Stuff we learned:

max duty cycle should be between 0.4 - 0.6 (we’ll spec it to be 0.5)
operating switching frequency should be between 50 to 140 kHz

(extra research)

For CCM flyback primary inductance:

Innoswitch Example:

Microsoft Word - DER-930 180W 20V9A 90-265VAC Battery Charger INN4177C-H189 CPZ1076M PFS5177F (power.com)

Nominal Primary Inductance of 430 microH
0.312 duty cycle
Vin max-410v
Vin min- 310v
100 kHz\
32 inductor turns
Po- 189 W
0.6 Z factor

Primary Side Inductance Calculation:

DCM formula: https://www.infineon.com/dgdl/an-1024.pdf?fileId=5546d462533600a401535591115e0f6d

If Cres doesn’t exist (which doesnt exist in the InnoSwitch design), then the following formula comes out (found on a different DCM source, https://www.monolithicpower.com/how-to-design-a-flyback-converter-in-seven-steps )

To check whether that formula is correct, let’s take the 180W innoswitch example and then plug in their design parameters and see if the inductance we get is close to what they used:

We don’t know how FACTOR_Z impacts the primary inductor calcs, so we’ll just ignore that.

(z-factor)

What is z-factor btw

https://piexpertonline.power.com/help/piexpert/en/topics/z_factor_definition.htm

Plugging in the values yields 414.1 uH, which is very close to the 430uH they used!

So now we can calculate with our parameters:

efficiency = 95% (the 180W example used 97% in calculation, but only achieved 95% irl)

Vin_min = 90V (min voltage of battery pack)

Pout_max = 70W (a little more than the 60W expected max power draw in requirements page)

K = 1 (because DCM)

f = 55 kHz (datasheet recommends 50-140kHz, but since lower freq is more efficient (less switching loss), we will also pick a low value. 55 is between 50 ('recommded min) and 57, which is the 180W example amount)

So the L we get from the above equation is the maximum primary side inductance we can use such that the converter is always in DCM, even in worst-case scenarios (which were the inputs to the above).

We will also spec the inductor to be slightly less than 250uH, just to be safe, so L = 240uH.

Using that inductance value, we can then reverse-calculate the duty cycle we expect to see at certain operating parameters (pack voltage and power consumption):

Nominal duty cycle of 0.23 is reasonable, so L = 240uH is reasonable!

Turns Ratio Calculation

Output voltage is a function of turns ratio, input voltage and duty cycle, in the following formula:

In our applications, since the input voltage is significantly higher than the output voltage, we expect N2 to be less than N1. Also, since it’s a ratio, we can define the ratio as ‘n:1’, where n is the number of coils on the primary side for a given number of coils on the secondary side (simplified to 1 for math’s purposes)

So then, the formula becomes:

Then, clearly, if n is very large, then D has to be larger to compensate in order to keep Vout the same. Likewise, is n is smaller, then D has to be smaller to compensate.

We will define the ‘worst case scenario’ as when D is as large as possible (0.5), and ask ‘at this point, what is the MAXIMUM value of n before the output voltage can no longer be the desired amount (12V). We will actually round 12V up to 12.5 just to be safe. In this worst-case scenario, we also will define the Vin to be the minimum (90V), since lower Vin means larger D.

So then, we define our worst-case parameters as follows:

Therefore, for an N:1 turns-ratio transformer, N must be less than 7.2. Conveniently, we will pick N=7, so our transformer should have a 7:1 primary to secondary turns ratio

Peak Primary Current, I_pk

Peak primary current is supposedly calculated using the following formula (from here):

Dmax = 0.5 (from above)

Vin_min = 90V (from above)

fsw = 55 kHz (from above)

Lp_max = 240uH (calculated from above)

But then why is there a 2 term in the denominator? We’re not using RMS, these are DC values

Why is there a 2 in the denominator

At aroudn 7:15 in this video:

https://www.youtube.com/watch?v=VGZyqmAHMw0

He mentions that the Duty Cycle of 0.48 should be multiplied by 2 for some reason?? IDk

From this other source, there’s no 2 in the equation:

When you also try to derive the equations yourself with V = L dI/dt and tON = D*1/f, you don’t get a 2

So we will assume there is no 2 involved:

I_pk = 3.4A, which is very high….

Looking at the InnoSwitch datasheet, these are the rated I_limit values

With this in mind, we should use INN4076 (or higher) with the 4.7uF BPP configuration

Transformer Core Selection

We will semi-blindly trust the area product method shown in this video:

https://www.youtube.com/watch?v=45kTQwZSLHc

Ac is the area of the core [mm^2]

A_W is the area of the winding space [mm^2]

E = max energy the inductor needs to store = 0.5*L*Ipk^2 =0.5 * 240uH * 3.4A = 1.3872 mJ

Bpk = saturation flux density of the core, which is determined by the material of the core. We will use a ferrite core (since it low losses), at which the common saturation limit is 300mT

J = current density in winding, which as a rule of thumb is given as 3A/mm for copper wire (from the video), so we will use that as well

PF = packing factor, which accounts for what percentage of the available wiring space is actually fillable by wire, since things like insulation take up space. The video estimates 50%, or 0.5, so we’ll use that too

Then 'A' (which is Aw*Ac) is 6200 mm^4, so we try to find a transformer where Ac (area of core) multiplied by Area of Winding (Aw) is similar

180W uses this: https://www.digikey.ca/en/products/detail/tdk-corporation/PC95PQ32-30Z-12/9607321

60W uses this: ATQ23.7 (not on digikey)

Alternatively, let’s follow the calculations from here:

https://coefs.charlotte.edu/mnoras/files/2013/03/Transformer-and-Inductor-Design-Handbook_Chapter_13.pdf

We need these parameters:

Skin depth, based on their formula, is:

where A_wire is in mm^2, so we want to pick a wire with a MINIMUM cross-sectional area of 0.25mm^2, or 0.0025cm^2, or 2.5cm^2*10^3

To verify this is correct, we use https://www.allaboutcircuits.com/tools/skin-depth-calculator/ , and see:

So then plug into area of circle formula:

We see minimum wire area is also ~0.25mm!

According to Table 4-9 in chapter 4, we the MINIMUM wire size we can use is 23-AWG

(details....)

https://coefs.charlotte.edu/mnoras/files/2013/03/Transformer-and-Inductor-Design-Handbook_Chapter_4.pdf

Then calculate Irms, which they give as:

We can simplify the equation to just:

Then, follow their steps for determining the right transformer to use

Assuming alpha (proportion of power loss due to copper resistance) is also their value of 1% (which is just 1 in their equation for some reason), then we find Kg = 0.0213 cm^5

So then go to chapter 3 of their guide, and find a core with a similar Kg (but slightly larger, just so that there’s more area to work with)

Click here to expand...

https://coefs.charlotte.edu/mnoras/files/2013/03/Transformer-and-Inductor-Design-Handbook_Chapter_3.pdf

For EE cores, nothing is within good range

Also nothing good for EC:

Nothing for ETD:

Nothing for ER:

For EFD ferrite cores, best is EFD-25 = 0.01911 or EFD-30 - 0.03047

For EPC ferrite cores, EPC-27 = 0.024036 is very close too!

For EP ferrit cores, EP-20 = 0.02892 works

For PQ ferrite cores (what they use in examples), we see PQ20/20 = 0.0227

So the ‘valid’ cores, in order of ascending Ku, are: PQ20/20, EPC-27, EP-20, EFD-30

Looked on Digikey and found the below options, then downloaded a spreadsheet format for all the valid options so it’s easier to filter and calculate stuff.