# Effective frequency of push-pull transformer

#### SiCEngineer

Joined May 22, 2019
209
I am reading a transformer design white paper which states that a push-pull transformer should be designed for a frequency of half the switching frequency. Eg, if my switches are running at 100kHz, the transformer frequency is 50kHz. Does this make sense? Personally, I would understand that although each primary winding is driven by one switch at 50kHz, the combined effect would be energy delivery at 100kHz. Am I wrong, and why? Thanks!

#### MrAl

Joined Jun 17, 2014
7,604
It depends on the volt seconds at the primary.
100v*10us vs 100v*20us for example.
Half the frequency implies twice the period meaning more turns required.
Why would each switch operate at 50kHz though?
100kHz has a period of 10us and that means for a square wave 5us per half period, which means if the transformer is really driven at 100kHz then it will be seeing 5us one polarity and 5us the opposite polarity so the output will be 100kHz also.
If each switch was at 50kHz that would mean 10us positive and 10us negative, which is not 100kHz.
Maybe you need to show your schematic if this does not make sense to you.

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• SiCEngineer

#### ronsimpson

Joined Oct 7, 2019
684
transformer design white paper
It is hard to know what is in some one's head, what they are thinking. Also it is hard to know where the frequency is measured.
I have built power supplies where (primary side) the top transistor is on for 10uS and feeds plus into the primary, then for 10uS the bottom transistor is on and feeds negative to the primary. (50khz) But looking at the secondary via full wave rectifying you get power every 10uS resulting in 100khz. What is the switching frequency 50 or 100khz?

This is the same thing as "60hz ac, what is the frequency of the light from a light bulb?" 60 or 120?

• SiCEngineer

#### SiCEngineer

Joined May 22, 2019
209
This is what I am doing I believe... the switching frequency of the switches are 100kHz, meaning that the period is 10uS, and each switch is on for ~5uS each - therefore each of the primary windings is driven for 5uS of the 10uS period. But I am confused. Here is the paper, if it helps: https://www.ti.com/lit/ml/slup126/slup126.pdf. The part which has what I am asking about is page 4-4, and continues on 4-5, under the subtitle "Frequency". Thanks for your help guys.

#### kubeek

Joined Sep 20, 2005
5,724
The way I see it with your example the switch frequency is 200khz - two switch actions per 10us, and the transformer frequency is 100khz - that is the frequency you design the tranformer for.

• SiCEngineer

#### MrAl

Joined Jun 17, 2014
7,604
I think i see why they are saying that.

In a push pull arrangement, one winding is driven to ground and so that puts -Vcc across winding A, and later the other winding is driven to ground, and that puts +Vcc on the A winding. So the A winding (as well as the B winding) goes from -Vcc to +Vcc, and that is 2 times Vcc. Therefore the volt seconds are twice as high as when you just drive the winding with only a level of Vcc. So to make the transformer suitable you design it AS IF it was driven at 1/2 the switch frequency because that is the same as designing it for 2 times the primary voltage. The operative phrase here is "AS IF" because it really is not 1/2 frequency.

• SiCEngineer

#### ronsimpson

Joined Oct 7, 2019
684
The part which has what I am asking about is page 4-4,
white paper said:
It is the frequency seen by the output filter, the frequency of the output ripple and input ripple current,
So the higher of the two numbers we are talking about.

#### SiCEngineer

Joined May 22, 2019
209
I think i see why they are saying that.

In a push pull arrangement, one winding is driven to ground and so that puts -Vcc across winding A, and later the other winding is driven to ground, and that puts +Vcc on the A winding. So the A winding (as well as the B winding) goes from -Vcc to +Vcc, and that is 2 times Vcc. Therefore the volt seconds are twice as high as when you just drive the winding with only a level of Vcc. So to make the transformer suitable you design it AS IF it was driven at 1/2 the switch frequency because that is the same as designing it for 2 times the primary voltage. The operative phrase here is "AS IF" because it really is not 1/2 frequency.
Okay, I think I understand! So, although the transformer is actually delivering power at 100kHz to the secondary side, for example, it is okay to do all the core loss calculations as if it was operating at 50kHz. However, does this mean the calculations of core loss etc will be accurate? Because although the transformer is driven at 100kHz, the core loss will still have to transfer energy at 100kHz - so it doesn't make sense to me to use the lower frequency value in the calculations?

#### SiCEngineer

Joined May 22, 2019
209
The way I see it with your example the switch frequency is 200khz - two switch actions per 10us, and the transformer frequency is 100khz - that is the frequency you design the tranformer for.
I understand where you are coming from, but I rarely see this being the case - in resonant half bridge for example. If the switching frequency is 100kHz, the resonant period is still 10uS..?

#### MrAl

Joined Jun 17, 2014
7,604
Okay, I think I understand! So, although the transformer is actually delivering power at 100kHz to the secondary side, for example, it is okay to do all the core loss calculations as if it was operating at 50kHz. However, does this mean the calculations of core loss etc will be accurate? Because although the transformer is driven at 100kHz, the core loss will still have to transfer energy at 100kHz - so it doesn't make sense to me to use the lower frequency value in the calculations?
Yeah i know what you mean. That is only valid for calculating the turns for the primary.

If you have doubts then just use 2x the voltage to do the calculations and leave the frequency alone.
Note that with 2x the voltage the turns will have to be 2x what it would be with 1x voltage.
Also remember the constant in the denominator of the flux density formula changes for square wave, and that means you can apply the full 1/2 cycle period pulse to each primary winding one at a time of course. PWM to less than that would mean less output and less power loss.