What I'm doing:

I am designing a DCDC converter to generate ±24v from an incoming supply of 18v - 36v. For this I am using the TI TPS54160, and following the document Create a Split Rail Power Supply with a Wide Input Voltage.


To save space, I designed a planar transformer, using a split transformer core. I put 12 turns on each side of the transformer, which according to the datasheet of the core should give 244uH (12x12x1700nH).

Split planar transformer core Split planar transformer core


I have been using an Excel based calculator provided by TI to calculate the correct component values. The calculator is specifically for designing this circuit topology with this IC.

The Problem:

The problem is that at 500kHz switching frequency, the transformer is getting very hot. If I reduce the switching frequency, I can get it a little cooler, but if I reduce too far, the circuit no longer has enough drive current.

My Question:

What should I try in version 2? Would a physically larger transformer core help? Should I try reducing the number of turns on the transformer? At 500kHz, I calculate that I only need 65uH, so I could certainly go down to 8 turns.

  • 2
    \$\begingroup\$ Do you have any way to measure coil current? You might be saturating. \$\endgroup\$
    – Daniel
    Nov 17, 2015 at 16:10
  • \$\begingroup\$ Tried to instead of calculating measure things? And what gets hot, the windings or the core? Do the waveforms look ok? \$\endgroup\$
    – PlasmaHH
    Nov 17, 2015 at 16:10
  • \$\begingroup\$ @PlasmaHH - I can't tell if it's the windings or the core, because the windings are completely enclosed by the core. My hunch is that it's the core, just because of the speed at which it heats up. \$\endgroup\$ Nov 17, 2015 at 16:25
  • 1
    \$\begingroup\$ @Daniel - Hmm, I've just measured the coil current, and it seems to be about 24A! (1.2v across a 0.05R sense resistor). That's way, way higher than it should be. \$\endgroup\$ Nov 17, 2015 at 16:36
  • 1
    \$\begingroup\$ How are your windings stacked up? (How many layers, are they interleaved, etc?) Did you measure the DC resistance of the windings? \$\endgroup\$ Nov 17, 2015 at 20:19

3 Answers 3


The problem is that the cores being used have no gap, so the inductor saturates. A topology with Flyback type operation stores energy in the core when the switch is on, moving the core up the BH curve. But, for the simple example of discontinuous conduction (DCM), when the switch turns off and current falls to zero, B does not return to zero but to an elevated residual point. So, the usable \$\Delta B\$ is very small and the core is walked into saturation. Continuous conduction (CCM) is even worse in this regard.

Adding a gap moves the residual point nearer to zero, giving a usable \$\Delta B\$. In the case of a gap, the inductance will be determined by the gap, not so much the core \$\mu\$. Consider the inductance of a gapped core inductor; with core cross section \$A_c\$ and gap length \$l_g\$ and winding turns \$n\$:

\$L_g\$ = \$\frac{n^2 A_c \mu _o}{l_g}\$

also relating maximum winding current (\$I_{\text{max}}\$) to gap flux (\$B_{\text{max}}\$):

\$n I_{\text{max}}\$ = \$\frac{B_{\max } l_g}{\mu _o}\$

By starting with a value for \$L_g\$, \$B_{\text{max}}\$, \$A_c\$, and \$I_{\text{max}}\$, it's possible to get an idea what \$l_g\$ and \$n\$ for the inductor would need to be. Let \$L_g\$=100\$\mu H\$, \$B_{\text{max}}\$=0.2T, \$A_c\$=20\$mm^2\$

\$l_g\$ = \$\frac{I_{\max }^2 L_g \mu _o}{A_c B_{\max }^2}\$ = \$\frac{1 Amp^2 100\mu H \mu _o}{20 mm^2 0.2T^2}\$ ~ \$0.16 mm\$


\$n\$ = \$\frac{i_{\max } L_g}{A_c B_{\max }}\$ = \$\frac{1Amp 100\mu H}{20 mm^2 0.2T}\$ = \$25 turns\$

This analysis is pretty simplified, leaving a lot out, but gives an idea of what to expect. Designing these types of inductors gets very involved. You might look at "Inductor and Flyback Transformer Design" as a reference.

  • \$\begingroup\$ In my answer I concluded that the magnetization current is about 73mA (I could have made an error of course) and this doesn't seem like anywhere near enough to create core saturation given the core dimensions. \$\endgroup\$
    – Andy aka
    Nov 18, 2015 at 8:08
  • \$\begingroup\$ @Andyaka, if this were a transformer, I'd agree with you. But it's an inductor with DC bias, which constrains the usable B. As the output voltages and load currents come up, the core will be walked into saturation. This is usually a problem even with a DCM Flyback due to the high remanence or residual level left in an ungapped core. \$\endgroup\$
    – gsills
    Nov 19, 2015 at 1:07
  • \$\begingroup\$ Not all flyback converters need a gap - plenty don't and, like I said, the mag current is only 74 mA which gives an MMF of 0.9 At. The length of the core is about 1.5 cm so the H field is about 60 At/m. Looking at N97 material it's only just starting to enter the saturation region but, given the TPS54160 can turn it's "ON" time down to 130ns I can't really see it walking into saturation because the overheating happens at no-load AND moderate load. \$\endgroup\$
    – Andy aka
    Nov 20, 2015 at 19:08

The hole for the center leg of the core on the PCB looks plated in the figure. Is it plated in the actual PCB? If it is, that explains why you might have large currents. You have a shorted turn that gets coupled thru the core.


I think you are using N87 material so I'm going to do a quick calculation of stuff. At 500 kHz the inductor current can rise to a certain value in 1 microsecond (50:50 duty cycle). You say it has an inductance of 244 uH so with 18V applied I expect the current to rise up to: -

18V x 1 us / 244 uH = 74mA - this is the magnetization current (it stores the enrgy that is released in the next half cycle) but it sounds really, really low. The energy stored up in the main winding has to transfer to the output and this energy is 0.66 uJ (still sounding very low). The power that can be transferred to a load is therefore 0.66 uJ x 500 kHz = 0.33 watts.

I think you need to look at other examples in that data sheet you linked. I see one that can work with voltages as high as 30V and operating at 300 kHz using an inductor of 150 uH so I think your main losses are copper losses in the windings - how did you fabricate these?

I'll also point out that N87 material is going to give you about 5 to 10% losses at 500 kHz too so it's probably not the best choice.

Added to this make sure that the output winding produces a negative voltage when positive is applied to the primary. In other words the phasing of the windings is fundamental to this type of flyback circuit.

My reasoning about this discontinuous mode assessment is that although you may be expecting to run in continuous conduction mode you can get a reasonable idea by looking at it in DCM and trying to work out whether DCM is in the right ballpark.


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