Flyback Converter Magnetic Flux Calculation

In summary, the core of the transformer saturates when the magnetic field density at the narrowest area of the core is greater than 217.032 mT. The peak current in the MOSFET becomes exponential, which means the core saturates. There might be something wrong with the calculation if the drain current waveform is not clean.
  • #1
BlackMelon
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Hi

I am designing a flyback converter. The calculation (.jpg file) is based on Daniel W. Hart Power Electronics book. The transformer model consists only of an ideal transformer and its magnetizing inductance (or "primary inductance" in some textbooks).
The datasheet of the transformer's core is the pdf file "pq32_30". The other pdf file has details about the core's material (N87). So, I calculate the magnetic field density (B) at the narrowest area of the core. From this B value (217.032 mT) I do not expect to have any core saturation issue.

However, when I test the real circuit, I can see the peak of the MOSFET's drain current becoming exponential, which means the core saturates.

So, is there something wrong with my calculation?
 

Attachments

  • SaturationTrf.jpg
    SaturationTrf.jpg
    48.9 KB · Views: 87
  • N87 Saturation Data.pdf
    402.3 KB · Views: 94
  • pq_32_30.pdf
    358.7 KB · Views: 80
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  • #2
Check the Drain current with the secondary open-circuit (no load, no filter cap). If the Drain current waveform is clean, your calculations did not include the load current.

edit: Or maybe the filter cap is shorted.
/edit;

Cheers,
Tom
 
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  • #3
I don't understand how you can expect us to figure this out without knowing the core dimensions (Ac & lm).Also, I calculated a quite different value of Ipk in the primary, 0.673A (assuming 100% efficiency). Although we nearly agree on the current change in the primary ##\Delta I_p##.

Anyway, I'll leave you with my favorite magnetics design equations, which are often all you need for this sort of problem:

## B=\mu H = \mu \frac{NI}{l_m} ##
## L= \mu \frac{N^2 A_c}{l_m} ##

You can combine these to show that ##B=\frac{LI}{NA_c}##.

##A_c## is the effective core area.
##l_m## is the effective magnetic path length.

PS: Oops, I see the core is specified in your attachments. You'll get better (more) feedback if you really show your work better without making us look for it. Anyway with ##A_c =153.8 mm^2## I calculated ##B_{max} = 0.16T## which should be fine for ferrite. Of course it's actually a bit higher since your efficiency isn't 100%.
 
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  • #4
How does the slope ##\frac{dI}{dt}## of the primary current compare to the expected value ##\frac{V_{in}}{L_m}## initially before the slope increases with saturation? This will tell you if the magnetizing inductance (i.e. air gap, turns, etc) is what you expected. How does the measured peak current compare to your calculations when you see saturation? Somewhere, your assumptions must be wrong.
 
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  • #5
Hi All,
Actually, I figure it out why it saturates. The center of the transformer core (area named Ae) got the highest flux density, not its wing (even the wing's area is smaller than the center, it has much lesser flux). Please see my calculation on the mediafire link below. First, let me summarize/answers major concerns of each guy in here.
Tom: I include the load current. (actually, the increasing load current will increase the mean value of Lm)
DaveE: For B = LI/(NAc) I use "flux = LI/N" where flux is B*Ac. Same equation.
dI/dt and Vin/Lm is normal before saturation
See my calculation on the images.

https://www.mediafire.com/file/xsmzea322c2lull/Page1FB.jpg/file
https://www.mediafire.com/file/fzw9wnw0r0k57fj/Page2FB.jpg/file
I have attached some pages from Daniel Hart Power Electronics to show the proof of the formula I used. See the pdf file on attachment
 

Attachments

  • Flyback Converter Daniel W. Hart.pdf
    327.9 KB · Views: 125
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1. What is a flyback converter and how does it work?

A flyback converter is a type of switching power supply that converts one DC voltage to another DC voltage. It works by using a transformer to store energy in its magnetic field and then releasing it to the output load when the switch is turned off.

2. Why is it important to calculate the magnetic flux in a flyback converter?

Calculating the magnetic flux is important because it helps determine the size and design of the transformer, which affects the overall performance and efficiency of the flyback converter. It also ensures that the transformer does not saturate, which could lead to malfunction or damage.

3. What factors affect the magnetic flux calculation in a flyback converter?

The main factors that affect the magnetic flux calculation in a flyback converter are the input voltage, output voltage, switching frequency, duty cycle, and transformer turns ratio. Other factors such as core material, winding resistance, and operating temperature can also have an impact.

4. How is the magnetic flux calculated in a flyback converter?

The magnetic flux in a flyback converter can be calculated using the formula: B = V * t / (N * A * μ), where B is the magnetic flux, V is the input voltage, t is the switching time, N is the number of turns in the primary winding, A is the cross-sectional area of the core, and μ is the permeability of the core material.

5. Are there any limitations or challenges in calculating the magnetic flux in a flyback converter?

Yes, there are some limitations and challenges in calculating the magnetic flux in a flyback converter. These include non-linear behavior of the core material, parasitic elements in the circuit, and the need for accurate measurements and calculations. Additionally, the design of the transformer may need to be adjusted to account for these limitations.

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