# Center tapped transformer equivalent circuit model

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.

EDIT1: circuit model is needed for the following center tapped transformer:

• can you add an image of what's your initial starting point? is the tap on primary or secondary? is the tapped terminal loaded? if so, are the other terminals loaded with something else? the model you posted works well for a regular single phase transformer, and under some specific circumstances it would work also for a tapped transformer – ppmbb May 6 '20 at 22:40
• @ppmbb I will add the image as an edit. In that image all I need is to replace the isolation transformer with its equivalent circuit. Thanks in advance – Ams May 7 '20 at 0:39
• @ppmbb What I more specifically need is an analytical solution for image in EDIT1. Could you please help with that? – Ams May 7 '20 at 0:51
• Ok, I'm assuming that your circuit input is on right hand and the secondary is on left hand of the diagram. As I can see here is that the primary of the transformer is operating alternating the winding for which current flows. In such case, the equivalent circuit is valid, you only have to consider that your primary will be Lp1 and Lp2 in each cycle. – ppmbb May 7 '20 at 0:54
• Also, seems you're working with low currents, if so, the $R_C$ and $X_M$ can be ignored because they won't represent 'significant' amount of power loss. – ppmbb May 7 '20 at 0:57

What is the correct equivalent circuit model of an arbitrary center tapped transformer?

Below is a low frequency transformer equivalent circuit from this site.

• LP is the primary leakage inductance
• RP is the primary copper loss
• RC is the core losses due to eddy currents and hysteresis
• LM is the magnetization inductance
• LS is the secondary leakage inductance
• RS is the secondary copper loss

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.

• Thank you for the answer. I will be driving the gates with 1MHz of speed and the transformer I use is Q1553-45. What my intention is to analytically understand why I get 9V peak to peak on the primary wingdings in SPICE simulation. – Ams May 9 '20 at 8:07
• Theoretically you would get 10 volt peak to peak. Reason: When one side is switched on it is pulled down to 0 volts with 5 volts on the centre tap. This has to induce 5 volts into the other winding due to transformer action solely on the primary. Given that the windings in the primary follow the same physical winding trajectory throughout, that 5 volts induced makes the open circuited end rise to 10 volts @Ams – Andy aka May 9 '20 at 8:11
• Think of it as a child's see-saw - the midpoint remains fixed at 5 volts and if one end goes down (below 5 volts), then the other end goes up (above 5 volts). – Andy aka May 9 '20 at 8:13
• Thank you that makes sense – Ams May 9 '20 at 8:18
• @Ams if you're all done with this question and answer, please do formally accept one of the answers so it can be completed - or raise a new comment. – Andy aka May 9 '20 at 8:20

Your drawing is correct, just make an electrical connection 50% down the secondary winding and bring it out. That is all there is to it. Also, recognize that load connected from one bushing to the center tap will have half the number of turns (Ns) when you go to reflect it to the primary.

At each half cycle only one of the primary winding is conducting, making a clockwise current for the up winding and counterclockwise current for the low winding.

Since both windings of the primary only conducts in alternated half cycles, there's no significant coupling effect between $$\L_{P1}\$$ and $$\L_{P2}\$$.

And the load reflected to the primary is given by the inductance ratio $$\\frac{150uH}{961.23uH} = 0.156\$$. So your reflected load is given by $$\R_L^{'} = 0.156*70\Omega = 10.92\Omega\$$

Edit: If you know your secondary resistance $$\R_s\$$ you can consider it as if it were in series with $$\R_1\$$. Besides that, $$\X_s\$$ and $$\X_p\$$ are the magnetic flow losses in the windings, you can ignore them in a first approximation to your equivalent circuit considering you're working with low currents.

• There is $\color{red}{\text{no}}$ counterclockwise current for the low winding - there is induction but no significant current. Also when you say this there's no significant coupling effect between LP1 and LP2 - you are wrong - there is a significant coupling effect via induction. – Andy aka May 7 '20 at 8:44
• would you elaborate on the significant coupling effect of a non current conducting inductor to help me improve my answer? – ppmbb May 7 '20 at 8:48
• No, don't make guesses. If you want to know, read up on stuff or ask a formal question. – Andy aka May 7 '20 at 8:49