come from by looking at the datasheet. Which method is correct for selecting Bpk?

Also, correction for above, I read the datasheet wront: Method 1 Bpk should be 500G (0.05T) with a core loss of 185mW/cm^3 and the number of turns works out to 23.5
 

Ian0

Joined Aug 7, 2020
5,140
. Which method is correct for selecting Bpk?
1.Work out the core losses from Bpeak.
2. work out the number of turns you need
3. work out the wire size*
4. Work out the wire resistance
5. work out the Conduction losses (for both primary and secondary)
6. Add the two together to get the total losses.
Then repeat for different Values of B until you find the lowest losses.

*aim for one complete layer for each winding, or perhaps a sandwich with primary-secondary-primary, otherwise you will run into problems with the proximity effect.
 

Ian0

Joined Aug 7, 2020
5,140
Okay, so i'm still not getting it sorry! How do I select Bpk to begin with?
Guess, based on experience!
For a first try, start with 0.2T as that is about the highest you can use with ferrite to avoid saturation.
It’s an iterative process, and more confusing if you load is not constant - do you design for maximum efficiency at peak load or at standby load?
 

Thread Starter

macafeeje

Joined Sep 9, 2021
6
Okay, so I made a spreadsheet with the formulas and values from the datasheet to work out the core/copper losses based on the number of turns. 0.2T gives 6 turns, a core loss of 2.7W which is obviously pretty poor. At 34 turns, 345G (0.0345T) the core loss is 71mW. But is there any restrictions on how low Bpk can be? Or is it as low as you want until you start increasing the copper losses because of the number of turns/length of copper (which in this example have been fairly negligable).

Google docs link if it helps; https://docs.google.com/spreadsheets/d/1qa7Jwav22smyHgdX0aTOFk01GFrf1ir5KeoKeu5rpcg/edit?usp=sharing

Oh, and fortunatly the load should be fairly constant, thank goodness!
 

Ian0

Joined Aug 7, 2020
5,140
Okay, so I made a spreadsheet with the formulas and values from the datasheet to work out the core/copper losses based on the number of turns. 0.2T gives 6 turns, a core loss of 2.7W which is obviously pretty poor. At 34 turns, 345G (0.0345T) the core loss is 71mW. But is there any restrictions on how low Bpk can be? Or is it as low as you
I use a spreadsheet as well.
The lower the better for peak flux, as it’s the one that varies most steeply.
You’re wasting you time with iron powder cores for a transformer.
Use ferrite for AC, iron powder for DC.
 

Thread Starter

macafeeje

Joined Sep 9, 2021
6
So a ferrite core would be better because it reduces the number of turns required, a higher Bpk before saturation, or is there another factor which means 34 turns on a powder core wouldn't be suitable?
 

Ian0

Joined Aug 7, 2020
5,140
So a ferrite core would be better because it reduces the number of turns required, a higher Bpk before saturation, or is there another factor which means 34 turns on a powder core wouldn't be suitable?
Inductance.
The inductance, especially on a #2 core would be much too low (9.7uH), resulting in far too much magnetisation current.
A comparable toroid would give about 3mH. Magnetisation current would be reduced by a factor of 300.
 

Thread Starter

macafeeje

Joined Sep 9, 2021
6
Ah, that's what i've been missing I think, that makes much more sense thank you. So total loss in the primary side is the sum of core loss, copper loss based on input power (0.74A^2 * copper_resistance) and magnetisation current (8.5V / Xl)^2 * Xl
Xl being 2 * pi * f * L, is that how you would calculate that?
 
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