Push pull converter transformerdesign following colonel mclyman

Thread Starter


Joined Mar 13, 2024
Hi all,

I have decided to post my question with my calcs following his example on page 7-21 of colonel mclymans book (he design a multi-output, I will design a single output).

It has been a while since I have looked into transformer design and I will most likely reread the book with more supporting material.

My goal here is to design a center tapped transformer but also gain more understanding to be able to design transformers and inductors for a range of smps.

Here are the initial design specs:

1. Input voltage, V(min) = 13volts
2. Output voltage , V(o) = 20.0 volts
3. Output current , I(o) = 0.5 amps
6. Frequency, f = 100kHz
7. Efficiency, η = 98%
8. Regulation, α = 0.5%
9. Diode voltage drop, Vd = 0.5 volt
10. Operating flux density, Bac = 0.05 teslas
11. Core Material = ferrite
12. Window utilization, Ku = 0.4
13. Temperature rise goal, Tr = 30°C
14. Maximum duty ratio, Dmax = 0.5
15. Notes:
Using a center-tapped winding, U = 1.41

Everything here is pretty straightforward for me, however I am unsure why he picked the operating flux density as 0,05. I do understand it is related to the material however I might have missed this in the text, this is question 1.

I apologize in advanced for my handwriting.
Fiirst thing is to select wire size

His example and mine operate at 100kHz so wire size is #26 AWG.

Step 1: Determine output power

Po = 10,25W, no questions here

Step No. 2: Calculate the total secondary apparent power, Pts

Pts = 14,45W, no questions here

Step No. 3: Calculate the total apparent power

Pt = 29,2W. I do however not know where he got Pa from, he has a value of 1,41 and I can only think it is related to U = 1,41 for a center-tapped winding?

Step No. 4: Calculate the electrical conditions, Ke.

ke = 5800, no questions here.

Step No. 5: Calculate the core geometry, Kg.

Kg = 0,005 no questions here.

Now he mentions the following.

*When operating at high frequencies, the engineer has to review the window utilization factor, Ku,in Chapter 4. When using small bobbin ferrites, use the ratio of the bobbin winding area to the core window area which is only about 0.6. Operating at 100kHz and having to use a #26 wire, because of the skin effect, the ratio of the bare copper area to the total area is 0.78. Therefore, the overall window utilization, Ku, is reduced. To return the design back to the norm, the core geometry, Kg, is to be multiplied by 1.35, and then, the current density, J, is calculated, using a window utilization factor of 0.29.*

Here is where things get a bit fuzzy for me, in the prior chapters he does explain the reduction in ku when using the small bobbin ferrites is due to shrinkage from the manufacturing process and gives a percentage of 15 to 30% for expected reduction but here he gives a solid 0,6. Where did this come from?

From his text here is the calculated reduction and multiplication factor of 1,35

Step No. 6: Select a PQ core from Chapter 3, comparable in core geometry Kg

For the core in his example he selected pq 2020 with pc44 material.

The closest core I found was pq2016 with my kg being 0,00675 and the pq2016 having a kq of 0.0166, All other cores provided in the book are much larger.

Now I am unsure from the image where he got the 2000um from, the copper weight and the millihenrys.

Might also be good to know how to select the correct material for the core.

For pq2016 I found the following datasheet: https://www.digikey.co.za/en/products/detail/epcos-tdk-electronics/B65875B0000R049/3914439

Step No. 7: Calculate the number of primary turns, Np, using Faraday’s Law

My input voltage will vary from 13V to 18V (stack of series connected 18650 batteries). I used 13V for Vp as he specs this in the design specs initially and I assume this will be the worst case scenario for turns and input current

For the rest of the steps, I dont have any questions.

Everything seems to work out well.
My temp rise is ~5 DegC
Regulation is 0,25%
Ku is 0,387.

Any inputs, questions and help will be deeply appreciated.

To summarize my questions

1. I am unsure why he picked the operating flux density as 0,05
2. I do not know where he got Pa from in step 3, he has a value of 1,41 and I can only think it is related to U = 1,41 for a center-tapped winding?
3. Where did the 0,6 factor come from in the reduction of ku value in step 5
4. After selecting the material where did he get the 2000um from, the copper weight and the millihenrys in step 6. Might also be good to know how to select the correct material for the core.
5. Am I correct to assume that Vp should be the minimum input voltage in step7.

Thanks in advance.


Joined Feb 24, 2006
To answer just one of your questions:
A flux density of 50 milli-Tesla is chosen based on the core material, ferrite, to avoid saturating the core. The reason you want to avoid the saturation region is that the transformer becomes highly nonlinear and can be damaged by repeated saturation cycles.

N.B. You should get in the habit of including the units with your numerical values, if for no other reason, so that you don't combine numbers with incompatible units and produce garbage results.


Joined Aug 7, 2020
50mT is nowhere near saturation for ferrite (which usually saturates at about 200-300mT)
50mT is used to minimise core loss which is proportional to the square root of the fifth power of peak-to-peak flux excursion.