Power Supply – What to Do About a Very Hot Planar Inductor?

dc/dc converterpower supplytransformer

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.

Best Answer

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.

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