0
\$\begingroup\$

I'm a VHDL guy for work and tinker with power electronics on the weekends, so please forgive if this question is dumb.

I'm designing a transformer with a large step down ratio. In case it matters, the end goal is to have a source of large DC currents.

Here is my question:

If I know the output current i, output power p, and switching frequency f, how much cross sectional area should the core have?

\$\endgroup\$
7
  • 1
    \$\begingroup\$ What power level are you targetting, 1 W, 100 W, 10 kW? The type of solution can depend on the scale. X-section sort of drops out of the power level, as the total weight of the transformer is more or less linearly proportional to power, half the weight is core and half copper, and the shape of the core is essentially constrained to be manufacturable. Once you have a core size, then you can design your turns per volt to match your frequency and supply voltage. \$\endgroup\$
    – Neil_UK
    Commented Sep 25, 2023 at 6:35
  • \$\begingroup\$ What type of transformer driver circuit are you proposing to use? What method of secondary rectification are you using? \$\endgroup\$
    – Andy aka
    Commented Sep 25, 2023 at 8:45
  • \$\begingroup\$ @Neil_UK This is the kind of dialog I was hoping to find. The answer > 100 kW. (Total energy is not especially high. Only delivering it in a short timeframe.) I had a hunch that weight and frequency impact each other. Can you say more? \$\endgroup\$ Commented Sep 25, 2023 at 22:34
  • \$\begingroup\$ OK, so start the dialog, what is a large current (amps) and a short (seconds) timeframe with not especially high total energy (joules)? The duty cycle would be useful as well, as cooling and output impedance are different limiting factors. This Q/A site format is not well suited to dialog, as you see in these comments, especially if getting info from you is like drawing teeth. You ask a question, you get an answer, so the question has to contain enough context for a sensible answer. Transformer theory for any conceivable application is not a good answer. \$\endgroup\$
    – Neil_UK
    Commented Sep 26, 2023 at 5:39
  • \$\begingroup\$ 100 kA. 30 us. 350 J. Duty cycle... no idea. The switching frequency needs to be > 150 kHz. In case it matters, I'll be delighted if I can get 50% efficiency out of this thing. \$\endgroup\$ Commented Sep 26, 2023 at 11:55

3 Answers 3

2
\$\begingroup\$

If I know the output current i, output power p, and switching frequency f, how much cross sectional area should the core have?

You have the wrong dimensions, so your thinking appears to be somewhat mixed up about transformers.

TL;DR There are so many variables, that attempting form one synthesis equation would be so complicated as to be uninstructive. In practice, it's probably best to pick a core size, pick a frequency, run through a design, and see what you get. And then iterate.

The power a transformer can shift is a rather woolly concept. It depends on the ratings of your application, what duty cycle you want, what cooling you apply, what regulation you can tolerate.

The transformer breaks down into two parts that, although coupled, can be designed entirely separately. The core, which is related to the operating voltage, and the hole in it that you fill with copper windings, related to the operating current.

The power shifted by a transformer depends on the product of the operating voltage and current. If you can increase one or the other, or both, you can increase the power. To take you back to your question, core x-section only gives you one factor, the voltage, and you need voltage and current for power.

There are several things that limit how much voltage and current can be used with a transformer. There are instantaneous (physics) limits, and long term (thermal) limits.

A continuously operating transformer will generally be thermally limited. A transformer operated with a very low duty cycle will usually hit the instantaneous limits.

Instantaneous limits

  • The core cannot exceed its saturation flux. Doing so reduces its permeability and so core inductance, by an order of magnitude or several, which absolutely limits the integral of the applied voltage.

  • The output current is limited by the regulation - how much voltage drop the winding resistance causes. In the case of short-circuiting the output, the resistance puts a maximum limit on the output current that can flow. Commercial transformers tend to be rated to around 5% regulation, but a special purpose low duty cycle transformer could be used at much higher voltage drop. Note that you will get maximum power output of the transformer at 50% regulation, when load resistance is equal to transformer resistance. A transformer can be used this way for short pulses, stopping before it overheats.

