circuit analysisinductancemodeltransformer
What is the correct equivalent circuit model of an arbitrary center tapped transformer? i.e. I simply want the equivalent circuit model with has to look something like this:
In center tapped the secondary (or primary) we have multiple couplings between wingdings and I wasn't sure my model was correct.
Any advice is appreciable.
EDIT1: circuit model is needed for the following center tapped transformer:
What you may be getting confused about is the "ideal transformer" in the equivalent circuit. You should not regard it as having any magnetic qualities at all. Try and see it like this: -
Whatever voltage you have on the input to the ideal transformer, \$V_P\$ is converted to \$V_S\$ on the output of this "theoretical" and perfect device. It converts power in to power out without loss or degradation such that: -
\$V_P\cdot I_P = V_S\cdot I_S\$
The ratio of \$V_P\$ to \$V_S\$ happens to be also called the turns ratio and almost quite literally it is on an unloaded transformer because there will be no volt-drop across R3 and L3 and only the tiniest of volt-drops on R1 and L1.
This means you now have a relatively easy way of constructing scenarios of load effects and recognizing the volt drops across the leakage components that are present.
The equivalent circuit of the transformer is really quite good once you accept that the ideal power converter is "untouchable" and should just be regarded as a black box. For instance you can measure both R1 and R3 and, by shorting the secondary you can get a pretty good idea what L3 and L1 are. With open circuit secondary you can measure the current into the primary and get a pretty good idea what \$L_M\$ is too.
Yes, a B-to-C connection would be the obvious thing to try first. After making that connection (and before you connect anything else to A and D), measure the voltage between A and D. If it's 12V, you're all set.
If it's a very low value, then one of the windings needs to be reversed. Connect B to D and use that as your center tap, and then A and C become your "end" connections.
Best Answer
Below is a low frequency transformer equivalent circuit from this site.
Then, because you are probably operating at a fairly high switching frequency compared to regular AC mains, you'll need to consider parasitic capacitance like this: -
With your primary inductances (LM in my diagrams) at 150 uH and K = 0.998 (unfeasibly close to 1 in my opinion), LP will be 0.3 uH but, in reality it will be more like 3 uH (K = 0.98 = more normal).
If you can avoid core saturation you can ignore core losses (RC).
I've also used dot notation to inform how you should wind and wire the primaries to operate a push-pull drive successfully. The interwinding capacitance (PR to SEC) can be quite significant and, to reduce high frequency common-mode noise coupling you should consider capacitors on each rectified secondary winding to ground (if you are intending to rectify).
Given also that you are operating from a 5 volt supply and at probably several tens of kHz, your primary inductance values of 150 uH might be a tad high and cause you unnecessary winding losses.
The IRF530 is also fairly unsuitable because you need significant gate-source voltage to properly activate it and you are using 3.3 volts gate drive according to your circuit. It's also rated at 100 volts and has a poor RDS(on) for such a low supply (5 volts) so, use a 40 volt rated MOSFET is my advice with mush lower on resistance.
Also watch out for leakage inductance back emfs - the natural voltage on the un-driven primary will flyback to 10 volts (due to proper transformer coupling) but, leakage flyback may cause a spike of several tens of volts above that. You might choose to use a snubber circuit or pick a 100 volt MOSFET (similar to the IRF530) but with significantly better on-characteristics.