Stuff we learned:
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Nominal duty cycle of 0.23 is reasonable, so L = 240uH is reasonable!
Turns Ratio Calculation
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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, Ipk
Peak primary current is supposedly calculated using the following formula (from here):
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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
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At aroudn 7:15 in this video: https://www.youtube.com/watch?v=VGZyqmAHMw0He 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:
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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:
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Ipk = 3.4A, which is very high….
Looking at the InnoSwitch datasheet, these are the rated I_limit values
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With this in mind, we should use INN4076 (or higher) with the 4.7uF BPP configuration
Transformer Core Selection
This is a very trial-and-error and convoluted process, but I will try to document the decisions and calculations
Calculations were performed on this spreadsheet: https://docs.google.com/spreadsheets/d/1O1wjWdbnPksTQmW_tAbk-rAPtSNWPjp_Yc3J71V07ag/edit#gid=1114770555 , and a Desmos page that Jonathan Xie used just for clarity of presentation
Skin Depth
Regardless of what core selection method we follow, skin depth is an important factor to consider, as it determines the minimum wire gauge that should be used at our operating frequency.
Skin depth, based on the formula from Charlotte EDU (see section below), is:
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where Awire 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:
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So then plug into area of circle formula:
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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. Since that is not a common AWG size, we will instead use22-AWG as our minimum wire size. 22 AWG will be used for the primary coils, since they carry low current and require many turns.
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Core Selection - Area Product Method
Previously, I tried following the area-product method from a video (shown below), but it didn’t yield such good results, due to lack of clear calculation steps and available resources for types of transformer cores and their properties.
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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] AW 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  |
However, the video is very useful for learning about the general procedure of finding a transformer, and the fundamentals behind the magnetics
Examples
180W uses this: https://www.digikey.ca/en/products/detail/tdk-corporation/PC95PQ32-30Z-12/9607321
60W uses this: ATQ23.7 (not on digikey)
Charlotte EDU - Ku Method
I found a series of lecture material regarding magnetics. The most important one is Chapter 13, Transformer and Inductor Design, which containts many examples (with calculations) of how to design a transformer for power converters:
This is a very long process, so I will summarize what was done below, and then put all the details in an expand:
Calculate parameters that we already knew such as peak primary current, average primary current, primary inductance, etc
Based on these input parameters, we use their magical equation to determine ‘Kg’, or the ‘core geometry factor’, which is a product of the Ac (core area), Aw (winding area) and MLT (mean length per turn)
Transformers come in standardized packages, each of which with varying values for Aw, Ac, etc, so they all have an associated Kg value. We then find transformers with similar (or slightly higher) Kg values as calculated above, and then we will use a spreadsheet on those transformers to pick one (next section)
For details, see below:
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We need these parameters:  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) 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.
 https://www.digikey.ca/en/products/detail/epcos-tdk-electronics/B65875A0000R087/3914436 EPC-27: (does not exist on digikey)  https://www.tdk-electronics.tdk.com/inf/80/db/fer/ep_20.pdf  https://www.digikey.ca/en/products/detail/epcos-tdk-electronics/B66423G0000X187/3914948 |
Number of Primary Windings
Now that we have a list of transformer core options, we can pick one that best suits our purposes, and then find the number of primary windings we should have. This was a semi trial-and-error process, but I got a reasonable conclusion in the end, so:
Both the Powerful Electronics and Charlotte EDU approach involve finding a balance between two things:
Finding a core area such as to not saturate the core during peak primary current
Adequate winding area (cross-sectional) so that there is enough space to fit all wire
The number of primary turns, which we will denote with ‘Np’, plays into both of the above parameters, because:
You need a MINIMUM amount of primary turns such as to not saturate your core during peak primary current, BUT
You cannot have too many turns, or else your wire becomes too small (and thus has too high of a resistance), or you will run out of transformer core space
So, whenever you are designing around 2 restrictions, you usually want to optimize for one restriction, and then calculate the corresponding effect on the other restriction
‘The dumb method:’
If you wish to base primary windings off the 2nd restriction, you have to:
estimate what percentage area of the transformer will be used for primary windings (typically 40-50%)
define a target current density, 'J' (typically 3A/mm^2 for copper)
define a factor that represents what percentage of the available space you actually have for copper, since you will have wasted space from wire insulation, inability to pack properly, etc (usually also 0.4 to 0.5)
THEN, you can calculate, based on the available winding area of the transformer, how much space you have available for primary windings, and then divide by the cross-sectional area of the wire you want to use, which will yield your desired number of turns. But this is kinda dumb, because it runs off raw assumptions.
'The smart method'
Instead, we will use the 1st restriction, which is must more simple, and more critical, because you never want to saturate your core.
Using this good resource, https://cettechnology.com/the-effects-of-an-air-gap-in-magnetics/#:~:text=Gapping%20a%20core%20with%20an,more%20heating%20at%20this%20location. and the Powerful Magnetics video, we can derive from Faraday’s Law and the properties of an inductor, that:
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Where IP is peak current, L is primary inductance, B is magnetic flux density, A is the core area, and N is the number of turns. Every core saturates at a certian value of Bmax, which is a material property. Since we will use a Ferrite core (it is more efficient at these high switching frequencies and is commonly available for cheap), we will define Bmax = 300mT.
So then, we can calculate the minimum number of turns, Np, such that the resulting B is less than 300mT!!