  • There is no practical limit to how a high current the windings can tolerate instantaneously - electromigration doesn't happen at any geometries bigger than IC internals, and magnetic forces only become troublesome at district-wide grid distribution levels.

Thermal limits

It's quite easy to calculate the heating power of the losses when operating a transformer. It's quite difficult to estimate the temperature rise that this power causes, especially with mixed materials, irregular geometries, passive convention all affecting the cooling. It's probably best to simply measure temperature rises, unless you have access to industrial strength thermal simulation software

  • Current in the windings causes I2R losses. These are quite easy to calculate.

  • Cycling the core round a hysteresis loop dumps a certain amount of energy into the core. Core heating power is therefore proportional to operating frequency. The amount of energy per hysteresis cycle varies at least as the square of the operating flux, and often a higher exponent. For an iron core at mains frequencies, this is often a minor contribution to the total losses. For a continuously operating ferrite transformer however, core heating tends to be the limiting factor rather than core saturation, as you push the flux and frequency upwards to get more power through your transformer.

  • If you have an intermittent application and a ferrite transformer, note that keeping the primary energised will keep generating heat in the core, even if you are not drawing current from the transformer. Energising only when output is required means it will run cooler, or at higher power when wanted.

\$\endgroup\$
3
  • \$\begingroup\$ I read in multiple places that using a transformer that is too massive for the switching frequency results in huge losses. For instance, if I make a core the size of a pickup truck and try to switch at 1 MHz, efficiency will be abysmal. Is that not true? \$\endgroup\$ Commented Sep 26, 2023 at 12:10
  • 1
    \$\begingroup\$ @JamesStrieter If you want to run >100 kW, I don't think that's going to be a problem. Getting a large volume of core up to flux will create a lot of losses. If the transformer is 'too big', then you can run at a lower flux for much lower core losses. Too massive for the switching frequency??? You're mixing dimensions again. Any size of transformer will have increased losses as the frequency rises. Where you want to put that efficiency knee is up to you. Choosing the best switching frequency FOR THE ENTIRE SYSTEM, including losses in semiconductor drivers, is a many-variable optimisation. \$\endgroup\$
    – Neil_UK
    Commented Sep 26, 2023 at 12:15
  • \$\begingroup\$ Alright cool! That's the kind of thing I needed. Sounds like the only thing to do is write a script that models a lot of inputs, a lot of outputs and optimizes according the weight that I put on each variable. I can do that! \$\endgroup\$ Commented Sep 26, 2023 at 13:46
0
\$\begingroup\$

Depends if weight or volume of core matters for your purpose.
You can use a smaller core but you would have to increase number of turns (overall) so you increase the copper losses and the design is not efficient.
On other hand using a bigger core the transformer will be more efficient.

So due to this balancing between core and copper losses the exact formula is hard to crack.

\$\endgroup\$
1
  • \$\begingroup\$ So there is a trade-off between core cross sectional area and number of turns. Are able to enlighten us ? \$\endgroup\$
    – D Duck
    Commented Sep 25, 2023 at 0:47
0
\$\begingroup\$

As Michal pointed out, there is no simple solution. Transformer design is an iterative process. You need to balance out core size against copper AC & DC losses, number of turns, wire diameter, magnetizing inductance, and winding geometries to improve coupling (minimizing leakage inductance).

This post will lead you to some of the design processes for transformers. Write a simple script with the equations so you can try different configurations.

Not covered in the post is core loss. Core data sheets will have the necessary graphs to figure out core loss.

\$\endgroup\$
1
  • \$\begingroup\$ Right now I'm trying to understand the relationship between frequency, power and cross sectional area. Put another way, how much flux is needed at a given frequency to convey a given amount of power? \$\endgroup\$ Commented Sep 25, 2023 at 3:07

Not the answer you're looking for? Browse other questions tagged or ask your own question.