One flavor of a high-frequency switching transformer is known as a push-pull transformer.
A push–pull transformer is a DC-to-DC switching transformer; it is often used to change a power supply’s DC voltage. These power supplies are known as switched-mode-power-supplies (SMPS).
Older, linear power supplies often used low frequency, heavy silicon steel type transformers to step down incoming line voltage to something smaller that was then rectified and filtered.
A high-frequency switching transformer can take a rectified and filtered AC line voltage and quickly “switch” it down to a lower DC voltage. The applied input voltage is only applied for very short periods of time due to the high frequency and as such the transformer size can be reduced as the turns and core area required to avoid saturation are much smaller. A lighter ferrite core is the choice at these
frequencies…
We’re going to design a fairly simple push-pull transformer. See the diagram below.
We’re going to concentrate on the transformer only and will ignore the output inductor and output capacitor used for filtering and ripple reduction…
Now let’s pick some design parameters…
That should be enough. Now we’ll choose a core that seems reasonably sized for this 24VA transformer.
Let’s go with an EE22 sized core. This core has an area of 36.26mm^2. A typical bobbin for this core has an available window area of 30.34mm^2.
This core area and bobbin window area are easily found in the published information for this core and bobbin. Obviously if this core and bobbin set are not big enough, we go to the next size up and try again. If they are too big to begin with we go down a size and try again…
Now let’s choose the wire sizes based on a safe ampacity of 2A/mm^2.
For the PRI we get:
So we’ll use 28AWG wire…
For the SEC we get:
Now this would be 20AWG wire….but the skin depth at 50kHz is about 0.29mm so instead of using Litz wire let’s instead use two strands of 0.55mm (radius of 0.275mm) wire wound bifilar. This is actually kind of close to the skin depth but it will reduce transformer cost and should be ok. The effective area of 2 strands of 0.55mm is about 94.5% of one strand of 0.80mm. This should be close enough…
Ok so the 150V is applied to one half of the PRI for a period of time, then the other half for an equal period of time. That time is not exactly 50% of the period of switching though as there must be a little bit of time when each transistor is off to ensure they are then never on at the same time.
There will also be times when the load is perhaps low, meaning the transformer is not required to deliver 24VA of power. During those times the duty cycle is decreased even further.
Nonetheless we are going to do the math in determining turns based on a “worst case” scenario of a 50% duty cycle. This will help ensure the core will not saturate…
The core we will use will be a fairly standard ferrite power material who’s MAX flux density is about 0.48T.
We are further going to be safe by using a flux density in our math of 0.3T. So, let’s find the turns needed:
Now we can find the SEC turns by the following:
We are going to take the ~ 0.7V diode drop into account and add this to the output voltage.
Now before we go further, lets compare the bare copper area to the available window area. We want the bare copper area to be less than about 40% of the total window area…
This works out to…
So this is good. The remaining 62.7% is left to cover unused space between wires (packing factor), the insulation of the wire, taping, tubing, etc…
If this were over 40% I’d go back and pick the next core/bobbin size up, if it were much under 40% then I go back and pick the next core/bobbin size down…
So this is what we have so far…..
Now our theoretical output voltage is:
This assumes a duty cycle of 50% but as we already said this is the MAX duty cycle and the actual duty cycle will likely be smaller.
Let’s find out about how much we’ll lose in voltage due to PRI & SEC DCR…
The length per turn around the bobbin (found on the bobbin spec) is about 0.0432m.
The DCR of the SEC is therefore ~ 9.27mΩ. Which means we will lose about 9.27mV @ 1A.
The DCR of the PRI is therefore ~ 0.32Ω. Which means we will lose about 0.051V @ 0.16A.
So the output should be about…
This seems close enough especially given this is not exact and the duty cycle will be less than 50% in operation. There are other losses which can/will reduce this further…