Selecting a Core
Using the spreadsheet, we see that the EFD-30 core requires at least 39.5 primary turns to not saturate, which we can conveniently round up to the next number divisible by 7, which is 42! This of course means the secondary number of turns will be 6
Also, 42 seems like a ‘reasonable’ number. The 180W example has 40 primary turns.
I then found a core on digikey: https://www.digikey.ca/en/products/detail/epcos-tdk-electronics/B66423G0000X187/3914948
and its associated bobbin: https://www.digikey.ca/en/products/detail/epcos-tdk-electronics/B66424W1012D001/3914953
which both happened to have CAD models, so I downloaded those and started hypothetically testing wire fits.
Determining the Gap
(The resource https://cettechnology.com/the-effects-of-an-air-gap-in-magnetics/#:~:text=Gapping%20a%20core%20with%20an,more%20heating%20at%20this%20location. is a very good read, and will be heavily cited for this portion)
Now that we have a core, all that’s left to do is roughly calculate the gap that will be required to achieve our desired inductance.
Every core has a specified ‘Al’ value, which is in units of H/turns^2. The EFD-30 core has an Al value of 2.05uH/turns^2, so multiply that by 42^2 yields 3616uH, which is far larger than our desired 240uH. As such, we will introduce an air gap in our transformer to reduce the magnetic permeability of the transformer, thus decreasing the primary side inductance.
The magnetic reluctance, R, of a transformer is given by:
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Where N is the number of primary turns, and L is the required inductance
According to https://cettechnology.com/the-effects-of-an-air-gap-in-magnetics/#:~:text=Gapping%20a%20core%20with%20an,more%20heating%20at%20this%20location. and other resources, it is a reasonable assumption to assume that almost all of the magnetic relucatance of a gapped transformer comes from the air gap (since the core itself has very low reluctance). So then, the magnetic relucance, R, of the transformer based on a given gap, g, is given by:
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where μ0 is the magnetic permeability constance of air (4π*10^-7 H/m), and A is the effective area of the core (Ac). Then, the gap, g, can be easily isolated for.
However, that is not the end, as there is something known as the fringing factor. You can read the linked resource about what that is, but basically, this fringing factor causes the true inductance of the transformer to be actually HIGHER than what it theoretifcally should be. This can be counteracted in two ways:
reducing number of turns
increasing the gap to be larger
Online resources to use option 1, as less wire = less ESR (equivalent series resistance) and you save money and space. However, we absolutely don’t want to saturate the core, and we have enough space, so we will go with option 2.
Also, the theoreitcal gap required to achieve 240uH will definitely be different than the actual gap, so this will be a trial-and-error thing anyway. Thus, we should not change the primary number of coils to account for fringing factor, which is what some resources will recommend.
Also, since the gap is actually distributed across 3 sections of the transformer (as opposed to one), the ‘true’ fringing factor will be different, as the fringing factor formula is non linear. That is,
f(g) != 3*f(1/3*g)
where f(g) is the fringing factor for a given gap, g. But again, we will trial and error the gap. The only thing to check with calculations is that the required gap is not an unreasonable number. The calculator says the required gap for EFD-30 is 0.635mm, which is very reasonable and consistant with other examples
Transformer Cross-Section
The EFD-30 core (orange) and its bobbin (green) are shown in the assembly below
I then made the following transformer cross-section diagram:
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(outer bounding box is taken from the true space available between bobbin and core)
Layer | Function | # strands x wire gauge | Turns | Reasoning |
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1 | Primary Layer 1 | 1 x 22 AWG | 21 | 22 AWG is smallest common wire size based on skin depth at f = 55kHz |
2 | Primary Bias | 3 x 30 AWG | 4? | Yes 30 AWG is less than skin depth, but this shouldn’t matter since this is just for the bias windings, which draw very little current. Also there will be 3 strands |
2 | Secondary Bias | 3 x 30 AWG | 3? | ^^ |
2 | Shield | 1 x 30 AWG | (until the end) | Wires that exist for shielding go brrrrrr (idk everyone else does it and we need to fill the leftover space in the bias layer) |
3 | Secondary | 3 x 20 AWG | Want more surface area for secondary layer since higher current will be experienced. 20 AWG is the max size that the height restriction of the core allows, and 3 strands is the most that the width restriction allows | |
4 | Primary Layer 2 | 1 x 22 AWG | 21 | Split primary windings into 2 layer |
PI expert
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(ignore this, gave useless information)
Bias Windings
Now that we know there will be 42 primary windings (and 6 secondary windings), we can determine the required number of windings for the primary bias (VBPP) and secondary bias (VBPS)
According to absolute maximums:
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But then for nominal voltages:
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But, both biases should provide operational voltages within the entire operating input voltage range. Doing the math, while factoring in the 1.25V drop across the rectifying diode:
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We see that the bounding regions overlap. That is, there is no number that perfectly satisfies both the max/min voltage conditions at max/min battery voltage, respectively (the min is greater than the max).
As always, we learn by example, so let’s see what the the 180W example did:
The title of the design states that the module accepts 90-265VAC input, but under the ‘transformer design spreadsheet’, the input voltage range is defined as 300-410VAC for some reason. Either way, using the numbers they have for primary, bias 1 and bias 2 windings, we see that in every single case, the voltage at both the BPP and BPS pins are much higher than 6V.
So, I will assume it is always better to have over-voltage than under-voltage. So we will pick bias windings of Bias 1 = 5 turnsand Bias 2 = 4 turns